3.11.16 \(\int \frac {(a+i a \tan (e+f x))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [1016]

Optimal. Leaf size=204 \[ \frac {15 i a^{7/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f} \]

[Out]

15*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))/f/c^(1/2)-15/2*I*a^3*(a
+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c/f-5/2*I*a^2*(c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2
)/c/f-2*I*a*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(1/2)

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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 = 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 {15 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 )}{\sqrt {c} f}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((15*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]])])/(Sqrt[c]*f)
- ((2*I)*a*(a + I*a*Tan[e + f*x])^(5/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (((15*I)/2)*a^3*Sqrt[a + I*a*Tan[e
+ f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(c*f) - (((5*I)/2)*a^2*(a + I*a*Tan[e + f*x])^(3/2)*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))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(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))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {\left (5 a^2\right ) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {\left (15 a^3\right ) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {\left (15 a^4\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 {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}+\frac {\left (15 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 )}{f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}+\frac {\left (15 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 )}{f}\\ &=\frac {15 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 )}{\sqrt {c} f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}\\ \end {align*}

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Mathematica [A]
time = 8.56, size = 340, normalized size = 1.67 \begin {gather*} \frac {15 i e^{-i (4 e+f x)} \sqrt {e^{i f x}} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \text {ArcTan}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2}}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}} f \sec ^{\frac {7}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2}}+\frac {\cos ^3(e+f x) \left (\cos (2 f x) \left (-\frac {4 i \cos (e)}{c}-\frac {4 \sin (e)}{c}\right )+\sec (e) (16 \cos (e)+i \sin (e)) \left (-\frac {i \cos (3 e)}{2 c}-\frac {\sin (3 e)}{2 c}\right )+\sec (e) \sec (e+f x) \left (\frac {\cos (3 e)}{2 c}-\frac {i \sin (3 e)}{2 c}\right ) \sin (f x)+\left (\frac {4 \cos (e)}{c}-\frac {4 i \sin (e)}{c}\right ) \sin (2 f x)\right ) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} (a+i a \tan (e+f x))^{7/2}}{f (\cos (f x)+i \sin (f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((15*I)*Sqrt[E^(I*f*x)]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*ArcTan[E^(I*(e + f*x))]*(a + I*a*Tan[e
+ f*x])^(7/2))/(E^(I*(4*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*Sec[e + f*x]^(7/2)*(Cos[f*x] + I*Sin[f*
x])^(7/2)) + (Cos[e + f*x]^3*(Cos[2*f*x]*(((-4*I)*Cos[e])/c - (4*Sin[e])/c) + Sec[e]*(16*Cos[e] + I*Sin[e])*((
(-1/2*I)*Cos[3*e])/c - Sin[3*e]/(2*c)) + Sec[e]*Sec[e + f*x]*(Cos[3*e]/(2*c) - ((I/2)*Sin[3*e])/c)*Sin[f*x] +
((4*Cos[e])/c - ((4*I)*Sin[e])/c)*Sin[2*f*x])*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*
Tan[e + f*x])^(7/2))/(f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]
time = 0.42, size = 328, normalized size = 1.61

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (30 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+6 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+15 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )-24 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}-15 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )-31 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{2 f c \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(328\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (30 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+6 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+15 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )-24 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}-15 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )-31 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{2 f c \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(328\)

