3.222 \(\int \frac{(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx\)

Optimal. Leaf size=187 \[ \frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}+\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{2 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 d e^8}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]

[Out]

(2*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(33*d*e^8) + (2*a^4*Sin[c + d*x])/(5
5*d*e^5*(e*Sec[c + d*x])^(5/2)) + (2*a^4*Sin[c + d*x])/(33*d*e^7*Sqrt[e*Sec[c + d*x]]) - (((4*I)/15)*a*(a + I*
a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(15/2)) - (((4*I)/55)*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x
])^(11/2))

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Rubi [A]  time = 0.166563, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3496, 3769, 3771, 2641} \[ \frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}+\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{2 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 d e^8}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(15/2),x]

[Out]

(2*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(33*d*e^8) + (2*a^4*Sin[c + d*x])/(5
5*d*e^5*(e*Sec[c + d*x])^(5/2)) + (2*a^4*Sin[c + d*x])/(33*d*e^7*Sqrt[e*Sec[c + d*x]]) - (((4*I)/15)*a*(a + I*
a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(15/2)) - (((4*I)/55)*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^2*(e*Sec[c + d*x
])^(11/2))

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))^4}{(e \sec (c+d x))^{15/2}} \, dx &=-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}+\frac{a^2 \int \frac{(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{11/2}} \, dx}{5 e^2}\\ &=-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (7 a^4\right ) \int \frac{1}{(e \sec (c+d x))^{7/2}} \, dx}{55 e^4}\\ &=\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{a^4 \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{11 e^6}\\ &=\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{a^4 \int \sqrt{e \sec (c+d x)} \, dx}{33 e^8}\\ &=\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}+\frac{\left (a^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{33 e^8}\\ &=\frac{2 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{33 d e^8}+\frac{2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac{2 a^4 \sin (c+d x)}{33 d e^7 \sqrt{e \sec (c+d x)}}-\frac{4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac{4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}}\\ \end{align*}

Mathematica [A]  time = 2.04776, size = 155, normalized size = 0.83 \[ -\frac{i a^4 \sqrt{e \sec (c+d x)} (\cos (4 (c+2 d x))+i \sin (4 (c+2 d x))) \left (-54 i \sin (2 (c+d x))-37 i \sin (4 (c+d x))+112 \cos (2 (c+d x))+48 \cos (4 (c+d x))+40 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\sin (4 (c+d x))+i \cos (4 (c+d x)))+64\right )}{660 d e^8 (\cos (d x)+i \sin (d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(15/2),x]

[Out]

((-I/660)*a^4*Sqrt[e*Sec[c + d*x]]*(64 + 112*Cos[2*(c + d*x)] + 48*Cos[4*(c + d*x)] - (54*I)*Sin[2*(c + d*x)]
- (37*I)*Sin[4*(c + d*x)] + 40*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(I*Cos[4*(c + d*x)] + Sin[4*(c + d
*x)]))*(Cos[4*(c + 2*d*x)] + I*Sin[4*(c + 2*d*x)]))/(d*e^8*(Cos[d*x] + I*Sin[d*x])^4)

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Maple [A]  time = 0.364, size = 232, normalized size = 1.2 \begin{align*}{\frac{2\,{a}^{4}}{165\,d \left ( \cos \left ( dx+c \right ) \right ) ^{8}} \left ( -88\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+88\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+60\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}-16\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +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{15}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(15/2),x)

[Out]

2/165*a^4/d*(-88*I*cos(d*x+c)^8+88*sin(d*x+c)*cos(d*x+c)^7+60*I*cos(d*x+c)^6-16*cos(d*x+c)^5*sin(d*x+c)+5*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)+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)^8/(e/cos(d*x+c))^(15/2)

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Maxima [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))^4/(e*sec(d*x+c))^(15/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1320 \, d e^{8}{\rm integral}\left (-\frac{i \, \sqrt{2} a^{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 )}}{33 \, d e^{8}}, x\right ) + \sqrt{2}{\left (-11 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 58 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 128 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 166 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 85 i \, a^{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 )}}{1320 \, d e^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(15/2),x, algorithm="fricas")

[Out]

1/1320*(1320*d*e^8*integral(-1/33*I*sqrt(2)*a^4*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^4*e^(8*I*d*x + 8*I*c) - 58*I*a^4*e^(6*I*d*x + 6*I*c) - 128*I*a^4*e^(4*I*d*x + 4*I*
c) - 166*I*a^4*e^(2*I*d*x + 2*I*c) - 85*I*a^4)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*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))**4/(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 )}^{4}}{\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))^4/(e*sec(d*x+c))^(15/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(15/2), x)