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

Optimal. Leaf size=121 \[ \frac{16 i \sqrt{e \sec (c+d x)}}{15 d e^2 \sqrt{a+i a \tan (c+d x)}}-\frac{8 i \sqrt{a+i a \tan (c+d x)}}{15 a d (e \sec (c+d x))^{3/2}}+\frac{2 i}{5 d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}} \]

[Out]

((2*I)/5)/(d*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/15)*Sqrt[e*Sec[c + d*x]])/(d*e^2*Sq
rt[a + I*a*Tan[c + d*x]]) - (((8*I)/15)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(e*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.220198, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3502, 3497, 3488} \[ \frac{16 i \sqrt{e \sec (c+d x)}}{15 d e^2 \sqrt{a+i a \tan (c+d x)}}-\frac{8 i \sqrt{a+i a \tan (c+d x)}}{15 a d (e \sec (c+d x))^{3/2}}+\frac{2 i}{5 d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)/5)/(d*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/15)*Sqrt[e*Sec[c + d*x]])/(d*e^2*Sq
rt[a + I*a*Tan[c + d*x]]) - (((8*I)/15)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(e*Sec[c + d*x])^(3/2))

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 3497

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] + Dist[(a*(m + n))/(m*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && 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}{(e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{2 i}{5 d (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}+\frac{4 \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{5 a}\\ &=\frac{2 i}{5 d (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}-\frac{8 i \sqrt{a+i a \tan (c+d x)}}{15 a d (e \sec (c+d x))^{3/2}}+\frac{8 \int \frac{\sqrt{e \sec (c+d x)}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 e^2}\\ &=\frac{2 i}{5 d (e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}}+\frac{16 i \sqrt{e \sec (c+d x)}}{15 d e^2 \sqrt{a+i a \tan (c+d x)}}-\frac{8 i \sqrt{a+i a \tan (c+d x)}}{15 a d (e \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.250152, size = 68, normalized size = 0.56 \[ -\frac{i \sec ^2(c+d x) (4 i \sin (2 (c+d x))+\cos (2 (c+d x))-15)}{15 d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/15)*Sec[c + d*x]^2*(-15 + Cos[2*(c + d*x)] + (4*I)*Sin[2*(c + d*x)]))/(d*(e*Sec[c + d*x])^(3/2)*Sqrt[a +
I*a*Tan[c + d*x]])

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Maple [A]  time = 0.319, size = 105, normalized size = 0.9 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( 3\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +4\,i\cos \left ( dx+c \right ) +8\,\sin \left ( dx+c \right ) \right ) }{15\,ad{e}^{3}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\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))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

2/15/d/a*(e/cos(d*x+c))^(3/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^2*(3*I*cos(d*x+c)^3+3*
cos(d*x+c)^2*sin(d*x+c)+4*I*cos(d*x+c)+8*sin(d*x+c))/e^3

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Maxima [A]  time = 1.93319, size = 176, normalized size = 1.45 \begin{align*} \frac{3 i \, \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) - 5 i \, \cos \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 30 i \, \cos \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 3 \, \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 30 \, \sin \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right )}{30 \, \sqrt{a} d e^{\frac{3}{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))^(1/2),x, algorithm="maxima")

[Out]

1/30*(3*I*cos(5/2*d*x + 5/2*c) - 5*I*cos(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*I*cos(1
/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 3*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin(5/2*d
*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))/(sqrt(a
)*d*e^(3/2))

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Fricas [A]  time = 2.06385, 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 (-5 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 25 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 33 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-\frac{5}{2} i \, d x - \frac{5}{2} i \, c\right )}}{30 \, 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))^(1/2),x, algorithm="fricas")

[Out]

1/30*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-5*I*e^(6*I*d*x + 6*I*c) + 25*I*e^(4
*I*d*x + 4*I*c) + 33*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5/2*I*d*x - 5/2*I*c)/(a*d*e^2)

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

[Out]

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

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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}} \sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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