3.221 \(\int \frac{(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx\)

Optimal. Leaf size=156 \[ \frac{2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}} \]

[Out]

(2*a^4*EllipticE[(c + d*x)/2, 2])/(39*d*e^6*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*a^4*Sin[c + d*x])/(1
17*d*e^5*(e*Sec[c + d*x])^(3/2)) - (((4*I)/13)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(13/2)) - (((4*
I)/117)*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(9/2))

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Rubi [A]  time = 0.146584, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3496, 3769, 3771, 2639} \[ \frac{2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(13/2),x]

[Out]

(2*a^4*EllipticE[(c + d*x)/2, 2])/(39*d*e^6*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*a^4*Sin[c + d*x])/(1
17*d*e^5*(e*Sec[c + d*x])^(3/2)) - (((4*I)/13)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(13/2)) - (((4*
I)/117)*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(9/2))

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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{(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx &=-\frac{4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}+\frac{a^2 \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{9/2}} \, dx}{13 e^2}\\ &=-\frac{4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{\left (5 a^4\right ) \int \frac{1}{(e \sec (c+d x))^{5/2}} \, dx}{117 e^4}\\ &=\frac{2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{a^4 \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{39 e^6}\\ &=\frac{2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}+\frac{a^4 \int \sqrt{\cos (c+d x)} \, dx}{39 e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{39 d e^6 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 a^4 \sin (c+d x)}{117 d e^5 (e \sec (c+d x))^{3/2}}-\frac{4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}\\ \end{align*}

Mathematica [C]  time = 6.92916, size = 450, normalized size = 2.88 \[ \frac{\sec ^3(c+d x) (a+i a \tan (c+d x))^4 \left (\left (-\frac{59 \sin (c)}{468}-\frac{59}{468} i \cos (c)\right ) \cos (3 d x)+\left (\frac{37 \sin (c)}{468}-\frac{37}{468} i \cos (c)\right ) \cos (5 d x)+\left (\frac{1}{52} \sin (3 c)-\frac{1}{52} i \cos (3 c)\right ) \cos (7 d x)+\left (\frac{55}{468} \cos (3 c)-\frac{55}{468} i \sin (3 c)\right ) \sin (d x)+\left (\frac{59 \cos (c)}{468}-\frac{59}{468} i \sin (c)\right ) \sin (3 d x)+\left (\frac{37 \cos (c)}{468}+\frac{37}{468} i \sin (c)\right ) \sin (5 d x)+\left (\frac{1}{52} \cos (3 c)+\frac{1}{52} i \sin (3 c)\right ) \sin (7 d x)+\csc (c) (24 \cos (c)+31 i \sin (c)) \left (-\frac{1}{468} \cos (3 c)+\frac{1}{468} i \sin (3 c)\right ) \cos (d x)\right )}{d (\cos (d x)+i \sin (d x))^4 (e \sec (c+d x))^{13/2}}-\frac{2 i \sqrt{2} e^{-i (3 c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sec ^{\frac{5}{2}}(c+d x) \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) (a+i a \tan (c+d x))^4}{117 \left (-1+e^{2 i c}\right ) d (\cos (d x)+i \sin (d x))^4 (e \sec (c+d x))^{13/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(13/2),x]

[Out]

(((-2*I)/117)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-3*Sqrt[1
+ E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x
))])*Sec[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^4)/(d*E^(I*(3*c + d*x))*(-1 + E^((2*I)*c))*(e*Sec[c + d*x])^(13
/2)*(Cos[d*x] + I*Sin[d*x])^4) + (Sec[c + d*x]^3*(Cos[3*d*x]*(((-59*I)/468)*Cos[c] - (59*Sin[c])/468) + Cos[5*
d*x]*(((-37*I)/468)*Cos[c] + (37*Sin[c])/468) + Cos[d*x]*Csc[c]*(24*Cos[c] + (31*I)*Sin[c])*(-Cos[3*c]/468 + (
I/468)*Sin[3*c]) + Cos[7*d*x]*((-I/52)*Cos[3*c] + Sin[3*c]/52) + ((55*Cos[3*c])/468 - ((55*I)/468)*Sin[3*c])*S
in[d*x] + ((59*Cos[c])/468 - ((59*I)/468)*Sin[c])*Sin[3*d*x] + ((37*Cos[c])/468 + ((37*I)/468)*Sin[c])*Sin[5*d
*x] + (Cos[3*c]/52 + (I/52)*Sin[3*c])*Sin[7*d*x])*(a + I*a*Tan[c + d*x])^4)/(d*(e*Sec[c + d*x])^(13/2)*(Cos[d*
x] + I*Sin[d*x])^4)

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Maple [B]  time = 0.357, size = 380, normalized size = 2.4 \begin{align*}{\frac{2\,{a}^{4}}{117\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}} \left ( -72\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}-72\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}+52\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+3\,i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) \cos \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}}}-3\,i\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 ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +88\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}+3\,i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \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}}}-3\,i\sin \left ( dx+c \right ){\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-17\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x)

[Out]

2/117*a^4/d*(-72*I*sin(d*x+c)*cos(d*x+c)^7-72*cos(d*x+c)^8+52*I*sin(d*x+c)*cos(d*x+c)^5+3*I*EllipticF(I*(cos(d
*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*I*El
lipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)+88*cos(d*x+c)^6+3*I*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*I*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2
)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-17*cos(d*x+c)^4-2*cos(d*x+c)^2+3*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)^7/(e/co
s(d*x+c))^(13/2)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-9 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 9 i \, a^{4} e^{\left (7 i \, d x + 7 i \, c\right )} - 37 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 37 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 59 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 59 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 55 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 31 i \, a^{4} e^{\left (i \, d x + i \, c\right )} - 24 i \, a^{4}\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 )} + 468 \,{\left (d e^{7} e^{\left (i \, d x + i \, c\right )} - d e^{7}\right )}{\rm integral}\left (\frac{\sqrt{2}{\left (-i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a^{4} e^{\left (i \, d x + i \, c\right )} - i \, a^{4}\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 )}}{39 \,{\left (d e^{7} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{7} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{7} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{468 \,{\left (d e^{7} e^{\left (i \, d x + i \, c\right )} - d e^{7}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x, algorithm="fricas")

[Out]

1/468*(sqrt(2)*(-9*I*a^4*e^(8*I*d*x + 8*I*c) + 9*I*a^4*e^(7*I*d*x + 7*I*c) - 37*I*a^4*e^(6*I*d*x + 6*I*c) + 37
*I*a^4*e^(5*I*d*x + 5*I*c) - 59*I*a^4*e^(4*I*d*x + 4*I*c) + 59*I*a^4*e^(3*I*d*x + 3*I*c) - 55*I*a^4*e^(2*I*d*x
+ 2*I*c) + 31*I*a^4*e^(I*d*x + I*c) - 24*I*a^4)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 4
68*(d*e^7*e^(I*d*x + I*c) - d*e^7)*integral(1/39*sqrt(2)*(-I*a^4*e^(2*I*d*x + 2*I*c) - 2*I*a^4*e^(I*d*x + I*c)
- I*a^4)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(d*e^7*e^(3*I*d*x + 3*I*c) - 2*d*e^7*e^(2*
I*d*x + 2*I*c) + d*e^7*e^(I*d*x + I*c)), x))/(d*e^7*e^(I*d*x + I*c) - d*e^7)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(13/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(13/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(13/2), x)