\(\int \frac {\tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\) [36]

Optimal result

Integrand size = 33, antiderivative size = 79 \[ \int \frac {\tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a-b+c} e} \] Output:

-1/2*arctanh(1/2*(2*a-b+(b-2*c)*tan(e*x+d)^2)/(a-b+c)^(1/2)/(a+b*tan(e*x+d 
)^2+c*tan(e*x+d)^4)^(1/2))/(a-b+c)^(1/2)/e
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a-b+c} e} \] Input:

Integrate[Tan[d + e*x]/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
 

Output:

-1/2*ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[ 
a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])]/(Sqrt[a - b + c]*e)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3042, 4183, 1576, 1154, 219}

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.

\(\displaystyle \int \frac {\tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (d+e x)}{\sqrt {a+b \tan (d+e x)^2+c \tan (d+e x)^4}}dx\)

\(\Big \downarrow \) 4183

\(\displaystyle \frac {\int \frac {\tan (d+e x)}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan (d+e x)}{e}\)

\(\Big \downarrow \) 1576

\(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(d+e x)+1\right ) \sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}d\tan ^2(d+e x)}{2 e}\)

\(\Big \downarrow \) 1154

\(\displaystyle -\frac {\int \frac {1}{4 (a-b+c)-\tan ^4(d+e x)}d\frac {(b-2 c) \tan ^2(d+e x)+2 a-b}{\sqrt {c \tan ^4(d+e x)+b \tan ^2(d+e x)+a}}}{e}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\text {arctanh}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt {a-b+c}}\)

Input:

Int[Tan[d + e*x]/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
 

Output:

-1/2*ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[ 
a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])]/(Sqrt[a - b + c]*e)
 

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1576

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( 
p_.), x_Symbol] :> Simp[1/2   Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] 
, x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
 

rule 3042

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

rule 4183

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*( 
x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] 
 :> Simp[f/e   Subst[Int[(x/f)^m*((a + b*x^n + c*x^(2*n))^p/(f^2 + x^2)), x 
], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n 
2, 2*n] && NeQ[b^2 - 4*a*c, 0]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.29

method	result	size
derivativedivides	\(-\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2 e \sqrt {a -b +c}}\)	\(102\)
default	\(-\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+2 \sqrt {a -b +c}\, \sqrt {c \left (1+\tan \left (e x +d \right )^{2}\right )^{2}+\left (b -2 c \right ) \left (1+\tan \left (e x +d \right )^{2}\right )+a -b +c}}{1+\tan \left (e x +d \right )^{2}}\right )}{2 e \sqrt {a -b +c}}\)	\(102\)

Input:

int(tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x,method=_RETURNVER 
BOSE)
 

Output:

-1/2/e/(a-b+c)^(1/2)*ln((2*a-2*b+2*c+(b-2*c)*(1+tan(e*x+d)^2)+2*(a-b+c)^(1 
/2)*(c*(1+tan(e*x+d)^2)^2+(b-2*c)*(1+tan(e*x+d)^2)+a-b+c)^(1/2))/(1+tan(e* 
x+d)^2))
 

Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 299, normalized size of antiderivative = 3.78 \[ \int \frac {\tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\left [\frac {\log \left (\frac {{\left (b^{2} + 4 \, {\left (a - 2 \, b\right )} c + 8 \, c^{2}\right )} \tan \left (e x + d\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2} - 4 \, {\left (a - b\right )} c\right )} \tan \left (e x + d\right )^{2} - 4 \, \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} {\left ({\left (b - 2 \, c\right )} \tan \left (e x + d\right )^{2} + 2 \, a - b\right )} \sqrt {a - b + c} + 8 \, a^{2} - 8 \, a b + b^{2} + 4 \, a c}{\tan \left (e x + d\right )^{4} + 2 \, \tan \left (e x + d\right )^{2} + 1}\right )}{4 \, \sqrt {a - b + c} e}, -\frac {\sqrt {-a + b - c} \arctan \left (-\frac {\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} {\left ({\left (b - 2 \, c\right )} \tan \left (e x + d\right )^{2} + 2 \, a - b\right )} \sqrt {-a + b - c}}{2 \, {\left ({\left ({\left (a - b\right )} c + c^{2}\right )} \tan \left (e x + d\right )^{4} + {\left (a b - b^{2} + b c\right )} \tan \left (e x + d\right )^{2} + a^{2} - a b + a c\right )}}\right )}{2 \, {\left (a - b + c\right )} e}\right ] \] Input:

integrate(tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm= 
"fricas")
 

Output:

[1/4*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 
- 4*(a - b)*c)*tan(e*x + d)^2 - 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 
 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b + c) + 8*a^2 - 8*a*b 
 + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1))/(sqrt(a - b + c)* 
e), -1/2*sqrt(-a + b - c)*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + 
d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(-a + b - c)/(((a - b)* 
c + c^2)*tan(e*x + d)^4 + (a*b - b^2 + b*c)*tan(e*x + d)^2 + a^2 - a*b + a 
*c))/((a - b + c)*e)]
 

Sympy [F]

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

integrate(tan(e*x+d)/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2),x)
 

Output:

Integral(tan(d + e*x)/sqrt(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4), x)
 

Maxima [F]

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

integrate(tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm= 
"maxima")
 

Output:

integrate(tan(e*x + d)/sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\text {Timed out} \] Input:

integrate(tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm= 
"giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx=\int \frac {\mathrm {tan}\left (d+e\,x\right )}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a}} \,d x \] Input:

int(tan(d + e*x)/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2),x)
 

Output:

int(tan(d + e*x)/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2), x)
 

Reduce [F]

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

int(tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)
 

Output:

int((sqrt(tan(d + e*x)**4*c + tan(d + e*x)**2*b + a)*tan(d + e*x))/(tan(d 
+ e*x)**4*c + tan(d + e*x)**2*b + a),x)