3.12.67 \(\int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1167]
Optimal. Leaf size=181 \[ -\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {4 i a^2 \sqrt {a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}} \]
[Out]
-4*I*a^(5/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))*2^(1/2)/(c
-I*d)^(5/2)/f+4*I*a^2*(a+I*a*tan(f*x+e))^(1/2)/(c-I*d)^2/f/(c+d*tan(f*x+e))^(1/2)-2/3*a*(a+I*a*tan(f*x+e))^(3/
2)/(I*c+d)/f/(c+d*tan(f*x+e))^(3/2)
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Rubi [A]
time = 0.26, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3626, 3625,
214} \begin {gather*} -\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}+\frac {4 i a^2 \sqrt {a+i a \tan (e+f x)}}{f (c-i d)^2 \sqrt {c+d \tan (e+f x)}}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + I*a*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(5/2),x]
[Out]
((-4*I)*Sqrt[2]*a^(5/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e +
f*x]])])/((c - I*d)^(5/2)*f) - (2*a*(a + I*a*Tan[e + f*x])^(3/2))/(3*(I*c + d)*f*(c + d*Tan[e + f*x])^(3/2))
+ ((4*I)*a^2*Sqrt[a + I*a*Tan[e + f*x]])/((c - I*d)^2*f*Sqrt[c + d*Tan[e + f*x]])
Rule 214
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]
Rule 3625
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-2*a*(b/f), Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Rule 3626
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[a*b*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m - 1)*(a*c - b*d))), x] + Dist[2*(a^2/(a
*c - b*d)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f},
x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {(2 a) \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx}{c-i d}\\ &=-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {4 i a^2 \sqrt {a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\left (4 a^2\right ) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{(c-i d)^2}\\ &=-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {4 i a^2 \sqrt {a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}}-\frac {\left (8 i a^4\right ) \text {Subst}\left (\int \frac {1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^2 f}\\ &=-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {4 i a^2 \sqrt {a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 7.05, size = 283, normalized size = 1.56 \begin {gather*} \frac {(a+i a \tan (e+f x))^{5/2} \left (-\frac {4 i \sqrt {2} e^{-3 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \log \left (2 e^{-i e} \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{(c-i d)^{5/2}}-\frac {2 \sqrt {\sec (e+f x)} (\cos (2 (e+f x))-i \sin (2 (e+f x))) (-7 i c-d+(c-7 i d) \tan (e+f x))}{3 (c-i d)^2 (c+d \tan (e+f x))^{3/2}}\right )}{f \sec ^{\frac {5}{2}}(e+f x)} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + I*a*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(5/2),x]
[Out]
((a + I*a*Tan[e + f*x])^(5/2)*(((-4*I)*Sqrt[2]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*
I)*(e + f*x))]*Log[(2*(Sqrt[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c - (I*d*(-1 + E^((2
*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]))/E^(I*e)])/((c - I*d)^(5/2)*E^((3*I)*(e + f*x))) - (2*Sqrt[Sec[e
+ f*x]]*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])*((-7*I)*c - d + (c - (7*I)*d)*Tan[e + f*x]))/(3*(c - I*d)^2*(c
+ d*Tan[e + f*x])^(3/2))))/(f*Sec[e + f*x]^(5/2))
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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. 2967 vs. \(2 (146 ) = 292\).
time = 0.62, size = 2968, normalized size = 16.40
| | |
| method |
result |
size |
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| derivativedivides |
\(\text {Expression too large to display}\) |
\(2968\) |
| default |
\(\text {Expression too large to display}\) |
\(2968\) |
| | |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
[Out]
1/3/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(6*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(
1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d*(-a*(I*d-c))^(1/2)*tan(f*x+e)-7*(I*a*d)^(1/2)*(
a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^4-3*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d
+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*
c^2*(I*a*d)^(1/2)+6*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(
I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2-18*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*
x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^4*(-a*(I*d-c)
)^(1/2)*tan(f*x+e)^2+12*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/
2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^4*d*(-a*(I*d-c))^(1/2)*tan(f*x+e)-36*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*
x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2*d^3*(-a*(I*d-
c))^(1/2)*tan(f*x+e)-18*I*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)
*c^2*d^2*tan(f*x+e)-3*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(
I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2*(-a*(I*d-c))^(1/2)+(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^
(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^4-3*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c)
)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*d^2*(I*a*d)^(1/2)*tan(f*x+e)^2-3*ln((3*
a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))
^(1/2))/(tan(f*x+e)+I))*a*c^2*(I*a*d)^(1/2)-6*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(
1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d*(-a*(I*d-c))^(1/2)*tan(f*x+e)-6*ln((3*a*c+I*a*t
an(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(
tan(f*x+e)+I))*a*c*d*(I*a*d)^(1/2)*tan(f*x+e)-3*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*
(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*d^2*(I*a*d)^(1/2)*tan(f*x+e)
^2+3*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*
