\(\int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1161]

Optimal result

Integrand size = 32, antiderivative size = 209 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 \sqrt [4]{-1} a^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}} \]

[Out]

2*(-1)^(1/4)*a^(5/2)*arctanh((-1)^(3/4)*d^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c+d*tan(f*x+e))^(1/2))/d^(3/
2)/f-4*I*a^(5/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))*2^(1/2
)/(c-I*d)^(3/2)/f+2*a^2*(c+I*d)*(a+I*a*tan(f*x+e))^(1/2)/(c-I*d)/d/f/(c+d*tan(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3634, 3682, 3625, 214, 3680, 65, 223, 212} \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 \sqrt [4]{-1} a^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{d f (c-i d) \sqrt {c+d \tan (e+f x)}} \]

[In]

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

[Out]

(2*(-1)^(1/4)*a^(5/2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]
])])/(d^(3/2)*f) - ((4*I)*Sqrt[2]*a^(5/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sq
rt[a + I*a*Tan[e + f*x]])])/((c - I*d)^(3/2)*f) + (2*a^2*(c + I*d)*Sqrt[a + I*a*Tan[e + f*x]])/((c - I*d)*d*f*
Sqrt[c + d*Tan[e + f*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 212

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

Rule 214

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

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 3625

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

Rule 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x]
+ Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(
m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3680

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

Rule 3682

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}-\frac {2 \int \frac {\sqrt {a+i a \tan (e+f x)} \left (-\frac {1}{2} a^2 (c+3 i d)+\frac {1}{2} a^2 (i c+d) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{d (i c+d)} \\ & = \frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}+\frac {\left (4 a^2\right ) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{c-i d}-\frac {(i a) \int \frac {(a-i a \tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{d} \\ & = \frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}-\frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{d f}+\frac {\left (8 a^4\right ) \text {Subst}\left (\int \frac {1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}\right )}{(i c+d) f} \\ & = -\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+i d-\frac {i d x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{d f} \\ & = -\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{1+\frac {i d x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{d f} \\ & = \frac {2 \sqrt [4]{-1} a^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1510\) vs. \(2(209)=418\).

Time = 7.22 (sec) , antiderivative size = 1510, normalized size of antiderivative = 7.22 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {2 d (a+i a \tan (e+f x))^{5/2}}{\left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 \left (-\frac {3 i (i a c-a d)^3 \sqrt {\frac {i a}{-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}}} \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )^3 \sqrt {\frac {i a (c+d \tan (e+f x))}{i a c-a d}} \sqrt {1+\frac {i a d (a+i a \tan (e+f x))}{(i a c-a d) \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )}} \left (\frac {2 i a d (a+i a \tan (e+f x))}{(i a c-a d) \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )}+\frac {4 a^2 d^2 (a+i a \tan (e+f x))^2}{3 (i a c-a d)^2 \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )^2}-\frac {2 \sqrt [4]{-1} \sqrt {a} \sqrt {d} \text {arcsinh}\left (\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a c-a d} \sqrt {-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}}}\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a c-a d} \sqrt {-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}} \sqrt {1+\frac {i a d (a+i a \tan (e+f x))}{(i a c-a d) \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )}}}\right )}{4 a^2 d^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {a \left (\frac {1}{2} i a^2 (c+5 i d)+2 a^2 d\right ) \left (2 a \left (-\frac {2 \sqrt {2} \arctan \left (\frac {\sqrt {-a c+i a d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a c+i a d}}-\frac {2 (-1)^{3/4} \sqrt {c+i d} \sqrt {\frac {1}{\frac {c}{c+i d}+\frac {i d}{c+i d}}} \sqrt {\frac {c}{c+i d}+\frac {i d}{c+i d}} \arcsin \left (\frac {\sqrt [4]{-1} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+i d} \sqrt {\frac {c}{c+i d}+\frac {i d}{c+i d}}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{\sqrt {a} \sqrt {d} \sqrt {c+d \tan (e+f x)}}\right )-\frac {a (c+i d)^2 \sqrt {\frac {c+d \tan (e+f x)}{c+i d}} \sqrt {1-\frac {i d (a+i a \tan (e+f x))}{a (c+i d) \left (\frac {c}{c+i d}+\frac {i d}{c+i d}\right )}} \left (-\frac {2 i d (a+i a \tan (e+f x))}{a (c+i d) \left (\frac {c}{c+i d}+\frac {i d}{c+i d}\right )}+\frac {2 \sqrt [4]{-1} \sqrt {d} \arcsin \left (\frac {\sqrt [4]{-1} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+i d} \sqrt {\frac {c}{c+i d}+\frac {i d}{c+i d}}}\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+i d} \sqrt {\frac {c}{c+i d}+\frac {i d}{c+i d}} \sqrt {1-\frac {i d (a+i a \tan (e+f x))}{a (c+i d) \left (\frac {c}{c+i d}+\frac {i d}{c+i d}\right )}}}\right )}{2 d^2 \left (\frac {1}{\frac {c}{c+i d}+\frac {i d}{c+i d}}\right )^{3/2} \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\right )}{f}\right )}{a \left (c^2+d^2\right )} \]

[In]

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

[Out]