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))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3/c*(30*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)+6*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2
+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)^2-(a*c)^(1/2)*(a*
c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3-24*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-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))-31*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)/(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. 843 vs. \(2 (160) = 320\).
time = 0.59, size = 843, normalized size = 4.13 \begin {gather*} -\frac {2 \, {\left (36 \, 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 ) + 36 i \, 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 ) - 30 \, {\left (a^{3} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, a^{3} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{3} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{3} \sin \left (2 \, f x + 2 \, e\right ) + a^{3}\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 ) - 30 \, {\left (a^{3} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, a^{3} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{3} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{3} \sin \left (2 \, f x + 2 \, e\right ) + a^{3}\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 (8 \, a^{3} \cos \left (4 \, f x + 4 \, e\right ) + 16 \, a^{3} \cos \left (2 \, f x + 2 \, e\right ) + 8 i \, a^{3} \sin \left (4 \, f x + 4 \, e\right ) + 16 i \, a^{3} \sin \left (2 \, f x + 2 \, e\right ) + 15 \, a^{3}\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 ) - 15 \, {\left (i \, a^{3} \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{3} \cos \left (2 \, f x + 2 \, e\right ) - a^{3} \sin \left (4 \, f x + 4 \, e\right ) - 2 \, a^{3} \sin \left (2 \, f x + 2 \, e\right ) + i \, a^{3}\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 ) - 15 \, {\left (-i \, a^{3} \cos \left (4 \, f x + 4 \, e\right ) - 2 i \, a^{3} \cos \left (2 \, f x + 2 \, e\right ) + a^{3} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, a^{3} \sin \left (2 \, f x + 2 \, e\right ) - i \, a^{3}\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 (-8 i \, a^{3} \cos \left (4 \, f x + 4 \, e\right ) - 16 i \, a^{3} \cos \left (2 \, f x + 2 \, e\right ) + 8 \, a^{3} \sin \left (4 \, f x + 4 \, e\right ) + 16 \, a^{3} \sin \left (2 \, f x + 2 \, e\right ) - 15 i \, a^{3}\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}}{-8 \, {\left (i \, c \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, c \cos \left (2 \, f x + 2 \, e\right ) - c \sin \left (4 \, f x + 4 \, e\right ) - 2 \, 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))^(7/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-2*(36*a^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 36*I*a^3*sin(3/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e))) - 30*(a^3*cos(4*f*x + 4*e) + 2*a^3*cos(2*f*x + 2*e) + I*a^3*sin(4*f*x + 4*e) + 2*I*a^3*sin(2
*f*x + 2*e) + 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*(a^3*cos(4*f*x + 4*e) + 2*a^3*cos(2*f*x + 2*e) + I*a^3*sin(4*f*x + 4*e) + 2*
I*a^3*sin(2*f*x + 2*e) + a^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 4*(8*a^3*cos(4*f*x + 4*e) + 16*a^3*cos(2*f*x + 2*e) + 8*I*a^3*sin(4
*f*x + 4*e) + 16*I*a^3*sin(2*f*x + 2*e) + 15*a^3)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 15*(I
*a^3*cos(4*f*x + 4*e) + 2*I*a^3*cos(2*f*x + 2*e) - a^3*sin(4*f*x + 4*e) - 2*a^3*sin(2*f*x + 2*e) + I*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*(-I*a^3*cos(4*f*x + 4*e) - 2*I*a^3*cos(2*
f*x + 2*e) + a^3*sin(4*f*x + 4*e) + 2*a^3*sin(2*f*x + 2*e) - I*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) - 4*(-8*I*a^3*cos(4*f*x + 4*e) - 16*I*a^3*cos(2*f*x + 2*e) + 8*a^3*sin(4*f*x + 4*e)
+ 16*a^3*sin(2*f*x + 2*e) - 15*I*a^3)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((
-8*I*c*cos(4*f*x + 4*e) - 16*I*c*cos(2*f*x + 2*e) + 8*c*sin(4*f*x + 4*e) + 16*c*sin(2*f*x + 2*e) - 8*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. 416 vs. \(2 (160) = 320\).
time = 1.01, size = 416, normalized size = 2.04 \begin {gather*} -\frac {15 \, \sqrt {\frac {a^{7}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \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}} - \sqrt {\frac {a^{7}}{c f^{2}}} {\left (i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c f\right )}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 15 \, \sqrt {\frac {a^{7}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \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}} - \sqrt {\frac {a^{7}}{c f^{2}}} {\left (-i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c f\right )}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) + 4 \, {\left (8 i \, a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} + 25 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}}}{4 \, {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \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))^(1/2),x, algorithm="fricas")

[Out]

-1/4*(15*sqrt(a^7/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) + c*f)*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)) - sqrt(a^7/(c*f^2))*(I*c*f*e^(2*I*f*
x + 2*I*e) - I*c*f))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) - 15*sqrt(a^7/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) + c*f)*l
og(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)) - sqrt(a^7/(c*f^2))*(-I*c*f*e^(2*I*f*x + 2*I*e) + I*c*f))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) + 4*(
8*I*a^3*e^(5*I*f*x + 5*I*e) + 25*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*f*e^(2*I*f*x + 2*I*e) + c*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))**(1/2),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))^(1/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(7/2)/sqrt(-I*c*tan(f*x + e) + c), 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}}{\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)^(7/2)/(c - c*tan(e + f*x)*1i)^(1/2),x)

[Out]

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

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