3.3.17 \(\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx\) [217]
Optimal. Leaf size=156 \[ -\frac {42 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {42 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e^3}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}} \]
[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 = 0.11, 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 = {3577, 3853,
3856, 2719} \begin {gather*} \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}} \end {gather*}
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 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 3577
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 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 \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] Result contains higher order function than in optimal. Order 5 vs. order 4 in
optimal.
time = 2.56, size = 110, normalized size = 0.71 \begin {gather*} -\frac {4 i a^4 e^{2 i (c+d x)} \left (7+2 e^{2 i (c+d x)}-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 )\right )}{5 d e^2 \left (1+e^{2 i (c+d x)}\right ) \sqrt {e \sec (c+d x)}} \end {gather*}
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] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 3761 vs. \(2 (160 ) = 320\).
time = 0.68, size = 3762, normalized size = 24.12
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result |
size |
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\(\text {Expression too large to display}\) |
\(3762\) |
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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,method=_RETURNVERBOSE)
[Out]
-1/10*a^4/d*(-1+cos(d*x+c))*(20*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+160*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*
cos(d*x+c)^7-288*cos(d*x+c)^5*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-36*cos(d*x+c)^4*(-cos(d*x+c)/(1+cos(d*x+c))
^2)^(3/2)+224*cos(d*x+c)^3*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-28*cos(d*x+c)^2*(-cos(d*x+c)/(1+cos(d*x+c))^2)
^(3/2)+144*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^5-32*I*sin(d*x+c)*cos(d*x+c)^8*(-cos(d
*x+c)/(1+cos(d*x+c))^2)^(3/2)-32*I*sin(d*x+c)*cos(d*x+c)^7*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+144*I*(-cos(d*
x+c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^6-192*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)*cos
(d*x+c)^4-192*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^3+20*I*cos(d*x+c)^4*ln(-(2*(-cos(d*
x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(1+cos(d*x+c))^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*(2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^2-cos(d*x+
c)^2-2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)+80*I*(-cos(d*x+c)/(1+cos(
d*x+c))^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^2-40*I*cos(d*x+c)^3*ln(-(2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x
+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(1+cos(d*x+c))^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*(2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(1+cos(d*x+c))^2
)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)+80*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)*cos(d*x+
c)+20*I*cos(d*x+c)^2*ln(-(2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(1+c
os(d*x+c))^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)-20*I*ln(-2*(2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2
)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*cos(d*x+c)^2*
sin(d*x+c)+76*cos(d*x+c)^6*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-64*cos(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(
3/2)+21*I*sin(d*x+c)*cos(d*x+c)^7*(-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)-21*I*sin(d*x+c)*cos(d*x+c)^7*(-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)*EllipticE(I*(-1+cos(d*x+c))/sin(d
*x+c),I)-420*I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^5*(1/(1+cos(d*x+c)))^(1/2)*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+420*I*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c
),I)*sin(d*x+c)*cos(d*x+c)^5*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-cos(d*x+c)/(1+cos(d*
x+c))^2)^(3/2)+42*I*sin(d*x+c)*cos(d*x+c)^6*(-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)-42*I*sin(d*x+c)*cos(d*x+c)^6*(-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)*EllipticE(I*(-1+cos(d*x
+c))/sin(d*x+c),I)-315*I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^4*(1/(1+cos(d*x+c)))^
(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+315*I*EllipticE(I*(-1+cos(d*x+c))
/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^4*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-cos(d*x+c)
/(1+cos(d*x+c))^2)^(3/2)-63*I*sin(d*x+c)*cos(d*x+c)^5*(-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)+63*I*sin(d*x+c)*cos(d*x+c)^5
*(-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)*EllipticE(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)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^3-336*I*(-cos(d*x+c)/(1+c
os(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)*cos(d*x+c)^3-168*I*sin(d*x+c)*cos(d*x+c)^4*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(1/(1+co
s(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)+168*I*sin(d*x+c)*
cos(d*x+c)^4*(-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)*E
llipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+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)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^2-504*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)*sin(d*x+c)*cos(d*x+c)^2+168*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)*cos(d*x+c)-168*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)*cos(d*x+c...
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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((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
e^(-5/2)*integrate((I*a*tan(d*x + c) + a)^4/sec(d*x + c)^(5/2), x)
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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.10, size = 86, normalized size = 0.55 \begin {gather*} -\frac {2 \, {\left (21 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \frac {\sqrt {2} {\left (2 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, a^{4} e^{\left (i \, d x + i \, c\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}}\right )} e^{\left (-\frac {5}{2}\right )}}{5 \, d} \end {gather*}
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]
-2/5*(21*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(2*I*a^4*
e^(3*I*d*x + 3*I*c) + 7*I*a^4*e^(I*d*x + I*c))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-5/2)
/d
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 ) \end {gather*}
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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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((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^(-5/2)/sec(d*x + c)^(5/2), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}
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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