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

Optimal. Leaf size=116 \[ \frac{2 a^2 \sin (c+d x)}{7 d e^3 \sqrt{e \sec (c+d x)}}+\frac{2 a^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 d e^4}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}} \]

[Out]

(2*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(7*d*e^4) + (2*a^2*Sin[c + d*x])/(7*
d*e^3*Sqrt[e*Sec[c + d*x]]) - (((4*I)/7)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0873508, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3496, 3769, 3771, 2641} \[ \frac{2 a^2 \sin (c+d x)}{7 d e^3 \sqrt{e \sec (c+d x)}}+\frac{2 a^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 d e^4}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(7*d*e^4) + (2*a^2*Sin[c + d*x])/(7*
d*e^3*Sqrt[e*Sec[c + d*x]]) - (((4*I)/7)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(7/2))

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{7/2}} \, dx &=-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}+\frac{\left (3 a^2\right ) \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{7 e^2}\\ &=\frac{2 a^2 \sin (c+d x)}{7 d e^3 \sqrt{e \sec (c+d x)}}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}+\frac{a^2 \int \sqrt{e \sec (c+d x)} \, dx}{7 e^4}\\ &=\frac{2 a^2 \sin (c+d x)}{7 d e^3 \sqrt{e \sec (c+d x)}}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}+\frac{\left (a^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{7 e^4}\\ &=\frac{2 a^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 d e^4}+\frac{2 a^2 \sin (c+d x)}{7 d e^3 \sqrt{e \sec (c+d x)}}-\frac{4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.877063, size = 133, normalized size = 1.15 \[ \frac{a^2 \sqrt{e \sec (c+d x)} (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x))) \left (-\sin (2 (c+d x))-2 i \cos (2 (c+d x))+2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (2 (c+d x))-i \sin (2 (c+d x)))-2 i\right )}{7 d e^4 (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[e*Sec[c + d*x]]*(-2*I - (2*I)*Cos[2*(c + d*x)] + 2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos
[2*(c + d*x)] - I*Sin[2*(c + d*x)]) - Sin[2*(c + d*x)])*(Cos[2*(c + 2*d*x)] + I*Sin[2*(c + 2*d*x)]))/(7*d*e^4*
(Cos[d*x] + I*Sin[d*x])^2)

________________________________________________________________________________________

Maple [A]  time = 0.203, size = 189, normalized size = 1.6 \begin{align*} -{\frac{2\,{a}^{2}}{7\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}} \left ( 2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-i\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \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))^2/(e*sec(d*x+c))^(7/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{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))^2/(e*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{14 \, d e^{4}{\rm integral}\left (-\frac{i \, \sqrt{2} a^{2} \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 )}}{7 \, d e^{4}}, x\right ) + \sqrt{2}{\left (-i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, a^{2}\right )} \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 )}}{14 \, d e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/14*(14*d*e^4*integral(-1/7*I*sqrt(2)*a^2*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)/(d*e^4),
x) + sqrt(2)*(-I*a^2*e^(4*I*d*x + 4*I*c) - 4*I*a^2*e^(2*I*d*x + 2*I*c) - 3*I*a^2)*sqrt(e/(e^(2*I*d*x + 2*I*c)
+ 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^4)

________________________________________________________________________________________

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))**2/(e*sec(d*x+c))**(7/2),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 )}^{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))^2/(e*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

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