\(\int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx\) [398]

Optimal result

Integrand size = 17, antiderivative size = 74 \[ \int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {b}}+\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {b}}+\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \] Input:

Integrate[Tan[x]^3/Sqrt[a + b*Tan[x]^4],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 4153, 1579, 605, 224, 219, 488, 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.

\(\displaystyle \int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (x)^3}{\sqrt {a+b \tan (x)^4}}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \int \frac {\tan ^3(x)}{\left (\tan ^2(x)+1\right ) \sqrt {a+b \tan ^4(x)}}d\tan (x)\)

\(\Big \downarrow \) 1579

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

\(\Big \downarrow \) 605

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{\sqrt {b}}+\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{\sqrt {a+b}}\right )\)

Input:

Int[Tan[x]^3/Sqrt[a + b*Tan[x]^4],x]
 

Output:

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

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 488

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

rule 605

Int[((x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] 
:> Simp[1/d   Int[x^(m - 1)*(a + b*x^2)^p, x], x] - Simp[c/d   Int[x^(m - 1 
)*((a + b*x^2)^p/(c + d*x)), x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 
 0] && LtQ[-1, p, 0]
 

rule 1579

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], 
 x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]
 

rule 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {\ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )}{2 \sqrt {b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \sqrt {a +b}}\)	\(91\)
default	\(\frac {\ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )}{2 \sqrt {b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \sqrt {a +b}}\)	\(91\)

Input:

int(tan(x)^3/(a+b*tan(x)^4)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*ln(b^(1/2)*tan(x)^2+(a+b*tan(x)^4)^(1/2))/b^(1/2)+1/2/(a+b)^(1/2)*ln(( 
2*a+2*b-2*b*(1+tan(x)^2)+2*(a+b)^(1/2)*(b*(1+tan(x)^2)^2-2*b*(1+tan(x)^2)+ 
a+b)^(1/2))/(1+tan(x)^2))
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 483, normalized size of antiderivative = 6.53 \[ \int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\left [\frac {{\left (a + b\right )} \sqrt {b} \log \left (-2 \, b \tan \left (x\right )^{4} - 2 \, \sqrt {b \tan \left (x\right )^{4} + a} \sqrt {b} \tan \left (x\right )^{2} - a\right ) + \sqrt {a + b} b \log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \tan \left (x\right )^{4} - 2 \, a b \tan \left (x\right )^{2} - 2 \, \sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {a + b} + 2 \, a^{2} + a b}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right )}{4 \, {\left (a b + b^{2}\right )}}, -\frac {2 \, {\left (a + b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (x\right )^{4} + a} \sqrt {-b}}{b \tan \left (x\right )^{2}}\right ) - \sqrt {a + b} b \log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \tan \left (x\right )^{4} - 2 \, a b \tan \left (x\right )^{2} - 2 \, \sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {a + b} + 2 \, a^{2} + a b}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right )}{4 \, {\left (a b + b^{2}\right )}}, \frac {2 \, \sqrt {-a - b} b \arctan \left (\frac {\sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \tan \left (x\right )^{4} + a^{2} + a b}\right ) + {\left (a + b\right )} \sqrt {b} \log \left (-2 \, b \tan \left (x\right )^{4} - 2 \, \sqrt {b \tan \left (x\right )^{4} + a} \sqrt {b} \tan \left (x\right )^{2} - a\right )}{4 \, {\left (a b + b^{2}\right )}}, \frac {\sqrt {-a - b} b \arctan \left (\frac {\sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \tan \left (x\right )^{4} + a^{2} + a b}\right ) - {\left (a + b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (x\right )^{4} + a} \sqrt {-b}}{b \tan \left (x\right )^{2}}\right )}{2 \, {\left (a b + b^{2}\right )}}\right ] \] Input:

integrate(tan(x)^3/(a+b*tan(x)^4)^(1/2),x, algorithm="fricas")
 

Output:

[1/4*((a + b)*sqrt(b)*log(-2*b*tan(x)^4 - 2*sqrt(b*tan(x)^4 + a)*sqrt(b)*t 
an(x)^2 - a) + sqrt(a + b)*b*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 
- 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(a + b) + 2*a^2 + a*b)/(tan( 
x)^4 + 2*tan(x)^2 + 1)))/(a*b + b^2), -1/4*(2*(a + b)*sqrt(-b)*arctan(sqrt 
(b*tan(x)^4 + a)*sqrt(-b)/(b*tan(x)^2)) - sqrt(a + b)*b*log(((a*b + 2*b^2) 
*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt( 
a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)))/(a*b + b^2), 1/4*(2*sq 
rt(-a - b)*b*arctan(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(-a - b)/((a 
*b + b^2)*tan(x)^4 + a^2 + a*b)) + (a + b)*sqrt(b)*log(-2*b*tan(x)^4 - 2*s 
qrt(b*tan(x)^4 + a)*sqrt(b)*tan(x)^2 - a))/(a*b + b^2), 1/2*(sqrt(-a - b)* 
b*arctan(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(-a - b)/((a*b + b^2)*t 
an(x)^4 + a^2 + a*b)) - (a + b)*sqrt(-b)*arctan(sqrt(b*tan(x)^4 + a)*sqrt( 
-b)/(b*tan(x)^2)))/(a*b + b^2)]
 

Sympy [F]

\[ \int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\int \frac {\tan ^{3}{\left (x \right )}}{\sqrt {a + b \tan ^{4}{\left (x \right )}}}\, dx \] Input:

integrate(tan(x)**3/(a+b*tan(x)**4)**(1/2),x)
 

Output:

Integral(tan(x)**3/sqrt(a + b*tan(x)**4), x)
 

Maxima [F]

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

integrate(tan(x)^3/(a+b*tan(x)^4)^(1/2),x, algorithm="maxima")
 

Output:

integrate(tan(x)^3/sqrt(b*tan(x)^4 + a), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(x)^3/(a+b*tan(x)^4)^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio 
n over extensionDegree mismatch inside factorisation over extensionError: 
Bad Argum
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\int \frac {{\mathrm {tan}\left (x\right )}^3}{\sqrt {b\,{\mathrm {tan}\left (x\right )}^4+a}} \,d x \] Input:

int(tan(x)^3/(a + b*tan(x)^4)^(1/2),x)
 

Output:

int(tan(x)^3/(a + b*tan(x)^4)^(1/2), x)
 

Reduce [F]

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

int(tan(x)^3/(a+b*tan(x)^4)^(1/2),x)
 

Output:

int((sqrt(tan(x)**4*b + a)*tan(x)**3)/(tan(x)**4*b + a),x)