3.518 \(\int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\)

Optimal. Leaf size=173 \[ -\frac {8 a^3 (21 A-23 i B)}{105 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 (7 A-11 i B) \left (a^3 \cot (c+d x)+i a^3\right )}{35 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {8 a^3 (B+i A)}{d \sqrt {\cot (c+d x)}}+\frac {8 \sqrt [4]{-1} a^3 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i a B (a \cot (c+d x)+i a)^2}{7 d \cot ^{\frac {7}{2}}(c+d x)} \]

[Out]

8*(-1)^(1/4)*a^3*(A-I*B)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d-8/105*a^3*(21*A-23*I*B)/d/cot(d*x+c)^(3/2)+2/7
*I*a*B*(I*a+a*cot(d*x+c))^2/d/cot(d*x+c)^(7/2)-2/35*(7*A-11*I*B)*(I*a^3+a^3*cot(d*x+c))/d/cot(d*x+c)^(5/2)+8*a
^3*(I*A+B)/d/cot(d*x+c)^(1/2)

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Rubi [A]  time = 0.54, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3581, 3593, 3591, 3529, 3533, 208} \[ -\frac {8 a^3 (21 A-23 i B)}{105 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 (7 A-11 i B) \left (a^3 \cot (c+d x)+i a^3\right )}{35 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {8 a^3 (B+i A)}{d \sqrt {\cot (c+d x)}}+\frac {8 \sqrt [4]{-1} a^3 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i a B (a \cot (c+d x)+i a)^2}{7 d \cot ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(-1)^(1/4)*a^3*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (8*a^3*(21*A - (23*I)*B))/(105*d*Cot[c
+ d*x]^(3/2)) + (8*a^3*(I*A + B))/(d*Sqrt[Cot[c + d*x]]) + (((2*I)/7)*a*B*(I*a + a*Cot[c + d*x])^2)/(d*Cot[c
+ d*x]^(7/2)) - (2*(7*A - (11*I)*B)*(I*a^3 + a^3*Cot[c + d*x]))/(35*d*Cot[c + d*x]^(5/2))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3581

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

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 \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx &=\int \frac {(i a+a \cot (c+d x))^3 (B+A \cot (c+d x))}{\cot ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 i a B (i a+a \cot (c+d x))^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(i a+a \cot (c+d x))^2 \left (\frac {1}{2} a (7 i A+11 B)+\frac {1}{2} a (7 A-3 i B) \cot (c+d x)\right )}{\cot ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 i a B (i a+a \cot (c+d x))^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}-\frac {2 (7 A-11 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {(i a+a \cot (c+d x)) \left (a^2 (21 i A+23 B)+2 a^2 (7 A-6 i B) \cot (c+d x)\right )}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {8 a^3 (21 A-23 i B)}{105 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 i a B (i a+a \cot (c+d x))^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}-\frac {2 (7 A-11 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {35 a^3 (i A+B)+35 a^3 (A-i B) \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {8 a^3 (21 A-23 i B)}{105 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {8 a^3 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i a B (i a+a \cot (c+d x))^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}-\frac {2 (7 A-11 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {35 a^3 (A-i B)-35 a^3 (i A+B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=-\frac {8 a^3 (21 A-23 i B)}{105 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {8 a^3 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i a B (i a+a \cot (c+d x))^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}-\frac {2 (7 A-11 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {\left (280 a^6 (A-i B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-35 a^3 (A-i B)-35 a^3 (i A+B) x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {8 \sqrt [4]{-1} a^3 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {8 a^3 (21 A-23 i B)}{105 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {8 a^3 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i a B (i a+a \cot (c+d x))^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}-\frac {2 (7 A-11 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d \cot ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 15.18, size = 298, normalized size = 1.72 \[ \frac {a^3 \sin ^4(c+d x) (\cot (c+d x)+i)^3 (A \cot (c+d x)+B) \left (\frac {(\sin (3 c)+i \cos (3 c)) \sec ^2(c+d x) (10 ((31 B+21 i A) \cos (2 (c+d x))+21 i A+25 B)+21 (59 A-57 i B) \cot (c+d x)+21 (21 A-23 i B) \cos (3 (c+d x)) \csc (c+d x))}{210 \cot ^{\frac {3}{2}}(c+d x)}-8 e^{-3 i c} (A-i B) \sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\frac {i \left (1+e^{2 i (c+d x)}\right )}{-1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*((-8*(A - I*B)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c
+ d*x)))]*Sqrt[(I*(1 + E^((2*I)*(c + d*x))))/(-1 + E^((2*I)*(c + d*x)))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x
)))/(1 + E^((2*I)*(c + d*x)))]])/E^((3*I)*c) + ((10*((21*I)*A + 25*B + ((21*I)*A + 31*B)*Cos[2*(c + d*x)]) + 2
1*(59*A - (57*I)*B)*Cot[c + d*x] + 21*(21*A - (23*I)*B)*Cos[3*(c + d*x)]*Csc[c + d*x])*Sec[c + d*x]^2*(I*Cos[3
*c] + Sin[3*c]))/(210*Cot[c + d*x]^(3/2)))*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Si
n[c + d*x]))

