3.66 \(\int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=219 \[ \frac {b \cot ^4(c+d x)}{2 a^3 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {2 b \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (\tan (c+d x))}{a^7 d}+\frac {2 b \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{a^7 d}-\frac {b \left (a^2+b^2\right )^2}{a^6 d (a+b \tan (c+d x))}-\frac {\left (a^2+b^2\right ) \left (a^2+5 b^2\right ) \cot (c+d x)}{a^6 d}+\frac {2 b \left (a^2+b^2\right ) \cot ^2(c+d x)}{a^5 d}-\frac {\left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 a^4 d} \]

[Out]

-(a^2+b^2)*(a^2+5*b^2)*cot(d*x+c)/a^6/d+2*b*(a^2+b^2)*cot(d*x+c)^2/a^5/d-1/3*(2*a^2+3*b^2)*cot(d*x+c)^3/a^4/d+
1/2*b*cot(d*x+c)^4/a^3/d-1/5*cot(d*x+c)^5/a^2/d-2*b*(a^2+b^2)*(a^2+3*b^2)*ln(tan(d*x+c))/a^7/d+2*b*(a^2+b^2)*(
a^2+3*b^2)*ln(a+b*tan(d*x+c))/a^7/d-b*(a^2+b^2)^2/a^6/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.20, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3516, 894} \[ -\frac {b \left (a^2+b^2\right )^2}{a^6 d (a+b \tan (c+d x))}-\frac {\left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 a^4 d}+\frac {2 b \left (a^2+b^2\right ) \cot ^2(c+d x)}{a^5 d}-\frac {\left (a^2+b^2\right ) \left (a^2+5 b^2\right ) \cot (c+d x)}{a^6 d}-\frac {2 b \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (\tan (c+d x))}{a^7 d}+\frac {2 b \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{a^7 d}+\frac {b \cot ^4(c+d x)}{2 a^3 d}-\frac {\cot ^5(c+d x)}{5 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 + b^2)*(a^2 + 5*b^2)*Cot[c + d*x])/(a^6*d)) + (2*b*(a^2 + b^2)*Cot[c + d*x]^2)/(a^5*d) - ((2*a^2 + 3*b
^2)*Cot[c + d*x]^3)/(3*a^4*d) + (b*Cot[c + d*x]^4)/(2*a^3*d) - Cot[c + d*x]^5/(5*a^2*d) - (2*b*(a^2 + b^2)*(a^
2 + 3*b^2)*Log[Tan[c + d*x]])/(a^7*d) + (2*b*(a^2 + b^2)*(a^2 + 3*b^2)*Log[a + b*Tan[c + d*x]])/(a^7*d) - (b*(
a^2 + b^2)^2)/(a^6*d*(a + b*Tan[c + d*x]))

