3.11.33 \(\int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [1033]

Optimal. Leaf size=304 \[ \frac {63 i a^{11/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac {42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {63 i a^5 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f} \]

[Out]

63*I*a^(11/2)*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-63/2*I*a^5*(
a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c^3/f-21/2*I*a^4*(c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^
(3/2)/c^3/f-42/5*I*a^3*(a+I*a*tan(f*x+e))^(5/2)/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-2/5*I*a*(a+I*a*tan(f*x+e))^(9/2
)/f/(c-I*c*tan(f*x+e))^(5/2)+6/5*I*a^2*(a+I*a*tan(f*x+e))^(7/2)/c/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.17, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 9, 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 {63 i a^{11/2} \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 {63 i a^5 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a (a+i a \tan (e+f x))^{9/2}}{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])^(11/2)/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((63*I)*a^(11/2)*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])^(9/2))/(f*(c - I*c*Tan[e + f*x])^(5/2)) + (((6*I)/5)*a^2*(a + I*a*Tan[
e + f*x])^(7/2))/(c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (((42*I)/5)*a^3*(a + I*a*Tan[e + f*x])^(5/2))/(c^2*f*Sqr
t[c - I*c*Tan[e + f*x]]) - (((63*I)/2)*a^5*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(c^3*f) - ((
(21*I)/2)*a^4*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(c^3*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))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{9/2}}{(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))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {\left (9 a^2\right ) \text {Subst}\left (\int \frac {(a+i a x)^{7/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}+\frac {\left (21 a^3\right ) \text {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{5 c f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac {42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {\left (21 a^4\right ) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{\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))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac {42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {\left (63 a^5\right ) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c^2 f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac {42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {63 i a^5 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {\left (63 a^6\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 c^2 f}\\ &=-\frac {2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac {42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {63 i a^5 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}+\frac {\left (63 i a^5\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))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac {42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {63 i a^5 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}+\frac {\left (63 i a^5\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 {63 i a^{11/2} \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))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac {42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {63 i a^5 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}\\ \end {align*}

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Mathematica [A]
time = 13.03, size = 459, normalized size = 1.51 \begin {gather*} \frac {63 i e^{-i (6 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))^{11/2}}{c^2 \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} f \sec ^{\frac {11}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{11/2}}+\frac {\cos ^5(e+f x) \left (\cos (6 f x) \left (-\frac {4 i \cos (e)}{5 c^3}+\frac {4 \sin (e)}{5 c^3}\right )+\cos (4 f x) \left (\frac {16 i \cos (e)}{5 c^3}+\frac {16 \sin (e)}{5 c^3}\right )+\cos (2 f x) \left (-\frac {20 i \cos (3 e)}{c^3}-\frac {20 \sin (3 e)}{c^3}\right )+\sec (e) (64 \cos (e)+i \sin (e)) \left (-\frac {i \cos (5 e)}{2 c^3}-\frac {\sin (5 e)}{2 c^3}\right )+\sec (e) \sec (e+f x) \left (\frac {\cos (5 e)}{2 c^3}-\frac {i \sin (5 e)}{2 c^3}\right ) \sin (f x)+\left (\frac {20 \cos (3 e)}{c^3}-\frac {20 i \sin (3 e)}{c^3}\right ) \sin (2 f x)+\left (-\frac {16 \cos (e)}{5 c^3}+\frac {16 i \sin (e)}{5 c^3}\right ) \sin (4 f x)+\left (\frac {4 \cos (e)}{5 c^3}+\frac {4 i \sin (e)}{5 c^3}\right ) \sin (6 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))^{11/2}}{f (\cos (f x)+i \sin (f x))^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(11/2)/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((63*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])^(11/2))/(c^2*E^(I*(6*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*Sec[e + f*x]^(11/2)*(Cos[f*x] + I*
Sin[f*x])^(11/2)) + (Cos[e + f*x]^5*(Cos[6*f*x]*((((-4*I)/5)*Cos[e])/c^3 + (4*Sin[e])/(5*c^3)) + Cos[4*f*x]*((
((16*I)/5)*Cos[e])/c^3 + (16*Sin[e])/(5*c^3)) + Cos[2*f*x]*(((-20*I)*Cos[3*e])/c^3 - (20*Sin[3*e])/c^3) + Sec[
e]*(64*Cos[e] + I*Sin[e])*(((-1/2*I)*Cos[5*e])/c^3 - Sin[5*e]/(2*c^3)) + Sec[e]*Sec[e + f*x]*(Cos[5*e]/(2*c^3)
- ((I/2)*Sin[5*e])/c^3)*Sin[f*x] + ((20*Cos[3*e])/c^3 - ((20*I)*Sin[3*e])/c^3)*Sin[2*f*x] + ((-16*Cos[e])/(5*
c^3) + (((16*I)/5)*Sin[e])/c^3)*Sin[4*f*x] + ((4*Cos[e])/(5*c^3) + (((4*I)/5)*Sin[e])/c^3)*Sin[6*f*x])*Sqrt[Se
c[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*Tan[e + f*x])^(11/2))/(f*(Cos[f*x] + I*Sin[f*x])^5)

