3.3.21 \(\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{13/2}} \, dx\) [221]

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

[Out]

2/117*a^4*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(3/2)+2/39*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^6/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-4/13*I*a*(a+I*a*tan(d*x+c))^3/d/(
e*sec(d*x+c))^(13/2)-4/117*I*(a^4+I*a^4*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(9/2)

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Rubi [A]
time = 0.12, antiderivative size = 156, normalized size of antiderivative = 1.00, 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 = {3577, 3854, 3856, 2719} \begin {gather*} \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 \left (a^4+i a^4 \tan (c+d x)\right )}{117 d e^2 (e \sec (c+d x))^{9/2}}-\frac {4 i a (a+i a \tan (c+d x))^3}{13 d (e \sec (c+d x))^{13/2}} \end {gather*}

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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 3577

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 3854

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 3856

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 {(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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.89, size = 450, normalized size = 2.88 \begin {gather*} -\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)}} \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right ) \sec ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^4}{117 d \left (-1+e^{2 i c}\right ) (e \sec (c+d x))^{13/2} (\cos (d x)+i \sin (d x))^4}+\frac {\sec ^3(c+d x) \left (\cos (3 d x) \left (-\frac {59}{468} i \cos (c)-\frac {59 \sin (c)}{468}\right )+\cos (5 d x) \left (-\frac {37}{468} i \cos (c)+\frac {37 \sin (c)}{468}\right )+\cos (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 (7 d x) \left (-\frac {1}{52} i \cos (3 c)+\frac {1}{52} \sin (3 c)\right )+\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)\right ) (a+i a \tan (c+d x))^4}{d (e \sec (c+d x))^{13/2} (\cos (d x)+i \sin (d x))^4} \end {gather*}

Antiderivative was successfully verified.

[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])*(-1/468*Cos[3*c] +
(I/468)*Sin[3*c]) + Cos[7*d*x]*((-1/52*I)*Cos[3*c] + Sin[3*c]/52) + ((55*Cos[3*c])/468 - ((55*I)/468)*Sin[3*c
])*Sin[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)*(Co
s[d*x] + I*Sin[d*x])^4)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (160 ) = 320\).
time = 0.64, size = 380, normalized size = 2.44

method	result	size
risch	\(-\frac {i \left (9 \,{\mathrm e}^{6 i \left (d x +c \right )}+28 \,{\mathrm e}^{4 i \left (d x +c \right )}+31 \,{\mathrm e}^{2 i \left (d x +c \right )}+24\right ) a^{4} \sqrt {2}}{468 d \,e^{6} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \left (-\frac {2 \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}{e \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (e \,{\mathrm e}^{2 i \left (d x +c \right )}+e \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \EllipticE \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \EllipticF \left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}}\right ) a^{4} \sqrt {2}\, \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{39 d \,e^{6} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(352\)
default	\(-\frac {2 a^{4} \left (72 i \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )+72 \left (\cos ^{8}\left (d x +c \right )\right )-52 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-88 \left (\cos ^{6}\left (d x +c \right )\right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+17 \left (\cos ^{4}\left (d x +c \right )\right )+2 \left (\cos ^{2}\left (d x +c \right )\right )-3 \cos \left (d x +c \right )\right )}{117 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{7} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {13}{2}}}\)	\(380\)

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,method=_RETURNVERBOSE)

[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*(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)*sin(d*x+c)*cos(d*x+c)+3*I*(
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)*sin(d*x+c)
*cos(d*x+c)-88*cos(d*x+c)^6-3*I*(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)*sin(d*x+c)+3*I*(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)*sin(d*x+c)+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/cos(d*x+c))^(13/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 114, normalized size = 0.73 \begin {gather*} \frac {{\left (24 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \frac {\sqrt {2} {\left (-9 i \, a^{4} e^{\left (7 i \, d x + 7 i \, c\right )} - 37 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 59 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 31 i \, a^{4} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-\frac {13}{2}\right )}}{468 \, d} \end {gather*}

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*(24*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(-9*I*a^
4*e^(7*I*d*x + 7*I*c) - 37*I*a^4*e^(5*I*d*x + 5*I*c) - 59*I*a^4*e^(3*I*d*x + 3*I*c) - 31*I*a^4*e^(I*d*x + I*c)
)*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-13/2)/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^(-13/2)/sec(d*x + c)^(13/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{13/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(13/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(13/2), x)

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