3.3.48 \(\int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^3} \, dx\) [248]

Optimal. Leaf size=116 \[ \frac {10 i e^4 \sqrt {e \sec (c+d x)}}{3 a^3 d}-\frac {10 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 a^3 d}+\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{3 a d (a+i a \tan (c+d x))^2} \]

[Out]

10/3*I*e^4*(e*sec(d*x+c))^(1/2)/a^3/d-10/3*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^3/d+4/3*I*e^2*(e*sec(d*x+c))^(5/2)/a/d/(a+I*a*t
an(d*x+c))^2

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3581, 3582, 3856, 2720} \begin {gather*} \frac {10 i e^4 \sqrt {e \sec (c+d x)}}{3 a^3 d}-\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)}}{3 a^3 d}+\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{3 a d (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^3,x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.50, size = 125, normalized size = 1.08 \begin {gather*} \frac {2 e^4 \sec ^3(c+d x) \sqrt {e \sec (c+d x)} \left (-7 i \cos (c+d x)+5 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (c+d x)+i \sin (c+d x))+3 \sin (c+d x)\right ) (-i \cos (2 (c+d x))+\sin (2 (c+d x)))}{3 a^3 d (-i+\tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.78, size = 203, normalized size = 1.75

method	result	size
default	\(\frac {2 \left (-5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+4 i \left (\cos ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 i\right ) \left (\cos ^{4}\left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}{3 a^{3} d \sin \left (d x +c \right )^{4}}\)	\(203\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

2/3/a^3/d*(-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)*EllipticF(I*(-1+cos(d*x+
c))/sin(d*x+c),I)-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))/s
in(d*x+c),I)+4*I*cos(d*x+c)^2+4*sin(d*x+c)*cos(d*x+c)+3*I)*cos(d*x+c)^4*(e/cos(d*x+c))^(9/2)*(-1+cos(d*x+c))^2
*(1+cos(d*x+c))^2/sin(d*x+c)^4

________________________________________________________________________________________

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))^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 88, normalized size = 0.76 \begin {gather*} -\frac {2 \, {\left (-5 i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c + \frac {9}{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (-2 i \, e^{\frac {9}{2}} - 5 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 (-2 i \, d x - 2 i \, c\right )}}{3 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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))**3,x)

[Out]

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

________________________________________________________________________________________

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))^3,x, algorithm="giac")

[Out]

integrate(e^(9/2)*sec(d*x + c)^(9/2)/(I*a*tan(d*x + c) + a)^3, x)

________________________________________________________________________________________

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 )}^3} \,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)^3,x)

[Out]

int((e/cos(c + d*x))^(9/2)/(a + a*tan(c + d*x)*1i)^3, x)

________________________________________________________________________________________