3.217 \(\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=156 \[ \frac {42 a^4 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 d e^3}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {42 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}} \]
[Out]
-42/5*a^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/e^2/cos(d*x+
c)^(1/2)/(e*sec(d*x+c))^(1/2)+42/5*a^4*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/d/e^3-4/5*I*a*(a+I*a*tan(d*x+c))^3/d/(e
*sec(d*x+c))^(5/2)+28/5*I*(a^4+I*a^4*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(1/2)
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Rubi [A] time = 0.14, antiderivative size = 156, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used =
{3496, 3768, 3771, 2639} \[ \frac {42 a^4 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 d e^3}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {42 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(5/2),x]
[Out]
(-42*a^4*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (42*a^4*Sqrt[e*Sec[c +
d*x]]*Sin[c + d*x])/(5*d*e^3) - (((4*I)/5)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(5/2)) + (((28*I)/
5)*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^2*Sqrt[e*Sec[c + d*x]])
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 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 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 \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx &=-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (7 a^2\right ) \int \frac {(a+i a \tan (c+d x))^2}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2}\\ &=-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}+\frac {\left (21 a^4\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 e^4}\\ &=\frac {42 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e^3}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {\left (21 a^4\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2}\\ &=\frac {42 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e^3}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {\left (21 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {42 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {42 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e^3}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 2.79, size = 110, normalized size = 0.71 \[ -\frac {4 i a^4 e^{2 i (c+d x)} \left (-7 \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+2 e^{2 i (c+d x)}+7\right )}{5 d e^2 \left (1+e^{2 i (c+d x)}\right ) \sqrt {e \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(5/2),x]
[Out]
(((-4*I)/5)*a^4*E^((2*I)*(c + d*x))*(7 + 2*E^((2*I)*(c + d*x)) - 7*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometri
c2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(d*e^2*(1 + E^((2*I)*(c + d*x)))*Sqrt[e*Sec[c + d*x]])
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-4 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} + 28 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 14 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + 42 i \, a^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 5 \, {\left (d e^{3} e^{\left (i \, d x + i \, c\right )} - d e^{3}\right )} {\rm integral}\left (\frac {\sqrt {2} {\left (21 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 42 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + 21 i \, a^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, {\left (d e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{5 \, {\left (d e^{3} e^{\left (i \, d x + i \, c\right )} - d e^{3}\right )}} \]
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))^(5/2),x, algorithm="fricas")
[Out]
1/5*(sqrt(2)*(-4*I*a^4*e^(4*I*d*x + 4*I*c) + 4*I*a^4*e^(3*I*d*x + 3*I*c) + 28*I*a^4*e^(2*I*d*x + 2*I*c) + 14*I
*a^4*e^(I*d*x + I*c) + 42*I*a^4)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 5*(d*e^3*e^(I*d*x
+ I*c) - d*e^3)*integral(1/5*sqrt(2)*(21*I*a^4*e^(2*I*d*x + 2*I*c) + 42*I*a^4*e^(I*d*x + I*c) + 21*I*a^4)*sqr
t(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(d*e^3*e^(3*I*d*x + 3*I*c) - 2*d*e^3*e^(2*I*d*x + 2*I*c
) + d*e^3*e^(I*d*x + I*c)), x))/(d*e^3*e^(I*d*x + I*c) - d*e^3)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]
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))^(5/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(5/2), x)
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maple [B] time = 1.07, size = 2196, normalized size = 14.08 \[ \text {Expression too large to display} \]
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))^(5/2),x)
[Out]
1/10*a^4/d*(-1+cos(d*x+c))^3*(32*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^6*sin(d*x+c)+96*I*(-cos(d*x
+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^5*sin(d*x+c)+16*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^4*sin
(d*x+c)-208*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^3*sin(d*x+c)-240*I*(-cos(d*x+c)/(1+cos(d*x+c))^2
)^(3/2)*cos(d*x+c)^2*sin(d*x+c)-80*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)*sin(d*x+c)-20*I*cos(d*x+c
)^2*ln(-(2*cos(d*x+c)^2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-cos(d*x+c)^2+2*cos(d*x+c)-2*(-cos(d*x+c)/(1+cos(d
*x+c))^2)^(1/2)-1)/sin(d*x+c)^2)*sin(d*x+c)+20*I*cos(d*x+c)^2*ln(-2*(2*cos(d*x+c)^2*(-cos(d*x+c)/(1+cos(d*x+c)
)^2)^(1/2)-cos(d*x+c)^2+2*cos(d*x+c)-2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-1)/sin(d*x+c)^2)*sin(d*x+c)+63*I*(
-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)-84*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)-21*I*(-cos(d*x+c)/(1+cos(d*x
+c))^2)^(3/2)*(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*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)^4*sin(d*x+c)+32*cos(d*x+c)^7*(-cos(d
*x+c)/(1+cos(d*x+c))^2)^(3/2)-20*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+96*cos(d*x+c)^6*(-cos(d*x+c)/(1+cos(d*x+
c))^2)^(3/2)-172*cos(d*x+c)^4*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-56*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos
(d*x+c)^3+96*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^2+24*cos(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3
/2)-84*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*Ellip
ticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)^4*sin(d*x+c)-21*I*cos(d*x+c)^5*sin(d*x+c)*(-cos(d*x+c)/(1+cos(
d*x+c))^2)^(1/2)*(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)-336*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)*cos(d*x+c)^3*sin(d*x+c)-84*I*cos(d*x+c)^4*sin(d*x+c)*(-cos(d*x+c)/(1
+cos(d*x+c))^2)^(1/2)*(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))/s
in(d*x+c),I)+189*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)^2*sin(d*x+c)-504*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3
/2)*(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)*cos(d
*x+c)^2*sin(d*x+c)-126*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(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)^3*sin(d*x+c)+210*I*(-cos(d*x+c)/(1+cos(d*x+c))
^2)^(3/2)*(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)*sin(d*x+c)-336*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)*cos(d*x+c)*sin(d*x+c)-84*I*(-cos(d*x+c)/(1+cos(d*x+c)
)^2)^(1/2)*(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)^2*sin(d*x+c)-84*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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)*cos(d*x+c)^3*sin(d*x+c)-21*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)/(1+cos(d*x+c))^2)^
(1/2)*cos(d*x+c)*sin(d*x+c)-126*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(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))/cos(d*x+c)^3/sin(d*x+c)
^7/(e/cos(d*x+c))^(5/2)/(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]
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))^(5/2),x, algorithm="maxima")
[Out]
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(5/2),x)
[Out]
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(5/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ a^{4} \left (\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {6 \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx + \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {4 i \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {4 i \tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx\right ) \]
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))**(5/2),x)
[Out]
a**4*(Integral((e*sec(c + d*x))**(-5/2), x) + Integral(-6*tan(c + d*x)**2/(e*sec(c + d*x))**(5/2), x) + Integr
al(tan(c + d*x)**4/(e*sec(c + d*x))**(5/2), x) + Integral(4*I*tan(c + d*x)/(e*sec(c + d*x))**(5/2), x) + Integ
ral(-4*I*tan(c + d*x)**3/(e*sec(c + d*x))**(5/2), x))
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