\(\int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx\) [40]

Optimal result

Integrand size = 21, antiderivative size = 167 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {a \left (a^2+6 b^2\right ) \cot (c+d x)}{d}-\frac {b \left (6 a^2+b^2\right ) \cot ^2(c+d x)}{2 d}-\frac {a \left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {3 a^2 b \cot ^4(c+d x)}{4 d}-\frac {a^3 \cot ^5(c+d x)}{5 d}+\frac {b \left (3 a^2+2 b^2\right ) \log (\tan (c+d x))}{d}+\frac {3 a b^2 \tan (c+d x)}{d}+\frac {b^3 \tan ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3597, 962} \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {a^3 \cot ^5(c+d x)}{5 d}-\frac {a \left (2 a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {b \left (6 a^2+b^2\right ) \cot ^2(c+d x)}{2 d}-\frac {a \left (a^2+6 b^2\right ) \cot (c+d x)}{d}+\frac {b \left (3 a^2+2 b^2\right ) \log (\tan (c+d x))}{d}-\frac {3 a^2 b \cot ^4(c+d x)}{4 d}+\frac {3 a b^2 \tan (c+d x)}{d}+\frac {b^3 \tan ^2(c+d x)}{2 d} \]

[In]

Int[Csc[c + d*x]^6*(a + b*Tan[c + d*x])^3,x]

[Out]

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

Rule 962

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] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rule 3597

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(515\) vs. \(2(167)=334\).

Time = 3.55 (sec) , antiderivative size = 515, normalized size of antiderivative = 3.08 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {\csc ^5(c+d x) \sec ^2(c+d x) \left (40 a \left (5 a^2+3 b^2\right ) \cos (c+d x)+8 \left (a^3+15 a b^2\right ) \cos (3 (c+d x))-24 a^3 \cos (5 (c+d x))-360 a b^2 \cos (5 (c+d x))+8 a^3 \cos (7 (c+d x))+120 a b^2 \cos (7 (c+d x))+360 a^2 b \sin (c+d x)-240 b^3 \sin (c+d x)+225 a^2 b \log (\cos (c+d x)) \sin (c+d x)+150 b^3 \log (\cos (c+d x)) \sin (c+d x)-225 a^2 b \log (\sin (c+d x)) \sin (c+d x)-150 b^3 \log (\sin (c+d x)) \sin (c+d x)+270 a^2 b \sin (3 (c+d x))+180 b^3 \sin (3 (c+d x))+45 a^2 b \log (\cos (c+d x)) \sin (3 (c+d x))+30 b^3 \log (\cos (c+d x)) \sin (3 (c+d x))-45 a^2 b \log (\sin (c+d x)) \sin (3 (c+d x))-30 b^3 \log (\sin (c+d x)) \sin (3 (c+d x))-90 a^2 b \sin (5 (c+d x))-60 b^3 \sin (5 (c+d x))-135 a^2 b \log (\cos (c+d x)) \sin (5 (c+d x))-90 b^3 \log (\cos (c+d x)) \sin (5 (c+d x))+135 a^2 b \log (\sin (c+d x)) \sin (5 (c+d x))+90 b^3 \log (\sin (c+d x)) \sin (5 (c+d x))+45 a^2 b \log (\cos (c+d x)) \sin (7 (c+d x))+30 b^3 \log (\cos (c+d x)) \sin (7 (c+d x))-45 a^2 b \log (\sin (c+d x)) \sin (7 (c+d x))-30 b^3 \log (\sin (c+d x)) \sin (7 (c+d x))\right )}{960 d} \]

[In]

Integrate[Csc[c + d*x]^6*(a + b*Tan[c + d*x])^3,x]

[Out]

