3.242 \(\int \frac{1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=150 \[ \frac{4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}+\frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 a^2 d e^2}+\frac{2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac{10 \sin (c+d x)}{33 a^2 d e \sqrt{e \sec (c+d x)}} \]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(33*a^2*d*e^2) + (2*e*Sin[c + d*x])/(11
*a^2*d*(e*Sec[c + d*x])^(5/2)) + (10*Sin[c + d*x])/(33*a^2*d*e*Sqrt[e*Sec[c + d*x]]) + (((4*I)/11)*e^2)/(d*(e*
Sec[c + d*x])^(7/2)*(a^2 + I*a^2*Tan[c + d*x]))

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Rubi [A]  time = 0.109988, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3500, 3769, 3771, 2641} \[ \frac{4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}+\frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 a^2 d e^2}+\frac{2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac{10 \sin (c+d x)}{33 a^2 d e \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(33*a^2*d*e^2) + (2*e*Sin[c + d*x])/(11
*a^2*d*(e*Sec[c + d*x])^(5/2)) + (10*Sin[c + d*x])/(33*a^2*d*e*Sqrt[e*Sec[c + d*x]]) + (((4*I)/11)*e^2)/(d*(e*
Sec[c + d*x])^(7/2)*(a^2 + I*a^2*Tan[c + d*x]))

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

Mathematica [A]  time = 0.484103, size = 134, normalized size = 0.89 \[ -\frac{\sec ^4(c+d x) \left (-6 \sin (2 (c+d x))+7 \sin (4 (c+d x))+24 i \cos (2 (c+d x))-4 i \cos (4 (c+d x))+40 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+28 i\right )}{132 a^2 d (\tan (c+d x)-i)^2 (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

-(Sec[c + d*x]^4*(28*I + (24*I)*Cos[2*(c + d*x)] - (4*I)*Cos[4*(c + d*x)] + 40*Sqrt[Cos[c + d*x]]*EllipticF[(c
+ d*x)/2, 2]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) - 6*Sin[2*(c + d*x)] + 7*Sin[4*(c + d*x)]))/(132*a^2*d*(
e*Sec[c + d*x])^(3/2)*(-I + Tan[c + d*x])^2)

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Maple [A]  time = 0.339, size = 234, normalized size = 1.6 \begin{align*}{\frac{2\,\cos \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{33\,{a}^{2}d{e}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}} \left ( 6\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+6\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +5\,i\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +5\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

2/33/a^2/d*cos(d*x+c)*(e/cos(d*x+c))^(3/2)*(cos(d*x+c)-1)^2*(cos(d*x+c)+1)^2*(6*I*cos(d*x+c)^6+6*cos(d*x+c)^5*
sin(d*x+c)+5*I*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1
)/sin(d*x+c),I)+5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(
d*x+c),I)+3*cos(d*x+c)^3*sin(d*x+c)+5*cos(d*x+c)*sin(d*x+c))/e^3/sin(d*x+c)^4

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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(1/(e*sec(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 (264 \, a^{2} d e^{2} e^{\left (6 i \, d x + 6 i \, c\right )}{\rm integral}\left (-\frac{5 i \, \sqrt{2} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac{1}{2} i \, d x - \frac{1}{2} i \, c\right )}}{33 \, a^{2} d e^{2}}, x\right ) + \sqrt{2} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-11 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 30 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 56 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{264 \, a^{2} d e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/264*(264*a^2*d*e^2*e^(6*I*d*x + 6*I*c)*integral(-5/33*I*sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*
d*x - 1/2*I*c)/(a^2*d*e^2), x) + sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-11*I*e^(8*I*d*x + 8*I*c) + 30*I*e
^(6*I*d*x + 6*I*c) + 56*I*e^(4*I*d*x + 4*I*c) + 18*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(1/2*I*d*x + 1/2*I*c))*e^(-6
*I*d*x - 6*I*c)/(a^2*d*e^2)

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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(1/(e*sec(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{1}{\left (e \sec \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(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)^2), x)