3.1.80 \(\int \frac {\sin ^5(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\) [80]

Optimal. Leaf size=264 \[ -\frac {\sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 (a-b)^{11/2} f}-\frac {\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 (a-b)^5 f}+\frac {(10 a-b) \cos ^3(e+f x)}{15 (a-b)^4 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )} \]

[Out]

-1/5*(5*a^2+20*a*b+2*b^2)*cos(f*x+e)/(a-b)^5/f+1/15*(10*a-b)*cos(f*x+e)^3/(a-b)^4/f-1/5*cos(f*x+e)^5/(a-b)/f/(
a-b+b*sec(f*x+e)^2)^2-1/20*b*(5*a^2+4*b^2)*sec(f*x+e)/(a-b)^4/f/(a-b+b*sec(f*x+e)^2)^2-1/40*b*(35*a^2+40*a*b+2
4*b^2)*sec(f*x+e)/(a-b)^5/f/(a-b+b*sec(f*x+e)^2)-1/8*(15*a^2+40*a*b+8*b^2)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/
2))*b^(1/2)/(a-b)^(11/2)/f

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Rubi [A]
time = 0.30, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3745, 473, 467, 1273, 1275, 211} \begin {gather*} -\frac {\sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 f (a-b)^{11/2}}-\frac {\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 f (a-b)^5}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 f (a-b)^5 \left (a+b \sec ^2(e+f x)-b\right )}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )^2}+\frac {(10 a-b) \cos ^3(e+f x)}{15 f (a-b)^4}-\frac {\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

-1/8*(Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/((a - b)^(11/2)*f) - ((5*a
^2 + 20*a*b + 2*b^2)*Cos[e + f*x])/(5*(a - b)^5*f) + ((10*a - b)*Cos[e + f*x]^3)/(15*(a - b)^4*f) - Cos[e + f*
x]^5/(5*(a - b)*f*(a - b + b*Sec[e + f*x]^2)^2) - (b*(5*a^2 + 4*b^2)*Sec[e + f*x])/(20*(a - b)^4*f*(a - b + b*
Sec[e + f*x]^2)^2) - (b*(35*a^2 + 40*a*b + 24*b^2)*Sec[e + f*x])/(40*(a - b)^5*f*(a - b + b*Sec[e + f*x]^2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 467

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 1273

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(m
/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*
p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x],
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 3745

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m
+ 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{x^6 \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac {\text {Subst}\left (\int \frac {-10 a+b+5 (a-b) x^2}{x^4 \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \text {Subst}\left (\int \frac {\frac {4 (10 a-b)}{(a-b) b}-\frac {4 \left (5 a^2+4 b^2\right ) x^2}{(a-b)^2 b}+\frac {3 \left (5 a^2+4 b^2\right ) x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{20 (a-b) f}\\ &=-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {8 (a-b) (10 a-b) b-8 b \left (5 a^2+10 a b+3 b^2\right ) x^2+\frac {b^2 \left (35 a^2+40 a b+24 b^2\right ) x^4}{a-b}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{40 (a-b)^4 b f}\\ &=-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \left (\frac {8 (10 a-b) b}{x^4}-\frac {8 b \left (5 a^2+20 a b+2 b^2\right )}{(a-b) x^2}+\frac {5 b^2 \left (15 a^2+40 a b+8 b^2\right )}{(a-b) \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{40 (a-b)^4 b f}\\ &=-\frac {\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 (a-b)^5 f}+\frac {(10 a-b) \cos ^3(e+f x)}{15 (a-b)^4 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\left (b \left (15 a^2+40 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{8 (a-b)^5 f}\\ &=-\frac {\sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 (a-b)^{11/2} f}-\frac {\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 (a-b)^5 f}+\frac {(10 a-b) \cos ^3(e+f x)}{15 (a-b)^4 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 3.69, size = 278, normalized size = 1.05 \begin {gather*} \frac {\frac {30 \sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{11/2}}+\frac {30 \sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{11/2}}+\frac {-30 \cos (e+f x) \left (11 b^2+16 a b \left (2+\frac {b}{a+b+(a-b) \cos (2 (e+f x))}\right )+a^2 \left (5-\frac {8 b^2}{(a+b+(a-b) \cos (2 (e+f x)))^2}+\frac {18 b}{a+b+(a-b) \cos (2 (e+f x))}\right )\right )+(a-b) (5 (5 a+7 b) \cos (3 (e+f x))+3 (-a+b) \cos (5 (e+f x)))}{(a-b)^5}}{240 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

