3.34 \(\int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=200 \[ \frac{a^4 (145 B+148 i A) \cot ^2(c+d x)}{60 d}-\frac{8 a^4 (A-i B) \cot (c+d x)}{d}+\frac{8 a^4 (B+i A) \log (\sin (c+d x))}{d}-\frac{(5 B+8 i A) \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{(28 A-25 i B) \cot ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{30 d}-8 a^4 x (A-i B)-\frac{a A \cot ^5(c+d x) (a+i a \tan (c+d x))^3}{5 d} \]

[Out]

-8*a^4*(A - I*B)*x - (8*a^4*(A - I*B)*Cot[c + d*x])/d + (a^4*((148*I)*A + 145*B)*Cot[c + d*x]^2)/(60*d) + (8*a
^4*(I*A + B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^3)/(5*d) - (((8*I)*A + 5*B)*Cot
[c + d*x]^4*(a^2 + I*a^2*Tan[c + d*x])^2)/(20*d) + ((28*A - (25*I)*B)*Cot[c + d*x]^3*(a^4 + I*a^4*Tan[c + d*x]
))/(30*d)

________________________________________________________________________________________

Rubi [A]  time = 0.591938, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3593, 3591, 3529, 3531, 3475} \[ \frac{a^4 (145 B+148 i A) \cot ^2(c+d x)}{60 d}-\frac{8 a^4 (A-i B) \cot (c+d x)}{d}+\frac{8 a^4 (B+i A) \log (\sin (c+d x))}{d}-\frac{(5 B+8 i A) \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{(28 A-25 i B) \cot ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{30 d}-8 a^4 x (A-i B)-\frac{a A \cot ^5(c+d x) (a+i a \tan (c+d x))^3}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

-8*a^4*(A - I*B)*x - (8*a^4*(A - I*B)*Cot[c + d*x])/d + (a^4*((148*I)*A + 145*B)*Cot[c + d*x]^2)/(60*d) + (8*a
^4*(I*A + B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^3)/(5*d) - (((8*I)*A + 5*B)*Cot
[c + d*x]^4*(a^2 + I*a^2*Tan[c + d*x])^2)/(20*d) + ((28*A - (25*I)*B)*Cot[c + d*x]^3*(a^4 + I*a^4*Tan[c + d*x]
))/(30*d)

Rule 3593

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] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3591

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

Rule 3529

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

Rule 3531

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 3475

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

Rubi steps

\begin{align*} \int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot ^5(c+d x) (a+i a \tan (c+d x))^3}{5 d}+\frac{1}{5} \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (a (8 i A+5 B)-a (2 A-5 i B) \tan (c+d x)) \, dx\\ &=-\frac{a A \cot ^5(c+d x) (a+i a \tan (c+d x))^3}{5 d}-\frac{(8 i A+5 B) \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{1}{20} \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \left (-2 a^2 (28 A-25 i B)-6 a^2 (4 i A+5 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{a A \cot ^5(c+d x) (a+i a \tan (c+d x))^3}{5 d}-\frac{(8 i A+5 B) \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{(28 A-25 i B) \cot ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{30 d}+\frac{1}{60} \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \left (-2 a^3 (148 i A+145 B)+2 a^3 (92 A-95 i B) \tan (c+d x)\right ) \, dx\\ &=\frac{a^4 (148 i A+145 B) \cot ^2(c+d x)}{60 d}-\frac{a A \cot ^5(c+d x) (a+i a \tan (c+d x))^3}{5 d}-\frac{(8 i A+5 B) \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{(28 A-25 i B) \cot ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{30 d}+\frac{1}{60} \int \cot ^2(c+d x) \left (480 a^4 (A-i B)+480 a^4 (i A+B) \tan (c+d x)\right ) \, dx\\ &=-\frac{8 a^4 (A-i B) \cot (c+d x)}{d}+\frac{a^4 (148 i A+145 B) \cot ^2(c+d x)}{60 d}-\frac{a A \cot ^5(c+d x) (a+i a \tan (c+d x))^3}{5 d}-\frac{(8 i A+5 B) \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{(28 A-25 i B) \cot ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{30 d}+\frac{1}{60} \int \cot (c+d x) \left (480 a^4 (i A+B)-480 a^4 (A-i B) \tan (c+d x)\right ) \, dx\\ &=-8 a^4 (A-i B) x-\frac{8 a^4 (A-i B) \cot (c+d x)}{d}+\frac{a^4 (148 i A+145 B) \cot ^2(c+d x)}{60 d}-\frac{a A \cot ^5(c+d x) (a+i a \tan (c+d x))^3}{5 d}-\frac{(8 i A+5 B) \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{(28 A-25 i B) \cot ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{30 d}+\left (8 a^4 (i A+B)\right ) \int \cot (c+d x) \, dx\\ &=-8 a^4 (A-i B) x-\frac{8 a^4 (A-i B) \cot (c+d x)}{d}+\frac{a^4 (148 i A+145 B) \cot ^2(c+d x)}{60 d}+\frac{8 a^4 (i A+B) \log (\sin (c+d x))}{d}-\frac{a A \cot ^5(c+d x) (a+i a \tan (c+d x))^3}{5 d}-\frac{(8 i A+5 B) \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac{(28 A-25 i B) \cot ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{30 d}\\ \end{align*}

