3.245 \(\int \frac{(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=178 \[ -\frac{22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac{22 e^7 \sin (c+d x) \sqrt{e \sec (c+d x)}}{5 a^3 d}+\frac{22 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{15 a^3 d}-\frac{22 e^8 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2} \]

[Out]

(-22*e^8*EllipticE[(c + d*x)/2, 2])/(5*a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (((22*I)/21)*e^4*(e*Se
c[c + d*x])^(7/2))/(a^3*d) + (22*e^7*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/(5*a^3*d) + (22*e^5*(e*Sec[c + d*x])^(
5/2)*Sin[c + d*x])/(15*a^3*d) - (((4*I)/3)*e^2*(e*Sec[c + d*x])^(11/2))/(a*d*(a + I*a*Tan[c + d*x])^2)

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Rubi [A]  time = 0.170492, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac{22 e^7 \sin (c+d x) \sqrt{e \sec (c+d x)}}{5 a^3 d}+\frac{22 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{15 a^3 d}-\frac{22 e^8 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(-22*e^8*EllipticE[(c + d*x)/2, 2])/(5*a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (((22*I)/21)*e^4*(e*Se
c[c + d*x])^(7/2))/(a^3*d) + (22*e^7*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/(5*a^3*d) + (22*e^5*(e*Sec[c + d*x])^(
5/2)*Sin[c + d*x])/(15*a^3*d) - (((4*I)/3)*e^2*(e*Sec[c + d*x])^(11/2))/(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))^{15/2}}{(a+i a \tan (c+d x))^3} \, dx &=-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}+\frac{\left (11 e^2\right ) \int \frac{(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx}{3 a^2}\\ &=-\frac{22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}+\frac{\left (11 e^4\right ) \int (e \sec (c+d x))^{7/2} \, dx}{3 a^3}\\ &=-\frac{22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac{22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^3 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}+\frac{\left (11 e^6\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 a^3}\\ &=-\frac{22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac{22 e^7 \sqrt{e \sec (c+d x)} \sin (c+d x)}{5 a^3 d}+\frac{22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^3 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}-\frac{\left (11 e^8\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac{22 e^7 \sqrt{e \sec (c+d x)} \sin (c+d x)}{5 a^3 d}+\frac{22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^3 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}-\frac{\left (11 e^8\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^3 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=-\frac{22 e^8 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{22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac{22 e^7 \sqrt{e \sec (c+d x)} \sin (c+d x)}{5 a^3 d}+\frac{22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^3 d}-\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 1.54052, size = 128, normalized size = 0.72 \[ -\frac{e^6 (\tan (c+d x)-i) (e \sec (c+d x))^{3/2} \left (77 e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-868 \cos (2 (c+d x))+143 i \tan (c+d x)+203 i \sin (3 (c+d x)) \sec (c+d x)-556\right )}{210 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^3,x]

[Out]

-(e^6*(e*Sec[c + d*x])^(3/2)*(-556 - 868*Cos[2*(c + d*x)] + (77*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric
2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((2*I)*(c + d*x)) + (203*I)*Sec[c + d*x]*Sin[3*(c + d*x)] + (143*I
)*Tan[c + d*x])*(-I + Tan[c + d*x]))/(210*a^3*d)

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Maple [B]  time = 0.323, size = 392, normalized size = 2.2 \begin{align*} -{\frac{2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{105\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ( 231\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}\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 ) -231\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +231\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \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 ) -231\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +231\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+140\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -294\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-15\,i\sin \left ( dx+c \right ) +63\,\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{15}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^3,x)

[Out]

-2/105/a^3/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(231*I*sin(d*x+c)*cos(d*x+c)^4*(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)-231*I*sin(d*x+c)*cos(d*x+c)^4*(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)+231*I*cos(d*x+c)^3*sin
(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)-23
1*I*cos(d*x+c)^3*sin(d*x+c)*(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)+231*cos(d*x+c)^4+140*I*cos(d*x+c)^2*sin(d*x+c)-294*cos(d*x+c)^3-15*I*sin(d*x+c)+63*cos(d*x+c
))*(e/cos(d*x+c))^(15/2)*cos(d*x+c)^4/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))^(15/2)/(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 (-462 i \, e^{7} e^{\left (7 i \, d x + 7 i \, c\right )} - 1694 i \, e^{7} e^{\left (5 i \, d x + 5 i \, c\right )} - 2266 i \, e^{7} e^{\left (3 i \, d x + 3 i \, c\right )} - 1274 i \, e^{7} e^{\left (i \, d x + i \, c\right )}\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 )} + 105 \,{\left (a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}{\rm integral}\left (\frac{11 i \, \sqrt{2} e^{7} \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 )}{105 \,{\left (a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/105*(sqrt(2)*(-462*I*e^7*e^(7*I*d*x + 7*I*c) - 1694*I*e^7*e^(5*I*d*x + 5*I*c) - 2266*I*e^7*e^(3*I*d*x + 3*I*
c) - 1274*I*e^7*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 105*(a^3*d*e^(6*I
*d*x + 6*I*c) + 3*a^3*d*e^(4*I*d*x + 4*I*c) + 3*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*integral(11/5*I*sqrt(2)*e^7
*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^(6*I*d*x + 6*I*c) + 3*a^3*d*e
^(4*I*d*x + 4*I*c) + 3*a^3*d*e^(2*I*d*x + 2*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))**(15/2)/(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{15}{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))^(15/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(15/2)/(I*a*tan(d*x + c) + a)^3, x)