d)/(I*a*d)^(1/2))*a*c^2*(-a*(I*d-c))^(1/2)+6*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1
+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^5*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2-18*2^(1/2)*ln(1/
2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*
c^4*d*(-a*(I*d-c))^(1/2)+6*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1
/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2*d^3*(-a*(I*d-c))^(1/2)-3*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+
2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^2*(-a*(I*d-c))^(1/2)*tan(f
*x+e)^2+18*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*d^2-18*2^(
1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)
^(1/2))*a*c^2*d^3*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2-4*2^(1/2)*c^3*d*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f
*x+e)))^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)+20*2^(1/2)*c*d^3*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)
))^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)-36*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*
tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)+12*2^(1/2)*ln(1/2
*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c
*d^4*(-a*(I*d-c))^(1/2)*tan(f*x+e)-18*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*ta
n(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3*d^2*(-a*(I*d-c))^(1/2)-I*2^(1/2)*c^4*(I*a*d)^(1/2)*(a
*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(-a*(I*d-c))^(1/2)*tan(f*x+e)+7*I*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*
(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^4*tan(f*x+e)+3*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c
+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^2*(-a*(I*d-c))^(1/2)*tan(
f*x+e)^2-20*I*(I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c^3*d+4*I*(
I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d^3-6*I*ln((3*a*c+I*a*ta
n(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(t
an(f*x+e)+I))*a*c*d*(I*a*d)^(1/2)*tan(f*x+e)+6*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e)
)*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^5*(-a*(I*d-c))^(1/2))*a^2/(c+d*tan(f*x+e))^(3/
2)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/...
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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 873 vs. \(2 (143) = 286\).
time = 1.22, size = 873, normalized size = 4.82 \begin {gather*} -\frac {8 \, \sqrt {2} {\left (4 \, {\left (-i \, a^{2} c - a^{2} d\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-7 i \, a^{2} c - a^{2} d\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 \, {\left (-i \, a^{2} c + a^{2} d\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - 3 \, {\left ({\left (c^{4} - 4 i \, c^{3} d - 6 \, c^{2} d^{2} + 4 i \, c d^{3} + d^{4}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} - 2 i \, c^{3} d - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} f\right )} \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} \log \left (\frac {{\left ({\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2}}\right ) + 3 \, {\left ({\left (c^{4} - 4 i \, c^{3} d - 6 \, c^{2} d^{2} + 4 i \, c d^{3} + d^{4}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} - 2 i \, c^{3} d - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} f\right )} \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} \log \left (\frac {{\left ({\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} \sqrt {-\frac {32 i \, a^{5}}{{\left (i \, c^{5} + 5 \, c^{4} d - 10 i \, c^{3} d^{2} - 10 \, c^{2} d^{3} + 5 i \, c d^{4} + d^{5}\right )} f^{2}}} f e^{\left (i \, f x + i \, e\right )} + 4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2}}\right )}{6 \, {\left ({\left (c^{4} - 4 i \, c^{3} d - 6 \, c^{2} d^{2} + 4 i \, c d^{3} + d^{4}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (c^{4} - 2 i \, c^{3} d - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} f\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
-1/6*(8*sqrt(2)*(4*(-I*a^2*c - a^2*d)*e^(5*I*f*x + 5*I*e) + (-7*I*a^2*c - a^2*d)*e^(3*I*f*x + 3*I*e) + 3*(-I*a
^2*c + a^2*d)*e^(I*f*x + I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(
a/(e^(2*I*f*x + 2*I*e) + 1)) - 3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4*I*e) + 2*(c
^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^4 + 2*c^2*d^2 + d^4)*f)*sqrt(-32*I*a^5/((I*c^5 +
5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(1/4*((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*sqrt
(-32*I*a^5/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2
)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sq
rt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2) + 3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*
e^(4*I*f*x + 4*I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^4 + 2*c^2*d^2 + d^4)*f)
*sqrt(-32*I*a^5/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(1/4*((-I*c^3 - 3*c^
2*d + 3*I*c*d^2 + d^3)*sqrt(-32*I*a^5/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*f
*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e
^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2))/((c^4 - 4*I*c^3*d - 6*c^2*d
^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c
^4 + 2*c^2*d^2 + d^4)*f)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))**(5/2)/(c+d*tan(f*x+e))**(5/2),x)
[Out]
Integral((I*a*(tan(e + f*x) - I))**(5/2)/(c + d*tan(e + f*x))**(5/2), x)
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Timed out
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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 (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*tan(e + f*x)*1i)^(5/2)/(c + d*tan(e + f*x))^(5/2),x)
[Out]
int((a + a*tan(e + f*x)*1i)^(5/2)/(c + d*tan(e + f*x))^(5/2), x)
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