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 3516

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

Rubi steps

\begin {align*} \int \frac {\csc ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac {b \operatorname {Subst}\left (\int \frac {\left (b^2+x^2\right )^2}{x^6 (a+x)^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac {b \operatorname {Subst}\left (\int \left (\frac {b^4}{a^2 x^6}-\frac {2 b^4}{a^3 x^5}+\frac {2 a^2 b^2+3 b^4}{a^4 x^4}-\frac {4 b^2 \left (a^2+b^2\right )}{a^5 x^3}+\frac {a^4+6 a^2 b^2+5 b^4}{a^6 x^2}-\frac {2 \left (a^4+4 a^2 b^2+3 b^4\right )}{a^7 x}+\frac {\left (a^2+b^2\right )^2}{a^6 (a+x)^2}+\frac {2 \left (a^4+4 a^2 b^2+3 b^4\right )}{a^7 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac {\left (a^2+b^2\right ) \left (a^2+5 b^2\right ) \cot (c+d x)}{a^6 d}+\frac {2 b \left (a^2+b^2\right ) \cot ^2(c+d x)}{a^5 d}-\frac {\left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 a^4 d}+\frac {b \cot ^4(c+d x)}{2 a^3 d}-\frac {\cot ^5(c+d x)}{5 a^2 d}-\frac {2 b \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (\tan (c+d x))}{a^7 d}+\frac {2 b \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{a^7 d}-\frac {b \left (a^2+b^2\right )^2}{a^6 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 6.25, size = 589, normalized size = 2.69 \[ \frac {b \csc ^4(c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{2 a^3 d (a+b \tan (c+d x))^2}-\frac {\csc ^5(c+d x) \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{5 a^2 d (a+b \tan (c+d x))^2}+\frac {b \left (a^2+2 b^2\right ) \csc ^2(c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{a^5 d (a+b \tan (c+d x))^2}+\frac {\csc ^3(c+d x) \sec ^2(c+d x) \left (-4 a^2 \cos (c+d x)-15 b^2 \cos (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{15 a^4 d (a+b \tan (c+d x))^2}-\frac {2 \left (a^4 b+4 a^2 b^3+3 b^5\right ) \sec ^2(c+d x) \log (\sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^2}{a^7 d (a+b \tan (c+d x))^2}+\frac {2 \left (a^4 b+4 a^2 b^3+3 b^5\right ) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \log (a \cos (c+d x)+b \sin (c+d x))}{a^7 d (a+b \tan (c+d x))^2}+\frac {\sec ^2(c+d x) \left (a^4 b^2 \sin (c+d x)+2 a^2 b^4 \sin (c+d x)+b^6 \sin (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))}{a^7 d (a+b \tan (c+d x))^2}+\frac {\csc (c+d x) \sec ^2(c+d x) \left (-8 a^4 \cos (c+d x)-75 a^2 b^2 \cos (c+d x)-75 b^4 \cos (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{15 a^6 d (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(Csc[c + d*x]^5*Sec[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(a^2*d*(a + b*Tan[c + d*x])^2) + ((-8*a
^4*Cos[c + d*x] - 75*a^2*b^2*Cos[c + d*x] - 75*b^4*Cos[c + d*x])*Csc[c + d*x]*Sec[c + d*x]^2*(a*Cos[c + d*x] +
b*Sin[c + d*x])^2)/(15*a^6*d*(a + b*Tan[c + d*x])^2) + (b*(a^2 + 2*b^2)*Csc[c + d*x]^2*Sec[c + d*x]^2*(a*Cos[
c + d*x] + b*Sin[c + d*x])^2)/(a^5*d*(a + b*Tan[c + d*x])^2) + ((-4*a^2*Cos[c + d*x] - 15*b^2*Cos[c + d*x])*Cs
c[c + d*x]^3*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(15*a^4*d*(a + b*Tan[c + d*x])^2) + (b*Csc[c
+ d*x]^4*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(2*a^3*d*(a + b*Tan[c + d*x])^2) - (2*(a^4*b + 4*
a^2*b^3 + 3*b^5)*Log[Sin[c + d*x]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(a^7*d*(a + b*Tan[c + d
*x])^2) + (2*(a^4*b + 4*a^2*b^3 + 3*b^5)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]*Sec[c + d*x]^2*(a*Cos[c + d*x] +
b*Sin[c + d*x])^2)/(a^7*d*(a + b*Tan[c + d*x])^2) + (Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])*(a^4*b^
2*Sin[c + d*x] + 2*a^2*b^4*Sin[c + d*x] + b^6*Sin[c + d*x]))/(a^7*d*(a + b*Tan[c + d*x])^2)