((30*Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(11/2
) + (30*Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(1
1/2) + (-30*Cos[e + f*x]*(11*b^2 + 16*a*b*(2 + b/(a + b + (a - b)*Cos[2*(e + f*x)])) + a^2*(5 - (8*b^2)/(a + b
+ (a - b)*Cos[2*(e + f*x)])^2 + (18*b)/(a + b + (a - b)*Cos[2*(e + f*x)]))) + (a - b)*(5*(5*a + 7*b)*Cos[3*(e
+ f*x)] + 3*(-a + b)*Cos[5*(e + f*x)]))/(a - b)^5)/(240*f)

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Maple [A]
time = 0.66, size = 282, normalized size = 1.07

method	result	size
derivativedivides	\(\frac {-\frac {\frac {a^{2} \left (\cos ^{5}\left (f x +e \right )\right )}{5}-\frac {2 a b \left (\cos ^{5}\left (f x +e \right )\right )}{5}+\frac {b^{2} \left (\cos ^{5}\left (f x +e \right )\right )}{5}-\frac {2 a^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{3}+\frac {a b \left (\cos ^{3}\left (f x +e \right )\right )}{3}+\frac {b^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{3}+a^{2} \cos \left (f x +e \right )+4 a b \cos \left (f x +e \right )+b^{2} \cos \left (f x +e \right )}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {b \left (\frac {-\frac {a \left (9 a^{2}-a b -8 b^{2}\right ) \left (\cos ^{3}\left (f x +e \right )\right )}{8}+\left (-\frac {7}{8} a^{2} b -a \,b^{2}\right ) \cos \left (f x +e \right )}{\left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {\left (15 a^{2}+40 a b +8 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{5}}}{f}\)	\(282\)
default	\(\frac {-\frac {\frac {a^{2} \left (\cos ^{5}\left (f x +e \right )\right )}{5}-\frac {2 a b \left (\cos ^{5}\left (f x +e \right )\right )}{5}+\frac {b^{2} \left (\cos ^{5}\left (f x +e \right )\right )}{5}-\frac {2 a^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{3}+\frac {a b \left (\cos ^{3}\left (f x +e \right )\right )}{3}+\frac {b^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{3}+a^{2} \cos \left (f x +e \right )+4 a b \cos \left (f x +e \right )+b^{2} \cos \left (f x +e \right )}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {b \left (\frac {-\frac {a \left (9 a^{2}-a b -8 b^{2}\right ) \left (\cos ^{3}\left (f x +e \right )\right )}{8}+\left (-\frac {7}{8} a^{2} b -a \,b^{2}\right ) \cos \left (f x +e \right )}{\left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {\left (15 a^{2}+40 a b +8 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{5}}}{f}\)	\(282\)
risch	\(\text {Expression too large to display}\)	\(1175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(-1/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*a*b+b^2)*(1/5*a^2*cos(f*x+e)^5-2/5*a*b*cos(f*x+e)^5+1/5*b^2*cos(f*x+e
)^5-2/3*a^2*cos(f*x+e)^3+1/3*a*b*cos(f*x+e)^3+1/3*b^2*cos(f*x+e)^3+a^2*cos(f*x+e)+4*a*b*cos(f*x+e)+b^2*cos(f*x
+e))+b/(a-b)^5*((-1/8*a*(9*a^2-a*b-8*b^2)*cos(f*x+e)^3+(-7/8*a^2*b-a*b^2)*cos(f*x+e))/(a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)^2+1/8*(15*a^2+40*a*b+8*b^2)/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

