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

Optimal. Leaf size=74 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{b}}+\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}} \]

[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])

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Rubi [A]  time = 0.128865, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {3670, 1252, 844, 217, 206, 725} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{b}}+\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}} \]

Antiderivative was successfully verified.

[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 3670

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]))

Rule 1252

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 844

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 217

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 725

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]

Rubi steps

\begin{align*} \int \frac{\tan ^3(x)}{\sqrt{a+b \tan ^4(x)}} \, dx &=\operatorname{Subst}\left (\int \frac{x^3}{\left (1+x^2\right ) \sqrt{a+b x^4}} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{(1+x) \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )\\ &=\frac{1}{2} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{b}}+\frac{\tanh ^{-1}\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]  time = 0.0532017, size = 74, normalized size = 1. \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{b}}+\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}} \]

Antiderivative was successfully verified.

[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])

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Maple [A]  time = 0.065, size = 91, normalized size = 1.2 \begin{align*}{\frac{1}{2}\ln \left ( \sqrt{b} \left ( \tan \left ( x \right ) \right ) ^{2}+\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}} \right ){\frac{1}{\sqrt{b}}}}+{\frac{1}{2}\ln \left ({\frac{1}{1+ \left ( \tan \left ( x \right ) \right ) ^{2}} \left ( 2\,a+2\,b-2\,b \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) +2\,\sqrt{a+b}\sqrt{b \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{2}-2\,b \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) +a+b} \right ) } \right ){\frac{1}{\sqrt{a+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (x\right )^{3}}{\sqrt{b \tan \left (x\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.97415, size = 1251, normalized size = 16.91 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\left (x \right )}}{\sqrt{a + b \tan ^{4}{\left (x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.16823, size = 101, normalized size = 1.36 \begin{align*} -\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 )}{\sqrt{-a - b}} - \frac{\log \left ({\left | -\sqrt{b} \tan \left (x\right )^{2} + \sqrt{b \tan \left (x\right )^{4} + a} \right |}\right )}{2 \, \sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(-(sqrt(b)*tan(x)^2 - sqrt(b*tan(x)^4 + a) + sqrt(b))/sqrt(-a - b))/sqrt(-a - b) - 1/2*log(abs(-sqrt(b)
*tan(x)^2 + sqrt(b*tan(x)^4 + a)))/sqrt(b)