3.1.86 \(\int \frac {\sin ^4(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\) [86]

Optimal. Leaf size=250 \[ \frac {3 \left (a^2+10 a b+5 b^2\right ) x}{8 (a-b)^5}-\frac {3 \sqrt {b} \left (5 a^2+10 a b+b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 \sqrt {a} (a-b)^5 f}-\frac {(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {3 b (a+b) \tan (e+f x)}{2 (a-b)^4 f \left (a+b \tan ^2(e+f x)\right )} \]

[Out]

3/8*(a^2+10*a*b+5*b^2)*x/(a-b)^5-3/8*(5*a^2+10*a*b+b^2)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))*b^(1/2)/(a-b)^5/f/a
^(1/2)-1/8*(5*a+3*b)*cos(f*x+e)*sin(f*x+e)/(a-b)^2/f/(a+b*tan(f*x+e)^2)^2+1/4*cos(f*x+e)^3*sin(f*x+e)/(a-b)/f/
(a+b*tan(f*x+e)^2)^2-1/8*b*(7*a+5*b)*tan(f*x+e)/(a-b)^3/f/(a+b*tan(f*x+e)^2)^2-3/2*b*(a+b)*tan(f*x+e)/(a-b)^4/
f/(a+b*tan(f*x+e)^2)

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Rubi [A]
time = 0.23, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3744, 481, 541, 536, 209, 211} \begin {gather*} -\frac {3 \sqrt {b} \left (5 a^2+10 a b+b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 \sqrt {a} f (a-b)^5}+\frac {3 x \left (a^2+10 a b+5 b^2\right )}{8 (a-b)^5}-\frac {3 b (a+b) \tan (e+f x)}{2 f (a-b)^4 \left (a+b \tan ^2(e+f x)\right )}-\frac {b (7 a+5 b) \tan (e+f x)}{8 f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(5 a+3 b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(3*(a^2 + 10*a*b + 5*b^2)*x)/(8*(a - b)^5) - (3*Sqrt[b]*(5*a^2 + 10*a*b + b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/S
qrt[a]])/(8*Sqrt[a]*(a - b)^5*f) - ((5*a + 3*b)*Cos[e + f*x]*Sin[e + f*x])/(8*(a - b)^2*f*(a + b*Tan[e + f*x]^
2)^2) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) - (b*(7*a + 5*b)*Tan[e + f*x])/(8
*(a - b)^3*f*(a + b*Tan[e + f*x]^2)^2) - (3*b*(a + b)*Tan[e + f*x])/(2*(a - b)^4*f*(a + b*Tan[e + f*x]^2))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 481

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

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 3744

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

Rubi steps

\begin {align*} \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac {\text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^3 \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\text {Subst}\left (\int \frac {a+(-4 a-3 b) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{4 (a-b) f}\\ &=-\frac {(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\text {Subst}\left (\int \frac {a (3 a+5 b)-5 b (5 a+3 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^2 f}\\ &=-\frac {(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\text {Subst}\left (\int \frac {12 a^2 (a+3 b)-12 a b (7 a+5 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{32 a (a-b)^3 f}\\ &=-\frac {(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {3 b (a+b) \tan (e+f x)}{2 (a-b)^4 f \left (a+b \tan ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {24 a^2 \left (a^2+6 a b+b^2\right )-96 a^2 b (a+b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{64 a^2 (a-b)^4 f}\\ &=-\frac {(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {3 b (a+b) \tan (e+f x)}{2 (a-b)^4 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\left (3 b \left (5 a^2+10 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^5 f}+\frac {\left (3 \left (a^2+10 a b+5 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^5 f}\\ &=\frac {3 \left (a^2+10 a b+5 b^2\right ) x}{8 (a-b)^5}-\frac {3 \sqrt {b} \left (5 a^2+10 a b+b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 \sqrt {a} (a-b)^5 f}-\frac {(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {3 b (a+b) \tan (e+f x)}{2 (a-b)^4 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.61, size = 194, normalized size = 0.78 \begin {gather*} \frac {12 \left (a^2+10 a b+5 b^2\right ) (e+f x)-\frac {12 \sqrt {b} \left (5 a^2+10 a b+b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{\sqrt {a}}-8 (a-b) (a+2 b) \sin (2 (e+f x))+\frac {16 a (a-b) b^2 \sin (2 (e+f x))}{(a+b+(a-b) \cos (2 (e+f x)))^2}-\frac {4 (a-b) b (9 a+5 b) \sin (2 (e+f x))}{a+b+(a-b) \cos (2 (e+f x))}+(a-b)^2 \sin (4 (e+f x))}{32 (a-b)^5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(12*(a^2 + 10*a*b + 5*b^2)*(e + f*x) - (12*Sqrt[b]*(5*a^2 + 10*a*b + b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a
]])/Sqrt[a] - 8*(a - b)*(a + 2*b)*Sin[2*(e + f*x)] + (16*a*(a - b)*b^2*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[
2*(e + f*x)])^2 - (4*(a - b)*b*(9*a + 5*b)*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)]) + (a - b)^2*Si
n[4*(e + f*x)])/(32*(a - b)^5*f)

