3.30 \(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=144 \[ \frac {a^4 (B+4 i A) \log (\sin (c+d x))}{d}+\frac {a^4 (7 B+4 i A) \log (\cos (c+d x))}{d}-8 a^4 x (A-i B)-\frac {3 B \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {(-B+2 i A) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d} \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 144, 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, 3594, 3589, 3475, 3531} \[ \frac {(-B+2 i A) \left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac {a^4 (B+4 i A) \log (\sin (c+d x))}{d}+\frac {a^4 (7 B+4 i A) \log (\cos (c+d x))}{d}-8 a^4 x (A-i B)-\frac {3 B \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

-8*a^4*(A - I*B)*x + (a^4*((4*I)*A + 7*B)*Log[Cos[c + d*x]])/d + (a^4*((4*I)*A + B)*Log[Sin[c + d*x]])/d - (a*
A*Cot[c + d*x]*(a + I*a*Tan[c + d*x])^3)/d + (((2*I)*A - B)*(a^2 + I*a^2*Tan[c + d*x])^2)/(2*d) - (3*B*(a^4 +
I*a^4*Tan[c + d*x]))/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 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 3589

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 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 3594

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

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Mathematica [B]  time = 11.53, size = 1122, normalized size = 7.79 \[ a^4 \left (\frac {x (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \left (-22 A \cos ^4(c)+\frac {17}{2} i B \cos ^4(c)-4 i A \cot (c) \cos ^4(c)-B \cot (c) \cos ^4(c)+50 i A \sin (c) \cos ^3(c)+\frac {55}{2} B \sin (c) \cos ^3(c)+60 A \sin ^2(c) \cos ^2(c)-45 i B \sin ^2(c) \cos ^2(c)+2 A \cos ^2(c)-\frac {7}{2} i B \cos ^2(c)-40 i A \sin ^3(c) \cos (c)-40 B \sin ^3(c) \cos (c)-6 i A \sin (c) \cos (c)-\frac {21}{2} B \sin (c) \cos (c)-14 A \sin ^4(c)+\frac {37}{2} i B \sin ^4(c)-6 A \sin ^2(c)+\frac {21}{2} i B \sin ^2(c)+(4 \cos (2 c) B-3 B+4 i A \cos (2 c)) \csc (c) \sec (c) (\cos (4 c)-i \sin (4 c))+2 i A \sin ^4(c) \tan (c)+\frac {7}{2} B \sin ^4(c) \tan (c)+2 i A \sin ^2(c) \tan (c)+\frac {7}{2} B \sin ^2(c) \tan (c)\right ) \sin ^5(c+d x)}{(\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (4 i A \cos (2 c)+B \cos (2 c)+4 A \sin (2 c)-i B \sin (2 c)) \left (-i \tan ^{-1}(\tan (5 c+d x)) \cos (2 c)-\tan ^{-1}(\tan (5 c+d x)) \sin (2 c)\right ) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (4 i A \cos (2 c)+7 B \cos (2 c)+4 A \sin (2 c)-7 i B \sin (2 c)) \left (\frac {1}{2} \cos (2 c) \log \left (\cos ^2(c+d x)\right )-\frac {1}{2} i \log \left (\cos ^2(c+d x)\right ) \sin (2 c)\right ) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (4 i A \cos (2 c)+B \cos (2 c)+4 A \sin (2 c)-i B \sin (2 c)) \left (\frac {1}{2} \cos (2 c) \log \left (\sin ^2(c+d x)\right )-\frac {1}{2} i \log \left (\sin ^2(c+d x)\right ) \sin (2 c)\right ) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(A-i B) (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) (8 i d x \sin (4 c)-8 d x \cos (4 c)) \sin ^5(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \sec (c) (\cos (4 c)-i \sin (4 c)) (A \sin (d x)-4 i B \sin (d x)) \tan (c+d x) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac {A (\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \csc (c) (\cos (4 c)-i \sin (4 c)) \sin (d x) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}+\frac {(\cot (c+d x)+i)^4 (B+A \cot (c+d x)) \left (\frac {1}{2} B \cos (4 c)-\frac {1}{2} i B \sin (4 c)\right ) \tan ^2(c+d x) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^4 (A \cos (c+d x)+B \sin (c+d x))}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

