3.237 \(\int \frac{(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx\)
Optimal. Leaf size=115 \[ -\frac{6 e^3 \sin (c+d x) \sqrt{e \sec (c+d x)}}{a^2 d}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{6 e^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]
[Out]
(6*e^4*EllipticE[(c + d*x)/2, 2])/(a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (6*e^3*Sqrt[e*Sec[c + d*x]
]*Sin[c + d*x])/(a^2*d) + ((4*I)*e^2*(e*Sec[c + d*x])^(3/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))
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Rubi [A] time = 0.0888349, antiderivative size = 115, 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 =
{3500, 3768, 3771, 2639} \[ -\frac{6 e^3 \sin (c+d x) \sqrt{e \sec (c+d x)}}{a^2 d}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{6 e^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^2,x]
[Out]
(6*e^4*EllipticE[(c + d*x)/2, 2])/(a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (6*e^3*Sqrt[e*Sec[c + d*x]
]*Sin[c + d*x])/(a^2*d) + ((4*I)*e^2*(e*Sec[c + d*x])^(3/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))
Rule 3500
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
Rule 3768
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rubi steps
\begin{align*} \int \frac{(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx &=\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac{\left (3 e^2\right ) \int (e \sec (c+d x))^{3/2} \, dx}{a^2}\\ &=-\frac{6 e^3 \sqrt{e \sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (3 e^4\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{a^2}\\ &=-\frac{6 e^3 \sqrt{e \sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (3 e^4\right ) \int \sqrt{\cos (c+d x)} \, dx}{a^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{6 e^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{6 e^3 \sqrt{e \sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 0.473509, size = 80, normalized size = 0.7 \[ \frac{2 i e^3 e^{-i (c+d x)} \left (-1+3 \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )\right ) \sqrt{e \sec (c+d x)}}{a^2 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^2,x]
[Out]
((2*I)*e^3*(-1 + 3*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])*Sqrt
[e*Sec[c + d*x]])/(a^2*d*E^(I*(c + d*x)))
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Maple [B] time = 0.239, size = 1093, normalized size = 9.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x)
[Out]
1/a^2/d*(e/cos(d*x+c))^(7/2)*(cos(d*x+c)-1)^4*(cos(d*x+c)+1)^7*(-12*I*cos(d*x+c)^3*sin(d*x+c)*(-cos(d*x+c)/(co
s(d*x+c)+1)^2)^(3/2)-12*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^2*sin(d*x+c)-4*I*(-cos(d*x+c)/(cos(d
*x+c)+1)^2)^(3/2)*cos(d*x+c)*sin(d*x+c)+4*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^5-6*I*(1/(cos(d*x+c)
+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*EllipticE(I*(cos(
d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)-12*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)
*(1/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^2*sin(d*x+c)+I*cos(d*x+c)^2*ln(-
(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos
(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)+6*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^4+6*I*(cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d
*x+c),I)*cos(d*x+c)^3*sin(d*x+c)+12*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(
1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^2*sin(d*x+c)-I*ln(-2*(2*(-cos(d*x+
c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/si
n(d*x+c)^2)*cos(d*x+c)^2*sin(d*x+c)-4*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^3+6*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)*(-cos(d*x+c)/(cos(d*x+c)+1)^
2)^(1/2)*cos(d*x+c)*sin(d*x+c)-6*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(
cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^3*sin(d*x+c)-4*I*cos(d*x+c)^4*sin(d*x
+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)-8*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^2+2*(-cos(d*x+c)/(c
os(d*x+c)+1)^2)^(3/2))*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)/sin(d*x+c)^9
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (a^{2} d e^{\left (i \, d x + i \, c\right )}{\rm integral}\left (-\frac{3 i \, \sqrt{2} e^{3} \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 )}}{a^{2} d}, x\right ) + \sqrt{2}{\left (6 i \, e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, e^{3}\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 )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
[Out]
(a^2*d*e^(I*d*x + I*c)*integral(-3*I*sqrt(2)*e^3*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(a^
2*d), x) + sqrt(2)*(6*I*e^3*e^(2*I*d*x + 2*I*c) + 4*I*e^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/
2*I*c))*e^(-I*d*x - I*c)/(a^2*d)
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**2,x)
[Out]
Exception raised: AttributeError
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec \left (d x + c\right )\right )^{\frac{7}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
[Out]
integrate((e*sec(d*x + c))^(7/2)/(I*a*tan(d*x + c) + a)^2, x)