3.433 \(\int \frac{(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=80 \[ \frac{4 i (e \sec (c+d x))^{3/2}}{21 a d (a+i a \tan (c+d x))^{3/2}}+\frac{2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(((2*I)/7)*(e*Sec[c + d*x])^(3/2))/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((4*I)/21)*(e*Sec[c + d*x])^(3/2))/(a*d
*(a + I*a*Tan[c + d*x])^(3/2))

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Rubi [A]  time = 0.165989, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3502, 3488} \[ \frac{4 i (e \sec (c+d x))^{3/2}}{21 a d (a+i a \tan (c+d x))^{3/2}}+\frac{2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((2*I)/7)*(e*Sec[c + d*x])^(3/2))/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((4*I)/21)*(e*Sec[c + d*x])^(3/2))/(a*d
*(a + I*a*Tan[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 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{(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}}+\frac{2 \int \frac{(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx}{7 a}\\ &=\frac{2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}}+\frac{4 i (e \sec (c+d x))^{3/2}}{21 a d (a+i a \tan (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.199161, size = 63, normalized size = 0.79 \[ \frac{2 (2 \tan (c+d x)-5 i) (e \sec (c+d x))^{3/2}}{21 a^2 d (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(2*(e*Sec[c + d*x])^(3/2)*(-5*I + 2*Tan[c + d*x]))/(21*a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.291, size = 112, normalized size = 1.4 \begin{align*}{\frac{-{\frac{2\,i}{21}} \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( 12\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -12\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+i\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +5\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) }{d{a}^{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((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-2/21*I/d/a^3*(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*(12*I*cos(d*x+c
)^3*sin(d*x+c)-12*cos(d*x+c)^4+I*sin(d*x+c)*cos(d*x+c)+5*cos(d*x+c)^2+2)

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Maxima [A]  time = 1.9025, size = 116, normalized size = 1.45 \begin{align*} \frac{{\left (3 i \, e \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 7 i \, e \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 ) + 3 \, e \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 7 \, e \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 )\right )} \sqrt{e}}{21 \, a^{\frac{5}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/21*(3*I*e*cos(7/2*d*x + 7/2*c) + 7*I*e*cos(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 3*e*si
n(7/2*d*x + 7/2*c) + 7*e*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*sqrt(e)/(a^(5/2)*d)

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Fricas [A]  time = 2.01614, size = 232, normalized size = 2.9 \begin{align*} \frac{{\left (7 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 10 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, e\right )} \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}} e^{\left (-\frac{7}{2} i \, d x - \frac{7}{2} i \, c\right )}}{21 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/21*(7*I*e*e^(4*I*d*x + 4*I*c) + 10*I*e*e^(2*I*d*x + 2*I*c) + 3*I*e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e
/(e^(2*I*d*x + 2*I*c) + 1))*e^(-7/2*I*d*x - 7/2*I*c)/(a^3*d)

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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((e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(3/2)/(I*a*tan(d*x + c) + a)^(5/2), x)