3.25 \(\int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=180 \[ \frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}+\frac {2 a^3 (B+i A) \cot ^2(c+d x)}{d}-\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {4 a^3 (B+i A) \log (\sin (c+d x))}{d}-\frac {(5 B+7 i A) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}-4 a^3 x (A-i B)-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

[Out]

-4*a^3*(A-I*B)*x-4*a^3*(A-I*B)*cot(d*x+c)/d+2*a^3*(I*A+B)*cot(d*x+c)^2/d+1/60*a^3*(47*A-45*I*B)*cot(d*x+c)^3/d
+4*a^3*(I*A+B)*ln(sin(d*x+c))/d-1/5*a*A*cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2/d-1/20*(7*I*A+5*B)*cot(d*x+c)^4*(a^3
+I*a^3*tan(d*x+c))/d

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Rubi [A]  time = 0.46, antiderivative size = 180, normalized size of antiderivative = 1.00, 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} \[ \frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}+\frac {2 a^3 (B+i A) \cot ^2(c+d x)}{d}-\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {4 a^3 (B+i A) \log (\sin (c+d x))}{d}-\frac {(5 B+7 i A) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}-4 a^3 x (A-i B)-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-4*a^3*(A - I*B)*x - (4*a^3*(A - I*B)*Cot[c + d*x])/d + (2*a^3*(I*A + B)*Cot[c + d*x]^2)/d + (a^3*(47*A - (45*
I)*B)*Cot[c + d*x]^3)/(60*d) + (4*a^3*(I*A + B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^5*(a + I*a*Tan[c + d*
x])^2)/(5*d) - (((7*I)*A + 5*B)*Cot[c + d*x]^4*(a^3 + I*a^3*Tan[c + d*x]))/(20*d)

Rule 3475

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

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 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 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]

Rubi steps

\begin {align*} \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (a (7 i A+5 B)-a (3 A-5 i B) \tan (c+d x)) \, dx\\ &=-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \left (-a^2 (47 A-45 i B)-a^2 (33 i A+35 B) \tan (c+d x)\right ) \, dx\\ &=\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \cot ^3(c+d x) \left (-80 a^3 (i A+B)+80 a^3 (A-i B) \tan (c+d x)\right ) \, dx\\ &=\frac {2 a^3 (i A+B) \cot ^2(c+d x)}{d}+\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \cot ^2(c+d x) \left (80 a^3 (A-i B)+80 a^3 (i A+B) \tan (c+d x)\right ) \, dx\\ &=-\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {2 a^3 (i A+B) \cot ^2(c+d x)}{d}+\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \cot (c+d x) \left (80 a^3 (i A+B)-80 a^3 (A-i B) \tan (c+d x)\right ) \, dx\\ &=-4 a^3 (A-i B) x-\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {2 a^3 (i A+B) \cot ^2(c+d x)}{d}+\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\left (4 a^3 (i A+B)\right ) \int \cot (c+d x) \, dx\\ &=-4 a^3 (A-i B) x-\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {2 a^3 (i A+B) \cot ^2(c+d x)}{d}+\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}+\frac {4 a^3 (i A+B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}\\ \end {align*}

