\(\int \frac {(d \tan (e+f x))^{3/2}}{a+a \tan (e+f x)} \, dx\) [359]

Optimal result

Integrand size = 25, antiderivative size = 87 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+a \tan (e+f x)} \, dx=\frac {d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}-\frac {d^{3/2} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(280\) vs. \(2(87)=174\).

Time = 0.38 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.22 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+a \tan (e+f x)} \, dx=\frac {8 d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-4 \left (-d^2\right )^{3/4} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right )+2 \sqrt {2} d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )-2 \sqrt {2} d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+4 \left (-d^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right )+\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )-\sqrt {2} d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{8 a f} \] Input:

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

Output:

(8*d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]] - 4*(-d^2)^(3/4)*ArcTan[Sq 
rt[d*Tan[e + f*x]]/(-d^2)^(1/4)] + 2*Sqrt[2]*d^(3/2)*ArcTan[1 - (Sqrt[2]*S 
qrt[d*Tan[e + f*x]])/Sqrt[d]] - 2*Sqrt[2]*d^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt 
[d*Tan[e + f*x]])/Sqrt[d]] + 4*(-d^2)^(3/4)*ArcTanh[Sqrt[d*Tan[e + f*x]]/( 
-d^2)^(1/4)] + Sqrt[2]*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2 
]*Sqrt[d*Tan[e + f*x]]] - Sqrt[2]*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f* 
x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(8*a*f)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4056, 25, 3042, 4015, 221, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4056

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4015

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 216

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

Input:

Int[(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]])/(a*f) - (d^(3/2)*ArcTanh[(a 
*d^2 + a*d^2*Tan[e + f*x])/(Sqrt[2]*a*d^(3/2)*Sqrt[d*Tan[e + f*x]])])/(Sqr 
t[2]*a*f)
 

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 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 221

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 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 4056

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2)   Int[Simp[a^2*c - b^2*c + 
2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e + f*x 
]], x], x] + Simp[(b*c - a*d)^2/(c^2 + d^2)   Int[(1 + Tan[e + f*x]^2)/(Sqr 
t[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(70)=140\).

Time = 1.39 (sec) , antiderivative size = 298, normalized size of antiderivative = 3.43

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 \left (d^{2}\right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}\right )}{f a}\)	\(298\)
default	\(\frac {2 d^{2} \left (-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 \left (d^{2}\right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}\right )}{f a}\)	\(298\)

Input:

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

Output:

2/f/a*d^2*(-1/16/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*ta 
n(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*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))+1/16/(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*arctan(-2 
^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))+1/2/d^(1/2)*arctan((d*tan(f*x+ 
e))^(1/2)/d^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.45 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+a \tan (e+f x)} \, dx=\left [\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {-d} d \arctan \left (\frac {\sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{d \tan \left (f x + e\right )}\right ) + \sqrt {-d} d \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right )}{2 \, a f}, \frac {\sqrt {\frac {1}{2}} d^{\frac {3}{2}} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 4 \, \sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right )}{2 \, a f}\right ] \] Input:

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

Output:

[1/2*(2*sqrt(1/2)*sqrt(-d)*d*arctan(sqrt(1/2)*sqrt(d*tan(f*x + e))*sqrt(-d 
)*(tan(f*x + e) + 1)/(d*tan(f*x + e))) + sqrt(-d)*d*log((d*tan(f*x + e) + 
2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)))/(a*f), 1/2*(sqrt 
(1/2)*d^(3/2)*log((d*tan(f*x + e)^2 - 4*sqrt(1/2)*sqrt(d*tan(f*x + e))*sqr 
t(d)*(tan(f*x + e) + 1) + 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) - 2* 
d^(3/2)*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)*tan(f*x + e))))/(a*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.29 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+a \tan (e+f x)} \, dx=-\frac {\frac {d^{3} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a} - \frac {4 \, d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a}}{4 \, d f} \] Input:

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

Output:

-1/4*(d^3*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt( 
d) + d)/sqrt(d) - sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e) 
)*sqrt(d) + d)/sqrt(d))/a - 4*d^(5/2)*arctan(sqrt(d*tan(f*x + e))/sqrt(d)) 
/a)/(d*f)
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90 \[ \int \frac {(d \tan (e+f x))^{3/2}}{a+a \tan (e+f x)} \, dx=\frac {d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,f}-\frac {\sqrt {2}\,d^{3/2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,d^{25/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{12\,d^{13}\,\mathrm {tan}\left (e+f\,x\right )+12\,d^{13}}\right )}{2\,a\,f} \] Input:

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

Output:

(d^(3/2)*atan((d*tan(e + f*x))^(1/2)/d^(1/2)))/(a*f) - (2^(1/2)*d^(3/2)*at 
anh((12*2^(1/2)*d^(25/2)*(d*tan(e + f*x))^(1/2))/(12*d^13*tan(e + f*x) + 1 
2*d^13)))/(2*a*f)
 

Reduce [F]

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

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