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]

-6/5*I*e^4/a^3/d/(e*sec(d*x+c))^(1/2)-6/5*e^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/
2*d*x+1/2*c),2^(1/2))/a^3/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)+4/5*I*e^2*(e*sec(d*x+c))^(3/2)/a/d/(a+I*a*ta
n(d*x+c))^2

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Rubi [A]  time = 0.13, antiderivative size = 116, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.65, size = 117, normalized size = 1.01 \[ \frac {2 e e^{-i d x} \left (-2+\frac {6 e^{2 i (c+d x)} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right ) (e \sec (c+d x))^{5/2} (\cos (c+2 d x)+i \sin (c+2 d x))}{5 a^3 d (\tan (c+d x)-i)^3} \]

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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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ \frac {{\left (5 \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} {\rm integral}\left (\frac {3 i \, \sqrt {2} e^{3} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, a^{3} d}, x\right ) + \sqrt {2} {\left (-6 i \, e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, e^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{5 \, a^{3} d} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

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)

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maple [B]  time = 1.32, size = 378, normalized size = 3.26 \[ -\frac {2 \left (-3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-4 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-3 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+4 \left (\cos ^{4}\left (d x +c \right )\right )+5 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-7 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \left (\cos ^{3}\left (d x +c \right )\right )}{5 a^{3} d \sin \left (d x +c \right )^{5}} \]

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]

-2/5/a^3/d*(-3*I*sin(d*x+c)*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*
(-1+cos(d*x+c))/sin(d*x+c),I)+3*I*sin(d*x+c)*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-4*I*cos(d*x+c)^3*sin(d*x+c)-3*I*sin(d*x+c)*(1/(1+cos(d*x+c)))^(
1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+3*I*sin(d*x+c)*(1/(1+cos(d*x+
c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+4*cos(d*x+c)^4+5*I*cos(
d*x+c)*sin(d*x+c)-7*cos(d*x+c)^2+3*cos(d*x+c))*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(e/cos(d*x+c))^(7/2)*cos(d*x
+c)^3/sin(d*x+c)^5

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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