3.212 \(\int \frac{(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx\)

Optimal. Leaf size=186 \[ \frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}+\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{2 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{11 d e^8}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]

[Out]

(2*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(11*d*e^8) + (6*a^3*Sin[c + d*x])/(5
5*d*e^5*(e*Sec[c + d*x])^(5/2)) + (2*a^3*Sin[c + d*x])/(11*d*e^7*Sqrt[e*Sec[c + d*x]]) - (((2*I)/15)*(a + I*a*
Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(15/2)) - (((12*I)/55)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x]
)^(11/2))

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Rubi [A]  time = 0.173392, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}+\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{2 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{11 d e^8}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(15/2),x]

[Out]

(2*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(11*d*e^8) + (6*a^3*Sin[c + d*x])/(5
5*d*e^5*(e*Sec[c + d*x])^(5/2)) + (2*a^3*Sin[c + d*x])/(11*d*e^7*Sqrt[e*Sec[c + d*x]]) - (((2*I)/15)*(a + I*a*
Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(15/2)) - (((12*I)/55)*(a^3 + I*a^3*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x]
)^(11/2))

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))^{15/2}} \, dx &=-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}+\frac{(3 a) \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{11/2}} \, dx}{5 e^2}\\ &=-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (21 a^3\right ) \int \frac{1}{(e \sec (c+d x))^{7/2}} \, dx}{55 e^4}\\ &=\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (3 a^3\right ) \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{11 e^6}\\ &=\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{a^3 \int \sqrt{e \sec (c+d x)} \, dx}{11 e^8}\\ &=\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (a^3 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{11 e^8}\\ &=\frac{2 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)}}{11 d e^8}+\frac{6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^3 \sin (c+d x)}{11 d e^7 \sqrt{e \sec (c+d x)}}-\frac{2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}\\ \end{align*}

Mathematica [A]  time = 1.85125, size = 170, normalized size = 0.91 \[ \frac{a^3 \sqrt{e \sec (c+d x)} (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x))) \left (-114 \sin (c+d x)-81 \sin (3 (c+d x))+33 \sin (5 (c+d x))-332 i \cos (c+d x)-154 i \cos (3 (c+d x))+22 i \cos (5 (c+d x))+240 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (3 (c+d x))-i \sin (3 (c+d x)))\right )}{1320 d e^8 (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(15/2),x]

[Out]

(a^3*Sqrt[e*Sec[c + d*x]]*((-332*I)*Cos[c + d*x] - (154*I)*Cos[3*(c + d*x)] + (22*I)*Cos[5*(c + d*x)] - 114*Si
n[c + d*x] + 240*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)]) - 81*Sin
[3*(c + d*x)] + 33*Sin[5*(c + d*x)])*(Cos[3*(c + 2*d*x)] + I*Sin[3*(c + 2*d*x)]))/(1320*d*e^8*(Cos[d*x] + I*Si
n[d*x])^3)

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Maple [A]  time = 0.378, size = 232, normalized size = 1.3 \begin{align*}{\frac{2\,{a}^{3}}{165\,d \left ( \cos \left ( dx+c \right ) \right ) ^{8}} \left ( -44\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+44\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+15\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+7\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +15\,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 ) +15\,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 ) +9\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +15\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{15}{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))^(15/2),x)

[Out]

2/165*a^3/d*(-44*I*cos(d*x+c)^8+44*sin(d*x+c)*cos(d*x+c)^7+15*I*cos(d*x+c)^6+7*cos(d*x+c)^5*sin(d*x+c)+15*I*co
s(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)+1
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)+9*cos(
d*x+c)^3*sin(d*x+c)+15*cos(d*x+c)*sin(d*x+c))/cos(d*x+c)^8/(e/cos(d*x+c))^(15/2)

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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{15}{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))^(15/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(15/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (2640 \, d e^{8} e^{\left (2 i \, d x + 2 i \, c\right )}{\rm integral}\left (-\frac{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 )}}{11 \, d e^{8}}, x\right ) + \sqrt{2}{\left (-11 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 73 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 218 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 446 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 235 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 55 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 )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2640 \, d e^{8}} \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))^(15/2),x, algorithm="fricas")

[Out]

1/2640*(2640*d*e^8*e^(2*I*d*x + 2*I*c)*integral(-1/11*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^8), x) + sqrt(2)*(-11*I*a^3*e^(10*I*d*x + 10*I*c) - 73*I*a^3*e^(8*I*d*x + 8*I*c) - 218*I
*a^3*e^(6*I*d*x + 6*I*c) - 446*I*a^3*e^(4*I*d*x + 4*I*c) - 235*I*a^3*e^(2*I*d*x + 2*I*c) + 55*I*a^3)*sqrt(e/(e
^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-2*I*d*x - 2*I*c)/(d*e^8)

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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))**(15/2),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{15}{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))^(15/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(15/2), x)