3.730 \(\int \cot ^{\frac{7}{2}}(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=106 \[ -\frac{8 i a^3 \cot ^{\frac{3}{2}}(c+d x)}{5 d}-\frac{2 \cot ^{\frac{3}{2}}(c+d x) \left (a^3 \cot (c+d x)+i a^3\right )}{5 d}+\frac{8 a^3 \sqrt{\cot (c+d x)}}{d}+\frac{8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

[Out]

(8*(-1)^(1/4)*a^3*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (8*a^3*Sqrt[Cot[c + d*x]])/d - (((8*I)/5)*a^3*Co
t[c + d*x]^(3/2))/d - (2*Cot[c + d*x]^(3/2)*(I*a^3 + a^3*Cot[c + d*x]))/(5*d)

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Rubi [A]  time = 0.201385, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3673, 3556, 3592, 3528, 3533, 208} \[ -\frac{8 i a^3 \cot ^{\frac{3}{2}}(c+d x)}{5 d}-\frac{2 \cot ^{\frac{3}{2}}(c+d x) \left (a^3 \cot (c+d x)+i a^3\right )}{5 d}+\frac{8 a^3 \sqrt{\cot (c+d x)}}{d}+\frac{8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(-1)^(1/4)*a^3*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (8*a^3*Sqrt[Cot[c + d*x]])/d - (((8*I)/5)*a^3*Co
t[c + d*x]^(3/2))/d - (2*Cot[c + d*x]^(3/2)*(I*a^3 + a^3*Cot[c + d*x]))/(5*d)

Rule 3673

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

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

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))^3 \, dx &=\int \sqrt{\cot (c+d x)} (i a+a \cot (c+d x))^3 \, dx\\ &=-\frac{2 \cot ^{\frac{3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}-\frac{1}{5} (2 i a) \int \sqrt{\cot (c+d x)} (-4 i a-6 a \cot (c+d x)) (i a+a \cot (c+d x)) \, dx\\ &=-\frac{8 i a^3 \cot ^{\frac{3}{2}}(c+d x)}{5 d}-\frac{2 \cot ^{\frac{3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}-\frac{1}{5} (2 i a) \int \sqrt{\cot (c+d x)} \left (10 a^2-10 i a^2 \cot (c+d x)\right ) \, dx\\ &=\frac{8 a^3 \sqrt{\cot (c+d x)}}{d}-\frac{8 i a^3 \cot ^{\frac{3}{2}}(c+d x)}{5 d}-\frac{2 \cot ^{\frac{3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}-\frac{1}{5} (2 i a) \int \frac{10 i a^2+10 a^2 \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{8 a^3 \sqrt{\cot (c+d x)}}{d}-\frac{8 i a^3 \cot ^{\frac{3}{2}}(c+d x)}{5 d}-\frac{2 \cot ^{\frac{3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}+\frac{\left (80 i a^5\right ) \operatorname{Subst}\left (\int \frac{1}{-10 i a^2+10 a^2 x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=\frac{8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{8 a^3 \sqrt{\cot (c+d x)}}{d}-\frac{8 i a^3 \cot ^{\frac{3}{2}}(c+d x)}{5 d}-\frac{2 \cot ^{\frac{3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}\\ \end{align*}

Mathematica [A]  time = 2.85173, size = 147, normalized size = 1.39 \[ -\frac{a^3 e^{-3 i c} \sqrt{\cot (c+d x)} (\cos (3 (c+d x))+i \sin (3 (c+d x))) \left (40 \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+\csc ^2(c+d x) (5 i \sin (2 (c+d x))+21 \cos (2 (c+d x))-19)\right )}{5 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sqrt[Cot[c + d*x]]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)])*(Csc[c + d*x]^2*(-19 + 21*Cos[2*(c + d*x)] +
(5*I)*Sin[2*(c + d*x)]) + 40*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Sqrt[I*Tan[c
+ d*x]]))/(5*d*E^((3*I)*c)*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [C]  time = 0.31, size = 1482, normalized size = 14. \begin{align*} \text{result 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))^3,x)

[Out]

1/5*a^3/d*2^(1/2)*(20*I*(-(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)^3-20*I*(-(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))
*cos(d*x+c)^3+20*I*(-(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)^2-20*I*(-(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))*cos(
d*x+c)^2+20*(-(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)^3-20*I*(-(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)*((co
s(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)+20*I*(-(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)*((c
os(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)
+20*(-(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)^
2-20*I*(-(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))+20*I*(-(
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))-5*I*2^(1/2)*cos(d*x+c)^2*sin(
d*x+c)-20*(-(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)-20*(-(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))-21*2^(
1/2)*cos(d*x+c)^3+20*2^(1/2)*cos(d*x+c))*(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 = 1.54392, size = 213, normalized size = 2.01 \begin{align*} \frac{5 \,{\left (\left (2 i - 2\right ) \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + \left (2 i - 2\right ) \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) - \left (i + 1\right ) \, \sqrt{2} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) + \left (i + 1\right ) \, \sqrt{2} \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{40 \, a^{3}}{\sqrt{\tan \left (d x + c\right )}} - \frac{10 i \, a^{3}}{\tan \left (d x + c\right )^{\frac{3}{2}}} - \frac{2 \, a^{3}}{\tan \left (d x + c\right )^{\frac{5}{2}}}}{5 \, 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))^3,x, algorithm="maxima")

[Out]

1/5*(5*((2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + (2*I - 2)*sqrt(2)*arctan(-1/2
*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - (I + 1)*sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) +
1) + (I + 1)*sqrt(2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 + 40*a^3/sqrt(tan(d*x + c)) -
10*I*a^3/tan(d*x + c)^(3/2) - 2*a^3/tan(d*x + c)^(5/2))/d

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Fricas [B]  time = 1.36915, size = 934, normalized size = 8.81 \begin{align*} -\frac{5 \, \sqrt{\frac{64 i \, 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{{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{64 i \, a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 )}}{4 \, a^{3}}\right ) - 5 \, \sqrt{\frac{64 i \, 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{{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{64 i \, a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 )}}{4 \, a^{3}}\right ) - 16 \,{\left (13 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 19 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, 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}}}{20 \,{\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))^3,x, algorithm="fricas")

[Out]

-1/20*(5*sqrt(64*I*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(1/4*(8*I*a^3*e^(2*I*d*x
+ 2*I*c) + sqrt(64*I*a^6/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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)/a^3) - 5*sqrt(64*I*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(1/4*(8*I*a^3*e^(2*I*d*x + 2*I*c) - sqrt(64*I*a^6/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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)/a^3) - 16*(13*a^3*e^(4*I*d*x + 4*I*c) - 19*a^3*e^(
2*I*d*x + 2*I*c) + 8*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)

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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))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \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))^3,x, algorithm="giac")

[Out]

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