3.11.8 \(\int (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2} \, dx\) [1008]

Optimal. Leaf size=159 \[ -\frac {i a^{3/2} c^{5/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f} \]

[Out]

-I*a^(3/2)*c^(5/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/f+1/2*a*c^2*(a+I*
a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)*tan(f*x+e)/f-1/3*I*c*(a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^
(3/2)/f

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Rubi [A]
time = 0.11, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3604, 51, 38, 65, 223, 209} \begin {gather*} -\frac {i a^{3/2} c^{5/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((-I)*a^(3/2)*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f + (
a*c^2*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*f) - ((I/3)*c*(a + I*a*Tan[e + f*
x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/f

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x*(a + b*x)^m*((c + d*x)^m/(2*m + 1))
, x] + Dist[2*a*c*(m/(2*m + 1)), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

Rule 51

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m
+ n + 1))), x] + Dist[2*c*(n/(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x]
&& EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0] && IGtQ[n + 1/2, 0] && LtQ[m, n]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[a*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2} \, dx &=\frac {(a c) \text {Subst}\left (\int \sqrt {a+i a x} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}+\frac {\left (a c^2\right ) \text {Subst}\left (\int \sqrt {a+i a x} \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}+\frac {\left (a^2 c^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac {a c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {\left (i a c^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{f}\\ &=\frac {a c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {\left (i a c^3\right ) \text {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{f}\\ &=-\frac {i a^{3/2} c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f}\\ \end {align*}

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Mathematica [A]
time = 3.12, size = 169, normalized size = 1.06 \begin {gather*} -\frac {i e^{-2 i (e+f x)} \left (\frac {c}{1+e^{2 i (e+f x)}}\right )^{5/2} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \left (e^{i (e+f x)} \left (-3+8 e^{2 i (e+f x)}+3 e^{4 i (e+f x)}\right )+3 \left (1+e^{2 i (e+f x)}\right )^3 \text {ArcTan}\left (e^{i (e+f x)}\right )\right ) (a+i a \tan (e+f x))^{3/2}}{3 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 - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((-1/3*I)*(c/(1 + E^((2*I)*(e + f*x))))^(5/2)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*(E^(I*(e + f*x))
*(-3 + 8*E^((2*I)*(e + f*x)) + 3*E^((4*I)*(e + f*x))) + 3*(1 + E^((2*I)*(e + f*x)))^3*ArcTan[E^(I*(e + f*x))])
*(a + I*a*Tan[e + f*x])^(3/2))/(E^((2*I)*(e + f*x))*f*Sec[e + f*x]^(3/2))

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Maple [A]
time = 0.33, size = 186, normalized size = 1.17