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Maple [A]
time = 0.54, size = 209, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (-\frac {1}{4} a b +\frac {7}{8} b^{2}-\frac {5}{8} a^{2}\right ) \left (\tan ^{3}\left (f x +e \right )\right )+\left (-\frac {3}{8} a^{2}-\frac {3}{4} a b +\frac {9}{8} b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}+\frac {3 \left (a^{2}+10 a b +5 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{5}}-\frac {b \left (\frac {\left (\frac {7}{8} a^{2} b -\frac {1}{4} a \,b^{2}-\frac {5}{8} b^{3}\right ) \left (\tan ^{3}\left (f x +e \right )\right )+\frac {3 a \left (3 a^{2}-2 a b -b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {3 \left (5 a^{2}+10 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{\left (a -b \right )^{5}}}{f}\)	\(209\)
default	\(\frac {\frac {\frac {\left (-\frac {1}{4} a b +\frac {7}{8} b^{2}-\frac {5}{8} a^{2}\right ) \left (\tan ^{3}\left (f x +e \right )\right )+\left (-\frac {3}{8} a^{2}-\frac {3}{4} a b +\frac {9}{8} b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}+\frac {3 \left (a^{2}+10 a b +5 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{5}}-\frac {b \left (\frac {\left (\frac {7}{8} a^{2} b -\frac {1}{4} a \,b^{2}-\frac {5}{8} b^{3}\right ) \left (\tan ^{3}\left (f x +e \right )\right )+\frac {3 a \left (3 a^{2}-2 a b -b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {3 \left (5 a^{2}+10 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{\left (a -b \right )^{5}}}{f}\)	\(209\)
risch	\(\frac {3 x \,a^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a -b \right )^{2}}+\frac {15 x a b}{4 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a -b \right )^{2}}+\frac {15 x \,b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a -b \right )^{2}}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{4 \left (a -b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )}}{64 \left (a -b \right )^{3} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} a}{8 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} a}{8 \left (a -b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}-\frac {i \left (-9 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}-9 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+13 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+5 b^{3} {\mathrm e}^{6 i \left (f x +e \right )}-27 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}-33 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}-37 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-15 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}-27 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}-11 a^{2} b \,{\mathrm e}^{2 i \left (f x +e \right )}+23 a \,b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+15 b^{3} {\mathrm e}^{2 i \left (f x +e \right )}-9 a^{3}+13 a^{2} b +a \,b^{2}-5 b^{3}\right ) b}{4 \left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) \left (-a \,{\mathrm e}^{4 i \left (f x +e \right )}+b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 a \,{\mathrm e}^{2 i \left (f x +e \right )}-2 b \,{\mathrm e}^{2 i \left (f x +e \right )}-a +b \right )^{2} f \left (a^{2}-2 a b +b^{2}\right )}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{4 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2} f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )}}{64 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right ) f}+\frac {15 a \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{16 \left (a -b \right )^{5} f}+\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{8 \left (a -b \right )^{5} f}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b^{2}}{16 a \left (a -b \right )^{5} f}-\frac {15 a \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{16 \left (a -b \right )^{5} f}-\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{8 \left (a -b \right )^{5} f}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b^{2}}{16 a \left (a -b \right )^{5} f}\)	\(948\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(1/(a-b)^5*(((-1/4*a*b+7/8*b^2-5/8*a^2)*tan(f*x+e)^3+(-3/8*a^2-3/4*a*b+9/8*b^2)*tan(f*x+e))/(1+tan(f*x+e)^
2)^2+3/8*(a^2+10*a*b+5*b^2)*arctan(tan(f*x+e)))-b/(a-b)^5*(((7/8*a^2*b-1/4*a*b^2-5/8*b^3)*tan(f*x+e)^3+3/8*a*(
3*a^2-2*a*b-b^2)*tan(f*x+e))/(a+b*tan(f*x+e)^2)^2+3/8*(5*a^2+10*a*b+b^2)/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)
^(1/2))))