Mathematica [B]  time = 8.23866, size = 542, normalized size = 2.71 \[ \frac{a^4 (\cot (c+d x)+i)^4 (A \cot (c+d x)+B) \left (-8 d x (A-i B) (\cos (4 c)-i \sin (4 c)) \sin ^5(c+d x)+4 (A-i B) (\sin (4 c)+i \cos (4 c)) \sin ^5(c+d x) \log \left (\sin ^2(c+d x)\right )+8 (A-i B) (\cos (4 c)-i \sin (4 c)) \sin ^5(c+d x) \tan ^{-1}(\tan (5 c+d x))+\frac{1}{120} \csc (c) (\cos (4 c)-i \sin (4 c)) (15 (20 A d x-14 i A-20 i B d x-11 B) \cos (2 c+d x)+15 \cos (d x) (A (-20 d x+14 i)+B (11+20 i d x))+345 A \sin (2 c+d x)-275 A \sin (2 c+3 d x)-120 A \sin (4 c+3 d x)+79 A \sin (4 c+5 d x)-90 i A \cos (2 c+3 d x)+150 A d x \cos (2 c+3 d x)+90 i A \cos (4 c+3 d x)-150 A d x \cos (4 c+3 d x)-30 A d x \cos (4 c+5 d x)+30 A d x \cos (6 c+5 d x)+445 A \sin (d x)-300 i B \sin (2 c+d x)+260 i B \sin (2 c+3 d x)+90 i B \sin (4 c+3 d x)-70 i B \sin (4 c+5 d x)-60 B \cos (2 c+3 d x)-150 i B d x \cos (2 c+3 d x)+60 B \cos (4 c+3 d x)+150 i B d x \cos (4 c+3 d x)+30 i B d x \cos (4 c+5 d x)-30 i B d x \cos (6 c+5 d x)-400 i B \sin (d x))\right )}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

(a^4*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*(-8*(A - I*B)*d*x*(Cos[4*c] - I*Sin[4*c])*Sin[c + d*x]^5 + 8*(A
 - I*B)*ArcTan[Tan[5*c + d*x]]*(Cos[4*c] - I*Sin[4*c])*Sin[c + d*x]^5 + 4*(A - I*B)*Log[Sin[c + d*x]^2]*(I*Cos
[4*c] + Sin[4*c])*Sin[c + d*x]^5 + (Csc[c]*(Cos[4*c] - I*Sin[4*c])*(15*(A*(14*I - 20*d*x) + B*(11 + (20*I)*d*x
))*Cos[d*x] + 15*((-14*I)*A - 11*B + 20*A*d*x - (20*I)*B*d*x)*Cos[2*c + d*x] - (90*I)*A*Cos[2*c + 3*d*x] - 60*
B*Cos[2*c + 3*d*x] + 150*A*d*x*Cos[2*c + 3*d*x] - (150*I)*B*d*x*Cos[2*c + 3*d*x] + (90*I)*A*Cos[4*c + 3*d*x] +
 60*B*Cos[4*c + 3*d*x] - 150*A*d*x*Cos[4*c + 3*d*x] + (150*I)*B*d*x*Cos[4*c + 3*d*x] - 30*A*d*x*Cos[4*c + 5*d*
x] + (30*I)*B*d*x*Cos[4*c + 5*d*x] + 30*A*d*x*Cos[6*c + 5*d*x] - (30*I)*B*d*x*Cos[6*c + 5*d*x] + 445*A*Sin[d*x
] - (400*I)*B*Sin[d*x] + 345*A*Sin[2*c + d*x] - (300*I)*B*Sin[2*c + d*x] - 275*A*Sin[2*c + 3*d*x] + (260*I)*B*
Sin[2*c + 3*d*x] - 120*A*Sin[4*c + 3*d*x] + (90*I)*B*Sin[4*c + 3*d*x] + 79*A*Sin[4*c + 5*d*x] - (70*I)*B*Sin[4
*c + 5*d*x]))/120))/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x]))