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Mathematica [B]  time = 9.51, size = 943, normalized size = 5.24 \[ a^3 \left (\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (i A \cos \left (\frac {3 c}{2}\right )+B \cos \left (\frac {3 c}{2}\right )+A \sin \left (\frac {3 c}{2}\right )-i B \sin \left (\frac {3 c}{2}\right )\right ) \left (-4 i \tan ^{-1}(\tan (4 c+d x)) \cos \left (\frac {3 c}{2}\right )-4 \tan ^{-1}(\tan (4 c+d x)) \sin \left (\frac {3 c}{2}\right )\right ) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (i A \cos \left (\frac {3 c}{2}\right )+B \cos \left (\frac {3 c}{2}\right )+A \sin \left (\frac {3 c}{2}\right )-i B \sin \left (\frac {3 c}{2}\right )\right ) \left (2 \cos \left (\frac {3 c}{2}\right ) \log \left (\sin ^2(c+d x)\right )-2 i \log \left (\sin ^2(c+d x)\right ) \sin \left (\frac {3 c}{2}\right )\right ) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {x (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (-16 A \cos ^3(c)+16 i B \cos ^3(c)-4 i A \cot (c) \cos ^3(c)-4 B \cot (c) \cos ^3(c)+24 i A \sin (c) \cos ^2(c)+24 B \sin (c) \cos ^2(c)+16 A \sin ^2(c) \cos (c)-16 i B \sin ^2(c) \cos (c)-4 i A \sin ^3(c)-4 B \sin ^3(c)+(i A+B) \cot (c) (4 \cos (3 c)-4 i \sin (3 c))\right ) \sin ^4(c+d x)}{(\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \csc (c) \csc (c+d x) \left (\frac {1}{240} \cos (3 c)-\frac {1}{240} i \sin (3 c)\right ) (225 i A \cos (d x)+195 B \cos (d x)-300 A d x \cos (d x)+300 i B d x \cos (d x)-225 i A \cos (2 c+d x)-195 B \cos (2 c+d x)+300 A d x \cos (2 c+d x)-300 i B d x \cos (2 c+d x)-105 i A \cos (2 c+3 d x)-75 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)+105 i A \cos (4 c+3 d x)+75 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)+470 A \sin (d x)-420 i B \sin (d x)+360 A \sin (2 c+d x)-330 i B \sin (2 c+d x)-280 A \sin (2 c+3 d x)+270 i B \sin (2 c+3 d x)-135 A \sin (4 c+3 d x)+105 i B \sin (4 c+3 d x)+83 A \sin (4 c+5 d x)-75 i B \sin (4 c+5 d x))}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

a^3*(((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*(I*A*Cos[(3*c)/2] + B*Cos[(3*c)/2] + A*Sin[(3*c)/2] - I*B*Sin[
(3*c)/2])*((-4*I)*ArcTan[Tan[4*c + d*x]]*Cos[(3*c)/2] - 4*ArcTan[Tan[4*c + d*x]]*Sin[(3*c)/2])*Sin[c + d*x]^4)
/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*
(I*A*Cos[(3*c)/2] + B*Cos[(3*c)/2] + A*Sin[(3*c)/2] - I*B*Sin[(3*c)/2])*(2*Cos[(3*c)/2]*Log[Sin[c + d*x]^2] -
(2*I)*Log[Sin[c + d*x]^2]*Sin[(3*c)/2])*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c
+ d*x])) + (x*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*(-16*A*Cos[c]^3 + (16*I)*B*Cos[c]^3 - (4*I)*A*Cos[c]^
3*Cot[c] - 4*B*Cos[c]^3*Cot[c] + (24*I)*A*Cos[c]^2*Sin[c] + 24*B*Cos[c]^2*Sin[c] + 16*A*Cos[c]*Sin[c]^2 - (16*
I)*B*Cos[c]*Sin[c]^2 - (4*I)*A*Sin[c]^3 - 4*B*Sin[c]^3 + (I*A + B)*Cot[c]*(4*Cos[3*c] - (4*I)*Sin[3*c]))*Sin[c
+ d*x]^4)/((Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c
+ d*x])*Csc[c]*Csc[c + d*x]*(Cos[3*c]/240 - (I/240)*Sin[3*c])*((225*I)*A*Cos[d*x] + 195*B*Cos[d*x] - 300*A*d*x
*Cos[d*x] + (300*I)*B*d*x*Cos[d*x] - (225*I)*A*Cos[2*c + d*x] - 195*B*Cos[2*c + d*x] + 300*A*d*x*Cos[2*c + d*x
] - (300*I)*B*d*x*Cos[2*c + d*x] - (105*I)*A*Cos[2*c + 3*d*x] - 75*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] + (105*I)*A*Cos[4*c + 3*d*x] + 75*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] + 470*A*Sin[d*x] - (420*I)*B*Sin[d*x] + 360*A*Sin[2*c + d
*x] - (330*I)*B*Sin[2*c + d*x] - 280*A*Sin[2*c + 3*d*x] + (270*I)*B*Sin[2*c + 3*d*x] - 135*A*Sin[4*c + 3*d*x]
+ (105*I)*B*Sin[4*c + 3*d*x] + 83*A*Sin[4*c + 5*d*x] - (75*I)*B*Sin[4*c + 5*d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^
3*(A*Cos[c + d*x] + B*Sin[c + d*x])))

