3.3.1 \(\int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3 \, dx\) [201]

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

[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 = 0.15, 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 = {3579, 3567, 3853, 3856, 2719} \begin {gather*} -\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} \end {gather*}

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 2719

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

Rule 3567

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 3579

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 3853

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 3856

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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.92, size = 442, normalized size = 2.19 \begin {gather*} \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 (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right ) (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^3}{3 d \left (-1+e^{2 i c}\right ) \sec ^{\frac {13}{2}}(c+d x) (\cos (d x)+i \sin (d x))^3}+\frac {\cos ^6(c+d x) (e \sec (c+d x))^{7/2} \left (\sec ^5(c+d x) \left (-\frac {2}{11} i \cos (3 c)-\frac {2}{11} \sin (3 c)\right )+\cos (d x) \csc (c) (2 \cos (3 c)-2 i \sin (3 c))+\sec (c) \sec ^3(c+d x) (12 \cos (c)+7 i \sin (c)) \left (\frac {2}{21} i \cos (3 c)+\frac {2}{21} \sin (3 c)\right )+\sec (c) \sec ^2(c+d x) \left (\frac {2}{3} \cos (3 c)-\frac {2}{3} i \sin (3 c)\right ) \sin (d x)+\sec (c) \sec ^4(c+d x) \left (-\frac {2}{3} \cos (3 c)+\frac {2}{3} i \sin (3 c)\right ) \sin (d x)+\sec (c+d x) \left (\frac {2}{3} \cos (3 c)-\frac {2}{3} i \sin (3 c)\right ) \tan (c)\right ) (a+i a \tan (c+d x))^3}{d (\cos (d x)+i \sin (d x))^3} \end {gather*}

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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Maple [A]
time = 0.61, size = 402, normalized size = 1.99

method	result	size
default	\(\frac {2 a^{3} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (231 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{6}\left (d x +c \right )\right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-231 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right )+231 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{5}\left (d x +c \right )\right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-231 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )-231 \left (\cos ^{6}\left (d x +c \right )\right )+154 \left (\cos ^{5}\left (d x +c \right )\right )+132 i \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+154 \left (\cos ^{3}\left (d x +c \right )\right )-21 i \sin \left (d x +c \right )-77 \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}}}{231 d \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{2}}\)	\(402\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 301, normalized size = 1.49 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (231 i \, a^{3} e^{\left (11 i \, d x + 11 i \, c + \frac {7}{2}\right )} + 1309 i \, a^{3} e^{\left (9 i \, d x + 9 i \, c + \frac {7}{2}\right )} + 946 i \, a^{3} e^{\left (7 i \, d x + 7 i \, c + \frac {7}{2}\right )} + 870 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c + \frac {7}{2}\right )} + 407 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c + \frac {7}{2}\right )} + 77 i \, a^{3} e^{\left (i \, d x + i \, c + \frac {7}{2}\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 231 \, {\left (i \, \sqrt {2} a^{3} e^{\frac {7}{2}} + i \, \sqrt {2} a^{3} e^{\left (10 i \, d x + 10 i \, c + \frac {7}{2}\right )} + 5 i \, \sqrt {2} a^{3} e^{\left (8 i \, d x + 8 i \, c + \frac {7}{2}\right )} + 10 i \, \sqrt {2} a^{3} e^{\left (6 i \, d x + 6 i \, c + \frac {7}{2}\right )} + 10 i \, \sqrt {2} a^{3} e^{\left (4 i \, d x + 4 i \, c + \frac {7}{2}\right )} + 5 i \, \sqrt {2} a^{3} e^{\left (2 i \, d x + 2 i \, c + \frac {7}{2}\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\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 )}} \end {gather*}

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]

-2/231*(sqrt(2)*(231*I*a^3*e^(11*I*d*x + 11*I*c + 7/2) + 1309*I*a^3*e^(9*I*d*x + 9*I*c + 7/2) + 946*I*a^3*e^(7
*I*d*x + 7*I*c + 7/2) + 870*I*a^3*e^(5*I*d*x + 5*I*c + 7/2) + 407*I*a^3*e^(3*I*d*x + 3*I*c + 7/2) + 77*I*a^3*e
^(I*d*x + I*c + 7/2))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1) + 231*(I*sqrt(2)*a^3*e^(7/2) + I*s
qrt(2)*a^3*e^(10*I*d*x + 10*I*c + 7/2) + 5*I*sqrt(2)*a^3*e^(8*I*d*x + 8*I*c + 7/2) + 10*I*sqrt(2)*a^3*e^(6*I*d
*x + 6*I*c + 7/2) + 10*I*sqrt(2)*a^3*e^(4*I*d*x + 4*I*c + 7/2) + 5*I*sqrt(2)*a^3*e^(2*I*d*x + 2*I*c + 7/2))*we
ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))/(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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 5987 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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