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

Optimal. Leaf size=128 \[ -\frac{2 a^2 (5 B+7 i A) \cot ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{4 a^2 (A-i B) \sqrt{\cot (c+d x)}}{d}+\frac{4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x) \left (a^2 \cot (c+d x)+i a^2\right )}{5 d} \]

[Out]

(4*(-1)^(1/4)*a^2*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (4*a^2*(A - I*B)*Sqrt[Cot[c + d*x]])/d
- (2*a^2*((7*I)*A + 5*B)*Cot[c + d*x]^(3/2))/(15*d) - (2*A*Cot[c + d*x]^(3/2)*(I*a^2 + a^2*Cot[c + d*x]))/(5*
d)

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Rubi [A]  time = 0.360165, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3581, 3594, 3592, 3528, 3533, 208} \[ -\frac{2 a^2 (5 B+7 i A) \cot ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{4 a^2 (A-i B) \sqrt{\cot (c+d x)}}{d}+\frac{4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x) \left (a^2 \cot (c+d x)+i a^2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1)^(1/4)*a^2*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (4*a^2*(A - I*B)*Sqrt[Cot[c + d*x]])/d
- (2*a^2*((7*I)*A + 5*B)*Cot[c + d*x]^(3/2))/(15*d) - (2*A*Cot[c + d*x]^(3/2)*(I*a^2 + a^2*Cot[c + d*x]))/(5*
d)

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 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]

Rule 3592

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 3528

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

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 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]

Rubi steps

\begin{align*} \int \cot ^{\frac{7}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\int \sqrt{\cot (c+d x)} (i a+a \cot (c+d x))^2 (B+A \cot (c+d x)) \, dx\\ &=-\frac{2 A \cot ^{\frac{3}{2}}(c+d x) \left (i a^2+a^2 \cot (c+d x)\right )}{5 d}-\frac{2}{5} \int \sqrt{\cot (c+d x)} (i a+a \cot (c+d x)) \left (\frac{1}{2} a (3 A-5 i B)-\frac{1}{2} a (7 i A+5 B) \cot (c+d x)\right ) \, dx\\ &=-\frac{2 a^2 (7 i A+5 B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x) \left (i a^2+a^2 \cot (c+d x)\right )}{5 d}-\frac{2}{5} \int \sqrt{\cot (c+d x)} \left (5 a^2 (i A+B)+5 a^2 (A-i B) \cot (c+d x)\right ) \, dx\\ &=\frac{4 a^2 (A-i B) \sqrt{\cot (c+d x)}}{d}-\frac{2 a^2 (7 i A+5 B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x) \left (i a^2+a^2 \cot (c+d x)\right )}{5 d}-\frac{2}{5} \int \frac{-5 a^2 (A-i B)+5 a^2 (i A+B) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{4 a^2 (A-i B) \sqrt{\cot (c+d x)}}{d}-\frac{2 a^2 (7 i A+5 B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x) \left (i a^2+a^2 \cot (c+d x)\right )}{5 d}-\frac{\left (20 a^4 (A-i B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{5 a^2 (A-i B)+5 a^2 (i A+B) x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=\frac{4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{4 a^2 (A-i B) \sqrt{\cot (c+d x)}}{d}-\frac{2 a^2 (7 i A+5 B) \cot ^{\frac{3}{2}}(c+d x)}{15 d}-\frac{2 A \cot ^{\frac{3}{2}}(c+d x) \left (i a^2+a^2 \cot (c+d x)\right )}{5 d}\\ \end{align*}

