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*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]])
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Rubi [A] time = 0.138728, antiderivative size = 156, normalized size of antiderivative = 1.,
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 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]
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 \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*}
Mathematica [C] time = 2.51534, size = 110, normalized size = 0.71 \[ -\frac{4 i a^4 e^{2 i (c+d x)} \left (-7 \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+2 e^{2 i (c+d x)}+7\right )}{5 d e^2 \left (1+e^{2 i (c+d x)}\right ) \sqrt{e \sec (c+d x)}} \]
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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Maple [B] time = 0.336, size = 3762, normalized size = 24.1 \begin{align*} \text{output too large to display} \end{align*}
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*(cos(d*x+c)-1)*(-64*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)+160*(-cos(d*x+c)/(cos(d*x+c)+1
)^2)^(3/2)*cos(d*x+c)^7+76*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^6-63*I*(-cos(d*x+c)/(cos(d*x+c)+1)^
2)^(1/2)*(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)*s
in(d*x+c)*cos(d*x+c)^5+63*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(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)*sin(d*x+c)*cos(d*x+c)^5+504*I*(-cos(d*x+c)/(cos(d*x+c)+
1)^2)^(3/2)*(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
)*sin(d*x+c)*cos(d*x+c)^2-504*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*cos(d*x+c)^2-168*I*(-cos(d*x+c)/(cos(d*x
+c)+1)^2)^(1/2)*(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)*sin(d*x+c)*cos(d*x+c)^4+168*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(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)*sin(d*x+c)*cos(d*x+c)^4+168*I*(-cos(d*x+c)/(cos
(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*cos(d*x+c)-168*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*cos(d*x+c)+168*I*(-cos(d*x+c)/(cos
(d*x+c)+1)^2)^(1/2)*(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)*sin(d*x+c)*cos(d*x+c)^2-168*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(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)*sin(d*x+c)*cos(d*x+c)^2+84*I*(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)*(-cos(d*x+c)/(cos(d*x+c)
+1)^2)^(1/2)*cos(d*x+c)*sin(d*x+c)-84*I*(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)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*sin(d*x+c)+21*I*(-cos(d*x+c)/(co
s(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*cos(d*x+c)^8-21*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*cos(d*x+c)^8+84*I*(-cos(d*x+c)/(
cos(d*x+c)+1)^2)^(3/2)*(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)/s
in(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^7-84*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*cos(d*x+c)^7+21*I*(-cos(d*x+c)
/(cos(d*x+c)+1)^2)^(1/2)*(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)*sin(d*x+c)*cos(d*x+c)^7-21*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(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)*sin(d*x+c)*cos(d*x+c)^7-42*I*(-cos(d*x+
c)/(cos(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*cos(d*x+c)^6+42*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^6-32*(-cos(d*x+
c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^9-32*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^8+20*(-cos(d*x+c)/(
cos(d*x+c)+1)^2)^(3/2)-32*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^8-32*I*(-cos(d*x+c)/(co
s(d*x+c)+1)^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^7+144*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^
6-192*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^3+80*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)
*sin(d*x+c)*cos(d*x+c)^2-20*I*cos(d*x+c)^4*ln(-2*(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+
c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)+20*I*cos(d*x+c)^4*ln(-(2*
(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*
x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)+80*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*sin(d*x+c)*cos(d*x+c)+40*I*cos(d*x+
c)^3*ln(-2*(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^
(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)-40*I*cos(d*x+c)^3*ln(-(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*c
os(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)-20*I*
cos(d*x+c)^2*ln(-2*(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c
)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)+20*I*cos(d*x+c)^2*ln(-(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)
^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+
c)-420*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellip
ticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^5+420*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*cos(d*
x+c)^5-315*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*E
llipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^4+315*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*co
s(d*x+c)^4+42*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(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)*sin(d*x+c)*cos(d*x+c)^6-42*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(
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)*sin(d*x+c)*
cos(d*x+c)^6+336*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(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)*sin(d*x+c)*cos(d*x+c)^3-336*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/
2)*(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)*sin(d*x
+c)*cos(d*x+c)^3-288*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^5-36*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)
*cos(d*x+c)^4+224*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*cos(d*x+c)^3-28*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*co
s(d*x+c)^2+144*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^5-192*I*(-cos(d*x+c)/(cos(d*x+c)+1
)^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^4)/(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)/sin(d*x+c)^7/cos(d*x+c)^3/(e/cos(d*x+
c))^(5/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}} \end{align*}
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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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((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
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)