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

Optimal. Leaf size=122 \[ \frac{8 i a}{15 d e^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{15 d e^2 \sqrt{e \sec (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}} \]

[Out]

(((8*I)/15)*a)/(d*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/5)*Sqrt[a + I*a*Tan[c + d*x]]
)/(d*(e*Sec[c + d*x])^(5/2)) - (((16*I)/15)*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^2*Sqrt[e*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.196973, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3497, 3502, 3488} \[ \frac{8 i a}{15 d e^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{15 d e^2 \sqrt{e \sec (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((8*I)/15)*a)/(d*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/5)*Sqrt[a + I*a*Tan[c + d*x]]
)/(d*(e*Sec[c + d*x])^(5/2)) - (((16*I)/15)*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^2*Sqrt[e*Sec[c + d*x]])

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 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{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx &=-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}+\frac{(4 a) \int \frac{1}{\sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{5 e^2}\\ &=\frac{8 i a}{15 d e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}} \, dx}{15 e^2}\\ &=\frac{8 i a}{15 d e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}-\frac{16 i \sqrt{a+i a \tan (c+d x)}}{15 d e^2 \sqrt{e \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.185985, size = 63, normalized size = 0.52 \[ \frac{i \sqrt{a+i a \tan (c+d x)} (-4 i \sin (2 (c+d x))+\cos (2 (c+d x))-15)}{15 d e^2 \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/15)*(-15 + Cos[2*(c + d*x)] - (4*I)*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^2*Sqrt[e*Sec[c + d*
x]])

________________________________________________________________________________________

Maple [A]  time = 0.35, size = 85, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -16\,i \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15\,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))^(1/2)/(e*sec(d*x+c))^(5/2),x)

[Out]

2/15/d*(I*cos(d*x+c)^2+4*cos(d*x+c)*sin(d*x+c)-8*I)*cos(d*x+c)^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2
)*(e/cos(d*x+c))^(5/2)/e^5

________________________________________________________________________________________

Maxima [A]  time = 1.94218, size = 176, normalized size = 1.44 \begin{align*} \frac{\sqrt{a}{\left (5 i \, \cos \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) - 3 i \, \cos \left (\frac{5}{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 ) - 30 i \, \cos \left (\frac{1}{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 ) + 5 \, \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 3 \, \sin \left (\frac{5}{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 ) + 30 \, \sin \left (\frac{1}{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 )}}{30 \, 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))^(1/2)/(e*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/30*sqrt(a)*(5*I*cos(3/2*d*x + 3/2*c) - 3*I*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 30
*I*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 5*sin(3/2*d*x + 3/2*c) + 3*sin(5/3*arctan2(s
in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 30*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))
/(d*e^(5/2))

________________________________________________________________________________________

Fricas [A]  time = 2.0929, size = 262, normalized size = 2.15 \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 (-3 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 33 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 25 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac{3}{2} i \, d x - \frac{3}{2} i \, c\right )}}{30 \, d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-3*I*e^(6*I*d*x + 6*I*c) - 33*I*e^(4
*I*d*x + 4*I*c) - 25*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-3/2*I*d*x - 3/2*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))**(1/2)/(e*sec(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{i \, a \tan \left (d x + c\right ) + a}}{\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))^(1/2)/(e*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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