\(\int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx\) [183]

Optimal result

Integrand size = 28, antiderivative size = 234 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \sqrt {d} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{8 \sqrt {2} a^3 f}+\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2} \] Output:

1/16*I*d^(1/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a^3/ 
f-1/16*I*d^(1/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a^ 
3/f-1/16*I*d^(1/2)*arctanh(2^(1/2)*(d*tan(f*x+e))^(1/2)/(d^(1/2)+d^(1/2)*t 
an(f*x+e)))*2^(1/2)/a^3/f+1/6*I*(d*tan(f*x+e))^(1/2)/f/(a+I*a*tan(f*x+e))^ 
3+1/12*I*(d*tan(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e))^2
 

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {6 i \sqrt {2} \sqrt {d} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a^3}-\frac {6 i \sqrt {2} \sqrt {d} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a^3}+\frac {3 i \sqrt {2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{a^3}-\frac {3 i \sqrt {2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{a^3}-\frac {24 i \sqrt {d \tan (e+f x)}}{a^3 (-i+\tan (e+f x))^2}+\frac {16 (d \tan (e+f x))^{3/2}}{d (a+i a \tan (e+f x))^3}}{96 f} \] Input:

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

Output:

(((6*I)*Sqrt[2]*Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] 
)/a^3 - ((6*I)*Sqrt[2]*Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/S 
qrt[d]])/a^3 + ((3*I)*Sqrt[2]*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - 
 Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/a^3 - ((3*I)*Sqrt[2]*Sqrt[d]*Log[Sqrt[d] + 
 Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/a^3 - ((24*I)*Sqrt[ 
d*Tan[e + f*x]])/(a^3*(-I + Tan[e + f*x])^2) + (16*(d*Tan[e + f*x])^(3/2)) 
/(d*(a + I*a*Tan[e + f*x])^3))/(96*f)
 

Rubi [A] (warning: unable to verify)

Time = 0.82 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.14, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.607, Rules used = {3042, 4040, 3042, 4079, 27, 2011, 3042, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3}dx\)

\(\Big \downarrow \) 4040

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {i a d-5 a d \tan (e+f x)}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)^2}dx}{12 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {i a d-5 a d \tan (e+f x)}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)^2}dx}{12 a^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {\int \frac {6 \left (i a^2 d^2-a^2 d^2 \tan (e+f x)\right )}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)}dx}{4 a^2 d}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 \int \frac {i a^2 d^2-a^2 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)}dx}{2 a^2 d}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 2011

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d \int \frac {1}{\sqrt {d \tan (e+f x)}}dx}{2 a}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d \int \frac {1}{\sqrt {d \tan (e+f x)}}dx}{2 a}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \int \frac {1}{\sqrt {d \tan (e+f x)} \left (\tan ^2(e+f x) d^2+d^2\right )}d(d \tan (e+f x))}{2 a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \int \frac {1}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\int \frac {d^2 \tan ^2(e+f x)+d}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\)

Input:

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

Output:

((I/6)*Sqrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^3) - (((3*I)*d^2*(( 
-(ArcTan[1 - Sqrt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + 
 Sqrt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d]))/(2*d) + (-1/2*Log[d - Sq 
rt[2]*d^(3/2)*Tan[e + f*x] + d^2*Tan[e + f*x]^2]/(Sqrt[2]*Sqrt[d]) + Log[d 
 + Sqrt[2]*d^(3/2)*Tan[e + f*x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d])) 
/(2*d)))/(a*f) - (I*a*Sqrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^2))/ 
(12*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4040

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*(a + b*Tan[e + f*x])^m*(Sqrt[c + d 
*Tan[e + f*x]]/(2*a*f*m)), x] + Simp[1/(4*a^2*m)   Int[(a + b*Tan[e + f*x]) 
^(m + 1)*(Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], x]/Sqrt[c + d*Ta 
n[e + f*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] && LtQ[m, 0] && IntegersQ[2*m]
 

rule 4079

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

Maple [A] (verified)

Time = 1.48 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.51

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{3} \sqrt {i d}}+\frac {\frac {-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{3}}\right )}{f \,a^{3}}\)	\(120\)
default	\(\frac {2 d^{4} \left (\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{3} \sqrt {i d}}+\frac {\frac {-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{3}}\right )}{f \,a^{3}}\)	\(120\)

Input:

int((d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

2/f/a^3*d^4*(1/16/d^3/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2)) 
+1/16/d^3*((-2/3*I*d*(d*tan(f*x+e))^(3/2)-2*d^2*(d*tan(f*x+e))^(1/2))/(d*t 
an(f*x+e)-I*d)^3-1/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(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. 547 vs. \(2 (174) = 348\).

