3.11.35 \(\int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [1035]

Optimal. Leaf size=204 \[ \frac {2 i a^{7/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a^3 \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

2*I*a^(7/2)*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))/c^(5/2)/f-2*I*a^3*(a+I*a
*tan(f*x+e))^(1/2)/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-2/5*I*a*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(5/2)+
2/3*I*a^2*(a+I*a*tan(f*x+e))^(3/2)/c/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.13, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3604, 49, 65, 223, 209} \begin {gather*} \frac {2 i a^{7/2} \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 )}{c^{5/2} f}-\frac {2 i a^3 \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(7/2)/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((2*I)*a^(7/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(c^(5/2)*f)
- (((2*I)/5)*a*(a + I*a*Tan[e + f*x])^(5/2))/(f*(c - I*c*Tan[e + f*x])^(5/2)) + (((2*I)/3)*a^2*(a + I*a*Tan[e
+ f*x])^(3/2))/(c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((2*I)*a^3*Sqrt[a + I*a*Tan[e + f*x]])/(c^2*f*Sqrt[c - I*c
*Tan[e + f*x]])

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 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))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {a^2 \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}+\frac {a^3 \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{c f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a^3 \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {a^4 \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{c^2 f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a^3 \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {\left (2 i a^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 )}{c^2 f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a^3 \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {\left (2 i a^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 )}{c^2 f}\\ &=\frac {2 i a^{7/2} \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 )}{c^{5/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a^3 \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 9.70, size = 205, normalized size = 1.00 \begin {gather*} \frac {2 a^3 \cos ^2(e+f x) \left (\cos \left (\frac {1}{2} (e-4 f x)\right )-i \sin \left (\frac {1}{2} (e-4 f x)\right )\right ) \left (4 i \cos (e+f x)+9 i \cos (3 (e+f x))+6 \sin (e+f x)-15 i \text {ArcTan}\left (e^{i (e+f x)}\right ) \cos (e+f x) (\cos (3 (e+f x))-i \sin (3 (e+f x)))+6 \sin (3 (e+f x))\right ) \left (i \cos \left (\frac {1}{2} (e+6 f x)\right )+\sin \left (\frac {1}{2} (e+6 f x)\right )\right ) (-i+\tan (e+f x))^3 \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(7/2)/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(2*a^3*Cos[e + f*x]^2*(Cos[(e - 4*f*x)/2] - I*Sin[(e - 4*f*x)/2])*((4*I)*Cos[e + f*x] + (9*I)*Cos[3*(e + f*x)]
+ 6*Sin[e + f*x] - (15*I)*ArcTan[E^(I*(e + f*x))]*Cos[e + f*x]*(Cos[3*(e + f*x)] - I*Sin[3*(e + f*x)]) + 6*Si
n[3*(e + f*x)])*(I*Cos[(e + 6*f*x)/2] + Sin[(e + 6*f*x)/2])*(-I + Tan[e + f*x])^3*Sqrt[a + I*a*Tan[e + f*x]])/
(15*c^2*f*Sqrt[c - I*c*Tan[e + f*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. 428 vs. \(2 (164 ) = 328\).
time = 0.35, size = 429, normalized size = 2.10

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (60 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+15 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )-60 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-90 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-94 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-46 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+15 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )+26 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+74 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{4} \sqrt {a c}}\)	\(429\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (60 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+15 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )-60 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-90 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-94 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-46 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+15 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )+26 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+74 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{4} \sqrt {a c}}\)	\(429\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3/c^3*(60*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)^3+15*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)^4-60*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)-90*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)^2-94*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-46*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/
2)*tan(f*x+e)^3+15*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))+26*I*(a*c*(1+
tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+74*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)/(tan(f*x+e)+I)^4/(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. 472 vs. \(2 (160) = 320\).
time = 0.59, size = 472, normalized size = 2.31 \begin {gather*} -\frac {{\left (-30 i \, a^{3} \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 ) - 30 i \, a^{3} \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 ) + 12 i \, a^{3} \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 ) - 20 i \, a^{3} \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 ) + 60 i \, a^{3} \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 ) + 15 \, a^{3} \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 ) - 15 \, a^{3} \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 ) - 12 \, a^{3} \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 ) + 20 \, a^{3} \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 ) - 60 \, a^{3} \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}}{30 \, c^{\frac {5}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-1/30*(-30*I*a^3*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) - 30*I*a^3*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 12*I*a^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
)) - 20*I*a^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 60*I*a^3*cos(1/2*arctan2(sin(2*f*x + 2*e)
, cos(2*f*x + 2*e))) + 15*a^3*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) - 15*a^3*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) - 12*a^3*sin(5/2*arctan2(sin(2*f*x + 2*e), co
s(2*f*x + 2*e))) + 20*a^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 60*a^3*sin(1/2*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(5/2)*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. 400 vs. \(2 (160) = 320\).
time = 1.22, size = 400, normalized size = 1.96 \begin {gather*} -\frac {15 \, c^{3} f \sqrt {\frac {a^{7}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} 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^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c^{3} f\right )} \sqrt {\frac {a^{7}}{c^{5} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 15 \, c^{3} f \sqrt {\frac {a^{7}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} 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^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{3} f\right )} \sqrt {\frac {a^{7}}{c^{5} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) + 4 \, {\left (3 i \, a^{3} e^{\left (7 i \, f x + 7 i \, e\right )} - 2 i \, a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} + 10 i \, a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + 15 i \, a^{3} 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}}}{30 \, c^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/30*(15*c^3*f*sqrt(a^7/(c^5*f^2))*log(4*(2*(a^3*e^(3*I*f*x + 3*I*e) + a^3*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^3*f*e^(2*I*f*x + 2*I*e) - I*c^3*f)*sqrt(a^7/(c^5*f^2
)))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) - 15*c^3*f*sqrt(a^7/(c^5*f^2))*log(4*(2*(a^3*e^(3*I*f*x + 3*I*e) + a^3*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^3*f*e^(2*I*f*x + 2*
I*e) + I*c^3*f)*sqrt(a^7/(c^5*f^2)))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) + 4*(3*I*a^3*e^(7*I*f*x + 7*I*e) - 2*I*a
^3*e^(5*I*f*x + 5*I*e) + 10*I*a^3*e^(3*I*f*x + 3*I*e) + 15*I*a^3*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)))/(c^3*f)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(7/2)/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3062 deep

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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))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{{\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)^(7/2)/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(7/2)/(c - c*tan(e + f*x)*1i)^(5/2), x)

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