-1/960*(Csc[c + d*x]^5*Sec[c + d*x]^2*(40*a*(5*a^2 + 3*b^2)*Cos[c + d*x] + 8*(a^3 + 15*a*b^2)*Cos[3*(c + d*x)]
- 24*a^3*Cos[5*(c + d*x)] - 360*a*b^2*Cos[5*(c + d*x)] + 8*a^3*Cos[7*(c + d*x)] + 120*a*b^2*Cos[7*(c + d*x)]
+ 360*a^2*b*Sin[c + d*x] - 240*b^3*Sin[c + d*x] + 225*a^2*b*Log[Cos[c + d*x]]*Sin[c + d*x] + 150*b^3*Log[Cos[c
+ d*x]]*Sin[c + d*x] - 225*a^2*b*Log[Sin[c + d*x]]*Sin[c + d*x] - 150*b^3*Log[Sin[c + d*x]]*Sin[c + d*x] + 27
0*a^2*b*Sin[3*(c + d*x)] + 180*b^3*Sin[3*(c + d*x)] + 45*a^2*b*Log[Cos[c + d*x]]*Sin[3*(c + d*x)] + 30*b^3*Log
[Cos[c + d*x]]*Sin[3*(c + d*x)] - 45*a^2*b*Log[Sin[c + d*x]]*Sin[3*(c + d*x)] - 30*b^3*Log[Sin[c + d*x]]*Sin[3
*(c + d*x)] - 90*a^2*b*Sin[5*(c + d*x)] - 60*b^3*Sin[5*(c + d*x)] - 135*a^2*b*Log[Cos[c + d*x]]*Sin[5*(c + d*x
)] - 90*b^3*Log[Cos[c + d*x]]*Sin[5*(c + d*x)] + 135*a^2*b*Log[Sin[c + d*x]]*Sin[5*(c + d*x)] + 90*b^3*Log[Sin
[c + d*x]]*Sin[5*(c + d*x)] + 45*a^2*b*Log[Cos[c + d*x]]*Sin[7*(c + d*x)] + 30*b^3*Log[Cos[c + d*x]]*Sin[7*(c
+ d*x)] - 45*a^2*b*Log[Sin[c + d*x]]*Sin[7*(c + d*x)] - 30*b^3*Log[Sin[c + d*x]]*Sin[7*(c + d*x)]))/d

Maple [A] (verified)

Time = 19.52 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+3 a^{2} b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\)	\(165\)
default	\(\frac {b^{3} \left (\frac {1}{2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {1}{\sin \left (d x +c \right )^{2}}+2 \ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+3 a^{2} b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\)	\(165\)
risch	\(\frac {6 a^{2} b \,{\mathrm e}^{12 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{12 i \left (d x +c \right )}-18 a^{2} b \,{\mathrm e}^{10 i \left (d x +c \right )}-12 b^{3} {\mathrm e}^{10 i \left (d x +c \right )}+\frac {16 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{5}+32 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-24 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-16 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+24 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-16 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-\frac {16 i a^{3}}{15}-16 i a \,b^{2}+18 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+12 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-16 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-48 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-6 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-\frac {16 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}}{15}-\frac {32 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}}{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(427\)

[In]

int(csc(d*x+c)^6*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(b^3*(1/2/sin(d*x+c)^2/cos(d*x+c)^2-1/sin(d*x+c)^2+2*ln(tan(d*x+c)))+3*a*b^2*(-1/3/sin(d*x+c)^3/cos(d*x+c)
+4/3/sin(d*x+c)/cos(d*x+c)-8/3*cot(d*x+c))+3*a^2*b*(-1/4/sin(d*x+c)^4-1/2/sin(d*x+c)^2+ln(tan(d*x+c)))+a^3*(-8
/15-1/5*csc(d*x+c)^4-4/15*csc(d*x+c)^2)*cot(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (157) = 314\).

