\(\int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx\) [396]

Optimal result

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

[Out]

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)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3751, 1266, 858, 223, 212, 739} \[ \int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {b}}+\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \]

[In]

Int[Tan[x]^3/Sqrt[a + b*Tan[x]^4],x]

[Out]

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])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 739

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

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1266

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[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 3751

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]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^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] && RationalQ[n]))

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x^3}{\left (1+x^2\right ) \sqrt {a+b x^4}} \, dx,x,\tan (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x}{(1+x) \sqrt {a+b x^2}} \, dx,x,\tan ^2(x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan ^2(x)\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x^2}} \, dx,x,\tan ^2(x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\frac {a-b \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right ) \\ & = \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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^3(x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {b}}+\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \]

[In]

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

[Out]

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])

Maple [A] (verified)

Time = 0.09 (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	\(\frac {\ln \left (\sqrt {b}\, \tan \left (x \right )^{2}+\sqrt {a +b \tan \left (x \right )^{4}}\right )}{2 \sqrt {b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \sqrt {a +b}}\)	\(91\)

[In]

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

[Out]

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)

none

Time = 0.40 (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 ] \]

[In]

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

[Out]

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

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

Mupad [F(-1)]

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

[In]

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

[Out]

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