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fricas [B]  time = 0.65, size = 561, normalized size = 3.24 \[ -\frac {105 \, \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {{\left (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 105 \, \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {{\left (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (64 i \, A^{2} + 128 \, A B - 64 i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 16 \, {\left ({\left (273 \, A - 319 i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, {\left (133 \, A - 109 i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 5 \, {\left (21 \, A - 19 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, {\left (133 \, A - 129 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, {\left (42 \, A - 41 i \, B\right )} a^{3}\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{420 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/420*(105*sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6
*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log(-(8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt((64*I*A
^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x
+ 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^3)) - 105*sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)
*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log
(-(8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt((64*I*A^2 + 128*A*B - 64*I*B^2)*a^6/d^2)*(-I*d*e^(2*I*d*x + 2*I*
c) + I*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^3
)) - 16*((273*A - 319*I*B)*a^3*e^(8*I*d*x + 8*I*c) + 3*(133*A - 109*I*B)*a^3*e^(6*I*d*x + 6*I*c) - 5*(21*A - 1
9*I*B)*a^3*e^(4*I*d*x + 4*I*c) - 3*(133*A - 129*I*B)*a^3*e^(2*I*d*x + 2*I*c) - 4*(42*A - 41*I*B)*a^3)*sqrt((I*
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^
(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\cot \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 1.90, size = 1043, normalized size = 6.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x)

[Out]

1/105*a^3/d*(-1+cos(d*x+c))*(15*I*B*sin(d*x+c)*2^(1/2)-441*I*A*2^(1/2)*cos(d*x+c)^3+155*I*B*sin(d*x+c)*2^(1/2)
*cos(d*x+c)^3+420*A*cos(d*x+c)^3*sin(d*x+c)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin
(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+
c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+420*B*cos(d*x+c)^3*sin(d*x+c)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*
((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+c
os(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-420*B*sin(d*x+c)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/si
n(d*x+c))^(1/2),1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*
(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-15*I*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)+441*I*A*2^(1/
2)*cos(d*x+c)^4-420*I*A*sin(d*x+c)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-1+
cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d
*x+c))^(1/2)*cos(d*x+c)^3-105*A*sin(d*x+c)*2^(1/2)*cos(d*x+c)^3-155*I*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)^2+483*B*
2^(1/2)*cos(d*x+c)^4-420*I*B*cos(d*x+c)^3*sin(d*x+c)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d
*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))
/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+105*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2+21*I*A*cos(d*x+c)*2^(1/2)-483*
B*2^(1/2)*cos(d*x+c)^3+420*I*A*sin(d*x+c)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,
1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1
+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-21*I*A*2^(1/2)*cos(d*x+c)^2-63*B*2^(1/2)*cos(d*x+c)^2+63*B*2^(1/2)
*cos(d*x+c))*(1+cos(d*x+c))^2/cos(d*x+c)^3/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/2)*2^(1/2)

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maxima [A]  time = 0.80, size = 219, normalized size = 1.27 \[ -\frac {2 \, {\left (15 i \, B a^{3} + \frac {{\left (21 i \, A + 63 \, B\right )} a^{3}}{\tan \left (d x + c\right )} + \frac {35 \, {\left (3 \, A - 4 i \, B\right )} a^{3}}{\tan \left (d x + c\right )^{2}} + \frac {{\left (-420 i \, A - 420 \, B\right )} a^{3}}{\tan \left (d x + c\right )^{3}}\right )} \tan \left (d x + c\right )^{\frac {7}{2}} + 105 \, {\left (\sqrt {2} {\left (-\left (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (-\left (2 i - 2\right ) \, A - \left (2 i + 2\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

-1/105*(2*(15*I*B*a^3 + (21*I*A + 63*B)*a^3/tan(d*x + c) + 35*(3*A - 4*I*B)*a^3/tan(d*x + c)^2 + (-420*I*A - 4
20*B)*a^3/tan(d*x + c)^3)*tan(d*x + c)^(7/2) + 105*(sqrt(2)*(-(2*I - 2)*A - (2*I + 2)*B)*arctan(1/2*sqrt(2)*(s
qrt(2) + 2/sqrt(tan(d*x + c)))) + sqrt(2)*(-(2*I - 2)*A - (2*I + 2)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(t
an(d*x + c)))) - sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(
2)*(-(I + 1)*A + (I - 1)*B)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3)/cot(c + d*x)^(1/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3)/cot(c + d*x)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i a^{3} \left (\int \frac {i A}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 A \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {A \tan ^{3}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 B \tan ^{2}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {B \tan ^{4}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 i A \tan ^{2}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {i B \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 i B \tan ^{3}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c))/cot(d*x+c)**(1/2),x)

[Out]

-I*a**3*(Integral(I*A/sqrt(cot(c + d*x)), x) + Integral(-3*A*tan(c + d*x)/sqrt(cot(c + d*x)), x) + Integral(A*
tan(c + d*x)**3/sqrt(cot(c + d*x)), x) + Integral(-3*B*tan(c + d*x)**2/sqrt(cot(c + d*x)), x) + Integral(B*tan
(c + d*x)**4/sqrt(cot(c + d*x)), x) + Integral(-3*I*A*tan(c + d*x)**2/sqrt(cot(c + d*x)), x) + Integral(I*B*ta
n(c + d*x)/sqrt(cot(c + d*x)), x) + Integral(-3*I*B*tan(c + d*x)**3/sqrt(cot(c + d*x)), x))

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