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fricas [A]  time = 0.67, size = 284, normalized size = 1.58 \[ \frac {{\left (-480 i \, A - 360 \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (1170 i \, A + 1050 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-1390 i \, A - 1230 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (770 i \, A + 690 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-166 i \, A - 150 \, B\right )} a^{3} + {\left ({\left (60 i \, A + 60 \, B\right )} a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + {\left (-300 i \, A - 300 \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (600 i \, A + 600 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-600 i \, A - 600 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (300 i \, A + 300 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-60 i \, A - 60 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/15*((-480*I*A - 360*B)*a^3*e^(8*I*d*x + 8*I*c) + (1170*I*A + 1050*B)*a^3*e^(6*I*d*x + 6*I*c) + (-1390*I*A -
1230*B)*a^3*e^(4*I*d*x + 4*I*c) + (770*I*A + 690*B)*a^3*e^(2*I*d*x + 2*I*c) + (-166*I*A - 150*B)*a^3 + ((60*I*
A + 60*B)*a^3*e^(10*I*d*x + 10*I*c) + (-300*I*A - 300*B)*a^3*e^(8*I*d*x + 8*I*c) + (600*I*A + 600*B)*a^3*e^(6*
I*d*x + 6*I*c) + (-600*I*A - 600*B)*a^3*e^(4*I*d*x + 4*I*c) + (300*I*A + 300*B)*a^3*e^(2*I*d*x + 2*I*c) + (-60
*I*A - 60*B)*a^3)*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)

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giac [B]  time = 5.82, size = 392, normalized size = 2.18 \[ \frac {6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 45 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 190 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 660 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 540 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2460 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2280 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1920 \, {\left (-4 i \, A a^{3} - 4 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 1920 \, {\left (2 i \, A a^{3} + 2 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {-8768 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8768 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2460 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2280 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 660 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 540 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 190 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/960*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 45*I*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 15*B*a^3*tan(1/2*d*x + 1/2*c)^4 -
190*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 120*I*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 660*I*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 5
40*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 2460*A*a^3*tan(1/2*d*x + 1/2*c) - 2280*I*B*a^3*tan(1/2*d*x + 1/2*c) + 1920*(
-4*I*A*a^3 - 4*B*a^3)*log(tan(1/2*d*x + 1/2*c) + I) + 1920*(2*I*A*a^3 + 2*B*a^3)*log(tan(1/2*d*x + 1/2*c)) + (
-8768*I*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 8768*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 2460*A*a^3*tan(1/2*d*x + 1/2*c)^4 +
2280*I*B*a^3*tan(1/2*d*x + 1/2*c)^4 + 660*I*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 540*B*a^3*tan(1/2*d*x + 1/2*c)^3 +
190*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 120*I*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 45*I*A*a^3*tan(1/2*d*x + 1/2*c) - 15*
B*a^3*tan(1/2*d*x + 1/2*c) - 6*A*a^3)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 0.48, size = 224, normalized size = 1.24 \[ \frac {4 a^{3} B \ln \left (\sin \left (d x +c \right )\right )}{d}-4 A \,a^{3} x +\frac {4 i A \,a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {4 i B \,a^{3} c}{d}+4 i B x \,a^{3}-\frac {3 i A \,a^{3} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {4 i B \cot \left (d x +c \right ) a^{3}}{d}-\frac {i a^{3} B \left (\cot ^{3}\left (d x +c \right )\right )}{d}-\frac {4 A \,a^{3} c}{d}+\frac {4 A \,a^{3} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {4 A \cot \left (d x +c \right ) a^{3}}{d}+\frac {2 a^{3} B \left (\cot ^{2}\left (d x +c \right )\right )}{d}-\frac {A \,a^{3} \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}-\frac {a^{3} B \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {2 i A \,a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.98, size = 153, normalized size = 0.85 \[ -\frac {240 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{3} - 60 \, {\left (-2 i \, A - 2 \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 \, {\left (4 i \, A + 4 \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {240 \, {\left (A - i \, B\right )} a^{3} \tan \left (d x + c\right )^{4} + {\left (-120 i \, A - 120 \, B\right )} a^{3} \tan \left (d x + c\right )^{3} - 20 \, {\left (4 \, A - 3 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + {\left (45 i \, A + 15 \, B\right )} a^{3} \tan \left (d x + c\right ) + 12 \, A a^{3}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(240*(d*x + c)*(A - I*B)*a^3 - 60*(-2*I*A - 2*B)*a^3*log(tan(d*x + c)^2 + 1) - 60*(4*I*A + 4*B)*a^3*log(
tan(d*x + c)) + (240*(A - I*B)*a^3*tan(d*x + c)^4 + (-120*I*A - 120*B)*a^3*tan(d*x + c)^3 - 20*(4*A - 3*I*B)*a
^3*tan(d*x + c)^2 + (45*I*A + 15*B)*a^3*tan(d*x + c) + 12*A*a^3)/tan(d*x + c)^5)/d

