3.249 \(\int \frac{(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^3} \, dx\)
Optimal. Leaf size=116 \[ -\frac{6 i e^4}{5 a^3 d \sqrt{e \sec (c+d x)}}-\frac{6 e^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{5 a d (a+i a \tan (c+d x))^2} \]
[Out]
(((-6*I)/5)*e^4)/(a^3*d*Sqrt[e*Sec[c + d*x]]) - (6*e^4*EllipticE[(c + d*x)/2, 2])/(5*a^3*d*Sqrt[Cos[c + d*x]]*
Sqrt[e*Sec[c + d*x]]) + (((4*I)/5)*e^2*(e*Sec[c + d*x])^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^2)
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Rubi [A] time = 0.132058, 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 =
{3500, 3501, 3771, 2639} \[ -\frac{6 i e^4}{5 a^3 d \sqrt{e \sec (c+d x)}}-\frac{6 e^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{5 a d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
[In]
Int[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^3,x]
[Out]
(((-6*I)/5)*e^4)/(a^3*d*Sqrt[e*Sec[c + d*x]]) - (6*e^4*EllipticE[(c + d*x)/2, 2])/(5*a^3*d*Sqrt[Cos[c + d*x]]*
Sqrt[e*Sec[c + d*x]]) + (((4*I)/5)*e^2*(e*Sec[c + d*x])^(3/2))/(a*d*(a + I*a*Tan[c + d*x])^2)
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 3501
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d^2*
(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(d^2*(m - 2))/(a*(m + n -
1)), 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] && LtQ[n, 0] && GtQ[m, 1] && !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 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))^3} \, dx &=\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{5 a d (a+i a \tan (c+d x))^2}-\frac{\left (3 e^2\right ) \int \frac{(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx}{5 a^2}\\ &=-\frac{6 i e^4}{5 a^3 d \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{5 a d (a+i a \tan (c+d x))^2}-\frac{\left (3 e^4\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{6 i e^4}{5 a^3 d \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{5 a d (a+i a \tan (c+d x))^2}-\frac{\left (3 e^4\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^3 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=-\frac{6 i e^4}{5 a^3 d \sqrt{e \sec (c+d x)}}-\frac{6 e^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{3/2}}{5 a d (a+i a \tan (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 0.575989, size = 117, normalized size = 1.01 \[ \frac{2 e e^{-i d x} \left (-2+\frac{6 e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right ) (e \sec (c+d x))^{5/2} (\cos (c+2 d x)+i \sin (c+2 d x))}{5 a^3 d (\tan (c+d x)-i)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^3,x]
[Out]
(2*e*(-2 + (6*E^((2*I)*(c + d*x))*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(
c + d*x))])*(e*Sec[c + d*x])^(5/2)*(Cos[c + 2*d*x] + I*Sin[c + 2*d*x]))/(5*a^3*d*E^(I*d*x)*(-I + Tan[c + d*x])
^3)
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Maple [B] time = 0.243, size = 1086, normalized size = 9.4 \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))^3,x)
[Out]
1/10/a^3/d*(e/cos(d*x+c))^(7/2)*(cos(d*x+c)-1)^2*cos(d*x+c)^3*(cos(d*x+c)+1)*(-12*I*(-cos(d*x+c)/(cos(d*x+c)+1
)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*EllipticF(I*(cos
(d*x+c)-1)/sin(d*x+c),I)-5*I*cos(d*x+c)*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)-20*I*cos(d*x+c)^2*sin(d*x+c)*(
-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+5*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)/sin(d*x+c)^2)*cos(d*x+c)*sin(d*x+c)-20*I*cos(d*x+c)*s
in(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+12*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1
/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)+16*I*co
s(d*x+c)^4*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-16*cos(d*x+c)^5*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2
)+12*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellipti
cE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)-16*cos(d*x+c)^4*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+16*I*cos(d*x
+c)^3*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+28*cos(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-24*
I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*s
in(d*x+c)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+24*I*cos(d*x+c)*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(
1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-12*I*
(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(c
os(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)+16*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-12*cos(d*x+c)*(-cos
(d*x+c)/(cos(d*x+c)+1)^2)^(1/2))/(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)/sin(d*x+c)^5
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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))^3,x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (5 \, a^{3} d e^{\left (3 i \, d x + 3 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 )}}{5 \, a^{3} d}, x\right ) + \sqrt{2}{\left (-6 i \, e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 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 (-3 i \, d x - 3 i \, c\right )}}{5 \, a^{3} 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))^3,x, algorithm="fricas")
[Out]
1/5*(5*a^3*d*e^(3*I*d*x + 3*I*c)*integral(3/5*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^3*d), x) + sqrt(2)*(-6*I*e^3*e^(4*I*d*x + 4*I*c) - 4*I*e^3*e^(2*I*d*x + 2*I*c) + 2*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^(-3*I*d*x - 3*I*c)/(a^3*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))**3,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 )}^{3}}\,{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))^3,x, algorithm="giac")
[Out]
integrate((e*sec(d*x + c))^(7/2)/(I*a*tan(d*x + c) + a)^3, x)