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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. 489 vs. \(2 (244 ) = 488\).
time = 0.34, size = 490, normalized size = 1.61

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{5} \left (1260 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+315 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )+60 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{4}\left (f x +e \right )\right )-5 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{5}\left (f x +e \right )\right )-1260 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-1890 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-1964 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-866 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+315 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )+496 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+1659 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{10 f \,c^{3} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{4}}\)	\(490\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{5} \left (1260 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+315 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )+60 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{4}\left (f x +e \right )\right )-5 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{5}\left (f x +e \right )\right )-1260 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-1890 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-1964 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-866 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+315 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )+496 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+1659 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{10 f \,c^{3} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{4}}\)	\(490\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(11/2)/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/10/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^5/c^3*(1260*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+315*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*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^4-5*(a*c
*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^5-1260*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)-1890*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)^2-1964*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-866*(a*c)^(1/2)*(a*c*(1+tan
(f*x+e)^2))^(1/2)*tan(f*x+e)^3+315*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
))+496*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+1659*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)^4

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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. 1159 vs. \(2 (238) = 476\).
time = 0.61, size = 1159, normalized size = 3.81 \begin {gather*} \frac {10 \, {\left (630 \, {\left (a^{5} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{5} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) + a^{5}\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 ) + 630 \, {\left (a^{5} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{5} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) + a^{5}\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 ) - 32 \, {\left (a^{5} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{5} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) + a^{5}\right )} \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 \, {\left (8 \, a^{5} \cos \left (4 \, f x + 4 \, e\right ) + 16 \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) + 8 i \, a^{5} \sin \left (4 \, f x + 4 \, e\right ) + 16 i \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) - 9 \, a^{5}\right )} \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 \, {\left (16 \, a^{5} \cos \left (4 \, f x + 4 \, e\right ) + 32 \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) + 16 i \, a^{5} \sin \left (4 \, f x + 4 \, e\right ) + 32 i \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) + 21 \, a^{5}\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 ) + 315 \, {\left (i \, a^{5} \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) - a^{5} \sin \left (4 \, f x + 4 \, e\right ) - 2 \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) + i \, a^{5}\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 ) + 315 \, {\left (-i \, a^{5} \cos \left (4 \, f x + 4 \, e\right ) - 2 i \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) + a^{5} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) - i \, a^{5}\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 ) + 32 \, {\left (-i \, a^{5} \cos \left (4 \, f x + 4 \, e\right ) - 2 i \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) + a^{5} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) - i \, a^{5}\right )} \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 \, {\left (8 i \, a^{5} \cos \left (4 \, f x + 4 \, e\right ) + 16 i \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) - 8 \, a^{5} \sin \left (4 \, f x + 4 \, e\right ) - 16 \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) - 9 i \, a^{5}\right )} \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 \, {\left (-16 i \, a^{5} \cos \left (4 \, f x + 4 \, e\right ) - 32 i \, a^{5} \cos \left (2 \, f x + 2 \, e\right ) + 16 \, a^{5} \sin \left (4 \, f x + 4 \, e\right ) + 32 \, a^{5} \sin \left (2 \, f x + 2 \, e\right ) - 21 i \, a^{5}\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}}{-200 \, {\left (i \, c^{3} \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, c^{3} \cos \left (2 \, f x + 2 \, e\right ) - c^{3} \sin \left (4 \, f x + 4 \, e\right ) - 2 \, c^{3} \sin \left (2 \, f x + 2 \, e\right ) + i \, c^{3}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(11/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