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mupad [B]  time = 6.94, size = 140, normalized size = 0.78 \[ -\frac {\frac {A\,a^3}{5}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {4\,A\,a^3}{3}-B\,a^3\,1{}\mathrm {i}\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (4\,A\,a^3-B\,a^3\,4{}\mathrm {i}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (2\,B\,a^3+A\,a^3\,2{}\mathrm {i}\right )+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^3}{4}+\frac {A\,a^3\,3{}\mathrm {i}}{4}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5}+\frac {a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^6*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

(a^3*atan(2*tan(c + d*x) + 1i)*(A*1i + B)*8i)/d - (tan(c + d*x)^4*(4*A*a^3 - B*a^3*4i) - tan(c + d*x)^2*((4*A*
a^3)/3 - B*a^3*1i) - tan(c + d*x)^3*(A*a^3*2i + 2*B*a^3) + (A*a^3)/5 + tan(c + d*x)*((A*a^3*3i)/4 + (B*a^3)/4)
)/(d*tan(c + d*x)^5)

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sympy [A]  time = 2.59, size = 296, normalized size = 1.64 \[ \frac {4 i a^{3} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 166 i A a^{3} - 150 B a^{3} + \left (770 i A a^{3} e^{2 i c} + 690 B a^{3} e^{2 i c}\right ) e^{2 i d x} + \left (- 1390 i A a^{3} e^{4 i c} - 1230 B a^{3} e^{4 i c}\right ) e^{4 i d x} + \left (1170 i A a^{3} e^{6 i c} + 1050 B a^{3} e^{6 i c}\right ) e^{6 i d x} + \left (- 480 i A a^{3} e^{8 i c} - 360 B a^{3} e^{8 i c}\right ) e^{8 i d x}}{15 d e^{10 i c} e^{10 i d x} - 75 d e^{8 i c} e^{8 i d x} + 150 d e^{6 i c} e^{6 i d x} - 150 d e^{4 i c} e^{4 i d x} + 75 d e^{2 i c} e^{2 i d x} - 15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

4*I*a**3*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-166*I*A*a**3 - 150*B*a**3 + (770*I*A*a**3*exp(2*I*c)
+ 690*B*a**3*exp(2*I*c))*exp(2*I*d*x) + (-1390*I*A*a**3*exp(4*I*c) - 1230*B*a**3*exp(4*I*c))*exp(4*I*d*x) + (1
170*I*A*a**3*exp(6*I*c) + 1050*B*a**3*exp(6*I*c))*exp(6*I*d*x) + (-480*I*A*a**3*exp(8*I*c) - 360*B*a**3*exp(8*
I*c))*exp(8*I*d*x))/(15*d*exp(10*I*c)*exp(10*I*d*x) - 75*d*exp(8*I*c)*exp(8*I*d*x) + 150*d*exp(6*I*c)*exp(6*I*
d*x) - 150*d*exp(4*I*c)*exp(4*I*d*x) + 75*d*exp(2*I*c)*exp(2*I*d*x) - 15*d)

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