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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.158046, 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} \[ -\frac{4 i a \sqrt{a+i a \tan (c+d x)}}{5 d e^2 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^{3/2}}{5 d (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.365108, size = 84, normalized size = 1.04 \[ -\frac{2 a (2 \tan (c+d x)+3 i) (\cos (d x)-i \sin (d x)) \sqrt{a+i a \tan (c+d x)} (\cos (c+2 d x)+i \sin (c+2 d x))}{5 d e (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.296, size = 86, normalized size = 1.1 \begin{align*} -{\frac{2\,a \left ( i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +2\,i \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5\,d{e}^{5}}\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{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.88918, size = 80, normalized size = 0.99 \begin{align*} \frac{{\left (-i \, a \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) - 5 i \, a \cos \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, a \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \sqrt{a}}{5 \, d e^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.11201, size = 227, normalized size = 2.8 \begin{align*} \frac{{\left (-i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, a\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 )}}{5 \, d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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{3}{2}}}{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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