3.247 \(\int \frac{(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx\)
Optimal. Leaf size=141 \[ \frac{14 i e^4 (e \sec (c+d x))^{3/2}}{3 a^3 d}-\frac{14 e^5 \sin (c+d x) \sqrt{e \sec (c+d x)}}{a^3 d}+\frac{14 e^6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2} \]
[Out]
(14*e^6*EllipticE[(c + d*x)/2, 2])/(a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((14*I)/3)*e^4*(e*Sec[c
+ d*x])^(3/2))/(a^3*d) - (14*e^5*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/(a^3*d) + ((4*I)*e^2*(e*Sec[c + d*x])^(7/2
))/(a*d*(a + I*a*Tan[c + d*x])^2)
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Rubi [A] time = 0.14939, antiderivative size = 141, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used =
{3500, 3501, 3768, 3771, 2639} \[ \frac{14 i e^4 (e \sec (c+d x))^{3/2}}{3 a^3 d}-\frac{14 e^5 \sin (c+d x) \sqrt{e \sec (c+d x)}}{a^3 d}+\frac{14 e^6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
[In]
Int[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x])^3,x]
[Out]
(14*e^6*EllipticE[(c + d*x)/2, 2])/(a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((14*I)/3)*e^4*(e*Sec[c
+ d*x])^(3/2))/(a^3*d) - (14*e^5*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/(a^3*d) + ((4*I)*e^2*(e*Sec[c + d*x])^(7/2
))/(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 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))^{11/2}}{(a+i a \tan (c+d x))^3} \, dx &=\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2}-\frac{\left (7 e^2\right ) \int \frac{(e \sec (c+d x))^{7/2}}{a+i a \tan (c+d x)} \, dx}{a^2}\\ &=\frac{14 i e^4 (e \sec (c+d x))^{3/2}}{3 a^3 d}+\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2}-\frac{\left (7 e^4\right ) \int (e \sec (c+d x))^{3/2} \, dx}{a^3}\\ &=\frac{14 i e^4 (e \sec (c+d x))^{3/2}}{3 a^3 d}-\frac{14 e^5 \sqrt{e \sec (c+d x)} \sin (c+d x)}{a^3 d}+\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2}+\frac{\left (7 e^6\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{a^3}\\ &=\frac{14 i e^4 (e \sec (c+d x))^{3/2}}{3 a^3 d}-\frac{14 e^5 \sqrt{e \sec (c+d x)} \sin (c+d x)}{a^3 d}+\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2}+\frac{\left (7 e^6\right ) \int \sqrt{\cos (c+d x)} \, dx}{a^3 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{14 e^6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{14 i e^4 (e \sec (c+d x))^{3/2}}{3 a^3 d}-\frac{14 e^5 \sqrt{e \sec (c+d x)} \sin (c+d x)}{a^3 d}+\frac{4 i e^2 (e \sec (c+d x))^{7/2}}{a d (a+i a \tan (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 0.918654, size = 93, normalized size = 0.66 \[ \frac{i e^4 (e \sec (c+d x))^{3/2} \left (-7 \left (1+e^{2 i (c+d x)}\right )^{3/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+9 i \sin (2 (c+d x))+33 \cos (2 (c+d x))+35\right )}{3 a^3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(e*Sec[c + d*x])^(11/2)/(a + I*a*Tan[c + d*x])^3,x]
[Out]
((I/3)*e^4*(e*Sec[c + d*x])^(3/2)*(35 + 33*Cos[2*(c + d*x)] - 7*(1 + E^((2*I)*(c + d*x)))^(3/2)*Hypergeometric
2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + (9*I)*Sin[2*(c + d*x)]))/(a^3*d)
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Maple [B] time = 0.298, size = 1562, normalized size = 11.1 \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))^(11/2)/(a+I*a*tan(d*x+c))^3,x)
[Out]
-2/3/a^3/d*(cos(d*x+c)+1)^7*(cos(d*x+c)-1)^4*(-3*I*cos(d*x+c)^3*sin(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)+3*I*co
s(d*x+c)^3*sin(d*x+c)*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)-84*I*cos(d*x+c)^4*sin(d*x+c)*EllipticE(I*(cos(d*x+c)-1)/s
in(d*x+c),I)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-8
4*I*cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)+37*I*cos(d*x+c)^3*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1
)^2)^(3/2)+126*I*cos(d*x+c)^3*sin(d*x+c)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(-cos(d*x+c)/(cos(d*x+c)+1)^
2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+21*I*cos(d*x+c)^5*sin(d*x+c)*EllipticF(I*(
cos(d*x+c)-1)/sin(d*x+c),I)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x
+c)+1))^(1/2)+84*I*cos(d*x+c)^4*sin(d*x+c)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(-cos(d*x+c)/(cos(d*x+c)+1
)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+12*I*cos(d*x+c)^5*sin(d*x+c)*(-cos(d*x+c
)/(cos(d*x+c)+1)^2)^(3/2)+I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*sin(d*x+c)+21*I*cos(d*x+c)*sin(d*x+c)*(-cos(d
*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)-12*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^6-126*I*cos(d*x+c)^3*sin(d*x+c)*Ellipti
cE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)-15*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^5+3*I*cos(d*x+c)*sin(d*x+c)*(-cos(d*x+c
)/(cos(d*x+c)+1)^2)^(3/2)+18*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^4+84*I*cos(d*x+c)^2*sin(d*x+c)*(-
cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)+24*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^3-21*I*cos(d*x+c)*sin(d*x+c)*(-cos
(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)-21*I*cos(d*x+c)^5*sin(d*x+c)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*(-cos(d*x+c)/(cos(
d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+36*I*cos(d*x+c)^4*sin(d*x+c)*(-c
os(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)-6*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^2+15*I*cos(d*x+c)^2*sin(d*
x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)-9*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2))*(e/cos(d*x+c))^(1
1/2)*cos(d*x+c)*(-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))^(11/2)/(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{\sqrt{2}{\left (42 i \, e^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 70 i \, e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 24 i \, e^{5}\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 )} + 3 \,{\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}\right )}{\rm integral}\left (-\frac{7 i \, \sqrt{2} e^{5} \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^{3} d}, x\right )}{3 \,{\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
[Out]
1/3*(sqrt(2)*(42*I*e^5*e^(4*I*d*x + 4*I*c) + 70*I*e^5*e^(2*I*d*x + 2*I*c) + 24*I*e^5)*sqrt(e/(e^(2*I*d*x + 2*I
*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 3*(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + I*c))*integral(-7*I*sqrt(2
)*e^5*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))/(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*
d*e^(I*d*x + I*c))
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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))**(11/2)/(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{11}{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))^(11/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((e*sec(d*x + c))^(11/2)/(I*a*tan(d*x + c) + a)^3, x)