\(\int \frac {a+a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx\) [339]

Optimal result

Integrand size = 23, antiderivative size = 50 \[ \int \frac {a+a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {d} (1-\tan (e+f x))}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {d} f} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.48 \[ \int \frac {a+a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {(1-i) \sqrt [4]{-1} a \left (\arctan \left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+i \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )\right ) \sqrt {\tan (e+f x)}}{f \sqrt {d \tan (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4015, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4015

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(271\) vs. \(2(41)=82\).

Time = 1.72 (sec) , antiderivative size = 272, normalized size of antiderivative = 5.44

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}\)	\(272\)
default	\(\frac {a \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}\)	\(272\)
parts	\(\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f d}+\frac {a \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}\)	\(275\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.50 \[ \int \frac {a+a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=\left [\frac {\sqrt {2} a \sqrt {-\frac {1}{d}} \log \left (\frac {2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-\frac {1}{d}} {\left (\tan \left (f x + e\right ) - 1\right )} + \tan \left (f x + e\right )^{2} - 4 \, \tan \left (f x + e\right ) + 1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, f}, \frac {\sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right )}{\sqrt {d} f}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.56 \[ \int \frac {a+a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=\frac {a {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{f} \] Input:

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

Output:

a*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/s 
qrt(d))/sqrt(d) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d* 
tan(f*x + e)))/sqrt(d))/sqrt(d))/f
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 1.49 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30 \[ \int \frac {a+a \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx=\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (1-\mathrm {i}\right )}{\sqrt {d}\,f}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (-1-\mathrm {i}\right )}{\sqrt {d}\,f} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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