3.79 \(\int \frac {\csc ^6(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=182 \[ -\frac {\sqrt {b} (3 a-7 b) (a-b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{9/2} f}-\frac {(10 a-7 b) \cot ^3(e+f x)}{15 a^3 f}-\frac {b \left (5 a^2-10 a b+7 b^2\right ) \tan (e+f x)}{10 a^4 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\left (5 a^2-20 a b+14 b^2\right ) \cot (e+f x)}{5 a^4 f}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )} \]

[Out]

-1/5*(5*a^2-20*a*b+14*b^2)*cot(f*x+e)/a^4/f-1/15*(10*a-7*b)*cot(f*x+e)^3/a^3/f-1/2*(3*a-7*b)*(a-b)*arctan(b^(1
/2)*tan(f*x+e)/a^(1/2))*b^(1/2)/a^(9/2)/f-1/5*cot(f*x+e)^5/a/f/(a+b*tan(f*x+e)^2)-1/10*b*(5*a^2-10*a*b+7*b^2)*
tan(f*x+e)/a^4/f/(a+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.22, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3663, 462, 456, 1261, 205} \[ -\frac {b \left (5 a^2-10 a b+7 b^2\right ) \tan (e+f x)}{10 a^4 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\left (5 a^2-20 a b+14 b^2\right ) \cot (e+f x)}{5 a^4 f}-\frac {\sqrt {b} (3 a-7 b) (a-b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{9/2} f}-\frac {(10 a-7 b) \cot ^3(e+f x)}{15 a^3 f}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

-((3*a - 7*b)*(a - b)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*a^(9/2)*f) - ((5*a^2 - 20*a*b + 14*b^
2)*Cot[e + f*x])/(5*a^4*f) - ((10*a - 7*b)*Cot[e + f*x]^3)/(15*a^3*f) - Cot[e + f*x]^5/(5*a*f*(a + b*Tan[e + f
*x]^2)) - (b*(5*a^2 - 10*a*b + 7*b^2)*Tan[e + f*x])/(10*a^4*f*(a + b*Tan[e + f*x]^2))

Rule 205

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

Rule 456

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 462

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 1261

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 3663

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 {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {10 a-7 b+5 a x^2}{x^4 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \left (5 a^2-10 a b+7 b^2\right ) \tan (e+f x)}{10 a^4 f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \operatorname {Subst}\left (\int \frac {2 \left (\frac {7}{a}-\frac {10}{b}\right )+2 \left (\frac {10}{a}-\frac {5}{b}-\frac {7 b}{a^2}\right ) x^2+\frac {\left (5 a^2-10 a b+7 b^2\right ) x^4}{a^3}}{x^4 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{10 a f}\\ &=-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \left (5 a^2-10 a b+7 b^2\right ) \tan (e+f x)}{10 a^4 f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {2 (10 a-7 b)}{a^2 b x^4}-\frac {2 \left (5 a^2-20 a b+14 b^2\right )}{a^3 b x^2}+\frac {5 (3 a-7 b) (a-b)}{a^3 \left (a+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{10 a f}\\ &=-\frac {\left (5 a^2-20 a b+14 b^2\right ) \cot (e+f x)}{5 a^4 f}-\frac {(10 a-7 b) \cot ^3(e+f x)}{15 a^3 f}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \left (5 a^2-10 a b+7 b^2\right ) \tan (e+f x)}{10 a^4 f \left (a+b \tan ^2(e+f x)\right )}-\frac {((3 a-7 b) (a-b) b) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^4 f}\\ &=-\frac {(3 a-7 b) (a-b) \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{9/2} f}-\frac {\left (5 a^2-20 a b+14 b^2\right ) \cot (e+f x)}{5 a^4 f}-\frac {(10 a-7 b) \cot ^3(e+f x)}{15 a^3 f}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \left (5 a^2-10 a b+7 b^2\right ) \tan (e+f x)}{10 a^4 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.90, size = 151, normalized size = 0.83 \[ \frac {\sqrt {a} \left (-2 \cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+8 a^2+2 a (2 a-5 b) \csc ^2(e+f x)-50 a b+45 b^2\right )-\frac {15 b (a-b)^2 \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}\right )-15 \sqrt {b} \left (3 a^2-10 a b+7 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{30 a^{9/2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

(-15*Sqrt[b]*(3*a^2 - 10*a*b + 7*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] + Sqrt[a]*(-2*Cot[e + f*x]*(8*a^2
- 50*a*b + 45*b^2 + 2*a*(2*a - 5*b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4) - (15*(a - b)^2*b*Sin[2*(e + f*x)]
)/(a + b + (a - b)*Cos[2*(e + f*x)])))/(30*a^(9/2)*f)