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fricas [B]  time = 0.53, size = 787, normalized size = 3.59 \[ \frac {4 \, {\left (4 \, a^{6} + 45 \, a^{4} b^{2} + 45 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{6} - 75 \, a^{4} b^{2} - 90 \, a^{2} b^{4} - 10 \, {\left (4 \, a^{6} + 45 \, a^{4} b^{2} + 45 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (2 \, a^{6} + 23 \, a^{4} b^{2} + 24 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left ({\left (a^{4} b^{2} + 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4} - 3 \, b^{6} - 3 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (a^{5} b + 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{5} - 2 \, {\left (a^{5} b + 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{5} b + 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 30 \, {\left ({\left (a^{4} b^{2} + 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4} - 3 \, b^{6} - 3 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (a^{5} b + 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{5} - 2 \, {\left (a^{5} b + 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{5} b + 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) + {\left (4 \, {\left (4 \, a^{5} b + 45 \, a^{3} b^{3} + 45 \, a b^{5}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (a^{5} b + 33 \, a^{3} b^{3} + 36 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (a^{5} b - 10 \, a^{3} b^{3} - 12 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{7} b d \cos \left (d x + c\right )^{6} - 3 \, a^{7} b d \cos \left (d x + c\right )^{4} + 3 \, a^{7} b d \cos \left (d x + c\right )^{2} - a^{7} b d - {\left (a^{8} d \cos \left (d x + c\right )^{5} - 2 \, a^{8} d \cos \left (d x + c\right )^{3} + a^{8} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(4*(4*a^6 + 45*a^4*b^2 + 45*a^2*b^4)*cos(d*x + c)^6 - 75*a^4*b^2 - 90*a^2*b^4 - 10*(4*a^6 + 45*a^4*b^2 +
45*a^2*b^4)*cos(d*x + c)^4 + 15*(2*a^6 + 23*a^4*b^2 + 24*a^2*b^4)*cos(d*x + c)^2 + 30*((a^4*b^2 + 4*a^2*b^4 +
3*b^6)*cos(d*x + c)^6 - a^4*b^2 - 4*a^2*b^4 - 3*b^6 - 3*(a^4*b^2 + 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^4 + 3*(a^4*
b^2 + 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^2 - ((a^5*b + 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^5 - 2*(a^5*b + 4*a^3*b^3
+ 3*a*b^5)*cos(d*x + c)^3 + (a^5*b + 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c))*sin(d*x + c))*log(2*a*b*cos(d*x + c)*
sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 30*((a^4*b^2 + 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^6 - a^4*b^2
- 4*a^2*b^4 - 3*b^6 - 3*(a^4*b^2 + 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^4 + 3*(a^4*b^2 + 4*a^2*b^4 + 3*b^6)*cos(d*x
+ c)^2 - ((a^5*b + 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^5 - 2*(a^5*b + 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 + (a^
5*b + 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c))*sin(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1/4) + (4*(4*a^5*b + 45*a^3*b
^3 + 45*a*b^5)*cos(d*x + c)^5 - 10*(a^5*b + 33*a^3*b^3 + 36*a*b^5)*cos(d*x + c)^3 - 15*(a^5*b - 10*a^3*b^3 - 1
2*a*b^5)*cos(d*x + c))*sin(d*x + c))/(a^7*b*d*cos(d*x + c)^6 - 3*a^7*b*d*cos(d*x + c)^4 + 3*a^7*b*d*cos(d*x +
c)^2 - a^7*b*d - (a^8*d*cos(d*x + c)^5 - 2*a^8*d*cos(d*x + c)^3 + a^8*d*cos(d*x + c))*sin(d*x + c))