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Maxima [A]
time = 0.52, size = 470, normalized size = 1.88 \begin {gather*} \frac {\frac {3 \, {\left (a^{2} + 10 \, a b + 5 \, b^{2}\right )} {\left (f x + e\right )}}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} - \frac {3 \, {\left (5 \, a^{2} b + 10 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\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}} - \frac {12 \, {\left (a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{7} + {\left (19 \, a^{2} b + 34 \, a b^{2} + 19 \, b^{3}\right )} \tan \left (f x + e\right )^{5} + {\left (5 \, a^{3} + 31 \, a^{2} b + 31 \, a b^{2} + 5 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{3} + 6 \, a^{2} b + a b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{4} b^{2} - 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} - 4 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{8} + 2 \, {\left (a^{5} b - 3 \, a^{4} b^{2} + 2 \, a^{3} b^{3} + 2 \, a^{2} b^{4} - 3 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{6} + a^{6} - 4 \, a^{5} b + 6 \, a^{4} b^{2} - 4 \, a^{3} b^{3} + a^{2} b^{4} + {\left (a^{6} - 9 \, a^{4} b^{2} + 16 \, a^{3} b^{3} - 9 \, a^{2} b^{4} + b^{6}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{6} - 3 \, a^{5} b + 2 \, a^{4} b^{2} + 2 \, a^{3} b^{3} - 3 \, a^{2} b^{4} + a b^{5}\right )} \tan \left (f x + e\right )^{2}}}{8 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