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fricas [B]  time = 0.58, size = 855, normalized size = 4.70 \[ \left [-\frac {4 \, {\left (16 \, a^{3} - 131 \, a^{2} b + 220 \, a b^{2} - 105 \, b^{3}\right )} \cos \left (f x + e\right )^{7} - 4 \, {\left (40 \, a^{3} - 321 \, a^{2} b + 590 \, a b^{2} - 315 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (6 \, a^{3} - 47 \, a^{2} b + 104 \, a b^{2} - 63 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (3 \, a^{3} - 13 \, a^{2} b + 17 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (6 \, a^{3} - 29 \, a^{2} b + 44 \, a b^{2} - 21 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b - 10 \, a b^{2} + 7 \, b^{3} + {\left (3 \, a^{3} - 19 \, a^{2} b + 37 \, a b^{2} - 21 \, b^{3}\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 ) \sin \left (f x + e\right ) + 60 \, {\left (3 \, a^{2} b - 10 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{120 \, {\left ({\left (a^{5} - a^{4} b\right )} f \cos \left (f x + e\right )^{6} + a^{4} b f - {\left (2 \, a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (16 \, a^{3} - 131 \, a^{2} b + 220 \, a b^{2} - 105 \, b^{3}\right )} \cos \left (f x + e\right )^{7} - 2 \, {\left (40 \, a^{3} - 321 \, a^{2} b + 590 \, a b^{2} - 315 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (6 \, a^{3} - 47 \, a^{2} b + 104 \, a b^{2} - 63 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (3 \, a^{3} - 13 \, a^{2} b + 17 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (6 \, a^{3} - 29 \, a^{2} b + 44 \, a b^{2} - 21 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b - 10 \, a b^{2} + 7 \, b^{3} + {\left (3 \, a^{3} - 19 \, a^{2} b + 37 \, a b^{2} - 21 \, b^{3}\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 ) \sin \left (f x + e\right ) + 30 \, {\left (3 \, a^{2} b - 10 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{60 \, {\left ({\left (a^{5} - a^{4} b\right )} f \cos \left (f x + e\right )^{6} + a^{4} b f - {\left (2 \, a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[-1/120*(4*(16*a^3 - 131*a^2*b + 220*a*b^2 - 105*b^3)*cos(f*x + e)^7 - 4*(40*a^3 - 321*a^2*b + 590*a*b^2 - 315
*b^3)*cos(f*x + e)^5 + 20*(6*a^3 - 47*a^2*b + 104*a*b^2 - 63*b^3)*cos(f*x + e)^3 - 15*((3*a^3 - 13*a^2*b + 17*
a*b^2 - 7*b^3)*cos(f*x + e)^6 - (6*a^3 - 29*a^2*b + 44*a*b^2 - 21*b^3)*cos(f*x + e)^4 + 3*a^2*b - 10*a*b^2 + 7
*b^3 + (3*a^3 - 19*a^2*b + 37*a*b^2 - 21*b^3)*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))*sin(f*x + e) + 60*(3*a^2*b
- 10*a*b^2 + 7*b^3)*cos(f*x + e))/(((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2*a^5 - 3*a^4*b)*f*cos(f*x +
e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2)*sin(f*x + e)), -1/60*(2*(16*a^3 - 131*a^2*b + 220*a*b^2 - 105*b^3)*co
s(f*x + e)^7 - 2*(40*a^3 - 321*a^2*b + 590*a*b^2 - 315*b^3)*cos(f*x + e)^5 + 10*(6*a^3 - 47*a^2*b + 104*a*b^2
- 63*b^3)*cos(f*x + e)^3 - 15*((3*a^3 - 13*a^2*b + 17*a*b^2 - 7*b^3)*cos(f*x + e)^6 - (6*a^3 - 29*a^2*b + 44*a
*b^2 - 21*b^3)*cos(f*x + e)^4 + 3*a^2*b - 10*a*b^2 + 7*b^3 + (3*a^3 - 19*a^2*b + 37*a*b^2 - 21*b^3)*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)))*sin(f*x + e)
+ 30*(3*a^2*b - 10*a*b^2 + 7*b^3)*cos(f*x + e))/(((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2*a^5 - 3*a^4*b)
*f*cos(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2)*sin(f*x + e))]

