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

Optimal. Leaf size=121 \[ \frac{16 i \sqrt{e \sec (c+d x)}}{45 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{8 i \sqrt{e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac{2 i \sqrt{e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(((2*I)/9)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((8*I)/45)*Sqrt[e*Sec[c + d*x]])/(a*d*(a
+ I*a*Tan[c + d*x])^(3/2)) + (((16*I)/45)*Sqrt[e*Sec[c + d*x]])/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.219756, 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} \[ \frac{16 i \sqrt{e \sec (c+d x)}}{45 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{8 i \sqrt{e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac{2 i \sqrt{e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/9)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((8*I)/45)*Sqrt[e*Sec[c + d*x]])/(a*d*(a
+ I*a*Tan[c + d*x])^(3/2)) + (((16*I)/45)*Sqrt[e*Sec[c + d*x]])/(a^2*d*Sqrt[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 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{\sqrt{e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{2 i \sqrt{e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac{4 \int \frac{\sqrt{e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx}{9 a}\\ &=\frac{2 i \sqrt{e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac{8 i \sqrt{e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac{8 \int \frac{\sqrt{e \sec (c+d x)}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{45 a^2}\\ &=\frac{2 i \sqrt{e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac{8 i \sqrt{e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac{16 i \sqrt{e \sec (c+d x)}}{45 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.305609, size = 85, normalized size = 0.7 \[ -\frac{i \sec ^2(c+d x) \sqrt{e \sec (c+d x)} (20 i \sin (2 (c+d x))+25 \cos (2 (c+d x))+9)}{45 a^2 d (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/45)*Sec[c + d*x]^2*Sqrt[e*Sec[c + d*x]]*(9 + 25*Cos[2*(c + d*x)] + (20*I)*Sin[2*(c + d*x)]))/(a^2*d*(-I +
Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.313, size = 128, normalized size = 1.1 \begin{align*}{\frac{-{\frac{2\,i}{45}}\cos \left ( dx+c \right ) \left ( 20\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) -20\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}+3\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +7\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+8\,i\sin \left ( dx+c \right ) -4\,\cos \left ( dx+c \right ) \right ) }{d{a}^{3}}\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}}\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))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-2/45*I/d/a^3*(e/cos(d*x+c))^(1/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)*(20*I*cos(d*x+c)^
4*sin(d*x+c)-20*cos(d*x+c)^5+3*I*cos(d*x+c)^2*sin(d*x+c)+7*cos(d*x+c)^3+8*I*sin(d*x+c)-4*cos(d*x+c))

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Maxima [A]  time = 1.88968, size = 176, normalized size = 1.45 \begin{align*} \frac{\sqrt{e}{\left (5 i \, \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 18 i \, \cos \left (\frac{5}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 45 i \, \cos \left (\frac{1}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 18 \, \sin \left (\frac{5}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right ) + 45 \, \sin \left (\frac{1}{9} \, \arctan \left (\sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ), \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right )\right )\right )\right )}}{90 \, a^{\frac{5}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/90*sqrt(e)*(5*I*cos(9/2*d*x + 9/2*c) + 18*I*cos(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 4
5*I*cos(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 5*sin(9/2*d*x + 9/2*c) + 18*sin(5/9*arctan2
(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 45*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))
))/(a^(5/2)*d)

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Fricas [A]  time = 1.87818, size = 262, normalized size = 2.17 \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 (45 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 63 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 23 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac{9}{2} i \, d x - \frac{9}{2} i \, c\right )}}{90 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/90*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(45*I*e^(6*I*d*x + 6*I*c) + 63*I*e^(4
*I*d*x + 4*I*c) + 23*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-9/2*I*d*x - 9/2*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))**(1/2)/(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{\sqrt{e \sec \left (d x + c\right )}}{{\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))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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