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

Optimal. Leaf size=114 \[ \frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{21 a d e^2}+\frac{10 \sin (c+d x)}{21 a d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{7 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{3/2}} \]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(21*a*d*e^2) + (10*Sin[c + d*x])/(21*a*
d*e*Sqrt[e*Sec[c + d*x]]) + ((2*I)/7)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.0930508, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3502, 3769, 3771, 2641} \[ \frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{21 a d e^2}+\frac{10 \sin (c+d x)}{21 a d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{7 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(21*a*d*e^2) + (10*Sin[c + d*x])/(21*a*
d*e*Sqrt[e*Sec[c + d*x]]) + ((2*I)/7)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x]))

Rule 3502

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

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))} \, dx &=\frac{2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))}+\frac{5 \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{7 a}\\ &=\frac{10 \sin (c+d x)}{21 a d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))}+\frac{5 \int \sqrt{e \sec (c+d x)} \, dx}{21 a e^2}\\ &=\frac{10 \sin (c+d x)}{21 a d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))}+\frac{\left (5 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 a 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)}}{21 a d e^2}+\frac{10 \sin (c+d x)}{21 a d e \sqrt{e \sec (c+d x)}}+\frac{2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.487852, size = 125, normalized size = 1.1 \[ -\frac{\sec ^3(c+d x) \left (5 i \sin (c+d x)+5 i \sin (3 (c+d x))-14 \cos (c+d x)+2 \cos (3 (c+d x))+20 i \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (c+d x)+i \sin (c+d x))\right )}{42 a d (\tan (c+d x)-i) (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^3*(-14*Cos[c + d*x] + 2*Cos[3*(c + d*x)] + (20*I)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*
(Cos[c + d*x] + I*Sin[c + d*x]) + (5*I)*Sin[c + d*x] + (5*I)*Sin[3*(c + d*x)]))/(42*a*d*(e*Sec[c + d*x])^(3/2)
*(-I + Tan[c + d*x]))

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Maple [A]  time = 0.336, size = 218, normalized size = 1.9 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}\cos \left ( dx+c \right ) }{21\,ad{e}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}} \left ( 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 ) +3\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+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)),x)

[Out]

2/21/a/d*(e/cos(d*x+c))^(3/2)*(cos(d*x+c)-1)^2*(cos(d*x+c)+1)^2*cos(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)+3*I*cos(d*x+c)^4+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)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (84 \, a d e^{2} e^{\left (4 i \, d x + 4 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 )}}{21 \, a d e^{2}}, x\right ) + \sqrt{2} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-7 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 9 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 19 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 (-4 i \, d x - 4 i \, c\right )}}{84 \, a 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)),x, algorithm="fricas")

[Out]

1/84*(84*a*d*e^2*e^(4*I*d*x + 4*I*c)*integral(-5/21*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*d*e^2), x) + sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-7*I*e^(6*I*d*x + 6*I*c) + 9*I*e^(4*I*d*
x + 4*I*c) + 19*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(1/2*I*d*x + 1/2*I*c))*e^(-4*I*d*x - 4*I*c)/(a*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)),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 )}}\,{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)),x, algorithm="giac")

[Out]

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