________________________________________________________________________________________

Maple [A]  time = 0.081, size = 224, normalized size = 1.1 \begin{align*} 8\,{\frac{B{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,iB{a}^{4}c}{d}}+{\frac{8\,iB\cot \left ( dx+c \right ){a}^{4}}{d}}-{\frac{iA{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{8\,iA{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{A{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{B{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{7\,A{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{7\,B{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-8\,{\frac{A{a}^{4}c}{d}}-8\,{\frac{A\cot \left ( dx+c \right ){a}^{4}}{d}}+8\,iBx{a}^{4}-{\frac{{\frac{4\,i}{3}}B{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{d}}-8\,A{a}^{4}x+{\frac{4\,iA{a}^{4} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 2.00317, size = 211, normalized size = 1.05 \begin{align*} -\frac{60 \,{\left (d x + c\right )}{\left (8 \, A - 8 i \, B\right )} a^{4} + 240 \,{\left (i \, A + B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 480 \,{\left (-i \, A - B\right )} a^{4} \log \left (\tan \left (d x + c\right )\right ) + \frac{{\left (480 \, A - 480 i \, B\right )} a^{4} \tan \left (d x + c\right )^{4} - 30 \,{\left (8 i \, A + 7 \, B\right )} a^{4} \tan \left (d x + c\right )^{3} -{\left (140 \, A - 80 i \, B\right )} a^{4} \tan \left (d x + c\right )^{2} - 15 \,{\left (-4 i \, A - B\right )} a^{4} \tan \left (d x + c\right ) + 12 \, A a^{4}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(60*(d*x + c)*(8*A - 8*I*B)*a^4 + 240*(I*A + B)*a^4*log(tan(d*x + c)^2 + 1) + 480*(-I*A - B)*a^4*log(tan
(d*x + c)) + ((480*A - 480*I*B)*a^4*tan(d*x + c)^4 - 30*(8*I*A + 7*B)*a^4*tan(d*x + c)^3 - (140*A - 80*I*B)*a^
4*tan(d*x + c)^2 - 15*(-4*I*A - B)*a^4*tan(d*x + c) + 12*A*a^4)/tan(d*x + c)^5)/d

________________________________________________________________________________________

Fricas [A]  time = 1.46433, size = 857, normalized size = 4.28 \begin{align*} \frac{{\left (-840 i \, A - 600 \, B\right )} a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (2220 i \, A + 1860 \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-2620 i \, A - 2260 \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (1460 i \, A + 1280 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-316 i \, A - 280 \, B\right )} a^{4} +{\left ({\left (120 i \, A + 120 \, B\right )} a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} +{\left (-600 i \, A - 600 \, B\right )} a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (1200 i \, A + 1200 \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-1200 i \, A - 1200 \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (600 i \, A + 600 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-120 i \, A - 120 \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{15 \,{\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/15*((-840*I*A - 600*B)*a^4*e^(8*I*d*x + 8*I*c) + (2220*I*A + 1860*B)*a^4*e^(6*I*d*x + 6*I*c) + (-2620*I*A -
2260*B)*a^4*e^(4*I*d*x + 4*I*c) + (1460*I*A + 1280*B)*a^4*e^(2*I*d*x + 2*I*c) + (-316*I*A - 280*B)*a^4 + ((120
*I*A + 120*B)*a^4*e^(10*I*d*x + 10*I*c) + (-600*I*A - 600*B)*a^4*e^(8*I*d*x + 8*I*c) + (1200*I*A + 1200*B)*a^4
*e^(6*I*d*x + 6*I*c) + (-1200*I*A - 1200*B)*a^4*e^(4*I*d*x + 4*I*c) + (600*I*A + 600*B)*a^4*e^(2*I*d*x + 2*I*c
) + (-120*I*A - 120*B)*a^4)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) +
 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) - d)

