\(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\) [31]

Optimal result

Integrand size = 34, antiderivative size = 156 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-8 a^4 (i A+B) x-\frac {a^4 (A-4 i B) \log (\cos (c+d x))}{d}-\frac {a^4 (7 A-4 i B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac {(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d} \]

[Out]

-8*a^4*(I*A+B)*x-a^4*(A-4*I*B)*ln(cos(d*x+c))/d-a^4*(7*A-4*I*B)*ln(sin(d*x+c))/d-1/2*a*A*cot(d*x+c)^2*(a+I*a*t
an(d*x+c))^3/d-1/2*(5*I*A+2*B)*cot(d*x+c)*(a^2+I*a^2*tan(d*x+c))^2/d-3*A*(a^4+I*a^4*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3674, 3675, 3670, 3556, 3612} \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {a^4 (7 A-4 i B) \log (\sin (c+d x))}{d}-\frac {a^4 (A-4 i B) \log (\cos (c+d x))}{d}-8 a^4 x (B+i A)-\frac {3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\frac {(2 B+5 i A) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d} \]

[In]

Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

-8*a^4*(I*A + B)*x - (a^4*(A - (4*I)*B)*Log[Cos[c + d*x]])/d - (a^4*(7*A - (4*I)*B)*Log[Sin[c + d*x]])/d - (a*
A*Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^3)/(2*d) - (((5*I)*A + 2*B)*Cot[c + d*x]*(a^2 + I*a^2*Tan[c + d*x])^2)
/(2*d) - (3*A*(a^4 + I*a^4*Tan[c + d*x]))/d

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3612

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c +
b*d)*(x/(a^2 + b^2)), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3670

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[B*(d/b), Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0]

Rule 3674

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x]
)^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c
+ d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m
- 1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3675

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}+\frac {1}{2} \int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 (a (5 i A+2 B)+a (A+2 i B) \tan (c+d x)) \, dx \\ & = -\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac {(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac {1}{2} \int \cot (c+d x) (a+i a \tan (c+d x))^2 \left (-2 a^2 (7 A-4 i B)+6 i a^2 A \tan (c+d x)\right ) \, dx \\ & = -\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac {(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {1}{2} \int \cot (c+d x) (a+i a \tan (c+d x)) \left (-2 a^3 (7 A-4 i B)-2 a^3 (i A+4 B) \tan (c+d x)\right ) \, dx \\ & = -\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac {(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {1}{2} \int \cot (c+d x) \left (-2 a^4 (7 A-4 i B)-16 a^4 (i A+B) \tan (c+d x)\right ) \, dx+\left (a^4 (A-4 i B)\right ) \int \tan (c+d x) \, dx \\ & = -8 a^4 (i A+B) x-\frac {a^4 (A-4 i B) \log (\cos (c+d x))}{d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac {(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\left (a^4 (7 A-4 i B)\right ) \int \cot (c+d x) \, dx \\ & = -8 a^4 (i A+B) x-\frac {a^4 (A-4 i B) \log (\cos (c+d x))}{d}-\frac {a^4 (7 A-4 i B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^3}{2 d}-\frac {(5 i A+2 B) \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 A \left (a^4+i a^4 \tan (c+d x)\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.85 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.78 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=a^4 \left (-\frac {4 i A \cot (c+d x)}{d}-\frac {B \cot (c+d x)}{d}-\frac {A \cot ^2(c+d x)}{2 d}-\frac {7 A \log (\tan (c+d x))}{d}+\frac {4 i B \log (\tan (c+d x))}{d}+\frac {8 A \log (i+\tan (c+d x))}{d}-\frac {8 i B \log (i+\tan (c+d x))}{d}+\frac {B \tan (c+d x)}{d}\right ) \]

[In]

Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

a^4*(((-4*I)*A*Cot[c + d*x])/d - (B*Cot[c + d*x])/d - (A*Cot[c + d*x]^2)/(2*d) - (7*A*Log[Tan[c + d*x]])/d + (
(4*I)*B*Log[Tan[c + d*x]])/d + (8*A*Log[I + Tan[c + d*x]])/d - ((8*I)*B*Log[I + Tan[c + d*x]])/d + (B*Tan[c +
d*x])/d)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(-\frac {a^{4} \left (16 i A x d +A \left (\cot ^{2}\left (d x +c \right )\right )+8 i A \cot \left (d x +c \right )-8 i B \ln \left (\tan \left (d x +c \right )\right )+8 i B \ln \left (\sec ^{2}\left (d x +c \right )\right )+16 B d x +14 A \ln \left (\tan \left (d x +c \right )\right )-8 A \ln \left (\sec ^{2}\left (d x +c \right )\right )+2 \cot \left (d x +c \right ) B -2 B \tan \left (d x +c \right )\right )}{2 d}\)	\(105\)
derivativedivides	\(\frac {a^{4} \left (-\frac {A \left (\cot ^{2}\left (d x +c \right )\right )}{2}-4 i A \cot \left (d x +c \right )-\cot \left (d x +c \right ) B +\frac {\left (-8 i B +8 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (8 i A +8 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )+\left (4 i B -A \right ) \ln \left (\cot \left (d x +c \right )\right )+\frac {B}{\cot \left (d x +c \right )}\right )}{d}\)	\(108\)
default	\(\frac {a^{4} \left (-\frac {A \left (\cot ^{2}\left (d x +c \right )\right )}{2}-4 i A \cot \left (d x +c \right )-\cot \left (d x +c \right ) B +\frac {\left (-8 i B +8 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (8 i A +8 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )+\left (4 i B -A \right ) \ln \left (\cot \left (d x +c \right )\right )+\frac {B}{\cot \left (d x +c \right )}\right )}{d}\)	\(108\)
norman	\(\frac {\left (-8 i A \,a^{4}-8 B \,a^{4}\right ) x \left (\tan ^{2}\left (d x +c \right )\right )+\frac {B \,a^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{d}-\frac {A \,a^{4}}{2 d}-\frac {\left (4 i A \,a^{4}+B \,a^{4}\right ) \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}-\frac {\left (-4 i B \,a^{4}+7 A \,a^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {4 \left (-i B \,a^{4}+A \,a^{4}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(140\)
risch	\(\frac {16 a^{4} B c}{d}+\frac {16 i a^{4} A c}{d}-\frac {2 i a^{4} \left (5 i A \,{\mathrm e}^{4 i \left (d x +c \right )}+i A \,{\mathrm e}^{2 i \left (d x +c \right )}+2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-4 i A -2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {4 i a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}-\frac {7 A \,a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {4 i a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{d}\)	\(190\)

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/2*a^4*(16*I*A*x*d+A*cot(d*x+c)^2+8*I*A*cot(d*x+c)-8*I*B*ln(tan(d*x+c))+8*I*B*ln(sec(d*x+c)^2)+16*B*d*x+14*A
*ln(tan(d*x+c))-8*A*ln(sec(d*x+c)^2)+2*cot(d*x+c)*B-2*B*tan(d*x+c))/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.63 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {10 \, A a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (A - 2 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, {\left (2 \, A - i \, B\right )} a^{4} - {\left ({\left (A - 4 i \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (A - 4 i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (A - 4 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - 4 i \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - {\left ({\left (7 \, A - 4 i \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (7 \, A - 4 i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (7 \, A - 4 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (7 \, A - 4 i \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (6 i \, d x + 6 i \, c\right )} - d e^{\left (4 i \, d x + 4 i \, c\right )} - d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

(10*A*a^4*e^(4*I*d*x + 4*I*c) + 2*(A - 2*I*B)*a^4*e^(2*I*d*x + 2*I*c) - 4*(2*A - I*B)*a^4 - ((A - 4*I*B)*a^4*e
^(6*I*d*x + 6*I*c) - (A - 4*I*B)*a^4*e^(4*I*d*x + 4*I*c) - (A - 4*I*B)*a^4*e^(2*I*d*x + 2*I*c) + (A - 4*I*B)*a
^4)*log(e^(2*I*d*x + 2*I*c) + 1) - ((7*A - 4*I*B)*a^4*e^(6*I*d*x + 6*I*c) - (7*A - 4*I*B)*a^4*e^(4*I*d*x + 4*I
*c) - (7*A - 4*I*B)*a^4*e^(2*I*d*x + 2*I*c) + (7*A - 4*I*B)*a^4)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(6*I*d*x +
6*I*c) - d*e^(4*I*d*x + 4*I*c) - d*e^(2*I*d*x + 2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.62 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=- \frac {a^{4} \left (A - 4 i B\right ) \log {\left (e^{2 i d x} + \frac {\left (4 A a^{4} - 4 i B a^{4} - a^{4} \left (A - 4 i B\right )\right ) e^{- 2 i c}}{3 A a^{4}} \right )}}{d} - \frac {a^{4} \cdot \left (7 A - 4 i B\right ) \log {\left (e^{2 i d x} + \frac {\left (4 A a^{4} - 4 i B a^{4} - a^{4} \cdot \left (7 A - 4 i B\right )\right ) e^{- 2 i c}}{3 A a^{4}} \right )}}{d} + \frac {10 A a^{4} e^{4 i c} e^{4 i d x} - 8 A a^{4} + 4 i B a^{4} + \left (2 A a^{4} e^{2 i c} - 4 i B a^{4} e^{2 i c}\right ) e^{2 i d x}}{d e^{6 i c} e^{6 i d x} - d e^{4 i c} e^{4 i d x} - d e^{2 i c} e^{2 i d x} + d} \]