Mathematica [B]  time = 5.66056, size = 272, normalized size = 2.12 \[ \frac{a^2 \sin ^3(c+d x) (\cot (c+d x)+i)^2 (A \cot (c+d x)+B) \left (-\frac{4 i e^{-2 i c} (A-i B) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{\frac{i \left (1+e^{2 i (c+d x)}\right )}{-1+e^{2 i (c+d x)}}}}-\frac{1}{15} (\cos (2 c)-i \sin (2 c)) \sqrt{\cot (c+d x)} \csc ^2(c+d x) (5 (B+2 i A) \sin (2 (c+d x))+(33 A-30 i B) \cos (2 (c+d x))-27 A+30 i B)\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(I + Cot[c + d*x])^2*(B + A*Cot[c + d*x])*Sin[c + d*x]^3*(((-4*I)*(A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(
c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/(E^((2*I)*c)*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)))]) - (Sqrt[Cot[c + d*x]]*Csc[c + d*x]^2*(Cos[2*
c] - I*Sin[2*c])*(-27*A + (30*I)*B + (33*A - (30*I)*B)*Cos[2*(c + d*x)] + 5*((2*I)*A + B)*Sin[2*(c + d*x)]))/1
5))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [C]  time = 0.446, size = 2947, normalized size = 23. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*a^2/d*2^(1/2)*(30*B*cos(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(
1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+
c))^(1/2)-33*A*cos(d*x+c)^3*2^(1/2)+30*A*cos(d*x+c)*2^(1/2)+30*B*cos(d*x+c)^2*((cos(d*x+c)-1+sin(d*x+c))/sin(d
*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos
(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+30*A*cos(d*x+c)^3*((cos(d*x+c)-1+sin(d*x+c))/si
n(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(
cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-30*B*cos(d*x+c)^3*((cos(d*x+c)-1+sin(d*x+c))
/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1
/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)+30*B*cos(d*x+c)^3*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x
+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d
*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+30*I*B*cos(d*x+c)^3*2^(1/2)-30*I*B*2^(1/2)*cos(d*
x+c)-30*B*cos(d*x+c)^2*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*Elliptic
F((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-5*B
*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)-30*A*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))
^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),
1/2+1/2*I,1/2*2^(1/2))-10*I*A*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+30*A*cos(d*x+c)^2*((cos(d*x+c)-1+sin(d*x+c))/sin
(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(c
os(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-30*B*cos(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/si
n(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(
cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-30*A*cos(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/s
in(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+30*I*A*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c
))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x
+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^3-30*I*A*((cos(d*x+c)-1+sin(d*x+c))/sin(
d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(cos
(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)^3-30*I*B*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c)
)^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+
c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^3+30*I*A*((cos(d*x+c)-1+sin(d*x+c))/sin(d
*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos
(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^2-30*I*A*((cos(d*x+c)-1+sin(d*x+c))/
sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)^2-30*I*B*((cos(d*x+c)-1+sin(d*x+c))/sin(d*
x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(
d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^2-30*I*A*((cos(d*x+c)-1+sin(d*x+c))/s
in(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)+30*I*A*cos(d*x+c)*(-(cos(d*x+c)-
1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)
*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+30*I*B*cos(d*x+c)*(-(cos(d*x+c)-1-sin(d*
x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*Ellipti
cPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+30*B*((cos(d*x+c)-1+sin(d*x+c))/sin(d
*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(
1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-30*B*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*
x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))
/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-30*I*A*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/
sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x
+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+30*I*A*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+
c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2
),1/2*2^(1/2))+30*I*B*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/
2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^
(1/2)))*(cos(d*x+c)/sin(d*x+c))^(7/2)*sin(d*x+c)/cos(d*x+c)^4

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Maxima [A]  time = 1.56812, size = 270, normalized size = 2.11 \begin{align*} -\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^{2} - \frac{4 \,{\left (30 \, A - 30 i \, B\right )} a^{2}}{\sqrt{\tan \left (d x + c\right )}} - \frac{20 \,{\left (-2 i \, A - B\right )} a^{2}}{\tan \left (d x + c\right )^{\frac{3}{2}}} + \frac{12 \, A a^{2}}{\tan \left (d x + c\right )^{\frac{5}{2}}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(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
(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^2 - 4*(30*A - 30*I*B)*a^2/sqrt(tan(d*x + c)) - 20*(-2*I*A - B)*a^2/tan
(d*x + c)^(3/2) + 12*A*a^2/tan(d*x + c)^(5/2))/d

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Fricas [B]  time = 1.59621, size = 1200, normalized size = 9.38 \begin{align*} -\frac{15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{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{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{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{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 8 \,{\left ({\left (43 \, A - 35 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 6 \,{\left (9 \, A - 10 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (23 \, A - 25 i \, B\right )} a^{2}\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}}}{60 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*l
og(-(4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/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)/((2*I*A + 2*B)*a^2
)) - 15*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log
(-(4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/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)/((2*I*A + 2*B)*a^2)
) - 8*((43*A - 35*I*B)*a^2*e^(4*I*d*x + 4*I*c) - 6*(9*A - 10*I*B)*a^2*e^(2*I*d*x + 2*I*c) + (23*A - 25*I*B)*a^
2)*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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \cot \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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