\(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx\) [362]

Optimal result

Integrand size = 25, antiderivative size = 111 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx=-\frac {\arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a d^{3/2} f}+\frac {\arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {2}{a d f \sqrt {d \tan (e+f x)}} \] Output:

-arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a/d^(3/2)/f+1/2*arctan(1/2*(d^(1/2)- 
d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/a/d^(3/2)/f-2/a/ 
d/f/(d*tan(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.39 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx=-\frac {2 \left (\frac {d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{2 f}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \sqrt [4]{-1} d^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt [4]{-1} d^{3/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}\right )}{a d^3}-\frac {2}{a d f \sqrt {d \tan (e+f x)}} \] Input:

Integrate[1/((d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])),x]
 

Output:

(-2*((d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(2*f) - ((1/4 - I/4)*( 
-1)^(1/4)*d^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - ( 
(1/4 + I/4)*(-1)^(1/4)*d^(3/2)*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/S 
qrt[d]])/f))/(a*d^3) - 2/(a*d*f*Sqrt[d*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4052, 27, 3042, 4136, 3042, 4015, 218, 4117, 27, 73, 216}

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.

\(\displaystyle \int \frac {1}{(a \tan (e+f x)+a) (d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \tan (e+f x)+a) (d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4015

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 216

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

Input:

Int[1/((d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])),x]
 

Output:

-(((d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f - (d^(3/2)*ArcTan[(a^2 
*d^2 - a^2*d^2*Tan[e + f*x])/(Sqrt[2]*a^2*d^(3/2)*Sqrt[d*Tan[e + f*x]])])/ 
(Sqrt[2]*f))/(a*d^3)) - 2/(a*d*f*Sqrt[d*Tan[e + f*x]])
 

Defintions of rubi rules used

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

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

rule 218

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

rule 3042

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

rule 4015

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

rule 4052

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c 
+ d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 
/((m + 1)*(a^2 + b^2)*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + 
d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - 
 a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / 
; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] 
 && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ 
erQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4117

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

rule 4136

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(318\) vs. \(2(93)=186\).

Time = 1.37 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.87

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{2 d^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {7}{2}}}-\frac {1}{d^{3} \sqrt {d \tan \left (f x +e \right )}}\right )}{f a}\)	\(319\)
default	\(\frac {2 d^{2} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{2 d^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {7}{2}}}-\frac {1}{d^{3} \sqrt {d \tan \left (f x +e \right )}}\right )}{f a}\)	\(319\)

Input:

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

Output:

2/f/a*d^2*(1/2/d^3*(-1/8/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/ 
4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d* 
tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*ta 
n(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))- 
1/8/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2) 
*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/ 
2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*ar 
ctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)))-1/2/d^(7/2)*arctan((d* 
tan(f*x+e))^(1/2)/d^(1/2))-1/d^3/(d*tan(f*x+e))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.51 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx=\left [-\frac {\sqrt {2} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right ) + 2 \, \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) \tan \left (f x + e\right ) + 8 \, \sqrt {d \tan \left (f x + e\right )}}{4 \, a d^{2} f \tan \left (f x + e\right )}, -\frac {\sqrt {2} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right ) - 2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right ) + 4 \, \sqrt {d \tan \left (f x + e\right )}}{2 \, a d^{2} f \tan \left (f x + e\right )}\right ] \] Input:

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

Output:

[-1/4*(sqrt(2)*sqrt(-d)*log((d*tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*tan(f*x + 
 e))*sqrt(-d)*(tan(f*x + e) - 1) - 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 
 1))*tan(f*x + e) + 2*sqrt(-d)*log((d*tan(f*x + e) + 2*sqrt(d*tan(f*x + e) 
)*sqrt(-d) - d)/(tan(f*x + e) + 1))*tan(f*x + e) + 8*sqrt(d*tan(f*x + e))) 
/(a*d^2*f*tan(f*x + e)), -1/2*(sqrt(2)*sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d*t 
an(f*x + e))*(tan(f*x + e) - 1)/(sqrt(d)*tan(f*x + e)))*tan(f*x + e) - 2*s 
qrt(d)*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)*tan(f*x + e)))*tan(f*x + e) + 
4*sqrt(d*tan(f*x + e)))/(a*d^2*f*tan(f*x + e))]
 

Sympy [F]

\[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx=\frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )} + \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx=-\frac {\frac {\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}}{a} + \frac {2 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a \sqrt {d}} + \frac {4}{\sqrt {d \tan \left (f x + e\right )} a}}{2 \, d f} \] Input:

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

Output:

-1/2*((sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e) 
))/sqrt(d))/sqrt(d) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqr 
t(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/a + 2*arctan(sqrt(d*tan(f*x + e))/sqr 
t(d))/(a*sqrt(d)) + 4/(sqrt(d*tan(f*x + e))*a))/(d*f)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 1.44 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx=-\frac {2}{a\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,d^{3/2}\,f}-\frac {\sqrt {2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{4\,a\,d^{3/2}\,f} \] Input:

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

Output:

- 2/(a*d*f*(d*tan(e + f*x))^(1/2)) - atan((d*tan(e + f*x))^(1/2)/d^(1/2))/ 
(a*d^(3/2)*f) - (2^(1/2)*(2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/ 
2))) + 2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2)) + (2^(1/2)*(d*t 
an(e + f*x))^(3/2))/(2*d^(3/2)))))/(4*a*d^(3/2)*f)
 

Reduce [F]

\[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{3}+\tan \left (f x +e \right )^{2}}d x \right )}{a \,d^{2}} \] Input:

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

Output:

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