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Fricas [A]
time = 4.29, size = 1040, normalized size = 3.94 \begin {gather*} \left [-\frac {48 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{9} - 16 \, {\left (10 \, a^{4} - 31 \, a^{3} b + 33 \, a^{2} b^{2} - 13 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{7} + 16 \, {\left (15 \, a^{4} + 10 \, a^{3} b - 57 \, a^{2} b^{2} + 24 \, a b^{3} + 8 \, b^{4}\right )} \cos \left (f x + e\right )^{5} + 50 \, {\left (15 \, a^{3} b + 25 \, a^{2} b^{2} - 32 \, a b^{3} - 8 \, b^{4}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left ({\left (15 \, a^{4} + 10 \, a^{3} b - 57 \, a^{2} b^{2} + 24 \, a b^{3} + 8 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 40 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 25 \, a^{2} b^{2} - 32 \, a b^{3} - 8 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, {\left (15 \, a^{2} b^{2} + 40 \, a b^{3} + 8 \, b^{4}\right )} \cos \left (f x + e\right )}{240 \, {\left ({\left (a^{7} - 7 \, a^{6} b + 21 \, a^{5} b^{2} - 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} - 21 \, a^{2} b^{5} + 7 \, a b^{6} - b^{7}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 6 \, a^{5} b^{2} + 15 \, a^{4} b^{3} - 20 \, a^{3} b^{4} + 15 \, a^{2} b^{5} - 6 \, a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} - 10 \, a^{2} b^{5} + 5 \, a b^{6} - b^{7}\right )} f\right )}}, -\frac {24 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{9} - 8 \, {\left (10 \, a^{4} - 31 \, a^{3} b + 33 \, a^{2} b^{2} - 13 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{7} + 8 \, {\left (15 \, a^{4} + 10 \, a^{3} b - 57 \, a^{2} b^{2} + 24 \, a b^{3} + 8 \, b^{4}\right )} \cos \left (f x + e\right )^{5} + 25 \, {\left (15 \, a^{3} b + 25 \, a^{2} b^{2} - 32 \, a b^{3} - 8 \, b^{4}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left ({\left (15 \, a^{4} + 10 \, a^{3} b - 57 \, a^{2} b^{2} + 24 \, a b^{3} + 8 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 40 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 25 \, a^{2} b^{2} - 32 \, a b^{3} - 8 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (15 \, a^{2} b^{2} + 40 \, a b^{3} + 8 \, b^{4}\right )} \cos \left (f x + e\right )}{120 \, {\left ({\left (a^{7} - 7 \, a^{6} b + 21 \, a^{5} b^{2} - 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} - 21 \, a^{2} b^{5} + 7 \, a b^{6} - b^{7}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 6 \, a^{5} b^{2} + 15 \, a^{4} b^{3} - 20 \, a^{3} b^{4} + 15 \, a^{2} b^{5} - 6 \, a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} - 10 \, a^{2} b^{5} + 5 \, a b^{6} - b^{7}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[-1/240*(48*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 16*(10*a^4 - 31*a^3*b + 33*a^2*b^2 -
13*a*b^3 + b^4)*cos(f*x + e)^7 + 16*(15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^5 + 50*(1
5*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^3 + 15*((15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b
^4)*cos(f*x + e)^4 + 15*a^2*b^2 + 40*a*b^3 + 8*b^4 + 2*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)
^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(
f*x + e)^2 + b)) + 30*(15*a^2*b^2 + 40*a*b^3 + 8*b^4)*cos(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3
+ 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x + e)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 +
15*a^2*b^5 - 6*a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^5*b^2 - 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7
)*f), -1/120*(24*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 8*(10*a^4 - 31*a^3*b + 33*a^2*b^
2 - 13*a*b^3 + b^4)*cos(f*x + e)^7 + 8*(15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^5 + 25
*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^3 + 15*((15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 +
8*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 + 40*a*b^3 + 8*b^4 + 2*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x +
e)^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 15*(15*a^2*b^2 + 40*a*b^3 + 8*b^4)*co
s(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3 + 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x + e
)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 + 15*a^2*b^5 - 6*a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^5*b^2
- 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7)*f)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**5/(a+b*tan(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 885 vs. \(2 (253) = 506\).
time = 1.19, size = 885, normalized size = 3.35 \begin {gather*} -\frac {\frac {15 \, {\left (15 \, a^{2} b + 40 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right ) - b}{\sqrt {a b - b^{2}} \cos \left (f x + e\right ) + \sqrt {a b - b^{2}}}\right )}{{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} \sqrt {a b - b^{2}}} + \frac {30 \, {\left (9 \, a^{3} b + 6 \, a^{2} b^{2} + \frac {27 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {32 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {40 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {27 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {54 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {24 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {48 \, b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {9 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {16 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {8 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} {\left (a + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{2}} - \frac {16 \, {\left (8 \, a^{2} + 59 \, a b + 23 \, b^{2} - \frac {40 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {250 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {70 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {80 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {320 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {140 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {270 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {90 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {45 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {45 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} {\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}}{120 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="giac")