(-2*d*(a + I*a*Tan[e + f*x])^(5/2))/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) + (2*((((-3*I)/4)*(I*a*c - a*d)^3
*Sqrt[(I*a)/(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))]*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c
- a*d))^3*Sqrt[(I*a*(c + d*Tan[e + f*x]))/(I*a*c - a*d)]*Sqrt[1 + (I*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*
d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)))]*(((2*I)*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(
-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))) + (4*a^2*d^2*(a + I*a*Tan[e + f*x])^2)/(3*(I*a*c - a*d)^2
*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))^2) - (2*(-1)^(1/4)*Sqrt[a]*Sqrt[d]*ArcSinh[((-1)^(1/4)*S
qrt[a]*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[I*a*c - a*d]*Sqrt[-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c
- a*d)])]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[I*a*c - a*d]*Sqrt[-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a
*d)]*Sqrt[1 + (I*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d
)))])))/(a^2*d^2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) + (a*((I/2)*a^2*(c + (5*I)*d) + 2*a^2*
d)*(2*a*((-2*Sqrt[2]*ArcTan[(Sqrt[-(a*c) + I*a*d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*
x]])])/Sqrt[-(a*c) + I*a*d] - (2*(-1)^(3/4)*Sqrt[c + I*d]*Sqrt[(c/(c + I*d) + (I*d)/(c + I*d))^(-1)]*Sqrt[c/(c
+ I*d) + (I*d)/(c + I*d)]*ArcSin[((-1)^(1/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + I*d]*Sqrt[
c/(c + I*d) + (I*d)/(c + I*d)])]*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)])/(Sqrt[a]*Sqrt[d]*Sqrt[c + d*Tan[e + f*x
]])) - (a*(c + I*d)^2*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)]*Sqrt[1 - (I*d*(a + I*a*Tan[e + f*x]))/(a*(c + I*d)*
(c/(c + I*d) + (I*d)/(c + I*d)))]*(((-2*I)*d*(a + I*a*Tan[e + f*x]))/(a*(c + I*d)*(c/(c + I*d) + (I*d)/(c + I*
d))) + (2*(-1)^(1/4)*Sqrt[d]*ArcSin[((-1)^(1/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + I*d]*Sqr
t[c/(c + I*d) + (I*d)/(c + I*d)])]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + I*d]*Sqrt[c/(c + I*d) + (I*d)
/(c + I*d)]*Sqrt[1 - (I*d*(a + I*a*Tan[e + f*x]))/(a*(c + I*d)*(c/(c + I*d) + (I*d)/(c + I*d)))])))/(2*d^2*((c
/(c + I*d) + (I*d)/(c + I*d))^(-1))^(3/2)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])))/f))/(a*(c^2 +
d^2))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1814 vs. \(2 (165 ) = 330\).

Time = 1.35 (sec) , antiderivative size = 1815, normalized size of antiderivative = 8.68

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1815\)
default	\(\text {Expression too large to display}\)	\(1815\)

[In]

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

[Out]

1/2/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(-2*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a
*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*c*d*(I*a*d)^(1/2)+2*I*2^(1/2)*l
n(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2)
)*a*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)-6*I*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*ta
n(f*x+e)))^(1/2)*c*d^2-3*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1
/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^4*(-a*(I*d-c))^(1/2)*tan(f*x+e)+ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(
a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^3*
d*tan(f*x+e)-7*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*
d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d^3*tan(f*x+e)+5*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2
*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2*d^2*(-a*(I*d-c))^(1/2)*ta
n(f*x+e)-2*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e)
)*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*d^2*(I*a*d)^(1/2)*tan(f*x+e)+5*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+
e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3*d*(-a*(I*d-c))^
(1/2)+2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d
)/(I*a*d)^(1/2))*a*c^4*(-a*(I*d-c))^(1/2)-7*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*
x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^2*d^2+2*I*2^(1/2)*c^3*(-a*(I*d-c
))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+2*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*
c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d*(-a*(I*d-c))^(1/2)-2*2
^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*
d)^(1/2))*a*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)-6*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1
/2)*(-a*(I*d-c))^(1/2)*c^2*d+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^
(1/2)*d^3-3*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(
1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^3*(-a*(I*d-c))^(1/2)-2*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(
c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d-2*ln((3*a*c+I*a*tan(
f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan
(f*x+e)+I))*a*d^2*(I*a*d)^(1/2)*tan(f*x+e)-2*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(
I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*c*d)*a^2/(c+d*tan(f
*x+e))^(1/2)/d/(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)/(c^2+d^2)/(I*c-d)/(I*a*d)^(1/2)/(-a*(I*d-c))^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 940 vs. \(2 (157) = 314\).

Time = 0.26 (sec) , antiderivative size = 940, normalized size of antiderivative = 4.50 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/2*(4*sqrt(2)*((a^2*c + I*a^2*d)*e^(3*I*f*x + 3*I*e) + (a^2*c + I*a^2*d)*e^(I*f*x + I*e))*sqrt(((c - I*d)*e^(
2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)) + ((c^2*d - 2*I*c*d^2
- d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(4*I*a^5/(d^3*f^2))*log((I*d^2*f*sqrt(4*I*a^5/(d^3*f^2))*
e^(I*f*x + I*e) + sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2
*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2) - ((c^2*d - 2*I*c*d^2 - d^3)*f*
e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(4*I*a^5/(d^3*f^2))*log((-I*d^2*f*sqrt(4*I*a^5/(d^3*f^2))*e^(I*f*x
+ I*e) + sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x +
2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2) + ((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*
x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(32*I*a^5/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/4*((I*c^2 + 2*c*
d - I*d^2)*sqrt(32*I*a^5/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I
*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*
f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2) - ((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f
)*sqrt(32*I*a^5/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/4*((-I*c^2 - 2*c*d + I*d^2)*sqrt(32*I*a^5/((
-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqrt((
(c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f
*x - I*e)/a^2))/((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((a+I*a*tan(f*x+e))**(5/2)/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Integral((I*a*(tan(e + f*x) - I))**(5/2)/(c + d*tan(e + f*x))**(3/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume((d^2-2*c*d-c^2)>0)', see `assu
me?` for mor

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Non regular value [0,0] was discarded and replaced randomly by 0=[41,-73]Warning, replacing 41 by -69, a su
bstitution

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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