3.413 \(\int \frac{(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=81 \[ -\frac{4 i a (a+i a \tan (c+d x))^{3/2}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{7 d (e \sec (c+d x))^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.153757, antiderivative size = 81, 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 = {3497, 3488} \[ -\frac{4 i a (a+i a \tan (c+d x))^{3/2}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{7 d (e \sec (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(7/2),x]

[Out]

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

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

Mathematica [A]  time = 0.39799, size = 92, normalized size = 1.14 \[ -\frac{2 a^2 (2 \tan (c+d x)+5 i) \sqrt{a+i a \tan (c+d x)} (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x)))}{21 d e^2 (\cos (d x)+i \sin (d x))^2 (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(7/2),x]

[Out]

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

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Maple [A]  time = 0.305, size = 105, normalized size = 1.3 \begin{align*} -{\frac{2\,{a}^{2} \left ( 6\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-6\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -i\cos \left ( dx+c \right ) -2\,\sin \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{21\,d{e}^{7}}\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{7}{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))^(7/2),x)

[Out]

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

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Maxima [A]  time = 2.07442, size = 127, normalized size = 1.57 \begin{align*} \frac{{\left (-7 i \, a^{2} \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) - 3 i \, a^{2} \cos \left (\frac{7}{3} \, \arctan \left (\sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ), \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )\right )\right ) + 7 \, a^{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3 \, a^{2} \sin \left (\frac{7}{3} \, \arctan \left (\sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ), \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right )\right )\right )\right )} \sqrt{a}}{21 \, d e^{\frac{7}{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))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.32796, size = 240, normalized size = 2.96 \begin{align*} \frac{{\left (-3 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 10 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 7 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{3}{2} i \, d x + \frac{3}{2} i \, c\right )}}{21 \, d e^{4}} \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))^(7/2),x, algorithm="fricas")

[Out]

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

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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))**(7/2),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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)/(e*sec(d*x + c))^(7/2), x)