3.201 \(\int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=202 \[ -\frac {2 a^3 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 a^3 e^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}+\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{3 d}+\frac {10 i \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}{33 d}+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{7/2}}{11 d} \]

[Out]

10/21*I*a^3*(e*sec(d*x+c))^(7/2)/d+2/3*a^3*e*(e*sec(d*x+c))^(5/2)*sin(d*x+c)/d-2*a^3*e^4*(cos(1/2*d*x+1/2*c)^2
)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)+2*a^3
*e^3*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/d+2/11*I*a*(e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^2/d+10/33*I*(e*sec(d*x
+c))^(7/2)*(a^3+I*a^3*tan(d*x+c))/d

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Rubi [A]  time = 0.20, antiderivative size = 202, normalized size of antiderivative = 1.00, 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} \[ \frac {2 a^3 e^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 a^3 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{3 d}+\frac {10 i \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}{33 d}+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{7/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(-2*a^3*e^4*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((10*I)/21)*a^3*(e*Sec[c
+ d*x])^(7/2))/d + (2*a^3*e^3*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/d + (2*a^3*e*(e*Sec[c + d*x])^(5/2)*Sin[c +
d*x])/(3*d) + (((2*I)/11)*a*(e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2)/d + (((10*I)/33)*(e*Sec[c + d*x])
^(7/2)*(a^3 + I*a^3*Tan[c + d*x]))/d

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]

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 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 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]

Rubi steps

\begin {align*} \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx &=\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {1}{11} (15 a) \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2 \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}+\frac {1}{3} \left (5 a^2\right ) \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x)) \, dx\\ &=\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}+\frac {1}{3} \left (5 a^3\right ) \int (e \sec (c+d x))^{7/2} \, dx\\ &=\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e (e \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}+\left (a^3 e^2\right ) \int (e \sec (c+d x))^{3/2} \, dx\\ &=\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 e (e \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}-\left (a^3 e^4\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx\\ &=\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 e (e \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}-\frac {\left (a^3 e^4\right ) \int \sqrt {\cos (c+d x)} \, dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {2 a^3 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {10 i a^3 (e \sec (c+d x))^{7/2}}{21 d}+\frac {2 a^3 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a^3 e (e \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {2 i a (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2}{11 d}+\frac {10 i (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )}{33 d}\\ \end {align*}

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Mathematica [C]  time = 7.85, size = 442, normalized size = 2.19 \[ \frac {2 i \sqrt {2} e^{-i (2 c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt {1+e^{2 i (c+d x)}}\right ) (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{7/2}}{3 \left (-1+e^{2 i c}\right ) d \sec ^{\frac {13}{2}}(c+d x) (\cos (d x)+i \sin (d x))^3}+\frac {\cos ^6(c+d x) (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{7/2} \left (\csc (c) (2 \cos (3 c)-2 i \sin (3 c)) \cos (d x)+\left (-\frac {2}{11} \sin (3 c)-\frac {2}{11} i \cos (3 c)\right ) \sec ^5(c+d x)+\sec (c) \left (-\frac {2}{3} \cos (3 c)+\frac {2}{3} i \sin (3 c)\right ) \sin (d x) \sec ^4(c+d x)+\sec (c) (12 \cos (c)+7 i \sin (c)) \left (\frac {2}{21} \sin (3 c)+\frac {2}{21} i \cos (3 c)\right ) \sec ^3(c+d x)+\sec (c) \left (\frac {2}{3} \cos (3 c)-\frac {2}{3} i \sin (3 c)\right ) \sin (d x) \sec ^2(c+d x)+\tan (c) \left (\frac {2}{3} \cos (3 c)-\frac {2}{3} i \sin (3 c)\right ) \sec (c+d x)\right )}{d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((2*I)/3)*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])^(7/2)*(a + I*a*Tan[c + d*x])^3)/(d*E^(I*(2*c + d*x))*(-1 + E^((2*I)*c))*Sec[c + d*x]^(13/2)
*(Cos[d*x] + I*Sin[d*x])^3) + (Cos[c + d*x]^6*(e*Sec[c + d*x])^(7/2)*(Sec[c + d*x]^5*(((-2*I)/11)*Cos[3*c] - (
2*Sin[3*c])/11) + Cos[d*x]*Csc[c]*(2*Cos[3*c] - (2*I)*Sin[3*c]) + Sec[c]*Sec[c + d*x]^3*(12*Cos[c] + (7*I)*Sin
[c])*(((2*I)/21)*Cos[3*c] + (2*Sin[3*c])/21) + Sec[c]*Sec[c + d*x]^2*((2*Cos[3*c])/3 - ((2*I)/3)*Sin[3*c])*Sin
[d*x] + Sec[c]*Sec[c + d*x]^4*((-2*Cos[3*c])/3 + ((2*I)/3)*Sin[3*c])*Sin[d*x] + Sec[c + d*x]*((2*Cos[3*c])/3 -
((2*I)/3)*Sin[3*c])*Tan[c])*(a + I*a*Tan[c + d*x])^3)/(d*(Cos[d*x] + I*Sin[d*x])^3)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-462 i \, a^{3} e^{3} e^{\left (11 i \, d x + 11 i \, c\right )} - 2618 i \, a^{3} e^{3} e^{\left (9 i \, d x + 9 i \, c\right )} - 1892 i \, a^{3} e^{3} e^{\left (7 i \, d x + 7 i \, c\right )} - 1740 i \, a^{3} e^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 814 i \, a^{3} e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 154 i \, a^{3} e^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 231 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} {\rm integral}\left (\frac {i \, \sqrt {2} a^{3} e^{3} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{d}, x\right )}{231 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/231*(sqrt(2)*(-462*I*a^3*e^3*e^(11*I*d*x + 11*I*c) - 2618*I*a^3*e^3*e^(9*I*d*x + 9*I*c) - 1892*I*a^3*e^3*e^(
7*I*d*x + 7*I*c) - 1740*I*a^3*e^3*e^(5*I*d*x + 5*I*c) - 814*I*a^3*e^3*e^(3*I*d*x + 3*I*c) - 154*I*a^3*e^3*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) + 231*(d*e^(10*I*d*x + 10*I*c) + 5*d*e^
(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)*integra
l(I*sqrt(2)*a^3*e^3*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^(10*I*d*x + 10*I*c)
+ 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a)^3, x)

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maple [A]  time = 1.09, size = 402, normalized size = 1.99 \[ -\frac {2 a^{3} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (231 i \left (\cos ^{6}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-231 i \left (\cos ^{6}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+231 i \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-231 i \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+231 \left (\cos ^{6}\left (d x +c \right )\right )-154 \left (\cos ^{5}\left (d x +c \right )\right )-132 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-154 \left (\cos ^{3}\left (d x +c \right )\right )+21 i \sin \left (d x +c \right )+77 \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}}}{231 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^3,x)

[Out]

-2/231*a^3/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(231*I*cos(d*x+c)^6*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-231*I*cos(d*x+c)^6*(1/(1+cos(d*x+c)))^(1
/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+231*I*cos(d*x+c)^5*
(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c
)-231*I*cos(d*x+c)^5*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/si
n(d*x+c),I)*sin(d*x+c)+231*cos(d*x+c)^6-154*cos(d*x+c)^5-132*I*cos(d*x+c)^2*sin(d*x+c)-154*cos(d*x+c)^3+21*I*s
in(d*x+c)+77*cos(d*x+c))*(e/cos(d*x+c))^(7/2)/cos(d*x+c)^2/sin(d*x+c)^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(7/2)*(a+I*a*tan(d*x+c))**3,x)

[Out]

Timed out

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