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

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

[Out]

1/4*f^x/a/(a+b*f^(2*x))^2/ln(f)+3/8*f^x/a^2/(a+b*f^(2*x))/ln(f)+3/8*arctan(f^x*b^(1/2)/a^(1/2))/a^(5/2)/ln(f)/
b^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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]

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*}

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Mathematica [A]  time = 0.05, size = 68, normalized size = 0.81 \[ \frac {\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {b}}+\frac {\sqrt {a} f^x \left (5 a+3 b f^{2 x}\right )}{\left (a+b f^{2 x}\right )^2}}{8 a^{5/2} \log (f)} \]

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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fricas [A]  time = 0.43, size = 258, normalized size = 3.07 \[ \left [\frac {6 \, a b^{2} f^{3 \, x} + 10 \, a^{2} b f^{x} - 3 \, {\left (\sqrt {-a b} b^{2} f^{4 \, x} + 2 \, \sqrt {-a b} a b f^{2 \, x} + \sqrt {-a b} a^{2}\right )} \log \left (\frac {b f^{2 \, x} - 2 \, \sqrt {-a b} f^{x} - a}{b f^{2 \, x} + a}\right )}{16 \, {\left (a^{3} b^{3} f^{4 \, x} \log \relax (f) + 2 \, a^{4} b^{2} f^{2 \, x} \log \relax (f) + a^{5} b \log \relax (f)\right )}}, \frac {3 \, a b^{2} f^{3 \, x} + 5 \, a^{2} b f^{x} - 3 \, {\left (\sqrt {a b} b^{2} f^{4 \, x} + 2 \, \sqrt {a b} a b f^{2 \, x} + \sqrt {a b} a^{2}\right )} \arctan \left (\frac {\sqrt {a b}}{b f^{x}}\right )}{8 \, {\left (a^{3} b^{3} f^{4 \, x} \log \relax (f) + 2 \, a^{4} b^{2} f^{2 \, x} \log \relax (f) + a^{5} b \log \relax (f)\right )}}\right ] \]

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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giac [A]  time = 0.40, size = 61, normalized size = 0.73 \[ \frac {3 \, \arctan \left (\frac {b f^{x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} \log \relax (f)} + \frac {3 \, b f^{3 \, x} + 5 \, a f^{x}}{8 \, {\left (b f^{2 \, x} + a\right )}^{2} a^{2} \log \relax (f)} \]

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)
)

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maple [A]  time = 0.06, size = 94, normalized size = 1.12 \[ \frac {\left (3 b \,f^{2 x}+5 a \right ) f^{x}}{8 \left (b \,f^{2 x}+a \right )^{2} a^{2} \ln \relax (f )}-\frac {3 \ln \left (-\frac {a}{\sqrt {-a b}}+f^{x}\right )}{16 \sqrt {-a b}\, a^{2} \ln \relax (f )}+\frac {3 \ln \left (\frac {a}{\sqrt {-a b}}+f^{x}\right )}{16 \sqrt {-a b}\, a^{2} \ln \relax (f )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.98, size = 76, normalized size = 0.90 \[ \frac {3 \, b f^{3 \, x} + 5 \, a f^{x}}{8 \, {\left (a^{2} b^{2} f^{4 \, x} + 2 \, a^{3} b f^{2 \, x} + a^{4}\right )} \log \relax (f)} + \frac {3 \, \arctan \left (\frac {b f^{x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*b*f^(3*x) + 5*a*f^x)/((a^2*b^2*f^(4*x) + 2*a^3*b*f^(2*x) + a^4)*log(f)) + 3/8*arctan(b*f^x/sqrt(a*b))/(
sqrt(a*b)*a^2*log(f))

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mupad [B]  time = 3.57, size = 79, normalized size = 0.94 \[ \frac {\frac {5\,f^x}{8\,a\,\ln \relax (f)}+\frac {3\,b\,f^{3\,x}}{8\,a^2\,\ln \relax (f)}}{b^2\,f^{4\,x}+a^2+2\,a\,b\,f^{2\,x}}+\frac {3\,\mathrm {atan}\left (\frac {b\,f^x}{\sqrt {a\,b}}\right )}{8\,a^2\,\ln \relax (f)\,\sqrt {a\,b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^x/(a + b*f^(2*x))^3,x)

[Out]

((5*f^x)/(8*a*log(f)) + (3*b*f^(3*x))/(8*a^2*log(f)))/(b^2*f^(4*x) + a^2 + 2*a*b*f^(2*x)) + (3*atan((b*f^x)/(a
*b)^(1/2)))/(8*a^2*log(f)*(a*b)^(1/2))

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sympy [A]  time = 0.35, size = 85, normalized size = 1.01 \[ \frac {5 a f^{x} + 3 b f^{3 x}}{8 a^{4} \log {\relax (f )} + 16 a^{3} b f^{2 x} \log {\relax (f )} + 8 a^{2} b^{2} f^{4 x} \log {\relax (f )}} + \frac {\operatorname {RootSum} {\left (256 z^{2} a^{5} b + 9, \left (i \mapsto i \log {\left (\frac {16 i a^{3}}{3} + f^{x} \right )} \right )\right )}}{\log {\relax (f )}} \]

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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