3.665 \(\int \frac{(e \cos (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=154 \[ \frac{4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{10 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{3/2}}{33 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) \cos (c+d x) (e \cos (c+d x))^{3/2}}{11 a^2 d}+\frac{10 \tan (c+d x) (e \cos (c+d x))^{3/2}}{33 a^2 d} \]

[Out]

(10*(e*Cos[c + d*x])^(3/2)*EllipticF[(c + d*x)/2, 2])/(33*a^2*d*Cos[c + d*x]^(3/2)) + (2*Cos[c + d*x]*(e*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(11*a^2*d) + (10*(e*Cos[c + d*x])^(3/2)*Tan[c + d*x])/(33*a^2*d) + (((4*I)/11)*Co
s[c + d*x]^2*(e*Cos[c + d*x])^(3/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.198362, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3515, 3500, 3769, 3771, 2641} \[ \frac{4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{10 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{3/2}}{33 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) \cos (c+d x) (e \cos (c+d x))^{3/2}}{11 a^2 d}+\frac{10 \tan (c+d x) (e \cos (c+d x))^{3/2}}{33 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*(e*Cos[c + d*x])^(3/2)*EllipticF[(c + d*x)/2, 2])/(33*a^2*d*Cos[c + d*x]^(3/2)) + (2*Cos[c + d*x]*(e*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(11*a^2*d) + (10*(e*Cos[c + d*x])^(3/2)*Tan[c + d*x])/(33*a^2*d) + (((4*I)/11)*Co
s[c + d*x]^2*(e*Cos[c + d*x])^(3/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3500

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx &=\left ((e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac{1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx\\ &=\frac{4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (7 e^2 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac{1}{(e \sec (c+d x))^{7/2}} \, dx}{11 a^2}\\ &=\frac{2 \cos (c+d x) (e \cos (c+d x))^{3/2} \sin (c+d x)}{11 a^2 d}+\frac{4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (5 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{11 a^2}\\ &=\frac{2 \cos (c+d x) (e \cos (c+d x))^{3/2} \sin (c+d x)}{11 a^2 d}+\frac{10 (e \cos (c+d x))^{3/2} \tan (c+d x)}{33 a^2 d}+\frac{4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (5 (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \sqrt{e \sec (c+d x)} \, dx}{33 a^2 e^2}\\ &=\frac{2 \cos (c+d x) (e \cos (c+d x))^{3/2} \sin (c+d x)}{11 a^2 d}+\frac{10 (e \cos (c+d x))^{3/2} \tan (c+d x)}{33 a^2 d}+\frac{4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (5 (e \cos (c+d x))^{3/2}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{33 a^2 \cos ^{\frac{3}{2}}(c+d x)}\\ &=\frac{10 (e \cos (c+d x))^{3/2} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{33 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \cos (c+d x) (e \cos (c+d x))^{3/2} \sin (c+d x)}{11 a^2 d}+\frac{10 (e \cos (c+d x))^{3/2} \tan (c+d x)}{33 a^2 d}+\frac{4 i \cos ^2(c+d x) (e \cos (c+d x))^{3/2}}{11 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.661062, size = 131, normalized size = 0.85 \[ \frac{(e \cos (c+d x))^{3/2} \left (\sqrt{\cos (c+d x)} (13 \sin (c+d x)-7 \sin (3 (c+d x))-28 i \cos (c+d x)+4 i \cos (3 (c+d x)))-20 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))\right )}{66 a^2 d \cos ^{\frac{7}{2}}(c+d x) (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cos[c + d*x])^(3/2)*(-20*EllipticF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + Sqrt[Cos[c +
d*x]]*((-28*I)*Cos[c + d*x] + (4*I)*Cos[3*(c + d*x)] + 13*Sin[c + d*x] - 7*Sin[3*(c + d*x)])))/(66*a^2*d*Cos[c
+ d*x]^(7/2)*(-I + Tan[c + d*x])^2)

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Maple [A]  time = 2.975, size = 315, normalized size = 2.1 \begin{align*}{\frac{2\,{e}^{2}}{33\,{a}^{2}d} \left ( 384\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{13}-384\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{12}\cos \left ( 1/2\,dx+c/2 \right ) -1152\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{11}+960\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}\cos \left ( 1/2\,dx+c/2 \right ) +1440\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9}-1008\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-960\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}+552\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +360\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}-176\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -72\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}-5\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +28\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +6\,i\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

2/33/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^2*(384*I*sin(1/2*d*x+1/2*c)^13-384*sin(1/2*d
*x+1/2*c)^12*cos(1/2*d*x+1/2*c)-1152*I*sin(1/2*d*x+1/2*c)^11+960*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+1440
*I*sin(1/2*d*x+1/2*c)^9-1008*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-960*I*sin(1/2*d*x+1/2*c)^7+552*sin(1/2*d*
x+1/2*c)^6*cos(1/2*d*x+1/2*c)+360*I*sin(1/2*d*x+1/2*c)^5-176*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-72*I*sin(
1/2*d*x+1/2*c)^3-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),
2^(1/2))+28*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+6*I*sin(1/2*d*x+1/2*c))/d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (132 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )}{\rm integral}\left (-\frac{10 i \, \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )}}{33 \,{\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}}, x\right ) + \sqrt{\frac{1}{2}}{\left (-11 i \, e e^{\left (6 i \, d x + 6 i \, c\right )} + 41 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, e\right )} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{132 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/132*(132*a^2*d*e^(5*I*d*x + 5*I*c)*integral(-10/33*I*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*e*e^(-1/2*I*d
*x - 1/2*I*c)/(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d), x) + sqrt(1/2)*(-11*I*e*e^(6*I*d*x + 6*I*c) + 41*I*e*e^(4*I
*d*x + 4*I*c) + 15*I*e*e^(2*I*d*x + 2*I*c) + 3*I*e)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*e^(-1/2*I*d*x - 1/2*I*c))*
e^(-5*I*d*x - 5*I*c)/(a^2*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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