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

Optimal. Leaf size=105 \[ -\frac{2 i e^2 (e \sec (c+d x))^{5/2}}{5 a d}+\frac{2 e^3 \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 a d}+\frac{2 e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.0891565, antiderivative size = 105, normalized size of antiderivative = 1., 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 = {3501, 3768, 3771, 2641} \[ -\frac{2 i e^2 (e \sec (c+d x))^{5/2}}{5 a d}+\frac{2 e^3 \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 a d}+\frac{2 e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3501

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 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]

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]

Rubi steps

\begin{align*} \int \frac{(e \sec (c+d x))^{9/2}}{a+i a \tan (c+d x)} \, dx &=-\frac{2 i e^2 (e \sec (c+d x))^{5/2}}{5 a d}+\frac{e^2 \int (e \sec (c+d x))^{5/2} \, dx}{a}\\ &=-\frac{2 i e^2 (e \sec (c+d x))^{5/2}}{5 a d}+\frac{2 e^3 (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 a d}+\frac{e^4 \int \sqrt{e \sec (c+d x)} \, dx}{3 a}\\ &=-\frac{2 i e^2 (e \sec (c+d x))^{5/2}}{5 a d}+\frac{2 e^3 (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 a d}+\frac{\left (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}\\ &=\frac{2 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 d}-\frac{2 i e^2 (e \sec (c+d x))^{5/2}}{5 a d}+\frac{2 e^3 (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.584932, size = 62, normalized size = 0.59 \[ \frac{e^2 (e \sec (c+d x))^{5/2} \left (5 \sin (2 (c+d x))+10 \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-6 i\right )}{15 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*(e*Sec[c + d*x])^(5/2)*(-6*I + 10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)]))/(15
*a*d)

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Maple [A]  time = 0.239, size = 202, normalized size = 1.9 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{15\,ad \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( 5\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{3}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +5\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -3\,i \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{9}{2}}}} \end{align*}

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)),x)

[Out]

2/15/a/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos
(d*x+c)^3*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*cos(d*x+c)^2*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+5*cos(d*x+c)*sin(d*x+c)-3*I)*(e/cos(d*x+c))^(9/2)*co
s(d*x+c)^2/sin(d*x+c)^4

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

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)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-10 i \, e^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 24 i \, e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 10 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 )} + 15 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}{\rm integral}\left (-\frac{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 d}, x\right )}{15 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \end{align*}

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)),x, algorithm="fricas")

[Out]

1/15*(sqrt(2)*(-10*I*e^4*e^(4*I*d*x + 4*I*c) - 24*I*e^4*e^(2*I*d*x + 2*I*c) + 10*I*e^4)*sqrt(e/(e^(2*I*d*x + 2
*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 15*(a*d*e^(4*I*d*x + 4*I*c) + 2*a*d*e^(2*I*d*x + 2*I*c) + a*d)*integral(
-1/3*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*d), x))/(a*d*e^(4*I*d*x + 4*I
*c) + 2*a*d*e^(2*I*d*x + 2*I*c) + a*d)

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

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)),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec \left (d x + c\right )\right )^{\frac{9}{2}}}{i \, a \tan \left (d x + c\right ) + a}\,{d x} \end{align*}

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)),x, algorithm="giac")

[Out]

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