3.213 \(\int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=215 \[ -\frac{22 a^4 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{22 i a^4 (e \sec (c+d x))^{3/2}}{9 d}+\frac{22 a^4 e \sin (c+d x) \sqrt{e \sec (c+d x)}}{3 d}+\frac{10 i \left (a^2+i a^2 \tan (c+d x)\right )^2 (e \sec (c+d x))^{3/2}}{21 d}+\frac{22 i \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{21 d}+\frac{2 i a (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}{9 d} \]

[Out]

(-22*a^4*e^2*EllipticE[(c + d*x)/2, 2])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((22*I)/9)*a^4*(e*Sec
[c + d*x])^(3/2))/d + (22*a^4*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (((2*I)/9)*a*(e*Sec[c + d*x])^(3/2)
*(a + I*a*Tan[c + d*x])^3)/d + (((10*I)/21)*(e*Sec[c + d*x])^(3/2)*(a^2 + I*a^2*Tan[c + d*x])^2)/d + (((22*I)/
21)*(e*Sec[c + d*x])^(3/2)*(a^4 + I*a^4*Tan[c + d*x]))/d

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Rubi [A]  time = 0.256976, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3498, 3486, 3768, 3771, 2639} \[ -\frac{22 a^4 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{22 i a^4 (e \sec (c+d x))^{3/2}}{9 d}+\frac{22 a^4 e \sin (c+d x) \sqrt{e \sec (c+d x)}}{3 d}+\frac{10 i \left (a^2+i a^2 \tan (c+d x)\right )^2 (e \sec (c+d x))^{3/2}}{21 d}+\frac{22 i \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{21 d}+\frac{2 i a (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-22*a^4*e^2*EllipticE[(c + d*x)/2, 2])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((22*I)/9)*a^4*(e*Sec
[c + d*x])^(3/2))/d + (22*a^4*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (((2*I)/9)*a*(e*Sec[c + d*x])^(3/2)
*(a + I*a*Tan[c + d*x])^3)/d + (((10*I)/21)*(e*Sec[c + d*x])^(3/2)*(a^2 + I*a^2*Tan[c + d*x])^2)/d + (((22*I)/
21)*(e*Sec[c + d*x])^(3/2)*(a^4 + I*a^4*Tan[c + d*x]))/d

Rule 3498

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 - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

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 2639

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

Rubi steps

\begin{align*} \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^4 \, dx &=\frac{2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}{9 d}+\frac{1}{3} (5 a) \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx\\ &=\frac{2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}{9 d}+\frac{10 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )^2}{21 d}+\frac{1}{21} \left (55 a^2\right ) \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx\\ &=\frac{2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}{9 d}+\frac{10 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )^2}{21 d}+\frac{22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{21 d}+\frac{1}{3} \left (11 a^3\right ) \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx\\ &=\frac{22 i a^4 (e \sec (c+d x))^{3/2}}{9 d}+\frac{2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}{9 d}+\frac{10 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )^2}{21 d}+\frac{22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{21 d}+\frac{1}{3} \left (11 a^4\right ) \int (e \sec (c+d x))^{3/2} \, dx\\ &=\frac{22 i a^4 (e \sec (c+d x))^{3/2}}{9 d}+\frac{22 a^4 e \sqrt{e \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}{9 d}+\frac{10 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )^2}{21 d}+\frac{22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{21 d}-\frac{1}{3} \left (11 a^4 e^2\right ) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx\\ &=\frac{22 i a^4 (e \sec (c+d x))^{3/2}}{9 d}+\frac{22 a^4 e \sqrt{e \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}{9 d}+\frac{10 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )^2}{21 d}+\frac{22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{21 d}-\frac{\left (11 a^4 e^2\right ) \int \sqrt{\cos (c+d x)} \, dx}{3 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=-\frac{22 a^4 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{22 i a^4 (e \sec (c+d x))^{3/2}}{9 d}+\frac{22 a^4 e \sqrt{e \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}{9 d}+\frac{10 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )^2}{21 d}+\frac{22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{21 d}\\ \end{align*}