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giac [A]  time = 1.01, size = 332, normalized size = 1.52 \[ -\frac {\frac {60 \, {\left (a^{4} b + 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{7}} - \frac {60 \, {\left (a^{4} b^{2} + 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{7} b} + \frac {30 \, {\left (2 \, a^{4} b^{2} \tan \left (d x + c\right ) + 8 \, a^{2} b^{4} \tan \left (d x + c\right ) + 6 \, b^{6} \tan \left (d x + c\right ) + 3 \, a^{5} b + 10 \, a^{3} b^{3} + 7 \, a b^{5}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )} a^{7}} - \frac {137 \, a^{4} b \tan \left (d x + c\right )^{5} + 548 \, a^{2} b^{3} \tan \left (d x + c\right )^{5} + 411 \, b^{5} \tan \left (d x + c\right )^{5} - 30 \, a^{5} \tan \left (d x + c\right )^{4} - 180 \, a^{3} b^{2} \tan \left (d x + c\right )^{4} - 150 \, a b^{4} \tan \left (d x + c\right )^{4} + 60 \, a^{4} b \tan \left (d x + c\right )^{3} + 60 \, a^{2} b^{3} \tan \left (d x + c\right )^{3} - 20 \, a^{5} \tan \left (d x + c\right )^{2} - 30 \, a^{3} b^{2} \tan \left (d x + c\right )^{2} + 15 \, a^{4} b \tan \left (d x + c\right ) - 6 \, a^{5}}{a^{7} \tan \left (d x + c\right )^{5}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(60*(a^4*b + 4*a^2*b^3 + 3*b^5)*log(abs(tan(d*x + c)))/a^7 - 60*(a^4*b^2 + 4*a^2*b^4 + 3*b^6)*log(abs(b*
tan(d*x + c) + a))/(a^7*b) + 30*(2*a^4*b^2*tan(d*x + c) + 8*a^2*b^4*tan(d*x + c) + 6*b^6*tan(d*x + c) + 3*a^5*
b + 10*a^3*b^3 + 7*a*b^5)/((b*tan(d*x + c) + a)*a^7) - (137*a^4*b*tan(d*x + c)^5 + 548*a^2*b^3*tan(d*x + c)^5
+ 411*b^5*tan(d*x + c)^5 - 30*a^5*tan(d*x + c)^4 - 180*a^3*b^2*tan(d*x + c)^4 - 150*a*b^4*tan(d*x + c)^4 + 60*
a^4*b*tan(d*x + c)^3 + 60*a^2*b^3*tan(d*x + c)^3 - 20*a^5*tan(d*x + c)^2 - 30*a^3*b^2*tan(d*x + c)^2 + 15*a^4*
b*tan(d*x + c) - 6*a^5)/(a^7*tan(d*x + c)^5))/d

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maple [A]  time = 0.52, size = 343, normalized size = 1.57 \[ -\frac {b}{a^{2} d \left (a +b \tan \left (d x +c \right )\right )}-\frac {2 b^{3}}{d \,a^{4} \left (a +b \tan \left (d x +c \right )\right )}-\frac {b^{5}}{d \,a^{6} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{3} d}+\frac {8 b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \,a^{5}}+\frac {6 b^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \,a^{7}}-\frac {1}{5 d \,a^{2} \tan \left (d x +c \right )^{5}}-\frac {2}{3 d \,a^{2} \tan \left (d x +c \right )^{3}}-\frac {b^{2}}{d \,a^{4} \tan \left (d x +c \right )^{3}}-\frac {1}{d \,a^{2} \tan \left (d x +c \right )}-\frac {6 b^{2}}{d \,a^{4} \tan \left (d x +c \right )}-\frac {5 b^{4}}{d \,a^{6} \tan \left (d x +c \right )}+\frac {b}{2 d \,a^{3} \tan \left (d x +c \right )^{4}}+\frac {2 b}{d \,a^{3} \tan \left (d x +c \right )^{2}}+\frac {2 b^{3}}{d \,a^{5} \tan \left (d x +c \right )^{2}}-\frac {2 b \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}-\frac {8 b^{3} \ln \left (\tan \left (d x +c \right )\right )}{d \,a^{5}}-\frac {6 b^{5} \ln \left (\tan \left (d x +c \right )\right )}{d \,a^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^6/(a+b*tan(d*x+c))^2,x)

[Out]

-b/a^2/d/(a+b*tan(d*x+c))-2/d*b^3/a^4/(a+b*tan(d*x+c))-1/d*b^5/a^6/(a+b*tan(d*x+c))+2*b*ln(a+b*tan(d*x+c))/a^3
/d+8/d*b^3/a^5*ln(a+b*tan(d*x+c))+6/d*b^5/a^7*ln(a+b*tan(d*x+c))-1/5/d/a^2/tan(d*x+c)^5-2/3/d/a^2/tan(d*x+c)^3
-1/d/a^4/tan(d*x+c)^3*b^2-1/d/a^2/tan(d*x+c)-6/d/a^4/tan(d*x+c)*b^2-5/d/a^6/tan(d*x+c)*b^4+1/2/d/a^3*b/tan(d*x
+c)^4+2/d/a^3*b/tan(d*x+c)^2+2/d*b^3/a^5/tan(d*x+c)^2-2*b*ln(tan(d*x+c))/a^3/d-8/d*b^3/a^5*ln(tan(d*x+c))-6/d*
b^5/a^7*ln(tan(d*x+c))

