3.12.68 \(\int \frac {(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1168]
Optimal. Leaf size=179 \[ -\frac {2 i \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}-\frac {2 d (a+i a \tan (e+f x))^{3/2}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}} \]
[Out]
-2*I*a^(3/2)*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+2*I*a*(a+I*a*tan(f*x+e))^(1/2)/(c-I*d)^2/f/(c+d*tan(f*x+e))^(1/2)-2/3*d*(a+I*a*tan(f*x+e))^(3/2)
/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)
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Rubi [A]
time = 0.23, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3629, 3626,
3625, 214} \begin {gather*} -\frac {2 i \sqrt {2} a^{3/2} \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 {2 d (a+i a \tan (e+f x))^{3/2}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{f (c-i d)^2 \sqrt {c+d \tan (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + I*a*Tan[e + f*x])^(3/2)/(c + d*Tan[e + f*x])^(5/2),x]
[Out]
((-2*I)*Sqrt[2]*a^(3/2)*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*d*(a + I*a*Tan[e + f*x])^(3/2))/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)
) + ((2*I)*a*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]
Rule 3629
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-d)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*m*(c^2 + d^2))), x] + Dist[a/(a*c - b*d), Int[(
a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c -
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] && !LtQ[m, -1]
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {2 d (a+i a \tan (e+f x))^{3/2}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {\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 d (a+i a \tan (e+f x))^{3/2}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}}+\frac {(2 a) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{(c-i d)^2}\\ &=-\frac {2 d (a+i a \tan (e+f x))^{3/2}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}}-\frac {\left (4 i a^3\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 {2 i \sqrt {2} a^{3/2} \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 d (a+i a \tan (e+f x))^{3/2}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a \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 = 6.41, size = 278, normalized size = 1.55 \begin {gather*} \frac {(a+i a \tan (e+f x))^{3/2} \left (-\frac {2 i \sqrt {2} \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} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (1+e^{2 i (e+f x)}\right )^{3/2}}+\frac {2 (i+\tan (e+f x)) \left (3 c^2+4 i c d+d^2+2 (c+2 i d) d \tan (e+f x)\right )}{3 (c-i d)^2 (c+i d) \sqrt {\sec (e+f x)} (c+d \tan (e+f x))^{3/2}}\right )}{f \sec ^{\frac {3}{2}}(e+f x)} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + I*a*Tan[e + f*x])^(3/2)/(c + d*Tan[e + f*x])^(5/2),x]
[Out]
((a + I*a*Tan[e + f*x])^(3/2)*(((-2*I)*Sqrt[2]*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^(I
*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(3/2)*(1 + E^((2*I)*(e + f*x)))^(3/2)) + (2*(I + Tan[e + f*x])*(3*c^2 +
(4*I)*c*d + d^2 + 2*(c + (2*I)*d)*d*Tan[e + f*x]))/(3*(c - I*d)^2*(c + I*d)*Sqrt[Sec[e + f*x]]*(c + d*Tan[e +
f*x])^(3/2))))/(f*Sec[e + f*x]^(3/2))
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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. 2906 vs. \(2 (145 ) = 290\).
time = 0.61, size = 2907, normalized size = 16.24
| | |
| method |
result |
size |
| | |
| derivativedivides |
\(\text {Expression too large to display}\) |
\(2907\) |
| default |
\(\text {Expression too large to display}\) |
\(2907\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
[Out]
1/6/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)-6*(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)-16*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)+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)-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)+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)+8*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)-6*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*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*t
an(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+20*(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*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)+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*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*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/(c+d*tan(f*x
+e))^(3/2)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(c^2+d^2)^2/(I*c-d)/(I*a*d)^(1/2)/(-a*(I*d-c))^(1/2)
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((-(2*c*d^4)/((c^2-d^2)^2>0)',
see `assume
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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. 1030 vs. \(2 (143) = 286\).