1/8*(3*(a^2 + 10*a*b + 5*b^2)*(f*x + e)/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5) - 3*(5*a^2*b
+ 10*a*b^2 + b^3)*arctan(b*tan(f*x + e)/sqrt(a*b))/((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)) - (12*(a*b^2 + b^3)*tan(f*x + e)^7 + (19*a^2*b + 34*a*b^2 + 19*b^3)*tan(f*x + e)^5 + (5*a^3 + 31*a
^2*b + 31*a*b^2 + 5*b^3)*tan(f*x + e)^3 + 3*(a^3 + 6*a^2*b + a*b^2)*tan(f*x + e))/((a^4*b^2 - 4*a^3*b^3 + 6*a^
2*b^4 - 4*a*b^5 + b^6)*tan(f*x + e)^8 + 2*(a^5*b - 3*a^4*b^2 + 2*a^3*b^3 + 2*a^2*b^4 - 3*a*b^5 + b^6)*tan(f*x
+ e)^6 + a^6 - 4*a^5*b + 6*a^4*b^2 - 4*a^3*b^3 + a^2*b^4 + (a^6 - 9*a^4*b^2 + 16*a^3*b^3 - 9*a^2*b^4 + b^6)*ta
n(f*x + e)^4 + 2*(a^6 - 3*a^5*b + 2*a^4*b^2 + 2*a^3*b^3 - 3*a^2*b^4 + a*b^5)*tan(f*x + e)^2))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 565 vs. \(2 (241) = 482\).
time = 2.97, size = 1223, normalized size = 4.89 \begin {gather*} \left [\frac {12 \, {\left (a^{4} + 8 \, a^{3} b - 14 \, a^{2} b^{2} + 5 \, b^{4}\right )} f x \cos \left (f x + e\right )^{4} + 24 \, {\left (a^{3} b + 9 \, a^{2} b^{2} - 5 \, a b^{3} - 5 \, b^{4}\right )} f x \cos \left (f x + e\right )^{2} + 12 \, {\left (a^{2} b^{2} + 10 \, a b^{3} + 5 \, b^{4}\right )} f x - 3 \, {\left ({\left (5 \, a^{4} - 14 \, a^{2} b^{2} + 8 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + 5 \, a^{2} b^{2} + 10 \, a b^{3} + b^{4} + 2 \, {\left (5 \, a^{3} b + 5 \, a^{2} b^{2} - 9 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 4 \, {\left (2 \, {\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 )^{7} - {\left (5 \, a^{4} - 12 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} - 3 \, b^{4}\right )} \cos \left (f x + e\right )^{5} - {\left (19 \, a^{3} b - 21 \, a^{2} b^{2} - 15 \, a b^{3} + 17 \, b^{4}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\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 {6 \, {\left (a^{4} + 8 \, a^{3} b - 14 \, a^{2} b^{2} + 5 \, b^{4}\right )} f x \cos \left (f x + e\right )^{4} + 12 \, {\left (a^{3} b + 9 \, a^{2} b^{2} - 5 \, a b^{3} - 5 \, b^{4}\right )} f x \cos \left (f x + e\right )^{2} + 6 \, {\left (a^{2} b^{2} + 10 \, a b^{3} + 5 \, b^{4}\right )} f x + 3 \, {\left ({\left (5 \, a^{4} - 14 \, a^{2} b^{2} + 8 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + 5 \, a^{2} b^{2} + 10 \, a b^{3} + b^{4} + 2 \, {\left (5 \, a^{3} b + 5 \, a^{2} b^{2} - 9 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 2 \, {\left (2 \, {\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 )^{7} - {\left (5 \, a^{4} - 12 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} - 3 \, b^{4}\right )} \cos \left (f x + e\right )^{5} - {\left (19 \, a^{3} b - 21 \, a^{2} b^{2} - 15 \, a b^{3} + 17 \, b^{4}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\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)^4/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/32*(12*(a^4 + 8*a^3*b - 14*a^2*b^2 + 5*b^4)*f*x*cos(f*x + e)^4 + 24*(a^3*b + 9*a^2*b^2 - 5*a*b^3 - 5*b^4)*f
*x*cos(f*x + e)^2 + 12*(a^2*b^2 + 10*a*b^3 + 5*b^4)*f*x - 3*((5*a^4 - 14*a^2*b^2 + 8*a*b^3 + b^4)*cos(f*x + e)
^4 + 5*a^2*b^2 + 10*a*b^3 + b^4 + 2*(5*a^3*b + 5*a^2*b^2 - 9*a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(-b/a)*log(((a^2
+ 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f*x + e)^2 - 4*((a^2 + a*b)*cos(f*x + e)^3 - a*b*cos(f*x
+ e))*sqrt(-b/a)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2)
) + 4*(2*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^7 - (5*a^4 - 12*a^3*b + 6*a^2*b^2 + 4*a*b^3
- 3*b^4)*cos(f*x + e)^5 - (19*a^3*b - 21*a^2*b^2 - 15*a*b^3 + 17*b^4)*cos(f*x + e)^3 - 12*(a^2*b^2 - b^4)*cos(
f*x + e))*sin(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), 1/16*(6*(a^4 + 8*a^3*b - 14*a^2*b^2 +
5*b^4)*f*x*cos(f*x + e)^4 + 12*(a^3*b + 9*a^2*b^2 - 5*a*b^3 - 5*b^4)*f*x*cos(f*x + e)^2 + 6*(a^2*b^2 + 10*a*b^
3 + 5*b^4)*f*x + 3*((5*a^4 - 14*a^2*b^2 + 8*a*b^3 + b^4)*cos(f*x + e)^4 + 5*a^2*b^2 + 10*a*b^3 + b^4 + 2*(5*a^
3*b + 5*a^2*b^2 - 9*a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(b/a)*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(b/a)/(
b*cos(f*x + e)*sin(f*x + e))) + 2*(2*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^7 - (5*a^4 - 12*
a^3*b + 6*a^2*b^2 + 4*a*b^3 - 3*b^4)*cos(f*x + e)^5 - (19*a^3*b - 21*a^2*b^2 - 15*a*b^3 + 17*b^4)*cos(f*x + e)
^3 - 12*(a^2*b^2 - b^4)*cos(f*x + e))*sin(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3 + 35*a^3*b^4 - 2
1*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)**4/(a+b*tan(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [A]
time = 1.18, size = 399, normalized size = 1.60 \begin {gather*} \frac {\frac {3 \, {\left (a^{2} + 10 \, a b + 5 \, b^{2}\right )} {\left (f x + e\right )}}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} - \frac {3 \, {\left (5 \, a^{2} b + 10 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\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}} - \frac {12 \, a b^{2} \tan \left (f x + e\right )^{7} + 12 \, b^{3} \tan \left (f x + e\right )^{7} + 19 \, a^{2} b \tan \left (f x + e\right )^{5} + 34 \, a b^{2} \tan \left (f x + e\right )^{5} + 19 \, b^{3} \tan \left (f x + e\right )^{5} + 5 \, a^{3} \tan \left (f x + e\right )^{3} + 31 \, a^{2} b \tan \left (f x + e\right )^{3} + 31 \, a b^{2} \tan \left (f x + e\right )^{3} + 5 \, b^{3} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 18 \, a^{2} b \tan \left (f x + e\right ) + 3 \, a b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + b \tan \left (f x + e\right )^{2} + a\right )}^{2} {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )}}}{8 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^3,x, algorithm="giac")

