\(\int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [514]

Optimal result

Integrand size = 36, antiderivative size = 146 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {8 \sqrt [4]{-1} a^3 (A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {16 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{15 d}-\frac {2 a A \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2}{5 d}-\frac {2 (9 i A+5 B) \sqrt {\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{15 d} \]

[Out]

8*(-1)^(1/4)*a^3*(A-I*B)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d+16/15*a^3*(6*A-5*I*B)*cot(d*x+c)^(1/2)/d-2/5*a
*A*(I*a+a*cot(d*x+c))^2*cot(d*x+c)^(1/2)/d-2/15*(9*I*A+5*B)*(I*a^3+a^3*cot(d*x+c))*cot(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3662, 3675, 3673, 3614, 214} \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {8 \sqrt [4]{-1} a^3 (A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {16 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{15 d}-\frac {2 (5 B+9 i A) \sqrt {\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{15 d}-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d} \]

[In]

Int[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(8*(-1)^(1/4)*a^3*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (16*a^3*(6*A - (5*I)*B)*Sqrt[Cot[c + d
*x]])/(15*d) - (2*a*A*Sqrt[Cot[c + d*x]]*(I*a + a*Cot[c + d*x])^2)/(5*d) - (2*((9*I)*A + 5*B)*Sqrt[Cot[c + d*x
]]*(I*a^3 + a^3*Cot[c + d*x]))/(15*d)

Rule 214

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

Rule 3614

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 3662

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 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -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}& = \int \frac {(i a+a \cot (c+d x))^3 (B+A \cot (c+d x))}{\sqrt {\cot (c+d x)}} \, dx \\ & = -\frac {2 a A \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2}{5 d}-\frac {2}{5} \int \frac {(i a+a \cot (c+d x))^2 \left (\frac {1}{2} a (A-5 i B)-\frac {1}{2} a (9 i A+5 B) \cot (c+d x)\right )}{\sqrt {\cot (c+d x)}} \, dx \\ & = -\frac {2 a A \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2}{5 d}-\frac {2 (9 i A+5 B) \sqrt {\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{15 d}+\frac {4}{15} \int \frac {(i a+a \cot (c+d x)) \left (-a^2 (3 i A+5 B)-2 a^2 (6 A-5 i B) \cot (c+d x)\right )}{\sqrt {\cot (c+d x)}} \, dx \\ & = \frac {16 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{15 d}-\frac {2 a A \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2}{5 d}-\frac {2 (9 i A+5 B) \sqrt {\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{15 d}+\frac {4}{15} \int \frac {15 a^3 (A-i B)-15 a^3 (i A+B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx \\ & = \frac {16 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{15 d}-\frac {2 a A \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2}{5 d}-\frac {2 (9 i A+5 B) \sqrt {\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{15 d}+\frac {\left (120 a^6 (A-i B)^2\right ) \text {Subst}\left (\int \frac {1}{-15 a^3 (A-i B)-15 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) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {16 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{15 d}-\frac {2 a A \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2}{5 d}-\frac {2 (9 i A+5 B) \sqrt {\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{15 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.77 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 a^3 \sqrt {\cot (c+d x)} \left (-15 i B-5 i A \cot (c+d x)-15 B \cot (c+d x)-9 A \cot ^2(c+d x)+12 A \cot ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},i \tan (c+d x)\right )+20 B \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (c+d x)\right )\right )}{15 d} \]

[In]

Integrate[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(-2*a^3*Sqrt[Cot[c + d*x]]*((-15*I)*B - (5*I)*A*Cot[c + d*x] - 15*B*Cot[c + d*x] - 9*A*Cot[c + d*x]^2 + 12*A*C
ot[c + d*x]^2*Hypergeometric2F1[-5/2, 1, -3/2, I*Tan[c + d*x]] + 20*B*Cot[c + d*x]*Hypergeometric2F1[-3/2, 1,
-1/2, I*Tan[c + d*x]]))/(15*d)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (123 ) = 246\).

Time = 1.24 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.73

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {2 A \cot \left (d x +c \right )^{\frac {5}{2}}}{5}-2 i A \cot \left (d x +c \right )^{\frac {3}{2}}-\frac {2 B \cot \left (d x +c \right )^{\frac {3}{2}}}{3}-6 i B \sqrt {\cot \left (d x +c \right )}+8 A \sqrt {\cot \left (d x +c \right )}-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(252\)
default	\(\frac {a^{3} \left (-\frac {2 A \cot \left (d x +c \right )^{\frac {5}{2}}}{5}-2 i A \cot \left (d x +c \right )^{\frac {3}{2}}-\frac {2 B \cot \left (d x +c \right )^{\frac {3}{2}}}{3}-6 i B \sqrt {\cot \left (d x +c \right )}+8 A \sqrt {\cot \left (d x +c \right )}-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(252\)

[In]

int(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

a^3/d*(-2/5*A*cot(d*x+c)^(5/2)-2*I*A*cot(d*x+c)^(3/2)-2/3*B*cot(d*x+c)^(3/2)-6*I*B*cot(d*x+c)^(1/2)+8*A*cot(d*
x+c)^(1/2)-1/4*(4*A-4*I*B)*2^(1/2)*(ln((1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(1+cot(d*x+c)-2^(1/2)*cot(d*x+c
)^(1/2)))+2*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2)))-1/4*(-4*I*A-4*B)*2^(1/2)
*(ln((1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2)))+2*arctan(1+2^(1/2)*cot(d
*x+c)^(1/2))+2*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (118) = 236\).

Time = 0.27 (sec) , antiderivative size = 449, normalized size of antiderivative = 3.08 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 2 \, {\left ({\left (39 \, A - 25 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, {\left (19 \, A - 15 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left (6 \, A - 5 i \, B\right )} a^{3}\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/15*(15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2*
((A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*c) - I*d)*s
qrt((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)/((-I*A - B)*a^3)) - 15*sqrt(-
(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2*((A - I*B)*a^3*e
^(2*I*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + 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)/((-I*A - B)*a^3)) - 2*((39*A - 25*I*B)*a^3*e
^(4*I*d*x + 4*I*c) - 3*(19*A - 15*I*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*(6*A - 5*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^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**(7/2)*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.37 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} + \frac {30 \, {\left (4 \, A - 3 i \, B\right )} a^{3}}{\sqrt {\tan \left (d x + c\right )}} + \frac {10 \, {\left (-3 i \, A - B\right )} a^{3}}{\tan \left (d x + c\right )^{\frac {3}{2}}} - \frac {6 \, A a^{3}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{15 \, d} \]

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/15*(15*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*(
(I - 1)*A + (I + 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(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(ta
n(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 + 30*(4*A - 3*I*B)*a^3/sqrt(tan(d*x + c)) + 10*(-3*I*A - B)*a^3/tan(d*x
+ c)^(3/2) - 6*A*a^3/tan(d*x + c)^(5/2))/d

Giac [F]

\[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \cot \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*cot(d*x + c)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

[In]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3, x)