________________________________________________________________________________________

Sympy [A]  time = 146.132, size = 272, normalized size = 1.36 \begin{align*} \frac{8 a^{4} \left (i A + B\right ) \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{- \frac{\left (56 i A a^{4} + 40 B a^{4}\right ) e^{- 2 i c} e^{8 i d x}}{d} + \frac{\left (148 i A a^{4} + 124 B a^{4}\right ) e^{- 4 i c} e^{6 i d x}}{d} + \frac{\left (292 i A a^{4} + 256 B a^{4}\right ) e^{- 8 i c} e^{2 i d x}}{3 d} - \frac{\left (316 i A a^{4} + 280 B a^{4}\right ) e^{- 10 i c}}{15 d} - \frac{\left (524 i A a^{4} + 452 B a^{4}\right ) e^{- 6 i c} e^{4 i d x}}{3 d}}{e^{10 i d x} - 5 e^{- 2 i c} e^{8 i d x} + 10 e^{- 4 i c} e^{6 i d x} - 10 e^{- 6 i c} e^{4 i d x} + 5 e^{- 8 i c} e^{2 i d x} - e^{- 10 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

8*a**4*(I*A + B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-(56*I*A*a**4 + 40*B*a**4)*exp(-2*I*c)*exp(8*I*d*x)/d +
(148*I*A*a**4 + 124*B*a**4)*exp(-4*I*c)*exp(6*I*d*x)/d + (292*I*A*a**4 + 256*B*a**4)*exp(-8*I*c)*exp(2*I*d*x)/
(3*d) - (316*I*A*a**4 + 280*B*a**4)*exp(-10*I*c)/(15*d) - (524*I*A*a**4 + 452*B*a**4)*exp(-6*I*c)*exp(4*I*d*x)
/(3*d))/(exp(10*I*d*x) - 5*exp(-2*I*c)*exp(8*I*d*x) + 10*exp(-4*I*c)*exp(6*I*d*x) - 10*exp(-6*I*c)*exp(4*I*d*x
) + 5*exp(-8*I*c)*exp(2*I*d*x) - exp(-10*I*c))

________________________________________________________________________________________

Giac [B]  time = 1.86528, size = 529, normalized size = 2.64 \begin{align*} \frac{6 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 60 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 15 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 310 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 160 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1200 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 900 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4740 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4320 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15360 \,{\left (i \, A a^{4} + B a^{4}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 7680 \,{\left (-i \, A a^{4} - B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{-17536 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 17536 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4740 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 4320 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 1200 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 900 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 310 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 160 i \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 60 i \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, A a^{4}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/960*(6*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 60*I*A*a^4*tan(1/2*d*x + 1/2*c)^4 - 15*B*a^4*tan(1/2*d*x + 1/2*c)^4 -
310*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 160*I*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 1200*I*A*a^4*tan(1/2*d*x + 1/2*c)^2 +
900*B*a^4*tan(1/2*d*x + 1/2*c)^2 + 4740*A*a^4*tan(1/2*d*x + 1/2*c) - 4320*I*B*a^4*tan(1/2*d*x + 1/2*c) - 15360
*(I*A*a^4 + B*a^4)*log(tan(1/2*d*x + 1/2*c) + I) - 7680*(-I*A*a^4 - B*a^4)*log(abs(tan(1/2*d*x + 1/2*c))) + (-
17536*I*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 17536*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 4740*A*a^4*tan(1/2*d*x + 1/2*c)^4
+ 4320*I*B*a^4*tan(1/2*d*x + 1/2*c)^4 + 1200*I*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 900*B*a^4*tan(1/2*d*x + 1/2*c)^3
 + 310*A*a^4*tan(1/2*d*x + 1/2*c)^2 - 160*I*B*a^4*tan(1/2*d*x + 1/2*c)^2 - 60*I*A*a^4*tan(1/2*d*x + 1/2*c) - 1
5*B*a^4*tan(1/2*d*x + 1/2*c) - 6*A*a^4)/tan(1/2*d*x + 1/2*c)^5)/d