3.210 \(\int \frac{(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx\)

Optimal. Leaf size=155 \[ \frac{10 a^3 \sin (c+d x)}{77 d e^5 \sqrt{e \sec (c+d x)}}-\frac{20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac{10 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{77 d e^6}-\frac{2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}} \]

[Out]

(10*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(77*d*e^6) + (10*a^3*Sin[c + d*x])/
(77*d*e^5*Sqrt[e*Sec[c + d*x]]) - (((2*I)/11)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(11/2)) - (((20*I)
/77)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(7/2))

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Rubi [A]  time = 0.15469, antiderivative size = 155, normalized size of antiderivative = 1., 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 = {3497, 3496, 3769, 3771, 2641} \[ \frac{10 a^3 \sin (c+d x)}{77 d e^5 \sqrt{e \sec (c+d x)}}-\frac{20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac{10 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{77 d e^6}-\frac{2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(11/2),x]

[Out]

(10*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(77*d*e^6) + (10*a^3*Sin[c + d*x])/
(77*d*e^5*Sqrt[e*Sec[c + d*x]]) - (((2*I)/11)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(11/2)) - (((20*I)
/77)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x])^(7/2))

Rule 3497

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d*
Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(a*f*m), x] + Dist[(a*(m + n))/(m*d^2), 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] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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{(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx &=-\frac{2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}+\frac{(5 a) \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{7/2}} \, dx}{11 e^2}\\ &=-\frac{2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac{20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac{\left (15 a^3\right ) \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{77 e^4}\\ &=\frac{10 a^3 \sin (c+d x)}{77 d e^5 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac{20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac{\left (5 a^3\right ) \int \sqrt{e \sec (c+d x)} \, dx}{77 e^6}\\ &=\frac{10 a^3 \sin (c+d x)}{77 d e^5 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac{20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac{\left (5 a^3 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{77 e^6}\\ &=\frac{10 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{77 d e^6}+\frac{10 a^3 \sin (c+d x)}{77 d e^5 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac{20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 1.25003, size = 148, normalized size = 0.95 \[ \frac{a^3 \sqrt{e \sec (c+d x)} (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x))) \left (-15 \sin (c+d x)-15 \sin (3 (c+d x))-46 i \cos (c+d x)-22 i \cos (3 (c+d x))+20 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (3 (c+d x))-i \sin (3 (c+d x)))\right )}{154 d e^6 (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(11/2),x]

[Out]

(a^3*Sqrt[e*Sec[c + d*x]]*((-46*I)*Cos[c + d*x] - (22*I)*Cos[3*(c + d*x)] - 15*Sin[c + d*x] + 20*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)]) - 15*Sin[3*(c + d*x)])*(Cos[3*(c + 2*d
*x)] + I*Sin[3*(c + 2*d*x)]))/(154*d*e^6*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [A]  time = 0.287, size = 216, normalized size = 1.4 \begin{align*} -{\frac{2\,{a}^{3}}{77\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}} \left ( 28\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}-28\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) -11\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-5\,i\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\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}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(11/2),x)

[Out]

-2/77*a^3/d*(28*I*cos(d*x+c)^6-28*cos(d*x+c)^5*sin(d*x+c)-11*I*cos(d*x+c)^4-5*I*cos(d*x+c)*(1/(cos(d*x+c)+1))^
(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*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)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)-3*cos(d*x+c)^3*sin(d*x+c)-5*cos(d*x
+c)*sin(d*x+c))/cos(d*x+c)^6/(e/cos(d*x+c))^(11/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(11/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(11/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{308 \, d e^{6}{\rm integral}\left (-\frac{5 i \, \sqrt{2} a^{3} \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 )}}{77 \, d e^{6}}, x\right ) + \sqrt{2}{\left (-7 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 31 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 61 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 37 i \, a^{3}\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 )}}{308 \, d e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(11/2),x, algorithm="fricas")

[Out]

1/308*(308*d*e^6*integral(-5/77*I*sqrt(2)*a^3*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)/(d*e^
6), x) + sqrt(2)*(-7*I*a^3*e^(6*I*d*x + 6*I*c) - 31*I*a^3*e^(4*I*d*x + 4*I*c) - 61*I*a^3*e^(2*I*d*x + 2*I*c) -
37*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(11/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(11/2), x)