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

-a**4*(A - 4*I*B)*log(exp(2*I*d*x) + (4*A*a**4 - 4*I*B*a**4 - a**4*(A - 4*I*B))*exp(-2*I*c)/(3*A*a**4))/d - a*
*4*(7*A - 4*I*B)*log(exp(2*I*d*x) + (4*A*a**4 - 4*I*B*a**4 - a**4*(7*A - 4*I*B))*exp(-2*I*c)/(3*A*a**4))/d + (
10*A*a**4*exp(4*I*c)*exp(4*I*d*x) - 8*A*a**4 + 4*I*B*a**4 + (2*A*a**4*exp(2*I*c) - 4*I*B*a**4*exp(2*I*c))*exp(
2*I*d*x))/(d*exp(6*I*c)*exp(6*I*d*x) - d*exp(4*I*c)*exp(4*I*d*x) - d*exp(2*I*c)*exp(2*I*d*x) + d)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.69 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {16 \, {\left (d x + c\right )} {\left (i \, A + B\right )} a^{4} - 8 \, {\left (A - i \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (7 \, A - 4 i \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )\right ) - 2 \, B a^{4} \tan \left (d x + c\right ) - \frac {2 \, {\left (-4 i \, A - B\right )} a^{4} \tan \left (d x + c\right ) - A a^{4}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(16*(d*x + c)*(I*A + B)*a^4 - 8*(A - I*B)*a^4*log(tan(d*x + c)^2 + 1) + 2*(7*A - 4*I*B)*a^4*log(tan(d*x +
c)) - 2*B*a^4*tan(d*x + c) - (2*(-4*I*A - B)*a^4*tan(d*x + c) - A*a^4)/tan(d*x + c)^2)/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (138) = 276\).

Time = 0.90 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.03 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, {\left (A a^{4} - 4 i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 128 \, {\left (A a^{4} - i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 8 \, {\left (A a^{4} - 4 i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + 8 \, {\left (7 \, A a^{4} - 4 i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {8 \, {\left (A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{4} + 4 i \, B a^{4}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {84 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{4}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(A*a^4*tan(1/2*d*x + 1/2*c)^2 - 16*I*A*a^4*tan(1/2*d*x + 1/2*c) - 4*B*a^4*tan(1/2*d*x + 1/2*c) + 8*(A*a^4
- 4*I*B*a^4)*log(tan(1/2*d*x + 1/2*c) + 1) - 128*(A*a^4 - I*B*a^4)*log(tan(1/2*d*x + 1/2*c) + I) + 8*(A*a^4 -
4*I*B*a^4)*log(tan(1/2*d*x + 1/2*c) - 1) + 8*(7*A*a^4 - 4*I*B*a^4)*log(tan(1/2*d*x + 1/2*c)) - 8*(A*a^4*tan(1
/2*d*x + 1/2*c)^2 - 4*I*B*a^4*tan(1/2*d*x + 1/2*c)^2 - 2*B*a^4*tan(1/2*d*x + 1/2*c) - A*a^4 + 4*I*B*a^4)/(tan(
1/2*d*x + 1/2*c)^2 - 1) - (84*A*a^4*tan(1/2*d*x + 1/2*c)^2 - 48*I*B*a^4*tan(1/2*d*x + 1/2*c)^2 - 16*I*A*a^4*ta
n(1/2*d*x + 1/2*c) - 4*B*a^4*tan(1/2*d*x + 1/2*c) - A*a^4)/tan(1/2*d*x + 1/2*c)^2)/d

Mupad [B] (verification not implemented)

Time = 7.67 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.65 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {B\,a^4\,\mathrm {tan}\left (c+d\,x\right )}{d}-\frac {a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (7\,A-B\,4{}\mathrm {i}\right )}{d}+\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {A\,a^4}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^4+A\,a^4\,4{}\mathrm {i}\right )\right )}{d} \]

[In]

int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(8*a^4*log(tan(c + d*x) + 1i)*(A - B*1i))/d - (a^4*log(tan(c + d*x))*(7*A - B*4i))/d - (cot(c + d*x)^2*((A*a^4
)/2 + tan(c + d*x)*(A*a^4*4i + B*a^4)))/d + (B*a^4*tan(c + d*x))/d