[Out]

-1/120*(15*(15*a^2*b + 40*a*b^2 + 8*b^3)*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*
x + e) + sqrt(a*b - b^2)))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*sqrt(a*b - b^2)) + 30*(9
*a^3*b + 6*a^2*b^2 + 27*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 32*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) - 40*a*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 27*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2
- 54*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 24*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 +
48*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 9*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 16*a^2
*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 8*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a^5 - 5*
a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x
+ e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)^2) - 16*(8*a^2 + 59*a*b + 23*b^2
- 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 250*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 70*b^2*(cos(f
*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 320*a*b*(cos(f*x + e) - 1
)^2/(cos(f*x + e) + 1)^2 + 140*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 270*a*b*(cos(f*x + e) - 1)^3/(c
os(f*x + e) + 1)^3 - 90*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 45*a*b*(cos(f*x + e) - 1)^4/(cos(f*x +
e) + 1)^4 + 45*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a
*b^4 - b^5)*((cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 1)^5))/f

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Mupad [B]
time = 16.33, size = 1536, normalized size = 5.82 \begin {gather*} \frac {\sqrt {b}\,\mathrm {atan}\left (\frac {{\left (a-b\right )}^{11}\,\left (2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {\sqrt {b}\,\left (15\,a^2+40\,a\,b+8\,b^2\right )\,\left (-240\,a^{14}\,b+1760\,a^{13}\,b^2-4528\,a^{12}\,b^3+1280\,a^{11}\,b^4+20640\,a^{10}\,b^5-58560\,a^9\,b^6+84000\,a^8\,b^7-73344\,a^7\,b^8+39120\,a^6\,b^9-11040\,a^5\,b^{10}+400\,a^4\,b^{11}+640\,a^3\,b^{12}-128\,a^2\,b^{13}\right )}{16\,a\,{\left (a-b\right )}^{21/2}}-\frac {\sqrt {b}\,\left (a-2\,b\right )\,{\left (15\,a^2+40\,a\,b+8\,b^2\right )}^2\,\left (128\,a^{18}-2176\,a^{17}\,b+17280\,a^{16}\,b^2-85120\,a^{15}\,b^3+291200\,a^{14}\,b^4-733824\,a^{13}\,b^5+1409408\,a^{12}\,b^6-2104960\,a^{11}\,b^7+2471040\,a^{10}\,b^8-2288000\,a^9\,b^9+1665664\,a^8\,b^{10}-943488\,a^7\,b^{11}+407680\,a^6\,b^{12}-129920\,a^5\,b^{13}+28800\,a^4\,b^{14}-3968\,a^3\,b^{15}+256\,a^2\,b^{16}\right )}{512\,a\,{\left (a-b\right )}^{33/2}}\right )-\frac {\sqrt {b}\,\left (a-2\,b\right )\,{\left (15\,a^2+40\,a\,b+8\,b^2\right )}^2\,\left (-128\,a^{18}+1920\,a^{17}\,b-13440\,a^{16}\,b^2+58240\,a^{15}\,b^3-174720\,a^{14}\,b^4+384384\,a^{13}\,b^5-640640\,a^{12}\,b^6+823680\,a^{11}\,b^7-823680\,a^{10}\,b^8+640640\,a^9\,b^9-384384\,a^8\,b^{10}+174720\,a^7\,b^{11}-58240\,a^6\,b^{12}+13440\,a^5\,b^{13}-1920\,a^4\,b^{14}+128\,a^3\,b^{15}\right )}{256\,a\,{\left (a-b\right )}^{33/2}}\right )}{225\,a^{16}\,b-1050\,a^{15}\,b^2-35\,a^{14}\,b^3+9240\,a^{13}\,b^4-20286\,a^{12}\,b^5+2660\,a^{11}\,b^6+57330\,a^{10}\,b^7-111960\,a^9\,b^8+104685\,a^8\,b^9-50778\,a^7\,b^{10}+7665\,a^6\,b^{11}+3920\,a^5\,b^{12}-1680\,a^4\,b^{13}+64\,a^2\,b^{15}}\right )\,\left (15\,a^2+40\,a\,b+8\,b^2\right )}{8\,f\,{\left (a-b\right )}^{11/2}}-\frac {\frac {64\,a^4+607\,a^3\,b+274\,a^2\,b^2}{60\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{14}\,\left (15\,a^3\,b+85\,a^2\,b^2+128\,a\,b^3+24\,b^4\right )}{2\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}\,\left (64\,a^4-365\,a^3\,b+1075\,a^2\,b^2+936\,a\,b^3+936\,b^4\right )}{6\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}\,\left (-224\,a^4+921\,a^3\,b-1545\,a^2\,b^2+4268\,a\,b^3+1872\,b^4\right )}{6\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (-128\,a^4-671\,a^3\,b+5973\,a^2\,b^2+6224\,a\,b^3+1832\,b^4\right )}{30\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (-448\,a^4+867\,a^3\,b-935\,a^2\,b^2+20696\,a\,b^3+6280\,b^4\right )}{30\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (1312\,a^4-4064\,a^3\,b+1527\,a^2\,b^2+21740\,a\,b^3+12560\,b^4\right )}{30\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (32\,a^4+447\,a^3\,b+2265\,a^2\,b^2+1036\,a\,b^3\right )}{30\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}+\frac {b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{16}\,\left (15\,a^3+40\,a^2\,b+8\,a\,b^2\right )}{4\,\left (a-b\right )\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )}}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{18}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (-4\,a^2+24\,a\,b+16\,b^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{14}\,\left (-4\,a^2+24\,a\,b+16\,b^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (-4\,a^2+8\,a\,b+80\,b^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}\,\left (-4\,a^2+8\,a\,b+80\,b^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (6\,a^2-40\,a\,b+160\,b^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}\,\left (6\,a^2-40\,a\,b+160\,b^2\right )+a^2+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (a^2+8\,b\,a\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{16}\,\left (a^2+8\,b\,a\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^5/(a + b*tan(e + f*x)^2)^3,x)

[Out]