10*(630*(a^5*cos(4*f*x + 4*e) + 2*a^5*cos(2*f*x + 2*e) + I*a^5*sin(4*f*x + 4*e) + 2*I*a^5*sin(2*f*x + 2*e) + a
^5)*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) + 630*(a^5*cos(4*f*x + 4*e) + 2*a^5*cos(2*f*x + 2*e) + I*a^5*sin(4*f*x + 4*e) + 2*I*a^5*sin(2*f*
x + 2*e) + a^5)*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) - 32*(a^5*cos(4*f*x + 4*e) + 2*a^5*cos(2*f*x + 2*e) + I*a^5*sin(4*f*x + 4*e) + 2*I*
a^5*sin(2*f*x + 2*e) + a^5)*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 20*(8*a^5*cos(4*f*x + 4*e)
+ 16*a^5*cos(2*f*x + 2*e) + 8*I*a^5*sin(4*f*x + 4*e) + 16*I*a^5*sin(2*f*x + 2*e) - 9*a^5)*cos(3/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) - 60*(16*a^5*cos(4*f*x + 4*e) + 32*a^5*cos(2*f*x + 2*e) + 16*I*a^5*sin(4*f*x
+ 4*e) + 32*I*a^5*sin(2*f*x + 2*e) + 21*a^5)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 315*(I*a^5
*cos(4*f*x + 4*e) + 2*I*a^5*cos(2*f*x + 2*e) - a^5*sin(4*f*x + 4*e) - 2*a^5*sin(2*f*x + 2*e) + I*a^5)*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) + 315*(-I*a^5*cos(4*f*x + 4*e) - 2*I*a^5*cos(2*f*x
+ 2*e) + a^5*sin(4*f*x + 4*e) + 2*a^5*sin(2*f*x + 2*e) - I*a^5)*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), c
os(2*f*x + 2*e))) + 1) + 32*(-I*a^5*cos(4*f*x + 4*e) - 2*I*a^5*cos(2*f*x + 2*e) + a^5*sin(4*f*x + 4*e) + 2*a^5
*sin(2*f*x + 2*e) - I*a^5)*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 20*(8*I*a^5*cos(4*f*x + 4*e)
+ 16*I*a^5*cos(2*f*x + 2*e) - 8*a^5*sin(4*f*x + 4*e) - 16*a^5*sin(2*f*x + 2*e) - 9*I*a^5)*sin(3/2*arctan2(sin
(2*f*x + 2*e), cos(2*f*x + 2*e))) + 60*(-16*I*a^5*cos(4*f*x + 4*e) - 32*I*a^5*cos(2*f*x + 2*e) + 16*a^5*sin(4*
f*x + 4*e) + 32*a^5*sin(2*f*x + 2*e) - 21*I*a^5)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)
*sqrt(c)/((-200*I*c^3*cos(4*f*x + 4*e) - 400*I*c^3*cos(2*f*x + 2*e) + 200*c^3*sin(4*f*x + 4*e) + 400*c^3*sin(2
*f*x + 2*e) - 200*I*c^3)*f)

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Fricas [A]
time = 1.26, size = 466, normalized size = 1.53 \begin {gather*} -\frac {315 \, {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )} \sqrt {\frac {a^{11}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a^{5} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{5} 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^{11}}{c^{5} f^{2}}}\right )}}{a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{5}}\right ) - 315 \, {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )} \sqrt {\frac {a^{11}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a^{5} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{5} 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^{11}}{c^{5} f^{2}}}\right )}}{a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{5}}\right ) + 4 \, {\left (8 i \, a^{5} e^{\left (9 i \, f x + 9 i \, e\right )} - 24 i \, a^{5} e^{\left (7 i \, f x + 7 i \, e\right )} + 168 i \, a^{5} e^{\left (5 i \, f x + 5 i \, e\right )} + 525 i \, a^{5} e^{\left (3 i \, f x + 3 i \, e\right )} + 315 i \, a^{5} 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}}}{20 \, {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(11/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/20*(315*(c^3*f*e^(2*I*f*x + 2*I*e) + c^3*f)*sqrt(a^11/(c^5*f^2))*log(4*(2*(a^5*e^(3*I*f*x + 3*I*e) + a^5*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^11/(c^5*f^2)))/(a^5*e^(2*I*f*x + 2*I*e) + a^5)) - 315*(c^3*f*e^(2*I*f*x + 2*I*e) + c^3*f
)*sqrt(a^11/(c^5*f^2))*log(4*(2*(a^5*e^(3*I*f*x + 3*I*e) + a^5*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^11/(c^5*f^2)))/(a^5*e^
(2*I*f*x + 2*I*e) + a^5)) + 4*(8*I*a^5*e^(9*I*f*x + 9*I*e) - 24*I*a^5*e^(7*I*f*x + 7*I*e) + 168*I*a^5*e^(5*I*f
*x + 5*I*e) + 525*I*a^5*e^(3*I*f*x + 3*I*e) + 315*I*a^5*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqr
t(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c^3*f*e^(2*I*f*x + 2*I*e) + c^3*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))**(11/2)/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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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))^(11/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(11/2)/(-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 )}^{11/2}}{{\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)^(11/2)/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(11/2)/(c - c*tan(e + f*x)*1i)^(5/2), x)

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