[Out]

1/8*(3*(a^2 + 10*a*b + 5*b^2)*(f*x + e)/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5) - 3*(5*a^2*b
+ 10*a*b^2 + b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))/((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)) - (12*a*b^2*tan(f*x + e)^7 + 12*b^3*tan(f*x + e)^7 + 19*a^
2*b*tan(f*x + e)^5 + 34*a*b^2*tan(f*x + e)^5 + 19*b^3*tan(f*x + e)^5 + 5*a^3*tan(f*x + e)^3 + 31*a^2*b*tan(f*x
+ e)^3 + 31*a*b^2*tan(f*x + e)^3 + 5*b^3*tan(f*x + e)^3 + 3*a^3*tan(f*x + e) + 18*a^2*b*tan(f*x + e) + 3*a*b^
2*tan(f*x + e))/((b*tan(f*x + e)^4 + a*tan(f*x + e)^2 + b*tan(f*x + e)^2 + a)^2*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4
*a*b^3 + b^4)))/f

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Mupad [B]
time = 16.39, size = 2500, normalized size = 10.00 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^4/(a + b*tan(e + f*x)^2)^3,x)

[Out]

(atan(((((tan(e + f*x)*(540*a*b^6 + 117*b^7 + 990*a^2*b^5 + 540*a^3*b^4 + 117*a^4*b^3))/(16*(a^8 - 8*a^7*b - 8
*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^2)) + (3*((6*a*b^13 - (3*b^14)/2 +
21*a^2*b^12 - 210*a^3*b^11 + (1395*a^4*b^10)/2 - 1332*a^5*b^9 + 1638*a^6*b^8 - 1332*a^7*b^7 + (1395*a^8*b^6)/
2 - 210*a^9*b^5 + 21*a^10*b^4 + 6*a^11*b^3 - (3*a^12*b^2)/2)/(a^12 - 12*a^11*b - 12*a*b^11 + b^12 + 66*a^2*b^1
0 - 220*a^3*b^9 + 495*a^4*b^8 - 792*a^5*b^7 + 924*a^6*b^6 - 792*a^7*b^5 + 495*a^8*b^4 - 220*a^9*b^3 + 66*a^10*
b^2) - (3*tan(e + f*x)*(10*a*b + a^2 + 5*b^2)*(1152*a*b^12 - 128*b^13 - 4480*a^2*b^11 + 9600*a^3*b^10 - 11520*
a^4*b^9 + 5376*a^5*b^8 + 5376*a^6*b^7 - 11520*a^7*b^6 + 9600*a^8*b^5 - 4480*a^9*b^4 + 1152*a^10*b^3 - 128*a^11
*b^2))/(256*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)*(a^8 - 8*a^7*b - 8*a*b^7 + b^8
+ 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^2)))*(10*a*b + a^2 + 5*b^2))/(16*(a*b^4*5i - a
^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)))*(10*a*b + a^2 + 5*b^2)*3i)/(16*(a*b^4*5i - a^4*b*5i +
a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)) + (((tan(e + f*x)*(540*a*b^6 + 117*b^7 + 990*a^2*b^5 + 540*a^3*
b^4 + 117*a^4*b^3))/(16*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 2
8*a^6*b^2)) - (3*((6*a*b^13 - (3*b^14)/2 + 21*a^2*b^12 - 210*a^3*b^11 + (1395*a^4*b^10)/2 - 1332*a^5*b^9 + 163
8*a^6*b^8 - 1332*a^7*b^7 + (1395*a^8*b^6)/2 - 210*a^9*b^5 + 21*a^10*b^4 + 6*a^11*b^3 - (3*a^12*b^2)/2)/(a^12 -
12*a^11*b - 12*a*b^11 + b^12 + 66*a^2*b^10 - 220*a^3*b^9 + 495*a^4*b^8 - 792*a^5*b^7 + 924*a^6*b^6 - 792*a^7*
b^5 + 495*a^8*b^4 - 220*a^9*b^3 + 66*a^10*b^2) + (3*tan(e + f*x)*(10*a*b + a^2 + 5*b^2)*(1152*a*b^12 - 128*b^1
3 - 4480*a^2*b^11 + 9600*a^3*b^10 - 11520*a^4*b^9 + 5376*a^5*b^8 + 5376*a^6*b^7 - 11520*a^7*b^6 + 9600*a^8*b^5
- 4480*a^9*b^4 + 1152*a^10*b^3 - 128*a^11*b^2))/(256*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a
