3.236 \(\int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=119 \[ \frac {10 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 a^2 d}+\frac {10 e^3 \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

10/3*e^3*(e*sec(d*x+c))^(3/2)*sin(d*x+c)/a^2/d+10/3*e^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt
icF(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^2/d-4*I*e^2*(e*sec(d*x+c))^(5/2)/d/(a^
2+I*a^2*tan(d*x+c))

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Rubi [A]  time = 0.09, antiderivative size = 119, 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 = {3500, 3768, 3771, 2641} \[ \frac {10 e^3 \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {10 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2641

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

Rule 3500

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 3768

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 3771

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))^2} \, dx &=-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 e^2\right ) \int (e \sec (c+d x))^{5/2} \, dx}{a^2}\\ &=\frac {10 e^3 (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 e^4\right ) \int \sqrt {e \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {10 e^3 (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \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}{3 a^2}\\ &=\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^2 d}+\frac {10 e^3 (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{5/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 67, normalized size = 0.56 \[ \frac {2 e^3 (e \sec (c+d x))^{3/2} \left (-\sin (c+d x)-6 i \cos (c+d x)+5 \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-10 i \, e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 14 i \, e^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 3 \, {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} {\rm integral}\left (-\frac {5 i \, \sqrt {2} e^{4} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{3 \, a^{2} d}, x\right )}{3 \, {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*(-10*I*e^4*e^(2*I*d*x + 2*I*c) - 14*I*e^4)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I
*c) + 3*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*integral(-5/3*I*sqrt(2)*e^4*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-
1/2*I*d*x - 1/2*I*c)/(a^2*d), x))/(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 1.18, size = 201, normalized size = 1.69 \[ \frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right )-6 i \cos \left (d x +c \right )-\sin \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \left (\cos ^{3}\left (d x +c \right )\right )}{3 a^{2} d \sin \left (d x +c \right )^{4}} \]

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))^2,x)

[Out]

2/3/a^2/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(5*I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),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)+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)*cos(d*x+c)-6*I*cos(d*x+c)-sin(d*x+c))*(e/cos(d*x+c))^(9/2)*co
s(d*x+c)^3/sin(d*x+c)^4

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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))^2,x, algorithm="maxima")

[Out]

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

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

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)^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**2,x)

[Out]

Timed out

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