Mathematica [C]  time = 7.48129, size = 429, normalized size = 2. \[ \frac{22 i \sqrt{2} e^{-i (3 c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) (a+i a \tan (c+d x))^4 (e \sec (c+d x))^{3/2}}{9 \left (-1+e^{2 i c}\right ) d \sec ^{\frac{11}{2}}(c+d x) (\cos (d x)+i \sin (d x))^4}+\frac{\cos ^5(c+d x) (a+i a \tan (c+d x))^4 (e \sec (c+d x))^{3/2} \left (\csc (c) \left (\frac{22}{3} \cos (4 c)-\frac{22}{3} i \sin (4 c)\right ) \cos (d x)+\sec (c) \left (\frac{2}{9} \cos (4 c)-\frac{2}{9} i \sin (4 c)\right ) \sin (d x) \sec ^4(c+d x)+\sec (c) (36 \cos (c)+7 i \sin (c)) \left (-\frac{2}{63} \sin (4 c)-\frac{2}{63} i \cos (4 c)\right ) \sec ^3(c+d x)+\sec (c) \left (-\frac{26}{9} \cos (4 c)+\frac{26}{9} i \sin (4 c)\right ) \sin (d x) \sec ^2(c+d x)+\sec (c) (24 \cos (c)+13 i \sin (c)) \left (\frac{2}{9} \sin (4 c)+\frac{2}{9} i \cos (4 c)\right ) \sec (c+d x)\right )}{d (\cos (d x)+i \sin (d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(((22*I)/9)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-3*Sqrt[1 +
E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))
])*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^4)/(d*E^(I*(3*c + d*x))*(-1 + E^((2*I)*c))*Sec[c + d*x]^(11/2
)*(Cos[d*x] + I*Sin[d*x])^4) + (Cos[c + d*x]^5*(e*Sec[c + d*x])^(3/2)*(Sec[c]*Sec[c + d*x]^3*(36*Cos[c] + (7*I
)*Sin[c])*(((-2*I)/63)*Cos[4*c] - (2*Sin[4*c])/63) + Cos[d*x]*Csc[c]*((22*Cos[4*c])/3 - ((22*I)/3)*Sin[4*c]) +
Sec[c]*Sec[c + d*x]*(24*Cos[c] + (13*I)*Sin[c])*(((2*I)/9)*Cos[4*c] + (2*Sin[4*c])/9) + Sec[c]*Sec[c + d*x]^4
*((2*Cos[4*c])/9 - ((2*I)/9)*Sin[4*c])*Sin[d*x] + Sec[c]*Sec[c + d*x]^2*((-26*Cos[4*c])/9 + ((26*I)/9)*Sin[4*c
])*Sin[d*x])*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4)

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Maple [A]  time = 0.325, size = 401, normalized size = 1.9 \begin{align*} -{\frac{2\,{a}^{4} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{63\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5} \left ( \cos \left ( dx+c \right ) \right ) ^{3}} \left ( 231\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}\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 ) -231\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) +231\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}\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 ) -231\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -168\,i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+231\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}-322\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+36\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +98\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-7 \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^4,x)

[Out]

-2/63*a^4/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(231*I*sin(d*x+c)*cos(d*x+c)^5*(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)-231*I*sin(d*x+c)*cos(d*x+c)^5*(1/(cos(d*x+c
)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)+231*I*sin(d*x+c)*cos(d*
x+c)^4*(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)-231
*I*sin(d*x+c)*cos(d*x+c)^4*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-
1)/sin(d*x+c),I)-168*I*sin(d*x+c)*cos(d*x+c)^3+231*cos(d*x+c)^5-322*cos(d*x+c)^4+36*I*cos(d*x+c)*sin(d*x+c)+98
*cos(d*x+c)^2-7)*(e/cos(d*x+c))^(3/2)/sin(d*x+c)^5/cos(d*x+c)^3

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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((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-462 i \, a^{4} e e^{\left (9 i \, d x + 9 i \, c\right )} - 812 i \, a^{4} e e^{\left (7 i \, d x + 7 i \, c\right )} - 1080 i \, a^{4} e e^{\left (5 i \, d x + 5 i \, c\right )} - 660 i \, a^{4} e e^{\left (3 i \, d x + 3 i \, c\right )} - 154 i \, a^{4} e e^{\left (i \, d x + i \, c\right )}\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 )} + 63 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}{\rm integral}\left (\frac{11 i \, \sqrt{2} a^{4} e \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 \, d}, x\right )}{63 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/63*(sqrt(2)*(-462*I*a^4*e*e^(9*I*d*x + 9*I*c) - 812*I*a^4*e*e^(7*I*d*x + 7*I*c) - 1080*I*a^4*e*e^(5*I*d*x +
5*I*c) - 660*I*a^4*e*e^(3*I*d*x + 3*I*c) - 154*I*a^4*e*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1
/2*I*d*x + 1/2*I*c) + 63*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2
*I*d*x + 2*I*c) + d)*integral(11/3*I*sqrt(2)*a^4*e*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/d
, x))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d
)

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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((e*sec(d*x+c))**(3/2)*(a+I*a*tan(d*x+c))**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)^4, x)