a^4*((A*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*Csc[c]*(Cos[4*c] - I*Sin[4*c])*Sin[d*x]*Sin[c + d*x]^4)/(d*(
Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*((4*I
)*A*Cos[2*c] + B*Cos[2*c] + 4*A*Sin[2*c] - I*B*Sin[2*c])*((-I)*ArcTan[Tan[5*c + d*x]]*Cos[2*c] - ArcTan[Tan[5*
c + d*x]]*Sin[2*c])*Sin[c + d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Co
t[c + d*x])^4*(B + A*Cot[c + d*x])*((4*I)*A*Cos[2*c] + 7*B*Cos[2*c] + 4*A*Sin[2*c] - (7*I)*B*Sin[2*c])*((Cos[2
*c]*Log[Cos[c + d*x]^2])/2 - (I/2)*Log[Cos[c + d*x]^2]*Sin[2*c])*Sin[c + d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*
(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*((4*I)*A*Cos[2*c] + B*Cos[2*c]
+ 4*A*Sin[2*c] - I*B*Sin[2*c])*((Cos[2*c]*Log[Sin[c + d*x]^2])/2 - (I/2)*Log[Sin[c + d*x]^2]*Sin[2*c])*Sin[c
+ d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((A - I*B)*(I + Cot[c + d*x])^4*(B
+ A*Cot[c + d*x])*(-8*d*x*Cos[4*c] + (8*I)*d*x*Sin[4*c])*Sin[c + d*x]^5)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[
c + d*x] + B*Sin[c + d*x])) + (x*(I + Cot[c + d*x])^4*(B + A*Cot[c + d*x])*Sin[c + d*x]^5*(2*A*Cos[c]^2 - ((7*
I)/2)*B*Cos[c]^2 - 22*A*Cos[c]^4 + ((17*I)/2)*B*Cos[c]^4 - (4*I)*A*Cos[c]^4*Cot[c] - B*Cos[c]^4*Cot[c] - (6*I)
*A*Cos[c]*Sin[c] - (21*B*Cos[c]*Sin[c])/2 + (50*I)*A*Cos[c]^3*Sin[c] + (55*B*Cos[c]^3*Sin[c])/2 - 6*A*Sin[c]^2
+ ((21*I)/2)*B*Sin[c]^2 + 60*A*Cos[c]^2*Sin[c]^2 - (45*I)*B*Cos[c]^2*Sin[c]^2 - (40*I)*A*Cos[c]*Sin[c]^3 - 40
*B*Cos[c]*Sin[c]^3 - 14*A*Sin[c]^4 + ((37*I)/2)*B*Sin[c]^4 + (-3*B + (4*I)*A*Cos[2*c] + 4*B*Cos[2*c])*Csc[c]*S
ec[c]*(Cos[4*c] - I*Sin[4*c]) + (2*I)*A*Sin[c]^2*Tan[c] + (7*B*Sin[c]^2*Tan[c])/2 + (2*I)*A*Sin[c]^4*Tan[c] +
(7*B*Sin[c]^4*Tan[c])/2))/((Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^
4*(B + A*Cot[c + d*x])*Sec[c]*(Cos[4*c] - I*Sin[4*c])*(A*Sin[d*x] - (4*I)*B*Sin[d*x])*Sin[c + d*x]^4*Tan[c + d
*x])/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^4*(B + A*Cot[c + d*
x])*((B*Cos[4*c])/2 - (I/2)*B*Sin[4*c])*Sin[c + d*x]^3*Tan[c + d*x]^2)/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c +
d*x] + B*Sin[c + d*x])))

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fricas [B]  time = 0.76, size = 254, normalized size = 1.76 \[ \frac {10 \, B a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-4 i \, A - 2 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-4 i \, A - 8 \, B\right )} a^{4} + {\left ({\left (4 i \, A + 7 \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (4 i \, A + 7 \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-4 i \, A - 7 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-4 i \, A - 7 \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left ({\left (4 i \, A + B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (4 i \, A + B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-4 i \, A - B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-4 i \, A - B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (6 i \, d x + 6 i \, c\right )} + d e^{\left (4 i \, d x + 4 i \, c\right )} - d e^{\left (2 i \, d x + 2 i \, c\right )} - d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

