3.414 \(\int \frac{(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=125 \[ -\frac{16 i a^2 \sqrt{a+i a \tan (c+d x)}}{45 d e^4 \sqrt{e \sec (c+d x)}}-\frac{8 i a (a+i a \tan (c+d x))^{3/2}}{45 d e^2 (e \sec (c+d x))^{5/2}}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}} \]

[Out]

(((-16*I)/45)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^4*Sqrt[e*Sec[c + d*x]]) - (((8*I)/45)*a*(a + I*a*Tan[c + d*
x])^(3/2))/(d*e^2*(e*Sec[c + d*x])^(5/2)) - (((2*I)/9)*(a + I*a*Tan[c + d*x])^(5/2))/(d*(e*Sec[c + d*x])^(9/2)
)

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Rubi [A]  time = 0.221543, antiderivative size = 125, 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 = {3497, 3488} \[ -\frac{16 i a^2 \sqrt{a+i a \tan (c+d x)}}{45 d e^4 \sqrt{e \sec (c+d x)}}-\frac{8 i a (a+i a \tan (c+d x))^{3/2}}{45 d e^2 (e \sec (c+d x))^{5/2}}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(9/2),x]

[Out]

(((-16*I)/45)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^4*Sqrt[e*Sec[c + d*x]]) - (((8*I)/45)*a*(a + I*a*Tan[c + d*
x])^(3/2))/(d*e^2*(e*Sec[c + d*x])^(5/2)) - (((2*I)/9)*(a + I*a*Tan[c + d*x])^(5/2))/(d*(e*Sec[c + d*x])^(9/2)
)

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{(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{9/2}} \, dx &=-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}}+\frac{(4 a) \int \frac{(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{5/2}} \, dx}{9 e^2}\\ &=-\frac{8 i a (a+i a \tan (c+d x))^{3/2}}{45 d e^2 (e \sec (c+d x))^{5/2}}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}}+\frac{\left (8 a^2\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}} \, dx}{45 e^4}\\ &=-\frac{16 i a^2 \sqrt{a+i a \tan (c+d x)}}{45 d e^4 \sqrt{e \sec (c+d x)}}-\frac{8 i a (a+i a \tan (c+d x))^{3/2}}{45 d e^2 (e \sec (c+d x))^{5/2}}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{9 d (e \sec (c+d x))^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.459157, size = 104, normalized size = 0.83 \[ \frac{a^2 \sqrt{a+i a \tan (c+d x)} (-20 i \sin (2 (c+d x))+25 \cos (2 (c+d x))+9) (\sin (2 (c+2 d x))-i \cos (2 (c+2 d x)))}{45 d e^4 (\cos (d x)+i \sin (d x))^2 \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(9/2),x]

[Out]

(a^2*(9 + 25*Cos[2*(c + d*x)] - (20*I)*Sin[2*(c + d*x)])*((-I)*Cos[2*(c + 2*d*x)] + Sin[2*(c + 2*d*x)])*Sqrt[a
+ I*a*Tan[c + d*x]])/(45*d*e^4*Sqrt[e*Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])^2)

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Maple [A]  time = 0.329, size = 115, normalized size = 0.9 \begin{align*} -{\frac{2\,{a}^{2} \left ( 10\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-10\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-4\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +8\,i \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{45\,d{e}^{9}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(9/2),x)

[Out]

-2/45/d*a^2*(10*I*cos(d*x+c)^4-10*cos(d*x+c)^3*sin(d*x+c)-I*cos(d*x+c)^2-4*cos(d*x+c)*sin(d*x+c)+8*I)*(a*(I*si
n(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(e/cos(d*x+c))^(9/2)*cos(d*x+c)^5/e^9

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Maxima [A]  time = 2.19263, size = 130, normalized size = 1.04 \begin{align*} \frac{{\left (-5 i \, a^{2} \cos \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) - 18 i \, a^{2} \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) - 45 i \, a^{2} \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, a^{2} \sin \left (\frac{9}{2} \, d x + \frac{9}{2} \, c\right ) + 18 \, a^{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 45 \, a^{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \sqrt{a}}{90 \, d e^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

1/90*(-5*I*a^2*cos(9/2*d*x + 9/2*c) - 18*I*a^2*cos(5/2*d*x + 5/2*c) - 45*I*a^2*cos(1/2*d*x + 1/2*c) + 5*a^2*si
n(9/2*d*x + 9/2*c) + 18*a^2*sin(5/2*d*x + 5/2*c) + 45*a^2*sin(1/2*d*x + 1/2*c))*sqrt(a)/(d*e^(9/2))

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Fricas [A]  time = 2.30069, size = 284, normalized size = 2.27 \begin{align*} \frac{{\left (-5 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 23 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 63 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 45 i \, a^{2}\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{1}{2} i \, d x + \frac{1}{2} i \, c\right )}}{90 \, d e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

1/90*(-5*I*a^2*e^(6*I*d*x + 6*I*c) - 23*I*a^2*e^(4*I*d*x + 4*I*c) - 63*I*a^2*e^(2*I*d*x + 2*I*c) - 45*I*a^2)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(d*e^5)

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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((a+I*a*tan(d*x+c))**(5/2)/(e*sec(d*x+c))**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\left (e \sec \left (d x + c\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)/(e*sec(d*x + c))^(9/2), x)