(b^(1/2)*atan(((a - b)^11*(2*tan(e/2 + (f*x)/2)^2*((b^(1/2)*(40*a*b + 15*a^2 + 8*b^2)*(640*a^3*b^12 - 128*a^2*
b^13 - 240*a^14*b + 400*a^4*b^11 - 11040*a^5*b^10 + 39120*a^6*b^9 - 73344*a^7*b^8 + 84000*a^8*b^7 - 58560*a^9*
b^6 + 20640*a^10*b^5 + 1280*a^11*b^4 - 4528*a^12*b^3 + 1760*a^13*b^2))/(16*a*(a - b)^(21/2)) - (b^(1/2)*(a - 2
*b)*(40*a*b + 15*a^2 + 8*b^2)^2*(128*a^18 - 2176*a^17*b + 256*a^2*b^16 - 3968*a^3*b^15 + 28800*a^4*b^14 - 1299
20*a^5*b^13 + 407680*a^6*b^12 - 943488*a^7*b^11 + 1665664*a^8*b^10 - 2288000*a^9*b^9 + 2471040*a^10*b^8 - 2104
960*a^11*b^7 + 1409408*a^12*b^6 - 733824*a^13*b^5 + 291200*a^14*b^4 - 85120*a^15*b^3 + 17280*a^16*b^2))/(512*a
*(a - b)^(33/2))) - (b^(1/2)*(a - 2*b)*(40*a*b + 15*a^2 + 8*b^2)^2*(1920*a^17*b - 128*a^18 + 128*a^3*b^15 - 19
20*a^4*b^14 + 13440*a^5*b^13 - 58240*a^6*b^12 + 174720*a^7*b^11 - 384384*a^8*b^10 + 640640*a^9*b^9 - 823680*a^
10*b^8 + 823680*a^11*b^7 - 640640*a^12*b^6 + 384384*a^13*b^5 - 174720*a^14*b^4 + 58240*a^15*b^3 - 13440*a^16*b
^2))/(256*a*(a - b)^(33/2))))/(225*a^16*b + 64*a^2*b^15 - 1680*a^4*b^13 + 3920*a^5*b^12 + 7665*a^6*b^11 - 5077
8*a^7*b^10 + 104685*a^8*b^9 - 111960*a^9*b^8 + 57330*a^10*b^7 + 2660*a^11*b^6 - 20286*a^12*b^5 + 9240*a^13*b^4
- 35*a^14*b^3 - 1050*a^15*b^2))*(40*a*b + 15*a^2 + 8*b^2))/(8*f*(a - b)^(11/2)) - ((607*a^3*b + 64*a^4 + 274*
a^2*b^2)/(60*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^14*(128*a*b^3 + 15*a^3
*b + 24*b^4 + 85*a^2*b^2))/(2*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^12*(9
36*a*b^3 - 365*a^3*b + 64*a^4 + 936*b^4 + 1075*a^2*b^2))/(6*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2
)) + (tan(e/2 + (f*x)/2)^10*(4268*a*b^3 + 921*a^3*b - 224*a^4 + 1872*b^4 - 1545*a^2*b^2))/(6*(a - b)*(a^4 - 4*
a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^4*(6224*a*b^3 - 671*a^3*b - 128*a^4 + 1832*b^4 + 597
3*a^2*b^2))/(30*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^6*(20696*a*b^3 + 86
7*a^3*b - 448*a^4 + 6280*b^4 - 935*a^2*b^2))/(30*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e
/2 + (f*x)/2)^8*(21740*a*b^3 - 4064*a^3*b + 1312*a^4 + 12560*b^4 + 1527*a^2*b^2))/(30*(a - b)*(a^4 - 4*a^3*b -
4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^2*(1036*a*b^3 + 447*a^3*b + 32*a^4 + 2265*a^2*b^2))/(30*(a
- b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (b*tan(e/2 + (f*x)/2)^16*(8*a*b^2 + 40*a^2*b + 15*a^3))/(4
*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))/(f*(a^2*tan(e/2 + (f*x)/2)^18 + tan(e/2 + (f*x)/2)^4*(2
4*a*b - 4*a^2 + 16*b^2) + tan(e/2 + (f*x)/2)^14*(24*a*b - 4*a^2 + 16*b^2) + tan(e/2 + (f*x)/2)^6*(8*a*b - 4*a^
2 + 80*b^2) + tan(e/2 + (f*x)/2)^12*(8*a*b - 4*a^2 + 80*b^2) + tan(e/2 + (f*x)/2)^8*(6*a^2 - 40*a*b + 160*b^2)
+ tan(e/2 + (f*x)/2)^10*(6*a^2 - 40*a*b + 160*b^2) + a^2 + tan(e/2 + (f*x)/2)^2*(8*a*b + a^2) + tan(e/2 + (f*
x)/2)^16*(8*a*b + a^2)))

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