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

Optimal. Leaf size=121 \[ -\frac{16 i \sqrt{a+i a \tan (c+d x)}}{21 a^2 d \sqrt{e \sec (c+d x)}}+\frac{8 i}{21 a d \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt{e \sec (c+d x)}} \]

[Out]

((2*I)/7)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + ((8*I)/21)/(a*d*Sqrt[e*Sec[c + d*x]]*Sqrt[a
+ I*a*Tan[c + d*x]]) - (((16*I)/21)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d*Sqrt[e*Sec[c + d*x]])

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Rubi [A]  time = 0.209232, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3502, 3488} \[ -\frac{16 i \sqrt{a+i a \tan (c+d x)}}{21 a^2 d \sqrt{e \sec (c+d x)}}+\frac{8 i}{21 a d \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)/7)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + ((8*I)/21)/(a*d*Sqrt[e*Sec[c + d*x]]*Sqrt[a
+ I*a*Tan[c + d*x]]) - (((16*I)/21)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d*Sqrt[e*Sec[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 3488

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{2 i}{7 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{4 \int \frac{1}{\sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{7 a}\\ &=\frac{2 i}{7 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{8 i}{21 a d \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}} \, dx}{21 a^2}\\ &=\frac{2 i}{7 d \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{8 i}{21 a d \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{21 a^2 d \sqrt{e \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.287931, size = 83, normalized size = 0.69 \[ -\frac{\sec ^2(c+d x) (12 i \sin (2 (c+d x))+9 \cos (2 (c+d x))-7)}{21 a d (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^2*(-7 + 9*Cos[2*(c + d*x)] + (12*I)*Sin[2*(c + d*x)]))/(21*a*d*Sqrt[e*Sec[c + d*x]]*(-I + Tan[c
+ d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.308, size = 106, normalized size = 0.9 \begin{align*} -{\frac{18\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-24\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -16\,i}{21\,{a}^{2}d \left ( 2\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1 \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21/d/a^2*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(9*I*cos(d*x+c)^2-12*cos(d*x+c)*sin(d*x+c)-8*I)/(2*
I*cos(d*x+c)*sin(d*x+c)+2*cos(d*x+c)^2-1)/(e/cos(d*x+c))^(1/2)

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Maxima [A]  time = 1.93716, size = 176, normalized size = 1.45 \begin{align*} \frac{3 i \, \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 14 i \, \cos \left (\frac{3}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right ) - 21 i \, \cos \left (\frac{1}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right ) + 3 \, \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 14 \, \sin \left (\frac{3}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right ) + 21 \, \sin \left (\frac{1}{7} \, \arctan \left (\sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ), \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right )\right )\right )}{42 \, a^{\frac{3}{2}} d \sqrt{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/42*(3*I*cos(7/2*d*x + 7/2*c) + 14*I*cos(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 21*I*cos(
1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 3*sin(7/2*d*x + 7/2*c) + 14*sin(3/7*arctan2(sin(7/2
*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 21*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))/(a^(3
/2)*d*sqrt(e))

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Fricas [A]  time = 2.04918, size = 265, normalized size = 2.19 \begin{align*} \frac{\sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-21 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 7 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 17 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-\frac{7}{2} i \, d x - \frac{7}{2} i \, c\right )}}{42 \, a^{2} d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/42*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-21*I*e^(6*I*d*x + 6*I*c) - 7*I*e^(4
*I*d*x + 4*I*c) + 17*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-7/2*I*d*x - 7/2*I*c)/(a^2*d*e)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{e \sec \left (d x + c\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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