^3*b^2*10i)*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^2)))
*(10*a*b + a^2 + 5*b^2))/(16*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)))*(10*a*b + a
^2 + 5*b^2)*3i)/(16*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)))/(((1755*a*b^7)/64 +
(135*b^8)/64 + (2511*a^2*b^6)/32 + (2511*a^3*b^5)/32 + (1755*a^4*b^4)/64 + (135*a^5*b^3)/64)/(a^12 - 12*a^11*b
- 12*a*b^11 + b^12 + 66*a^2*b^10 - 220*a^3*b^9 + 495*a^4*b^8 - 792*a^5*b^7 + 924*a^6*b^6 - 792*a^7*b^5 + 495*
a^8*b^4 - 220*a^9*b^3 + 66*a^10*b^2) - (3*((tan(e + f*x)*(540*a*b^6 + 117*b^7 + 990*a^2*b^5 + 540*a^3*b^4 + 11
7*a^4*b^3))/(16*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^
2)) + (3*((6*a*b^13 - (3*b^14)/2 + 21*a^2*b^12 - 210*a^3*b^11 + (1395*a^4*b^10)/2 - 1332*a^5*b^9 + 1638*a^6*b^
8 - 1332*a^7*b^7 + (1395*a^8*b^6)/2 - 210*a^9*b^5 + 21*a^10*b^4 + 6*a^11*b^3 - (3*a^12*b^2)/2)/(a^12 - 12*a^11
*b - 12*a*b^11 + b^12 + 66*a^2*b^10 - 220*a^3*b^9 + 495*a^4*b^8 - 792*a^5*b^7 + 924*a^6*b^6 - 792*a^7*b^5 + 49
5*a^8*b^4 - 220*a^9*b^3 + 66*a^10*b^2) - (3*tan(e + f*x)*(10*a*b + a^2 + 5*b^2)*(1152*a*b^12 - 128*b^13 - 4480
*a^2*b^11 + 9600*a^3*b^10 - 11520*a^4*b^9 + 5376*a^5*b^8 + 5376*a^6*b^7 - 11520*a^7*b^6 + 9600*a^8*b^5 - 4480*
a^9*b^4 + 1152*a^10*b^3 - 128*a^11*b^2))/(256*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*1
0i)*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^2)))*(10*a*b
+ a^2 + 5*b^2))/(16*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)))*(10*a*b + a^2 + 5*b
^2))/(16*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)) + (3*((tan(e + f*x)*(540*a*b^6 +
117*b^7 + 990*a^2*b^5 + 540*a^3*b^4 + 117*a^4*b^3))/(16*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*
b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^2)) - (3*((6*a*b^13 - (3*b^14)/2 + 21*a^2*b^12 - 210*a^3*b^11 + (1395
*a^4*b^10)/2 - 1332*a^5*b^9 + 1638*a^6*b^8 - 1332*a^7*b^7 + (1395*a^8*b^6)/2 - 210*a^9*b^5 + 21*a^10*b^4 + 6*a
^11*b^3 - (3*a^12*b^2)/2)/(a^12 - 12*a^11*b - 12*a*b^11 + b^12 + 66*a^2*b^10 - 220*a^3*b^9 + 495*a^4*b^8 - 792
*a^5*b^7 + 924*a^6*b^6 - 792*a^7*b^5 + 495*a^8*b^4 - 220*a^9*b^3 + 66*a^10*b^2) + (3*tan(e + f*x)*(10*a*b + a^
2 + 5*b^2)*(1152*a*b^12 - 128*b^13 - 4480*a^2*b^11 + 9600*a^3*b^10 - 11520*a^4*b^9 + 5376*a^5*b^8 + 5376*a^6*b
^7 - 11520*a^7*b^6 + 9600*a^8*b^5 - 4480*a^9*b^4 + 1152*a^10*b^3 - 128*a^11*b^2))/(256*(a*b^4*5i - a^4*b*5i +
a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4
*b^4 - 56*a^5*b^3 + 28*a^6*b^2)))*(10*a*b + a^2 + 5*b^2))/(16*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3
*10i + a^3*b^2*10i)))*(10*a*b + a^2 + 5*b^2))/(16*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b
^2*10i))))*(10*a*b + a^2 + 5*b^2)*3i)/(8*f*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)
) - ((3*tan(e + f*x)*(a*b^2 + 6*a^2*b + a^3))/(...

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