(10*B*a^4*e^(4*I*d*x + 4*I*c) + (-4*I*A - 2*B)*a^4*e^(2*I*d*x + 2*I*c) + (-4*I*A - 8*B)*a^4 + ((4*I*A + 7*B)*a
^4*e^(6*I*d*x + 6*I*c) + (4*I*A + 7*B)*a^4*e^(4*I*d*x + 4*I*c) + (-4*I*A - 7*B)*a^4*e^(2*I*d*x + 2*I*c) + (-4*
I*A - 7*B)*a^4)*log(e^(2*I*d*x + 2*I*c) + 1) + ((4*I*A + B)*a^4*e^(6*I*d*x + 6*I*c) + (4*I*A + B)*a^4*e^(4*I*d
*x + 4*I*c) + (-4*I*A - B)*a^4*e^(2*I*d*x + 2*I*c) + (-4*I*A - 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)

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giac [B]  time = 3.06, size = 337, normalized size = 2.34 \[ \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, {\left (4 i \, A a^{4} + 7 \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 4 \, {\left (-8 i \, A a^{4} - 8 \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 2 \, {\left (-4 i \, A a^{4} - 7 \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - 2 \, {\left (-4 i \, A a^{4} - B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {8 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{4}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {12 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 46 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 i \, A a^{4} + 21 \, B a^{4}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(A*a^4*tan(1/2*d*x + 1/2*c) + 2*(4*I*A*a^4 + 7*B*a^4)*log(tan(1/2*d*x + 1/2*c) + 1) + 4*(-8*I*A*a^4 - 8*B*
a^4)*log(tan(1/2*d*x + 1/2*c) + I) - 2*(-4*I*A*a^4 - 7*B*a^4)*log(tan(1/2*d*x + 1/2*c) - 1) - 2*(-4*I*A*a^4 -
B*a^4)*log(tan(1/2*d*x + 1/2*c)) - (8*I*A*a^4*tan(1/2*d*x + 1/2*c) + 2*B*a^4*tan(1/2*d*x + 1/2*c) + A*a^4)/tan
(1/2*d*x + 1/2*c) - (12*I*A*a^4*tan(1/2*d*x + 1/2*c)^4 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^4 + 4*A*a^4*tan(1/2*d*x
+ 1/2*c)^3 - 16*I*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 24*I*A*a^4*tan(1/2*d*x + 1/2*c)^2 - 46*B*a^4*tan(1/2*d*x + 1
/2*c)^2 - 4*A*a^4*tan(1/2*d*x + 1/2*c) + 16*I*B*a^4*tan(1/2*d*x + 1/2*c) + 12*I*A*a^4 + 21*B*a^4)/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^2)/d

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maple [A]  time = 0.40, size = 165, normalized size = 1.15 \[ -8 A \,a^{4} x +\frac {A \,a^{4} \tan \left (d x +c \right )}{d}-\frac {8 A \,a^{4} c}{d}+\frac {a^{4} B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {7 a^{4} B \ln \left (\cos \left (d x +c \right )\right )}{d}+8 i B x \,a^{4}+\frac {4 i A \,a^{4} \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {4 i A \,a^{4} \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {4 i a^{4} B \tan \left (d x +c \right )}{d}+\frac {8 i B \,a^{4} c}{d}-\frac {A \cot \left (d x +c \right ) a^{4}}{d}+\frac {a^{4} B \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.88, size = 104, normalized size = 0.72 \[ \frac {B a^{4} \tan \left (d x + c\right )^{2} - 16 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{4} + 2 \, {\left (-4 i \, A - 4 \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (4 i \, A + B\right )} a^{4} \log \left (\tan \left (d x + c\right )\right ) + 2 \, {\left (A - 4 i \, B\right )} a^{4} \tan \left (d x + c\right ) - \frac {2 \, A a^{4}}{\tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.56, size = 110, normalized size = 0.76 \[ \frac {B\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}+\frac {a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B+A\,4{}\mathrm {i}\right )}{d}-\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d}-\frac {A\,a^4\,\mathrm {cot}\left (c+d\,x\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^4\,1{}\mathrm {i}+a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotInvertible} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

Exception raised: NotInvertible

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