3.255 \(\int \frac{(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=192 \[ -\frac{154 e^7 \sin (c+d x) \sqrt{e \sec (c+d x)}}{5 a^4 d}-\frac{154 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{15 a^4 d}+\frac{44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{154 e^8 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3} \]

[Out]

(154*e^8*EllipticE[(c + d*x)/2, 2])/(5*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (154*e^7*Sqrt[e*Sec[c
+ d*x]]*Sin[c + d*x])/(5*a^4*d) - (154*e^5*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(15*a^4*d) + ((4*I)*e^2*(e*Sec
[c + d*x])^(11/2))/(a*d*(a + I*a*Tan[c + d*x])^3) + (((44*I)/3)*e^4*(e*Sec[c + d*x])^(7/2))/(d*(a^4 + I*a^4*Ta
n[c + d*x]))

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Rubi [A]  time = 0.174935, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3500, 3768, 3771, 2639} \[ -\frac{154 e^7 \sin (c+d x) \sqrt{e \sec (c+d x)}}{5 a^4 d}-\frac{154 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{15 a^4 d}+\frac{44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{154 e^8 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(154*e^8*EllipticE[(c + d*x)/2, 2])/(5*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (154*e^7*Sqrt[e*Sec[c
+ d*x]]*Sin[c + d*x])/(5*a^4*d) - (154*e^5*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(15*a^4*d) + ((4*I)*e^2*(e*Sec
[c + d*x])^(11/2))/(a*d*(a + I*a*Tan[c + d*x])^3) + (((44*I)/3)*e^4*(e*Sec[c + d*x])^(7/2))/(d*(a^4 + I*a^4*Ta
n[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))^{15/2}}{(a+i a \tan (c+d x))^4} \, dx &=\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}-\frac{\left (11 e^2\right ) \int \frac{(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^2} \, dx}{a^2}\\ &=\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac{44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{\left (77 e^4\right ) \int (e \sec (c+d x))^{7/2} \, dx}{3 a^4}\\ &=-\frac{154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac{44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac{\left (77 e^6\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 a^4}\\ &=-\frac{154 e^7 \sqrt{e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}-\frac{154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac{44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\left (77 e^8\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 a^4}\\ &=-\frac{154 e^7 \sqrt{e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}-\frac{154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac{44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\left (77 e^8\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{154 e^8 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{154 e^7 \sqrt{e \sec (c+d x)} \sin (c+d x)}{5 a^4 d}-\frac{154 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^4 d}+\frac{4 i e^2 (e \sec (c+d x))^{11/2}}{a d (a+i a \tan (c+d x))^3}+\frac{44 i e^4 (e \sec (c+d x))^{7/2}}{3 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 1.33692, size = 124, normalized size = 0.65 \[ -\frac{i e^5 (e \sec (c+d x))^{5/2} \left (77 e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-1133 \cos (c+d x)-3 (33 i \sin (c+d x)+37 i \sin (3 (c+d x))+117 \cos (3 (c+d x)))\right )}{30 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((-I/30)*e^5*(e*Sec[c + d*x])^(5/2)*(-1133*Cos[c + d*x] + (77*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F
1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) - 3*(117*Cos[3*(c + d*x)] + (33*I)*Sin[c + d*x] + (37*
I)*Sin[3*(c + d*x)])))/(a^4*d)

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Maple [B]  time = 0.339, size = 1628, normalized size = 8.5 \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))^(15/2)/(a+I*a*tan(d*x+c))^4,x)

[Out]

2/15/a^4/d*(cos(d*x+c)+1)^7*(cos(d*x+c)-1)^4*(-9*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)+120*(-cos(d*x
+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^7+129*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^6-60*I*(-cos(d*x+
c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^2*sin(d*x+c)-30*I*cos(d*x+c)^4*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)*sin(d*x+c)
+30*I*cos(d*x+c)^4*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)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c
)*sin(d*x+c)-120*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^6*sin(d*x+c)-360*I*(-cos(d*x+c)/(cos(d*x+c)
+1)^2)^(3/2)*cos(d*x+c)^5*sin(d*x+c)-380*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^4*sin(d*x+c)-180*I*
(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^3*sin(d*x+c)-3*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)+1386*I*(-c
os(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)*sin(d*x+c)*cos(d*
x+c)^4*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-1386*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)*sin(d*x+c)*cos(d*x+c)^4*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+231*
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)*sin(d*x+c)*c
os(d*x+c)^6*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-219*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^5-231
*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^4+108*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^3+105*(
-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^2-231*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)*sin(d*x+c)*cos(d*x+c)^6*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+924*
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)*sin(d*x+c)*c
os(d*x+c)^5*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-924*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)*sin(d*x+c)*cos(d*x+c)^5*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+
924*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)*sin(d*x+
c)*cos(d*x+c)^3*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-924*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)*sin(d*x+c)*cos(d*x+c)^3*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c)
,I)+231*I*sin(d*x+c)*cos(d*x+c)^2*(-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)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-231*I*sin(d*x+c)*cos(d*x+c)^2*(-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))*(e/cos(d*x+c))^(15/2)*cos(d*x+c)^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))^(15/2)/(a+I*a*tan(d*x+c))^4,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 (6 i \, d x + 6 i \, c\right )} + 1232 i \, e^{7} e^{\left (4 i \, d x + 4 i \, c\right )} + 1034 i \, e^{7} e^{\left (2 i \, d x + 2 i \, c\right )} + 240 i \, e^{7}\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 )} + 15 \,{\left (a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )}{\rm integral}\left (-\frac{77 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^{4} d}, x\right )}{15 \,{\left (a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} 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))^(15/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/15*(sqrt(2)*(462*I*e^7*e^(6*I*d*x + 6*I*c) + 1232*I*e^7*e^(4*I*d*x + 4*I*c) + 1034*I*e^7*e^(2*I*d*x + 2*I*c)
+ 240*I*e^7)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 15*(a^4*d*e^(5*I*d*x + 5*I*c) + 2*a^
4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*integral(-77/5*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^4*d), x))/(a^4*d*e^(5*I*d*x + 5*I*c) + 2*a^4*d*e^(3*I*d*x + 3*I*c) + a^4*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))**(15/2)/(a+I*a*tan(d*x+c))**4,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 )}^{4}}\,{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))^4,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(15/2)/(I*a*tan(d*x + c) + a)^4, x)