time = 1.57, size = 1030, normalized size = 5.75 \begin {gather*} -\frac {4 \, \sqrt {2} {\left ({\left (3 \, a c^{2} + 2 i \, a c d + 5 \, a d^{2}\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 2 \, {\left (3 \, a c^{2} + 4 i \, a c d + a d^{2}\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 \, {\left (a c^{2} + 2 i \, a c d - a d^{2}\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 (i \, c^{5} + 3 \, c^{4} d - 2 i \, c^{3} d^{2} + 2 \, c^{2} d^{3} - 3 i \, c d^{4} - d^{5}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (i \, c^{5} + c^{4} d + 2 i \, c^{3} d^{2} + 2 \, c^{2} d^{3} + i \, c d^{4} + d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c^{5} - c^{4} d + 2 i \, c^{3} d^{2} - 2 \, c^{2} d^{3} + i \, c d^{4} - d^{5}\right )} f\right )} \sqrt {-\frac {8 i \, a^{3}}{{\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 )} f \sqrt {-\frac {8 i \, a^{3}}{{\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}}} e^{\left (i \, f x + i \, e\right )} + 2 \, \sqrt {2} {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\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 )}}{2 \, a}\right ) - 3 \, {\left ({\left (-i \, c^{5} - 3 \, c^{4} d + 2 i \, c^{3} d^{2} - 2 \, c^{2} d^{3} + 3 i \, c d^{4} + d^{5}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-i \, c^{5} - c^{4} d - 2 i \, c^{3} d^{2} - 2 \, c^{2} d^{3} - i \, c d^{4} - d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c^{5} + c^{4} d - 2 i \, c^{3} d^{2} + 2 \, c^{2} d^{3} - i \, c d^{4} + d^{5}\right )} f\right )} \sqrt {-\frac {8 i \, a^{3}}{{\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 )} f \sqrt {-\frac {8 i \, a^{3}}{{\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}}} e^{\left (i \, f x + i \, e\right )} + 2 \, \sqrt {2} {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\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 )}}{2 \, a}\right )}{6 \, {\left ({\left (i \, c^{5} + 3 \, c^{4} d - 2 i \, c^{3} d^{2} + 2 \, c^{2} d^{3} - 3 i \, c d^{4} - d^{5}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (i \, c^{5} + c^{4} d + 2 i \, c^{3} d^{2} + 2 \, c^{2} d^{3} + i \, c d^{4} + d^{5}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c^{5} - c^{4} d + 2 i \, c^{3} d^{2} - 2 \, c^{2} d^{3} + i \, c d^{4} - d^{5}\right )} f\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
-1/6*(4*sqrt(2)*((3*a*c^2 + 2*I*a*c*d + 5*a*d^2)*e^(5*I*f*x + 5*I*e) + 2*(3*a*c^2 + 4*I*a*c*d + a*d^2)*e^(3*I*
f*x + 3*I*e) + 3*(a*c^2 + 2*I*a*c*d - a*d^2)*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*((I*c^5 + 3*c^4*d - 2*I*c^3*d^2 + 2*c^2*d^3 -
3*I*c*d^4 - d^5)*f*e^(4*I*f*x + 4*I*e) + 2*(I*c^5 + c^4*d + 2*I*c^3*d^2 + 2*c^2*d^3 + I*c*d^4 + d^5)*f*e^(2*I*
f*x + 2*I*e) + (I*c^5 - c^4*d + 2*I*c^3*d^2 - 2*c^2*d^3 + I*c*d^4 - d^5)*f)*sqrt(-8*I*a^3/((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(1/2*((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f*sqrt(-8*I*a^
3/((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))*e^(I*f*x + I*e) + 2*sqrt(2)*(a*e^(2*I
*f*x + 2*I*e) + a)*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) - 3*((-I*c^5 - 3*c^4*d + 2*I*c^3*d^2 - 2*c^2*d^3 + 3*I*c*d^4 + d^5)*f*e^
(4*I*f*x + 4*I*e) + 2*(-I*c^5 - c^4*d - 2*I*c^3*d^2 - 2*c^2*d^3 - I*c*d^4 - d^5)*f*e^(2*I*f*x + 2*I*e) + (-I*c
^5 + c^4*d - 2*I*c^3*d^2 + 2*c^2*d^3 - I*c*d^4 + d^5)*f)*sqrt(-8*I*a^3/((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(1/2*((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f*sqrt(-8*I*a^3/((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))*e^(I*f*x + I*e) + 2*sqrt(2)*(a*e^(2*I*f*x + 2*I*e) + a)
*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))/((I*c^5 + 3*c^4*d - 2*I*c^3*d^2 + 2*c^2*d^3 - 3*I*c*d^4 - d^5)*f*e^(4*I*f*x + 4*I*e) + 2*
(I*c^5 + c^4*d + 2*I*c^3*d^2 + 2*c^2*d^3 + I*c*d^4 + d^5)*f*e^(2*I*f*x + 2*I*e) + (I*c^5 - c^4*d + 2*I*c^3*d^2
- 2*c^2*d^3 + I*c*d^4 - d^5)*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 {3}{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))**(3/2)/(c+d*tan(f*x+e))**(5/2),x)
[Out]
Integral((I*a*(tan(e + f*x) - I))**(3/2)/(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))^(3/2)/(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 )}^{3/2}}{{\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)^(3/2)/(c + d*tan(e + f*x))^(5/2),x)
[Out]
int((a + a*tan(e + f*x)*1i)^(3/2)/(c + d*tan(e + f*x))^(5/2), x)
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