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maxima [A]  time = 0.60, size = 225, normalized size = 1.03 \[ \frac {\frac {9 \, a^{4} b \tan \left (d x + c\right ) - 60 \, {\left (a^{4} b + 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \tan \left (d x + c\right )^{5} - 6 \, a^{5} - 30 \, {\left (a^{5} + 4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \tan \left (d x + c\right )^{4} + 10 \, {\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \tan \left (d x + c\right )^{3} - 5 \, {\left (4 \, a^{5} + 3 \, a^{3} b^{2}\right )} \tan \left (d x + c\right )^{2}}{a^{6} b \tan \left (d x + c\right )^{6} + a^{7} \tan \left (d x + c\right )^{5}} + \frac {60 \, {\left (a^{4} b + 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{7}} - \frac {60 \, {\left (a^{4} b + 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{7}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*((9*a^4*b*tan(d*x + c) - 60*(a^4*b + 4*a^2*b^3 + 3*b^5)*tan(d*x + c)^5 - 6*a^5 - 30*(a^5 + 4*a^3*b^2 + 3*
a*b^4)*tan(d*x + c)^4 + 10*(4*a^4*b + 3*a^2*b^3)*tan(d*x + c)^3 - 5*(4*a^5 + 3*a^3*b^2)*tan(d*x + c)^2)/(a^6*b
*tan(d*x + c)^6 + a^7*tan(d*x + c)^5) + 60*(a^4*b + 4*a^2*b^3 + 3*b^5)*log(b*tan(d*x + c) + a)/a^7 - 60*(a^4*b
+ 4*a^2*b^3 + 3*b^5)*log(tan(d*x + c))/a^7)/d

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mupad [B]  time = 5.01, size = 237, normalized size = 1.08 \[ \frac {4\,b\,\mathrm {atanh}\left (\frac {2\,b\,\left (a^2+3\,b^2\right )\,\left (a^2+b^2\right )\,\left (a+2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,\left (2\,a^4\,b+8\,a^2\,b^3+6\,b^5\right )}\right )\,\left (a^2+3\,b^2\right )\,\left (a^2+b^2\right )}{a^7\,d}-\frac {\frac {1}{5\,a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^4+4\,a^2\,b^2+3\,b^4\right )}{a^5}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (4\,a^2+3\,b^2\right )}{6\,a^3}-\frac {3\,b\,\mathrm {tan}\left (c+d\,x\right )}{10\,a^2}+\frac {2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (a^4+4\,a^2\,b^2+3\,b^4\right )}{a^6}-\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (4\,a^2+3\,b^2\right )}{3\,a^4}}{d\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^6+a\,{\mathrm {tan}\left (c+d\,x\right )}^5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*b*atanh((2*b*(a^2 + 3*b^2)*(a^2 + b^2)*(a + 2*b*tan(c + d*x)))/(a*(2*a^4*b + 6*b^5 + 8*a^2*b^3)))*(a^2 + 3*
b^2)*(a^2 + b^2))/(a^7*d) - (1/(5*a) + (tan(c + d*x)^4*(a^4 + 3*b^4 + 4*a^2*b^2))/a^5 + (tan(c + d*x)^2*(4*a^2
+ 3*b^2))/(6*a^3) - (3*b*tan(c + d*x))/(10*a^2) + (2*b*tan(c + d*x)^5*(a^4 + 3*b^4 + 4*a^2*b^2))/a^6 - (b*tan
(c + d*x)^3*(4*a^2 + 3*b^2))/(3*a^4))/(d*(a*tan(c + d*x)^5 + b*tan(c + d*x)^6))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{6}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**6/(a+b*tan(d*x+c))**2,x)

[Out]

Integral(csc(c + d*x)**6/(a + b*tan(c + d*x))**2, x)

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