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giac [A]  time = 3.07, size = 212, normalized size = 1.16 \[ -\frac {\frac {15 \, {\left (3 \, a^{2} b - 10 \, a b^{2} + 7 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{\sqrt {a b} a^{4}} + \frac {15 \, {\left (a^{2} b \tan \left (f x + e\right ) - 2 \, a b^{2} \tan \left (f x + e\right ) + b^{3} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )} a^{4}} + \frac {2 \, {\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 60 \, a b \tan \left (f x + e\right )^{4} + 45 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} - 10 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )}}{a^{4} \tan \left (f x + e\right )^{5}}}{30 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/30*(15*(3*a^2*b - 10*a*b^2 + 7*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b))
)/(sqrt(a*b)*a^4) + 15*(a^2*b*tan(f*x + e) - 2*a*b^2*tan(f*x + e) + b^3*tan(f*x + e))/((b*tan(f*x + e)^2 + a)*
a^4) + 2*(15*a^2*tan(f*x + e)^4 - 60*a*b*tan(f*x + e)^4 + 45*b^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2 - 10*a
*b*tan(f*x + e)^2 + 3*a^2)/(a^4*tan(f*x + e)^5))/f

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maple [A]  time = 0.58, size = 281, normalized size = 1.54 \[ -\frac {b \tan \left (f x +e \right )}{2 f \,a^{2} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {b^{2} \tan \left (f x +e \right )}{f \,a^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {b^{3} \tan \left (f x +e \right )}{2 f \,a^{4} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \,a^{2} \sqrt {a b}}+\frac {5 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{f \,a^{3} \sqrt {a b}}-\frac {7 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \,a^{4} \sqrt {a b}}-\frac {1}{5 f \,a^{2} \tan \left (f x +e \right )^{5}}-\frac {2}{3 f \,a^{2} \tan \left (f x +e \right )^{3}}+\frac {2 b}{3 f \,a^{3} \tan \left (f x +e \right )^{3}}-\frac {1}{f \,a^{2} \tan \left (f x +e \right )}+\frac {4 b}{f \,a^{3} \tan \left (f x +e \right )}-\frac {3 b^{2}}{f \,a^{4} \tan \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^2,x)

[Out]

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

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maxima [A]  time = 0.79, size = 161, normalized size = 0.88 \[ -\frac {\frac {15 \, {\left (3 \, a^{2} b - 10 \, a b^{2} + 7 \, b^{3}\right )} \tan \left (f x + e\right )^{6} + 10 \, {\left (3 \, a^{3} - 10 \, a^{2} b + 7 \, a b^{2}\right )} \tan \left (f x + e\right )^{4} + 6 \, a^{3} + 2 \, {\left (10 \, a^{3} - 7 \, a^{2} b\right )} \tan \left (f x + e\right )^{2}}{a^{4} b \tan \left (f x + e\right )^{7} + a^{5} \tan \left (f x + e\right )^{5}} + \frac {15 \, {\left (3 \, a^{2} b - 10 \, a b^{2} + 7 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}}}{30 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

-1/30*((15*(3*a^2*b - 10*a*b^2 + 7*b^3)*tan(f*x + e)^6 + 10*(3*a^3 - 10*a^2*b + 7*a*b^2)*tan(f*x + e)^4 + 6*a^
3 + 2*(10*a^3 - 7*a^2*b)*tan(f*x + e)^2)/(a^4*b*tan(f*x + e)^7 + a^5*tan(f*x + e)^5) + 15*(3*a^2*b - 10*a*b^2
+ 7*b^3)*arctan(b*tan(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^4))/f

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mupad [B]  time = 12.37, size = 178, normalized size = 0.98 \[ -\frac {\frac {1}{5\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (3\,a^2-10\,a\,b+7\,b^2\right )}{3\,a^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (10\,a-7\,b\right )}{15\,a^2}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (3\,a^2-10\,a\,b+7\,b^2\right )}{2\,a^4}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^7+a\,{\mathrm {tan}\left (e+f\,x\right )}^5\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a-b\right )\,\left (3\,a-7\,b\right )}{\sqrt {a}\,\left (3\,a^2-10\,a\,b+7\,b^2\right )}\right )\,\left (a-b\right )\,\left (3\,a-7\,b\right )}{2\,a^{9/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^6*(a + b*tan(e + f*x)^2)^2),x)

[Out]

- (1/(5*a) + (tan(e + f*x)^4*(3*a^2 - 10*a*b + 7*b^2))/(3*a^3) + (tan(e + f*x)^2*(10*a - 7*b))/(15*a^2) + (b*t
an(e + f*x)^6*(3*a^2 - 10*a*b + 7*b^2))/(2*a^4))/(f*(a*tan(e + f*x)^5 + b*tan(e + f*x)^7)) - (b^(1/2)*atan((b^
(1/2)*tan(e + f*x)*(a - b)*(3*a - 7*b))/(a^(1/2)*(3*a^2 - 10*a*b + 7*b^2)))*(a - b)*(3*a - 7*b))/(2*a^(9/2)*f)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**6/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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