3.3.46 \(\int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx\) [246]

Optimal. Leaf size=141 \[ \frac {6 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{a^3 d}-\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2} \]

[Out]

-18/5*I*e^4*(e*sec(d*x+c))^(5/2)/a^3/d+6*e^5*(e*sec(d*x+c))^(3/2)*sin(d*x+c)/a^3/d+6*e^6*(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^3/d-4
*I*e^2*(e*sec(d*x+c))^(9/2)/a/d/(a+I*a*tan(d*x+c))^2

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Rubi [A]
time = 0.13, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3581, 3582, 3853, 3856, 2720} \begin {gather*} \frac {6 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{a^3 d}+\frac {6 e^5 \sin (c+d x) (e \sec (c+d x))^{3/2}}{a^3 d}-\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(6*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(a^3*d) - (((18*I)/5)*e^4*(e*Sec[c +
d*x])^(5/2))/(a^3*d) + (6*e^5*(e*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(a^3*d) - ((4*I)*e^2*(e*Sec[c + d*x])^(9/2
))/(a*d*(a + I*a*Tan[c + d*x])^2)

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 3582

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 3853

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 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))^{13/2}}{(a+i a \tan (c+d x))^3} \, dx &=-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2}+\frac {\left (9 e^2\right ) \int \frac {(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx}{a^2}\\ &=-\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2}+\frac {\left (9 e^4\right ) \int (e \sec (c+d x))^{5/2} \, dx}{a^3}\\ &=-\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2}+\frac {\left (3 e^6\right ) \int \sqrt {e \sec (c+d x)} \, dx}{a^3}\\ &=-\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2}+\frac {\left (3 e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{a^3}\\ &=\frac {6 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{a^3 d}-\frac {18 i e^4 (e \sec (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{a d (a+i a \tan (c+d x))^2}\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 74, normalized size = 0.52 \begin {gather*} \frac {e^4 (e \sec (c+d x))^{5/2} \left (-18 i-20 i \cos (2 (c+d x))+30 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-5 \sin (2 (c+d x))\right )}{5 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(13/2)/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(e^4*(e*Sec[c + d*x])^(5/2)*(-18*I - (20*I)*Cos[2*(c + d*x)] + 30*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2]
- 5*Sin[2*(c + d*x)]))/(5*a^3*d)

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Maple [A]
time = 0.76, size = 213, normalized size = 1.51

method	result	size
default	\(\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{3}\left (d x +c \right )\right )+15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{2}\left (d x +c \right )\right )-20 i \left (\cos ^{2}\left (d x +c \right )\right )-5 \sin \left (d x +c \right ) \cos \left (d x +c \right )+i\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {13}{2}} \left (\cos ^{4}\left (d x +c \right )\right )}{5 a^{3} d \sin \left (d x +c \right )^{4}}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

2/5/a^3/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(15*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)*cos(d*x+c)^3+15*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)*cos(d*x+c)^2-20*I*cos(d*x+c)^2-5*sin(d*x+c)*cos(d*x+c)+I)*(
e/cos(d*x+c))^(13/2)*cos(d*x+c)^4/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))^(13/2)/(a+I*a*tan(d*x+c))^3,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 = 0.10, size = 147, normalized size = 1.04 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (25 i \, e^{\frac {13}{2}} + 15 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {13}{2}\right )} + 36 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {13}{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}} + 15 \, {\left (i \, \sqrt {2} e^{\frac {13}{2}} + i \, \sqrt {2} e^{\left (4 i \, d x + 4 i \, c + \frac {13}{2}\right )} + 2 i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c + \frac {13}{2}\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{5 \, {\left (a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-2/5*(sqrt(2)*(25*I*e^(13/2) + 15*I*e^(4*I*d*x + 4*I*c + 13/2) + 36*I*e^(2*I*d*x + 2*I*c + 13/2))*e^(1/2*I*d*x
+ 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1) + 15*(I*sqrt(2)*e^(13/2) + I*sqrt(2)*e^(4*I*d*x + 4*I*c + 13/2) + 2*
I*sqrt(2)*e^(2*I*d*x + 2*I*c + 13/2))*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(a^3*d*e^(4*I*d*x + 4*I*c)
+ 2*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*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((e*sec(d*x+c))**(13/2)/(a+I*a*tan(d*x+c))**3,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((e*sec(d*x+c))^(13/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(e^(13/2)*sec(d*x + c)^(13/2)/(I*a*tan(d*x + c) + a)^3, 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 )}^{13/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(13/2)/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int((e/cos(c + d*x))^(13/2)/(a + a*tan(c + d*x)*1i)^3, x)

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