3.3.13 \(\int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^4 \, dx\) [213]

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

[Out]

22/9*I*a^4*(e*sec(d*x+c))^(3/2)/d-22/3*a^4*e^2*(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)+22/3*a^4*e*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/d+2/
9*I*a*(e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^3/d+10/21*I*(e*sec(d*x+c))^(3/2)*(a^2+I*a^2*tan(d*x+c))^2/d+22/2
1*I*(e*sec(d*x+c))^(3/2)*(a^4+I*a^4*tan(d*x+c))/d

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Rubi [A]
time = 0.19, antiderivative size = 215, 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 {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 {22 i \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{21 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 {2 i a (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}{9 d} \end {gather*}

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 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))^{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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.89, size = 429, normalized size = 2.00 \begin {gather*} \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 (-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))^{3/2} (a+i a \tan (c+d x))^4}{9 d \left (-1+e^{2 i c}\right ) \sec ^{\frac {11}{2}}(c+d x) (\cos (d x)+i \sin (d x))^4}+\frac {\cos ^5(c+d x) (e \sec (c+d x))^{3/2} \left (\sec (c) \sec ^3(c+d x) (36 \cos (c)+7 i \sin (c)) \left (-\frac {2}{63} i \cos (4 c)-\frac {2}{63} \sin (4 c)\right )+\cos (d x) \csc (c) \left (\frac {22}{3} \cos (4 c)-\frac {22}{3} i \sin (4 c)\right )+\sec (c) \sec (c+d x) (24 \cos (c)+13 i \sin (c)) \left (\frac {2}{9} i \cos (4 c)+\frac {2}{9} \sin (4 c)\right )+\sec (c) \sec ^4(c+d x) \left (\frac {2}{9} \cos (4 c)-\frac {2}{9} i \sin (4 c)\right ) \sin (d x)+\sec (c) \sec ^2(c+d x) \left (-\frac {26}{9} \cos (4 c)+\frac {26}{9} i \sin (4 c)\right ) \sin (d x)\right ) (a+i a \tan (c+d x))^4}{d (\cos (d x)+i \sin (d x))^4} \end {gather*}

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 = 0.56, size = 401, normalized size = 1.87

method	result	size
default	\(\frac {2 a^{4} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (231 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{5}\left (d x +c \right )\right ) \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-231 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+231 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-231 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+168 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-231 \left (\cos ^{5}\left (d x +c \right )\right )+322 \left (\cos ^{4}\left (d x +c \right )\right )-36 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-98 \left (\cos ^{2}\left (d x +c \right )\right )+7\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{63 d \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{3}}\)	\(401\)

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,method=_RETURNVERBOSE)

[Out]

2/63*a^4/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)^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)*cos(d*x+c)^5*sin(d*x+c)+231*I*cos(d*x+c)^4*(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)^4*(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)+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]
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))^(3/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 256, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (231 i \, a^{4} e^{\left (9 i \, d x + 9 i \, c + \frac {3}{2}\right )} + 406 i \, a^{4} e^{\left (7 i \, d x + 7 i \, c + \frac {3}{2}\right )} + 540 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c + \frac {3}{2}\right )} + 330 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c + \frac {3}{2}\right )} + 77 i \, a^{4} e^{\left (i \, d x + i \, c + \frac {3}{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^{4} e^{\frac {3}{2}} + i \, \sqrt {2} a^{4} e^{\left (8 i \, d x + 8 i \, c + \frac {3}{2}\right )} + 4 i \, \sqrt {2} a^{4} e^{\left (6 i \, d x + 6 i \, c + \frac {3}{2}\right )} + 6 i \, \sqrt {2} a^{4} e^{\left (4 i \, d x + 4 i \, c + \frac {3}{2}\right )} + 4 i \, \sqrt {2} a^{4} e^{\left (2 i \, d x + 2 i \, c + \frac {3}{2}\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\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 {gather*}

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]

-2/63*(sqrt(2)*(231*I*a^4*e^(9*I*d*x + 9*I*c + 3/2) + 406*I*a^4*e^(7*I*d*x + 7*I*c + 3/2) + 540*I*a^4*e^(5*I*d
*x + 5*I*c + 3/2) + 330*I*a^4*e^(3*I*d*x + 3*I*c + 3/2) + 77*I*a^4*e^(I*d*x + I*c + 3/2))*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^4*e^(3/2) + I*sqrt(2)*a^4*e^(8*I*d*x + 8*I*c + 3/2) + 4*I
*sqrt(2)*a^4*e^(6*I*d*x + 6*I*c + 3/2) + 6*I*sqrt(2)*a^4*e^(4*I*d*x + 4*I*c + 3/2) + 4*I*sqrt(2)*a^4*e^(2*I*d*
x + 2*I*c + 3/2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))/(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{4} \left (\int \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx + \int \left (- 6 \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (c + d x \right )}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{4}{\left (c + d x \right )}\, dx + \int 4 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )}\, dx + \int \left (- 4 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{3}{\left (c + d x \right )}\right )\, dx\right ) \end {gather*}

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]

a**4*(Integral((e*sec(c + d*x))**(3/2), x) + Integral(-6*(e*sec(c + d*x))**(3/2)*tan(c + d*x)**2, x) + Integra
l((e*sec(c + d*x))**(3/2)*tan(c + d*x)**4, x) + Integral(4*I*(e*sec(c + d*x))**(3/2)*tan(c + d*x), x) + Integr
al(-4*I*(e*sec(c + d*x))**(3/2)*tan(c + d*x)**3, x))

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

[Out]

integrate((I*a*tan(d*x + c) + a)^4*e^(3/2)*sec(d*x + c)^(3/2), 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 )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

int((e/cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^4, x)

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