3.11.17 \(\int \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [1017]

Optimal. Leaf size=153 \[ \frac {6 i a^{5/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 i a (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 i a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c f} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 153, 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, 49, 52, 65, 223, 209} \begin {gather*} \frac {6 i a^{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 )}{\sqrt {c} f}-\frac {3 i a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {2 i a (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 52

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[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 \frac {(a+i a \tan (e+f x))^{5/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 i a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {\left (3 a^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 )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 i a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c f}+\frac {\left (6 i a^2\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 {2 i a (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 i a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c f}+\frac {\left (6 i a^2\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 {6 i a^{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 )}{\sqrt {c} f}-\frac {2 i a (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 i a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c f}\\ \end {align*}

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Mathematica [A]
time = 3.24, size = 155, normalized size = 1.01 \begin {gather*} -\frac {2 i e^{-3 i (e+f x)} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \left (e^{i (e+f x)} \left (3+2 e^{2 i (e+f x)}\right )-3 \left (1+e^{2 i (e+f x)}\right ) \text {ArcTan}\left (e^{i (e+f x)}\right )\right ) (a+i a \tan (e+f x))^{5/2}}{c 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)/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

((-2*I)*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*(E^(I*(e + f*x))*(3
+ 2*E^((2*I)*(e + f*x))) - 3*(1 + E^((2*I)*(e + f*x)))*ArcTan[E^(I*(e + f*x))])*(a + I*a*Tan[e + f*x])^(5/2))/
(c*E^((3*I)*(e + f*x))*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. 298 vs. \(2 (124 ) = 248\).
time = 0.43, size = 299, normalized size = 1.95

method	result	size
derivativedivides	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (3 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-3 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -6 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )-6 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+5 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{f c \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(299\)
default	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (3 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-3 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -6 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )-6 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+5 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{f c \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(299\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^2/c*(3*I*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))*a*c*tan(f*x+e)^2-3*I*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))*a*c-6*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)-6*ln((c*a*tan(f*x+e)+(a*c*(1+ta
n(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)-(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e
)^2+5*(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)/(tan(f*x+e)+I)^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. 578 vs. \(2 (121) = 242\).
time = 0.59, size = 578, normalized size = 3.78 \begin {gather*} \frac {{\left (6 \, {\left (a^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + a^{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^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + a^{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 ) - 4 \, {\left (2 \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) + 2 i \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + 3 \, a^{2}\right )} \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 ) + 3 \, {\left (i \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) - a^{2} \sin \left (2 \, f x + 2 \, e\right ) + i \, a^{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^{2} \cos \left (2 \, f x + 2 \, e\right ) + a^{2} \sin \left (2 \, f x + 2 \, e\right ) - i \, a^{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 ) + 4 \, {\left (-2 i \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) + 2 \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) - 3 i \, a^{2}\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 )\right )} \sqrt {a} \sqrt {c}}{-2 \, {\left (i \, c \cos \left (2 \, f x + 2 \, e\right ) - c \sin \left (2 \, f x + 2 \, e\right ) + i \, c\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

(6*(a^2*cos(2*f*x + 2*e) + I*a^2*sin(2*f*x + 2*e) + a^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^2*cos(2*f*x + 2*e) + I*a^2*sin(2*f*x
+ 2*e) + a^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) - 4*(2*a^2*cos(2*f*x + 2*e) + 2*I*a^2*sin(2*f*x + 2*e) + 3*a^2)*cos(1/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) + 3*(I*a^2*cos(2*f*x + 2*e) - a^2*sin(2*f*x + 2*e) + I*a^2)*log(cos(1/2*arcta
n2(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^2*cos(2*f*x + 2*e) + a^2*sin(2*f*x + 2*e) - I*a^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) + 4*(-2*I*a^2*cos(2*f*x + 2*e) + 2*a^2*s
in(2*f*x + 2*e) - 3*I*a^2)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-2*I*c*cos(
2*f*x + 2*e) + 2*c*sin(2*f*x + 2*e) - 2*I*c)*f)

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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. 358 vs. \(2 (121) = 242\).
time = 1.11, size = 358, normalized size = 2.34 \begin {gather*} -\frac {3 \, \sqrt {\frac {a^{5}}{c f^{2}}} c f \log \left (\frac {4 \, {\left (2 \, {\left (a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{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}} - {\left (i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c f\right )} \sqrt {\frac {a^{5}}{c f^{2}}}\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right ) - 3 \, \sqrt {\frac {a^{5}}{c f^{2}}} c f \log \left (\frac {4 \, {\left (2 \, {\left (a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{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}} - {\left (-i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c f\right )} \sqrt {\frac {a^{5}}{c f^{2}}}\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right ) + 4 \, {\left (2 i \, a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 i \, a^{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}}}{2 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/2*(3*sqrt(a^5/(c*f^2))*c*f*log(4*(2*(a^2*e^(3*I*f*x + 3*I*e) + a^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)) - (I*c*f*e^(2*I*f*x + 2*I*e) - I*c*f)*sqrt(a^5/(c*f^2)))/(a^2*e^(
2*I*f*x + 2*I*e) + a^2)) - 3*sqrt(a^5/(c*f^2))*c*f*log(4*(2*(a^2*e^(3*I*f*x + 3*I*e) + a^2*e^(I*f*x + I*e))*sq
rt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - (-I*c*f*e^(2*I*f*x + 2*I*e) + I*c*f)*sqrt(
a^5/(c*f^2)))/(a^2*e^(2*I*f*x + 2*I*e) + a^2)) + 4*(2*I*a^2*e^(3*I*f*x + 3*I*e) + 3*I*a^2*e^(I*f*x + I*e))*sqr
t(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c*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}}}{\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, 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-I*c*tan(f*x+e))**(1/2),x)

[Out]

Integral((I*a*(tan(e + f*x) - I))**(5/2)/sqrt(-I*c*(tan(e + f*x) + I)), 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(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

[Out]

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

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