3.3.56 \(\int \frac {(e \sec (c+d x))^{13/2}}{(a+i a \tan (c+d x))^4} \, dx\) [256]

Optimal. Leaf size=157 \[ -\frac {10 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{a^4 d}-\frac {10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

-10*e^5*(e*sec(d*x+c))^(3/2)*sin(d*x+c)/a^4/d-10*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^4/d+4/3*I*e^2*(e*sec(d*x+c))^(9/2)/a/d/(a
+I*a*tan(d*x+c))^3+12*I*e^4*(e*sec(d*x+c))^(5/2)/d/(a^4+I*a^4*tan(d*x+c))

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Rubi [A]
time = 0.12, antiderivative size = 157, 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 = {3581, 3853, 3856, 2720} \begin {gather*} -\frac {10 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^4 d}-\frac {10 e^5 \sin (c+d x) (e \sec (c+d x))^{3/2}}{a^4 d}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(a^4*d) - (10*e^5*(e*Sec[c + d*x])
^(3/2)*Sin[c + d*x])/(a^4*d) + (((4*I)/3)*e^2*(e*Sec[c + d*x])^(9/2))/(a*d*(a + I*a*Tan[c + d*x])^3) + ((12*I)
*e^4*(e*Sec[c + d*x])^(5/2))/(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 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))^4} \, dx &=\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}-\frac {\left (3 e^2\right ) \int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^2} \, dx}{a^2}\\ &=\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (15 e^4\right ) \int (e \sec (c+d x))^{5/2} \, dx}{a^4}\\ &=-\frac {10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (5 e^6\right ) \int \sqrt {e \sec (c+d x)} \, dx}{a^4}\\ &=-\frac {10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (5 e^6 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{a^4}\\ &=-\frac {10 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^4 d}-\frac {10 e^5 (e \sec (c+d x))^{3/2} \sin (c+d x)}{a^4 d}+\frac {4 i e^2 (e \sec (c+d x))^{9/2}}{3 a d (a+i a \tan (c+d x))^3}+\frac {12 i e^4 (e \sec (c+d x))^{5/2}}{d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.64, size = 134, normalized size = 0.85 \begin {gather*} \frac {i e^6 \sec ^5(c+d x) \sqrt {e \sec (c+d x)} \left (21+19 \cos (2 (c+d x))+30 i \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (c+d x)+i \sin (c+d x))+11 i \sin (2 (c+d x))\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))}{3 a^4 d (-i+\tan (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*e^6*Sec[c + d*x]^5*Sqrt[e*Sec[c + d*x]]*(21 + 19*Cos[2*(c + d*x)] + (30*I)*Cos[c + d*x]^(3/2)*EllipticF
[(c + d*x)/2, 2]*(Cos[c + d*x] + I*Sin[c + d*x]) + (11*I)*Sin[2*(c + d*x)])*(Cos[3*(c + d*x)] + I*Sin[3*(c + d
*x)]))/(a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]
time = 0.88, size = 226, normalized size = 1.44

method	result	size
default	\(\frac {2 \left (-15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{2}\left (d x +c \right )\right )-15 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+8 i \left (\cos ^{3}\left (d x +c \right )\right )+8 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+12 i \cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {13}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}{3 a^{4} d \sin \left (d x +c \right )^{4}}\)	\(226\)

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))^4,x,method=_RETURNVERBOSE)

[Out]

2/3/a^4/d*(-15*I*cos(d*x+c)^2*(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)-15*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*EllipticF(I*(-
1+cos(d*x+c))/sin(d*x+c),I)+8*I*cos(d*x+c)^3+8*cos(d*x+c)^2*sin(d*x+c)+12*I*cos(d*x+c)+sin(d*x+c))*cos(d*x+c)^
5*(e/cos(d*x+c))^(13/2)*(-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))^(13/2)/(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 = 0.10, size = 134, normalized size = 0.85 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (-4 i \, e^{\frac {13}{2}} - 15 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {13}{2}\right )} - 21 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^{\left (4 i \, d x + 4 i \, c + \frac {13}{2}\right )} - 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 )}}{3 \, {\left (a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )}\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))^4,x, algorithm="fricas")

[Out]

-2/3*(sqrt(2)*(-4*I*e^(13/2) - 15*I*e^(4*I*d*x + 4*I*c + 13/2) - 21*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^(4*I*d*x + 4*I*c + 13/2) - I*sqrt(2)*e^(2*I*d*x +
2*I*c + 13/2))*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(a^4*d*e^(4*I*d*x + 4*I*c) + a^4*d*e^(2*I*d*x + 2
*I*c))

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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))**4,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))^4,x, algorithm="giac")

[Out]

integrate(e^(13/2)*sec(d*x + c)^(13/2)/(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 )}^{13/2}}{{\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))^(13/2)/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

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

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