\(\int \frac {\tan (x)}{(a+b \tan ^4(x))^{3/2}} \, dx\) [403]

Optimal result

Integrand size = 15, antiderivative size = 74 \[ \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 (a+b)^{3/2}}+\frac {a+b \tan ^2(x)}{2 a (a+b) \sqrt {a+b \tan ^4(x)}} \] Output:

-1/2*arctanh((a-b*tan(x)^2)/(a+b)^(1/2)/(a+b*tan(x)^4)^(1/2))/(a+b)^(3/2)+ 
1/2*(a+b*tan(x)^2)/a/(a+b)/(a+b*tan(x)^4)^(1/2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99 \[ \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\frac {1}{2} \left (-\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{(a+b)^{3/2}}+\frac {a+b \tan ^2(x)}{a (a+b) \sqrt {a+b \tan ^4(x)}}\right ) \] Input:

Integrate[Tan[x]/(a + b*Tan[x]^4)^(3/2),x]
 

Output:

(-(ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(a + b)^(3 
/2)) + (a + b*Tan[x]^2)/(a*(a + b)*Sqrt[a + b*Tan[x]^4]))/2
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 4153, 1577, 496, 25, 27, 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.

\(\displaystyle \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (x)}{\left (a+b \tan (x)^4\right )^{3/2}}dx\)

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 1577

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

\(\Big \downarrow \) 496

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {a+b \tan ^2(x)}{a (a+b) \sqrt {a+b \tan ^4(x)}}-\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{(a+b)^{3/2}}\right )\)

Input:

Int[Tan[x]/(a + b*Tan[x]^4)^(3/2),x]
 

Output:

(-(ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(a + b)^(3 
/2)) + (a + b*Tan[x]^2)/(a*(a + b)*Sqrt[a + b*Tan[x]^4]))/2
 

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 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 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 496

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 
 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a 
 + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 
*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad 
raticQ[a, 0, b, c, d, n, p, x]
 

rule 1577

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

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(247\) vs. \(2(62)=124\).

Time = 0.15 (sec) , antiderivative size = 248, normalized size of antiderivative = 3.35

method	result	size
derivativedivides	\(\frac {b \ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \left (b +\sqrt {-a b}\right ) \left (-b +\sqrt {-a b}\right ) \sqrt {a +b}}+\frac {\sqrt {\left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )}}{4 a \left (b +\sqrt {-a b}\right ) \left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )}-\frac {\sqrt {\left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )}}{4 a \left (-b +\sqrt {-a b}\right ) \left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )}\)	\(248\)
default	\(\frac {b \ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \left (b +\sqrt {-a b}\right ) \left (-b +\sqrt {-a b}\right ) \sqrt {a +b}}+\frac {\sqrt {\left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )}}{4 a \left (b +\sqrt {-a b}\right ) \left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )}-\frac {\sqrt {\left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )}}{4 a \left (-b +\sqrt {-a b}\right ) \left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )}\)	\(248\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (64) = 128\).

Time = 0.30 (sec) , antiderivative size = 319, normalized size of antiderivative = 4.31 \[ \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\left [\frac {{\left (a b \tan \left (x\right )^{4} + a^{2}\right )} \sqrt {a + b} \log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \tan \left (x\right )^{4} - 2 \, a b \tan \left (x\right )^{2} + 2 \, \sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {a + b} + 2 \, a^{2} + a b}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right ) + 2 \, \sqrt {b \tan \left (x\right )^{4} + a} {\left ({\left (a b + b^{2}\right )} \tan \left (x\right )^{2} + a^{2} + a b\right )}}{4 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (x\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )}}, -\frac {{\left (a b \tan \left (x\right )^{4} + a^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \tan \left (x\right )^{4} + a^{2} + a b}\right ) - \sqrt {b \tan \left (x\right )^{4} + a} {\left ({\left (a b + b^{2}\right )} \tan \left (x\right )^{2} + a^{2} + a b\right )}}{2 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (x\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )}}\right ] \] Input:

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

Output:

[1/4*((a*b*tan(x)^4 + a^2)*sqrt(a + 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)) + 2*sqrt(b*tan(x)^4 + a)*((a*b + b^2)*ta 
n(x)^2 + a^2 + a*b))/((a^3*b + 2*a^2*b^2 + a*b^3)*tan(x)^4 + a^4 + 2*a^3*b 
 + a^2*b^2), -1/2*((a*b*tan(x)^4 + a^2)*sqrt(-a - 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)) - 
 sqrt(b*tan(x)^4 + a)*((a*b + b^2)*tan(x)^2 + a^2 + a*b))/((a^3*b + 2*a^2* 
b^2 + a*b^3)*tan(x)^4 + a^4 + 2*a^3*b + a^2*b^2)]
 

Sympy [F]

\[ \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\int \frac {\tan {\left (x \right )}}{\left (a + b \tan ^{4}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\int { \frac {\tan \left (x\right )}{{\left (b \tan \left (x\right )^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.61 \[ \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\frac {\frac {{\left (a b + b^{2}\right )} \tan \left (x\right )^{2}}{a^{3} + 2 \, a^{2} b + a b^{2}} + \frac {a^{2} + a b}{a^{3} + 2 \, a^{2} b + a b^{2}}}{2 \, \sqrt {b \tan \left (x\right )^{4} + a}} - \frac {\arctan \left (\frac {\sqrt {b} \tan \left (x\right )^{2} - \sqrt {b \tan \left (x\right )^{4} + a} + \sqrt {b}}{\sqrt {-a - b}}\right )}{{\left (a + b\right )} \sqrt {-a - b}} \] Input:

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

Output:

1/2*((a*b + b^2)*tan(x)^2/(a^3 + 2*a^2*b + a*b^2) + (a^2 + a*b)/(a^3 + 2*a 
^2*b + a*b^2))/sqrt(b*tan(x)^4 + a) - arctan((sqrt(b)*tan(x)^2 - sqrt(b*ta 
n(x)^4 + a) + sqrt(b))/sqrt(-a - b))/((a + b)*sqrt(-a - b))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\int \frac {\mathrm {tan}\left (x\right )}{{\left (b\,{\mathrm {tan}\left (x\right )}^4+a\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\tan (x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )}{\tan \left (x \right )^{8} b^{2}+2 \tan \left (x \right )^{4} a b +a^{2}}d x \] Input:

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

Output:

int((sqrt(tan(x)**4*b + a)*tan(x))/(tan(x)**8*b**2 + 2*tan(x)**4*a*b + a** 
2),x)