3.51 \(\int \frac{f^x}{(a+b f^{2 x})^3} \, dx\)

Optimal. Leaf size=84 \[ \frac{3 f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{b} \log (f)}+\frac{f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2} \]

[Out]

f^x/(4*a*(a + b*f^(2*x))^2*Log[f]) + (3*f^x)/(8*a^2*(a + b*f^(2*x))*Log[f]) + (3*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]
)/(8*a^(5/2)*Sqrt[b]*Log[f])

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Rubi [A]  time = 0.0510675, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2249, 199, 205} \[ \frac{3 f^x}{8 a^2 \log (f) \left (a+b f^{2 x}\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{b} \log (f)}+\frac{f^x}{4 a \log (f) \left (a+b f^{2 x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[f^x/(a + b*f^(2*x))^3,x]

[Out]

f^x/(4*a*(a + b*f^(2*x))^2*Log[f]) + (3*f^x)/(8*a^2*(a + b*f^(2*x))*Log[f]) + (3*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]]
)/(8*a^(5/2)*Sqrt[b]*Log[f])

Rule 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{f^x}{\left (a+b f^{2 x}\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^3} \, dx,x,f^x\right )}{\log (f)}\\ &=\frac{f^x}{4 a \left (a+b f^{2 x}\right )^2 \log (f)}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^2} \, dx,x,f^x\right )}{4 a \log (f)}\\ &=\frac{f^x}{4 a \left (a+b f^{2 x}\right )^2 \log (f)}+\frac{3 f^x}{8 a^2 \left (a+b f^{2 x}\right ) \log (f)}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,f^x\right )}{8 a^2 \log (f)}\\ &=\frac{f^x}{4 a \left (a+b f^{2 x}\right )^2 \log (f)}+\frac{3 f^x}{8 a^2 \left (a+b f^{2 x}\right ) \log (f)}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{b} \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.0501984, size = 68, normalized size = 0.81 \[ \frac{\frac{\sqrt{a} f^x \left (5 a+3 b f^{2 x}\right )}{\left (a+b f^{2 x}\right )^2}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{b}}}{8 a^{5/2} \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^x/(a + b*f^(2*x))^3,x]

[Out]

((Sqrt[a]*f^x*(5*a + 3*b*f^(2*x)))/(a + b*f^(2*x))^2 + (3*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/Sqrt[b])/(8*a^(5/2)*L
og[f])

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Maple [A]  time = 0.043, size = 94, normalized size = 1.1 \begin{align*}{\frac{{f}^{x} \left ( 3\,b \left ({f}^{x} \right ) ^{2}+5\,a \right ) }{8\,\ln \left ( f \right ){a}^{2} \left ( a+b \left ({f}^{x} \right ) ^{2} \right ) ^{2}}}-{\frac{3}{16\,\ln \left ( f \right ){a}^{2}}\ln \left ({f}^{x}-{a{\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{3}{16\,\ln \left ( f \right ){a}^{2}}\ln \left ({f}^{x}+{a{\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*f^x*(3*b*(f^x)^2+5*a)/ln(f)/a^2/(a+b*(f^x)^2)^2-3/16/(-a*b)^(1/2)/a^2/ln(f)*ln(f^x-1/(-a*b)^(1/2)*a)+3/16/
(-a*b)^(1/2)/a^2/ln(f)*ln(f^x+1/(-a*b)^(1/2)*a)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56869, size = 593, normalized size = 7.06 \begin{align*} \left [\frac{6 \, a b^{2} f^{3 \, x} + 10 \, a^{2} b f^{x} - 3 \,{\left (\sqrt{-a b} b^{2} f^{4 \, x} + 2 \, \sqrt{-a b} a b f^{2 \, x} + \sqrt{-a b} a^{2}\right )} \log \left (\frac{b f^{2 \, x} - 2 \, \sqrt{-a b} f^{x} - a}{b f^{2 \, x} + a}\right )}{16 \,{\left (a^{3} b^{3} f^{4 \, x} \log \left (f\right ) + 2 \, a^{4} b^{2} f^{2 \, x} \log \left (f\right ) + a^{5} b \log \left (f\right )\right )}}, \frac{3 \, a b^{2} f^{3 \, x} + 5 \, a^{2} b f^{x} - 3 \,{\left (\sqrt{a b} b^{2} f^{4 \, x} + 2 \, \sqrt{a b} a b f^{2 \, x} + \sqrt{a b} a^{2}\right )} \arctan \left (\frac{\sqrt{a b}}{b f^{x}}\right )}{8 \,{\left (a^{3} b^{3} f^{4 \, x} \log \left (f\right ) + 2 \, a^{4} b^{2} f^{2 \, x} \log \left (f\right ) + a^{5} b \log \left (f\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(6*a*b^2*f^(3*x) + 10*a^2*b*f^x - 3*(sqrt(-a*b)*b^2*f^(4*x) + 2*sqrt(-a*b)*a*b*f^(2*x) + sqrt(-a*b)*a^2)
*log((b*f^(2*x) - 2*sqrt(-a*b)*f^x - a)/(b*f^(2*x) + a)))/(a^3*b^3*f^(4*x)*log(f) + 2*a^4*b^2*f^(2*x)*log(f) +
a^5*b*log(f)), 1/8*(3*a*b^2*f^(3*x) + 5*a^2*b*f^x - 3*(sqrt(a*b)*b^2*f^(4*x) + 2*sqrt(a*b)*a*b*f^(2*x) + sqrt
(a*b)*a^2)*arctan(sqrt(a*b)/(b*f^x)))/(a^3*b^3*f^(4*x)*log(f) + 2*a^4*b^2*f^(2*x)*log(f) + a^5*b*log(f))]

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Sympy [A]  time = 0.46923, size = 85, normalized size = 1.01 \begin{align*} \frac{5 a f^{x} + 3 b f^{3 x}}{8 a^{4} \log{\left (f \right )} + 16 a^{3} b f^{2 x} \log{\left (f \right )} + 8 a^{2} b^{2} f^{4 x} \log{\left (f \right )}} + \frac{\operatorname{RootSum}{\left (256 z^{2} a^{5} b + 9, \left ( i \mapsto i \log{\left (\frac{16 i a^{3}}{3} + f^{x} \right )} \right )\right )}}{\log{\left (f \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a*f**x + 3*b*f**(3*x))/(8*a**4*log(f) + 16*a**3*b*f**(2*x)*log(f) + 8*a**2*b**2*f**(4*x)*log(f)) + RootSum(
256*_z**2*a**5*b + 9, Lambda(_i, _i*log(16*_i*a**3/3 + f**x)))/log(f)

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Giac [A]  time = 1.22519, size = 82, normalized size = 0.98 \begin{align*} \frac{3 \, \arctan \left (\frac{b f^{x}}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{2} \log \left (f\right )} + \frac{3 \, b f^{3 \, x} + 5 \, a f^{x}}{8 \,{\left (b f^{2 \, x} + a\right )}^{2} a^{2} \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*arctan(b*f^x/sqrt(a*b))/(sqrt(a*b)*a^2*log(f)) + 1/8*(3*b*f^(3*x) + 5*a*f^x)/((b*f^(2*x) + a)^2*a^2*log(f)
)