3.3.58 \(\int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^4} \, dx\) [258]

Optimal. Leaf size=132 \[ \frac {10 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{7 a d (a+i a \tan (c+d x))^3}-\frac {20 i e^4 \sqrt {e \sec (c+d x)}}{21 d \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

10/21*e^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/
2)*(e*sec(d*x+c))^(1/2)/a^4/d+4/7*I*e^2*(e*sec(d*x+c))^(5/2)/a/d/(a+I*a*tan(d*x+c))^3-20/21*I*e^4*(e*sec(d*x+c
))^(1/2)/d/(a^4+I*a^4*tan(d*x+c))

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Rubi [A]
time = 0.10, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3581, 3856, 2720} \begin {gather*} -\frac {20 i e^4 \sqrt {e \sec (c+d x)}}{21 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {10 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{7 a d (a+i a \tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(10*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(21*a^4*d) + (((4*I)/7)*e^2*(e*Sec[
c + d*x])^(5/2))/(a*d*(a + I*a*Tan[c + d*x])^3) - (((20*I)/21)*e^4*Sqrt[e*Sec[c + d*x]])/(d*(a^4 + I*a^4*Tan[c
+ d*x]))

Rule 2720

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

Rule 3581

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 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 {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^4} \, dx &=\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{7 a d (a+i a \tan (c+d x))^3}-\frac {\left (5 e^2\right ) \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx}{7 a^2}\\ &=\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{7 a d (a+i a \tan (c+d x))^3}-\frac {20 i e^4 \sqrt {e \sec (c+d x)}}{21 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\left (5 e^4\right ) \int \sqrt {e \sec (c+d x)} \, dx}{21 a^4}\\ &=\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{7 a d (a+i a \tan (c+d x))^3}-\frac {20 i e^4 \sqrt {e \sec (c+d x)}}{21 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\left (5 e^4 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 a^4}\\ &=\frac {10 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{7 a d (a+i a \tan (c+d x))^3}-\frac {20 i e^4 \sqrt {e \sec (c+d x)}}{21 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.60, size = 137, normalized size = 1.04 \begin {gather*} \frac {2 e^4 \sec ^4(c+d x) \sqrt {e \sec (c+d x)} \left (5 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-2 i (1+\cos (2 (c+d x))+4 i \sin (2 (c+d x)))\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))}{21 a^4 d (-i+\tan (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(2*e^4*Sec[c + d*x]^4*Sqrt[e*Sec[c + d*x]]*(5*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] +
I*Sin[2*(c + d*x)]) - (2*I)*(1 + Cos[2*(c + d*x)] + (4*I)*Sin[2*(c + d*x)]))*(Cos[2*(c + d*x)] + I*Sin[2*(c +
d*x)]))/(21*a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]
time = 0.79, size = 228, normalized size = 1.73

method	result	size
default	\(\frac {2 \left (24 i \left (\cos ^{4}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+24 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-28 i \left (\cos ^{2}\left (d x +c \right )\right )-16 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \left (\cos ^{4}\left (d x +c \right )\right ) \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}{21 a^{4} d \sin \left (d x +c \right )^{4}}\)	\(228\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

2/21/a^4/d*(24*I*cos(d*x+c)^4+5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*Ellipt
icF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+24*sin(d*x+c)*cos(d*x+c)^3+5*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)-28*I*cos(d*x+c)^2-16*sin(d*x+c)*cos(d*x+c))*(e/cos(d
*x+c))^(9/2)*cos(d*x+c)^4*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2/sin(d*x+c)^4

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 100, normalized size = 0.76 \begin {gather*} -\frac {2 \, {\left (5 i \, \sqrt {2} e^{\left (4 i \, d x + 4 i \, c + \frac {9}{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (-3 i \, e^{\frac {9}{2}} + 5 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {9}{2}\right )} + 2 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {9}{2}\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 (-4 i \, d x - 4 i \, c\right )}}{21 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-2/21*(5*I*sqrt(2)*e^(4*I*d*x + 4*I*c + 9/2)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(-3*I*e^(9/
2) + 5*I*e^(4*I*d*x + 4*I*c + 9/2) + 2*I*e^(2*I*d*x + 2*I*c + 9/2))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x +
2*I*c) + 1))*e^(-4*I*d*x - 4*I*c)/(a^4*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(9/2)/(a+I*a*tan(d*x+c))**4,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 7317 deep

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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((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate(e^(9/2)*sec(d*x + c)^(9/2)/(I*a*tan(d*x + c) + a)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{9/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(9/2)/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

int((e/cos(c + d*x))^(9/2)/(a + a*tan(c + d*x)*1i)^4, x)

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