Time = 0.28 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.05 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {32 \, {\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} - 80 \, {\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 180 \, a b^{2} \cos \left (d x + c\right ) + 60 \, {\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \, {\left ({\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left ({\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) - 15 \, {\left (2 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 2 \, b^{3} - 3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{6} - 2 \, d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(32*(a^3 + 15*a*b^2)*cos(d*x + c)^7 - 80*(a^3 + 15*a*b^2)*cos(d*x + c)^5 - 180*a*b^2*cos(d*x + c) + 60*(
a^3 + 15*a*b^2)*cos(d*x + c)^3 + 30*((3*a^2*b + 2*b^3)*cos(d*x + c)^6 - 2*(3*a^2*b + 2*b^3)*cos(d*x + c)^4 + (
3*a^2*b + 2*b^3)*cos(d*x + c)^2)*log(cos(d*x + c)^2)*sin(d*x + c) - 30*((3*a^2*b + 2*b^3)*cos(d*x + c)^6 - 2*(
3*a^2*b + 2*b^3)*cos(d*x + c)^4 + (3*a^2*b + 2*b^3)*cos(d*x + c)^2)*log(-1/4*cos(d*x + c)^2 + 1/4)*sin(d*x + c
) - 15*(2*(3*a^2*b + 2*b^3)*cos(d*x + c)^4 + 2*b^3 - 3*(3*a^2*b + 2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((d*cos
(d*x + c)^6 - 2*d*cos(d*x + c)^4 + d*cos(d*x + c)^2)*sin(d*x + c))

Sympy [F]

\[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \csc ^{6}{\left (c + d x \right )}\, dx \]

[In]

integrate(csc(d*x+c)**6*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*csc(c + d*x)**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.85 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {30 \, b^{3} \tan \left (d x + c\right )^{2} + 180 \, a b^{2} \tan \left (d x + c\right ) + 60 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {60 \, {\left (a^{3} + 6 \, a b^{2}\right )} \tan \left (d x + c\right )^{4} + 45 \, a^{2} b \tan \left (d x + c\right ) + 30 \, {\left (6 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{3} + 12 \, a^{3} + 20 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(30*b^3*tan(d*x + c)^2 + 180*a*b^2*tan(d*x + c) + 60*(3*a^2*b + 2*b^3)*log(tan(d*x + c)) - (60*(a^3 + 6*a
*b^2)*tan(d*x + c)^4 + 45*a^2*b*tan(d*x + c) + 30*(6*a^2*b + b^3)*tan(d*x + c)^3 + 12*a^3 + 20*(2*a^3 + 3*a*b^
2)*tan(d*x + c)^2)/tan(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.77 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.13 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {30 \, b^{3} \tan \left (d x + c\right )^{2} + 180 \, a b^{2} \tan \left (d x + c\right ) + 60 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {411 \, a^{2} b \tan \left (d x + c\right )^{5} + 274 \, b^{3} \tan \left (d x + c\right )^{5} + 60 \, a^{3} \tan \left (d x + c\right )^{4} + 360 \, a b^{2} \tan \left (d x + c\right )^{4} + 180 \, a^{2} b \tan \left (d x + c\right )^{3} + 30 \, b^{3} \tan \left (d x + c\right )^{3} + 40 \, a^{3} \tan \left (d x + c\right )^{2} + 60 \, a b^{2} \tan \left (d x + c\right )^{2} + 45 \, a^{2} b \tan \left (d x + c\right ) + 12 \, a^{3}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*b^3*tan(d*x + c)^2 + 180*a*b^2*tan(d*x + c) + 60*(3*a^2*b + 2*b^3)*log(abs(tan(d*x + c))) - (411*a^2*
b*tan(d*x + c)^5 + 274*b^3*tan(d*x + c)^5 + 60*a^3*tan(d*x + c)^4 + 360*a*b^2*tan(d*x + c)^4 + 180*a^2*b*tan(d
*x + c)^3 + 30*b^3*tan(d*x + c)^3 + 40*a^3*tan(d*x + c)^2 + 60*a*b^2*tan(d*x + c)^2 + 45*a^2*b*tan(d*x + c) +
12*a^3)/tan(d*x + c)^5)/d

Mupad [B] (verification not implemented)

Time = 4.32 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.87 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (3\,a^2\,b+2\,b^3\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^5\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {2\,a^3}{3}+a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,a^2\,b+\frac {b^3}{2}\right )+\frac {a^3}{5}+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^3+6\,a\,b^2\right )+\frac {3\,a^2\,b\,\mathrm {tan}\left (c+d\,x\right )}{4}\right )}{d}+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}+\frac {3\,a\,b^2\,\mathrm {tan}\left (c+d\,x\right )}{d} \]

[In]

int((a + b*tan(c + d*x))^3/sin(c + d*x)^6,x)

[Out]

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