\(\int \frac {\tan ^3(x)}{(a+b \tan ^4(x))^{3/2}} \, dx\) [402]

Optimal result

Integrand size = 17, antiderivative size = 71 \[ \int \frac {\tan ^3(x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 (a+b)^{3/2}}-\frac {1-\tan ^2(x)}{2 (a+b) \sqrt {a+b \tan ^4(x)}} \] 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*(1-tan(x)^2)/(a+b)/(a+b*tan(x)^4)^(1/2)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.94 \[ \int \frac {\tan ^3(x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\frac {1}{2} \left (\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{(a+b)^{3/2}}+\frac {-1+\tan ^2(x)}{(a+b) \sqrt {a+b \tan ^4(x)}}\right ) \] Input:

Integrate[Tan[x]^3/(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 
) + (-1 + Tan[x]^2)/((a + b)*Sqrt[a + b*Tan[x]^4]))/2
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 4153, 1579, 593, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 1579

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

\(\Big \downarrow \) 593

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

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

Input:

Int[Tan[x]^3/(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 
) - (1 - Tan[x]^2)/((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 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 593

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

Leaf count of result is larger than twice the leaf count of optimal. \(266\) vs. \(2(59)=118\).

Time = 29.78 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.76

method	result	size
derivativedivides	\(\frac {\tan \left (x \right )^{2}}{2 \sqrt {a +b \tan \left (x \right )^{4}}\, a}-\frac {b \ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \left (b +\sqrt {-a b}\right ) \left (-b +\sqrt {-a b}\right ) \sqrt {a +b}}-\frac {\sqrt {\left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )}}{4 a \left (b +\sqrt {-a b}\right ) \left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )}}{4 a \left (-b +\sqrt {-a b}\right ) \left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )}\)	\(267\)
default	\(\frac {\tan \left (x \right )^{2}}{2 \sqrt {a +b \tan \left (x \right )^{4}}\, a}-\frac {b \ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \left (b +\sqrt {-a b}\right ) \left (-b +\sqrt {-a b}\right ) \sqrt {a +b}}-\frac {\sqrt {\left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )}}{4 a \left (b +\sqrt {-a b}\right ) \left (\tan \left (x \right )^{2}-\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )}}{4 a \left (-b +\sqrt {-a b}\right ) \left (\tan \left (x \right )^{2}+\frac {\sqrt {-a b}}{b}\right )}\)	\(267\)

Input:

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

Output:

1/2/(a+b*tan(x)^4)^(1/2)/a*tan(x)^2-1/2*b/(b+(-a*b)^(1/2))/(-b+(-a*b)^(1/2 
))/(a+b)^(1/2)*ln((2*a+2*b-2*b*(1+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))-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. 136 vs. \(2 (59) = 118\).

Time = 0.24 (sec) , antiderivative size = 292, normalized size of antiderivative = 4.11 \[ \int \frac {\tan ^3(x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\left [\frac {{\left (b \tan \left (x\right )^{4} + a\right )} \sqrt {a + b} \log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \tan \left (x\right )^{4} - 2 \, a b \tan \left (x\right )^{2} - 2 \, \sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {a + b} + 2 \, a^{2} + a b}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right ) + 2 \, \sqrt {b \tan \left (x\right )^{4} + a} {\left ({\left (a + b\right )} \tan \left (x\right )^{2} - a - b\right )}}{4 \, {\left ({\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \tan \left (x\right )^{4} + a^{3} + 2 \, a^{2} b + a b^{2}\right )}}, \frac {{\left (b \tan \left (x\right )^{4} + a\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \tan \left (x\right )^{4} + a^{2} + a b}\right ) + \sqrt {b \tan \left (x\right )^{4} + a} {\left ({\left (a + b\right )} \tan \left (x\right )^{2} - a - b\right )}}{2 \, {\left ({\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \tan \left (x\right )^{4} + a^{3} + 2 \, a^{2} b + a b^{2}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.45 \[ \int \frac {\tan ^3(x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\frac {\frac {{\left (a + b\right )} \tan \left (x\right )^{2}}{a^{2} + 2 \, a b + b^{2}} - \frac {a + b}{a^{2} + 2 \, a b + b^{2}}}{2 \, \sqrt {b \tan \left (x\right )^{4} + a}} + \frac {\arctan \left (\frac {\sqrt {b} \tan \left (x\right )^{2} - \sqrt {b \tan \left (x\right )^{4} + a} + \sqrt {b}}{\sqrt {-a - b}}\right )}{{\left (a + b\right )} \sqrt {-a - b}} \] Input:

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

Output:

1/2*((a + b)*tan(x)^2/(a^2 + 2*a*b + b^2) - (a + b)/(a^2 + 2*a*b + b^2))/s 
qrt(b*tan(x)^4 + a) + arctan((sqrt(b)*tan(x)^2 - sqrt(b*tan(x)^4 + a) + sq 
rt(b))/sqrt(-a - b))/((a + b)*sqrt(-a - b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\tan ^3(x)}{\left (a+b \tan ^4(x)\right )^{3/2}} \, dx=\frac {-\sqrt {\tan \left (x \right )^{4} b +a}-2 \left (\int \frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )^{5}}{\tan \left (x \right )^{8} b^{2}+2 \tan \left (x \right )^{4} a b +a^{2}}d x \right ) \tan \left (x \right )^{4} b^{2}-2 \left (\int \frac {\sqrt {\tan \left (x \right )^{4} b +a}\, \tan \left (x \right )^{5}}{\tan \left (x \right )^{8} b^{2}+2 \tan \left (x \right )^{4} a b +a^{2}}d x \right ) a b}{2 b \left (\tan \left (x \right )^{4} b +a \right )} \] Input:

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

Output:

( - sqrt(tan(x)**4*b + a) - 2*int((sqrt(tan(x)**4*b + a)*tan(x)**5)/(tan(x 
)**8*b**2 + 2*tan(x)**4*a*b + a**2),x)*tan(x)**4*b**2 - 2*int((sqrt(tan(x) 
**4*b + a)*tan(x)**5)/(tan(x)**8*b**2 + 2*tan(x)**4*a*b + a**2),x)*a*b)/(2 
*b*(tan(x)**4*b + a))