Time = 0.10 (sec) , antiderivative size = 547, normalized size of antiderivative = 2.34 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

-1/48*(12*a^3*f*sqrt(1/64*I*d/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(8*(I* 
a^3*f*e^(2*I*f*x + 2*I*e) + I*a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d) 
/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/64*I*d/(a^6*f^2)) + I*d*e^(2*I*f*x + 2* 
I*e))*e^(-2*I*f*x - 2*I*e)) - 12*a^3*f*sqrt(1/64*I*d/(a^6*f^2))*e^(6*I*f*x 
 + 6*I*e)*log(-2*(8*(-I*a^3*f*e^(2*I*f*x + 2*I*e) - I*a^3*f)*sqrt((-I*d*e^ 
(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/64*I*d/(a^6*f^2 
)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 12*a^3*f*sqrt(-1/64* 
I*d/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*(8*(a^3*f*e^(2*I*f*x + 2*I*e) + 
 a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*s 
qrt(-1/64*I*d/(a^6*f^2)) + d)*e^(-2*I*f*x - 2*I*e)/(a^3*f)) + 12*a^3*f*sqr 
t(-1/64*I*d/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/8*(8*(a^3*f*e^(2*I*f*x + 
 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e 
) + 1))*sqrt(-1/64*I*d/(a^6*f^2)) - d)*e^(-2*I*f*x - 2*I*e)/(a^3*f)) - sqr 
t((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(2*I*e^(6*I* 
f*x + 6*I*e) + 5*I*e^(4*I*f*x + 4*I*e) + 4*I*e^(2*I*f*x + 2*I*e) + I))*e^( 
-6*I*f*x - 6*I*e)/(a^3*f)
 

Sympy [F]

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

integrate((d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**3,x)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i \, {\left (\frac {3 i \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\frac {i \, d}{{\left | d \right |}} + 1} - \frac {3 i \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{-\frac {i \, d}{{\left | d \right |}} + 1} + \frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right )} d^{4} \tan \left (f x + e\right ) - 3 i \, \sqrt {d \tan \left (f x + e\right )} d^{4}\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3}}\right )}}{24 \, a^{3} d f} \] Input:

integrate((d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

-1/24*I*(3*I*sqrt(2)*d^(3/2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt( 
2)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/(I*d/abs(d) + 1) - 3*I*sqrt(2)*d^(3/ 
2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(-I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt 
(d)*abs(d)))/(-I*d/abs(d) + 1) + 2*(sqrt(d*tan(f*x + e))*d^4*tan(f*x + e) 
- 3*I*sqrt(d*tan(f*x + e))*d^4)/(d*tan(f*x + e) - I*d)^3)/(a^3*d*f)
 

Mupad [B] (verification not implemented)

Time = 1.13 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,a^3\,f}+\frac {d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,1{}\mathrm {i}}{12\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}}-\frac {{\left (-1\right )}^{1/4}\,\sqrt {d}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {{\left (-1\right )}^{1/4}\,\sqrt {d}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f} \] Input:

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

Output:

((d^3*(d*tan(e + f*x))^(1/2))/(4*a^3*f) + (d^2*(d*tan(e + f*x))^(3/2)*1i)/ 
(12*a^3*f))/(3*d^3*tan(e + f*x) - d^3*1i + d^3*tan(e + f*x)^2*3i - d^3*tan 
(e + f*x)^3) - ((-1)^(1/4)*d^(1/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2) 
)/d^(1/2)))/(8*a^3*f) - ((-1)^(1/4)*d^(1/2)*atanh(((-1)^(1/4)*(d*tan(e + f 
*x))^(1/2))/d^(1/2)))/(8*a^3*f)
 

Reduce [F]

\[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\sqrt {d}\, \left (-2 \sqrt {\tan \left (f x +e \right )}\, i -\left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{4}-3 \tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}+\tan \left (f x +e \right ) i}d x \right ) f +\left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{3}-3 \tan \left (f x +e \right )^{2} i -3 \tan \left (f x +e \right )+i}d x \right ) f i +3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{3}-3 \tan \left (f x +e \right )^{2} i -3 \tan \left (f x +e \right )+i}d x \right ) f -2 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3}-3 \tan \left (f x +e \right )^{2} i -3 \tan \left (f x +e \right )+i}d x \right ) f i +2 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3}-3 \tan \left (f x +e \right )^{2} i -3 \tan \left (f x +e \right )+i}d x \right ) f \right )}{3 a^{3} f} \] Input:

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

Output:

(sqrt(d)*( - 2*sqrt(tan(e + f*x))*i - int(sqrt(tan(e + f*x))/(tan(e + f*x) 
**4 - 3*tan(e + f*x)**3*i - 3*tan(e + f*x)**2 + tan(e + f*x)*i),x)*f + int 
((sqrt(tan(e + f*x))*tan(e + f*x)**4)/(tan(e + f*x)**3 - 3*tan(e + f*x)**2 
*i - 3*tan(e + f*x) + i),x)*f*i + 3*int((sqrt(tan(e + f*x))*tan(e + f*x)** 
3)/(tan(e + f*x)**3 - 3*tan(e + f*x)**2*i - 3*tan(e + f*x) + i),x)*f - 2*i 
nt((sqrt(tan(e + f*x))*tan(e + f*x)**2)/(tan(e + f*x)**3 - 3*tan(e + f*x)* 
*2*i - 3*tan(e + f*x) + i),x)*f*i + 2*int((sqrt(tan(e + f*x))*tan(e + f*x) 
)/(tan(e + f*x)**3 - 3*tan(e + f*x)**2*i - 3*tan(e + f*x) + i),x)*f))/(3*a 
**3*f)