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} a \left (-2 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )-2 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+3 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{6 f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\)	\(186\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} a \left (-2 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )-2 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+3 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{6 f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*c^2*a*(-2*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1
/2)*tan(f*x+e)^2+3*a*c*ln((c*a*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))-2*I*(a*c*(1+t
an(f*x+e)^2))^(1/2)*(a*c)^(1/2)+3*tan(f*x+e)*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^
(1/2)/(a*c)^(1/2)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 949 vs. \(2 (126) = 252\).
time = 0.65, size = 949, normalized size = 5.97 \begin {gather*} -\frac {{\left (12 \, a c^{2} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 32 \, a c^{2} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 12 \, a c^{2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 12 i \, a c^{2} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 32 i \, a c^{2} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 12 i \, a c^{2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 6 \, {\left (a c^{2} \cos \left (6 \, f x + 6 \, e\right ) + 3 \, a c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 3 \, a c^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, a c^{2} \sin \left (6 \, f x + 6 \, e\right ) + 3 i \, a c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 3 i \, a c^{2} \sin \left (2 \, f x + 2 \, e\right ) + a c^{2}\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 6 \, {\left (a c^{2} \cos \left (6 \, f x + 6 \, e\right ) + 3 \, a c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 3 \, a c^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, a c^{2} \sin \left (6 \, f x + 6 \, e\right ) + 3 i \, a c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 3 i \, a c^{2} \sin \left (2 \, f x + 2 \, e\right ) + a c^{2}\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), -\sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 3 \, {\left (i \, a c^{2} \cos \left (6 \, f x + 6 \, e\right ) + 3 i \, a c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 3 i \, a c^{2} \cos \left (2 \, f x + 2 \, e\right ) - a c^{2} \sin \left (6 \, f x + 6 \, e\right ) - 3 \, a c^{2} \sin \left (4 \, f x + 4 \, e\right ) - 3 \, a c^{2} \sin \left (2 \, f x + 2 \, e\right ) + i \, a c^{2}\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 3 \, {\left (-i \, a c^{2} \cos \left (6 \, f x + 6 \, e\right ) - 3 i \, a c^{2} \cos \left (4 \, f x + 4 \, e\right ) - 3 i \, a c^{2} \cos \left (2 \, f x + 2 \, e\right ) + a c^{2} \sin \left (6 \, f x + 6 \, e\right ) + 3 \, a c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 3 \, a c^{2} \sin \left (2 \, f x + 2 \, e\right ) - i \, a c^{2}\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right )\right )} \sqrt {a} \sqrt {c}}{-12 \, f {\left (i \, \cos \left (6 \, f x + 6 \, e\right ) + 3 i \, \cos \left (4 \, f x + 4 \, e\right ) + 3 i \, \cos \left (2 \, f x + 2 \, e\right ) - \sin \left (6 \, f x + 6 \, e\right ) - 3 \, \sin \left (4 \, f x + 4 \, e\right ) - 3 \, \sin \left (2 \, f x + 2 \, e\right ) + i\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-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-(12*a*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 32*a*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e))) - 12*a*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*I*a*c^2*sin(5/2*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 32*I*a*c^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*I
*a*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 6*(a*c^2*cos(6*f*x + 6*e) + 3*a*c^2*cos(4*f*x +
4*e) + 3*a*c^2*cos(2*f*x + 2*e) + I*a*c^2*sin(6*f*x + 6*e) + 3*I*a*c^2*sin(4*f*x + 4*e) + 3*I*a*c^2*sin(2*f*x
+ 2*e) + a*c^2)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e)
, cos(2*f*x + 2*e))) + 1) + 6*(a*c^2*cos(6*f*x + 6*e) + 3*a*c^2*cos(4*f*x + 4*e) + 3*a*c^2*cos(2*f*x + 2*e) +
I*a*c^2*sin(6*f*x + 6*e) + 3*I*a*c^2*sin(4*f*x + 4*e) + 3*I*a*c^2*sin(2*f*x + 2*e) + a*c^2)*arctan2(cos(1/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 3*(I*
a*c^2*cos(6*f*x + 6*e) + 3*I*a*c^2*cos(4*f*x + 4*e) + 3*I*a*c^2*cos(2*f*x + 2*e) - a*c^2*sin(6*f*x + 6*e) - 3*
a*c^2*sin(4*f*x + 4*e) - 3*a*c^2*sin(2*f*x + 2*e) + I*a*c^2)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e))) + 1) + 3*(-I*a*c^2*cos(6*f*x + 6*e) - 3*I*a*c^2*cos(4*f*x + 4*e) - 3*I*a*c^2*cos(2*f*x + 2*e) +
a*c^2*sin(6*f*x + 6*e) + 3*a*c^2*sin(4*f*x + 4*e) + 3*a*c^2*sin(2*f*x + 2*e) - I*a*c^2)*log(cos(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(a)*sqrt(c)/(f*(-12*I*cos(6*f*x + 6*e) - 36*I*cos(4*f*x + 4
*e) - 36*I*cos(2*f*x + 2*e) + 12*sin(6*f*x + 6*e) + 36*sin(4*f*x + 4*e) + 36*sin(2*f*x + 2*e) - 12*I))

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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. 455 vs. \(2 (126) = 252\).
time = 1.14, size = 455, normalized size = 2.86 \begin {gather*} \frac {3 \, \sqrt {\frac {a^{3} c^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{3} c^{5}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2}}\right ) - 3 \, \sqrt {\frac {a^{3} c^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{3} c^{5}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2}}\right ) - 4 \, {\left (3 i \, a c^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + 8 i \, a c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} - 3 i \, a c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\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-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(a^3*c^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2*(a*c^2*e^(3*I*f*x + 3
*I*e) + a*c^2*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(a^3*
c^5/f^2)*(I*f*e^(2*I*f*x + 2*I*e) - I*f))/(a*c^2*e^(2*I*f*x + 2*I*e) + a*c^2)) - 3*sqrt(a^3*c^5/f^2)*(f*e^(4*I
*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2*(a*c^2*e^(3*I*f*x + 3*I*e) + a*c^2*e^(I*f*x + I*e))*sqrt
(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(a^3*c^5/f^2)*(-I*f*e^(2*I*f*x + 2*I*e)
+ I*f))/(a*c^2*e^(2*I*f*x + 2*I*e) + a*c^2)) - 4*(3*I*a*c^2*e^(5*I*f*x + 5*I*e) + 8*I*a*c^2*e^(3*I*f*x + 3*I*e
) - 3*I*a*c^2*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(4*I*
f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [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-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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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-I*c*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 {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\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 - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(5/2), x)

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