3.33 \(\int \frac{f^{a+b x}}{c+d f^{e+2 b x}} \, dx\)

Optimal. Leaf size=50 \[ \frac{f^{a-\frac{e}{2}} \tan ^{-1}\left (\frac{\sqrt{d} f^{b x+\frac{e}{2}}}{\sqrt{c}}\right )}{b \sqrt{c} \sqrt{d} \log (f)} \]

[Out]

(f^(a - e/2)*ArcTan[(Sqrt[d]*f^(e/2 + b*x))/Sqrt[c]])/(b*Sqrt[c]*Sqrt[d]*Log[f])

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Rubi [A]  time = 0.077644, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {2249, 205} \[ \frac{f^{a-\frac{e}{2}} \tan ^{-1}\left (\frac{\sqrt{d} f^{b x+\frac{e}{2}}}{\sqrt{c}}\right )}{b \sqrt{c} \sqrt{d} \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

(f^(a - e/2)*ArcTan[(Sqrt[d]*f^(e/2 + b*x))/Sqrt[c]])/(b*Sqrt[c]*Sqrt[d]*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 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^{a+b x}}{c+d f^{e+2 b x}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{c+d f^{-2 a+e} x^2} \, dx,x,f^{a+b x}\right )}{b \log (f)}\\ &=\frac{f^{a-\frac{e}{2}} \tan ^{-1}\left (\frac{\sqrt{d} f^{\frac{e}{2}+b x}}{\sqrt{c}}\right )}{b \sqrt{c} \sqrt{d} \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.0250429, size = 50, normalized size = 1. \[ \frac{f^{a-\frac{e}{2}} \tan ^{-1}\left (\frac{\sqrt{d} f^{b x+\frac{e}{2}}}{\sqrt{c}}\right )}{b \sqrt{c} \sqrt{d} \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

(f^(a - e/2)*ArcTan[(Sqrt[d]*f^(e/2 + b*x))/Sqrt[c]])/(b*Sqrt[c]*Sqrt[d]*Log[f])

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Maple [B]  time = 0.046, size = 91, normalized size = 1.8 \begin{align*} -{\frac{{f}^{a}}{2\,b\ln \left ( f \right ) }\ln \left ({f}^{bx+a}-{{f}^{a}c{\frac{1}{\sqrt{-{f}^{e}cd}}}} \right ){\frac{1}{\sqrt{-{f}^{e}cd}}}}+{\frac{{f}^{a}}{2\,b\ln \left ( f \right ) }\ln \left ({f}^{bx+a}+{{f}^{a}c{\frac{1}{\sqrt{-{f}^{e}cd}}}} \right ){\frac{1}{\sqrt{-{f}^{e}cd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x+a)/(c+d*f^(2*b*x+e)),x)

[Out]

-1/2/(-f^e*c*d)^(1/2)*f^a/b/ln(f)*ln(f^(b*x+a)-1/(-f^e*c*d)^(1/2)*f^a*c)+1/2/(-f^e*c*d)^(1/2)*f^a/b/ln(f)*ln(f
^(b*x+a)+1/(-f^e*c*d)^(1/2)*f^a*c)

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Maxima [A]  time = 1.59135, size = 100, normalized size = 2. \begin{align*} \frac{f^{a} \log \left (\frac{d f^{b x + a + e} - \sqrt{-c d f^{e}} f^{a}}{d f^{b x + a + e} + \sqrt{-c d f^{e}} f^{a}}\right )}{2 \, \sqrt{-c d f^{e}} b \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f^a*log((d*f^(b*x + a + e) - sqrt(-c*d*f^e)*f^a)/(d*f^(b*x + a + e) + sqrt(-c*d*f^e)*f^a))/(sqrt(-c*d*f^e)
*b*log(f))

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Fricas [A]  time = 1.79711, size = 397, normalized size = 7.94 \begin{align*} \left [-\frac{\sqrt{-c d f^{-2 \, a + e}} \log \left (\frac{d f^{2 \, b x + 2 \, a} f^{-2 \, a + e} - 2 \, \sqrt{-c d f^{-2 \, a + e}} f^{b x + a} - c}{d f^{2 \, b x + 2 \, a} f^{-2 \, a + e} + c}\right )}{2 \, b c d f^{-2 \, a + e} \log \left (f\right )}, -\frac{\sqrt{c d f^{-2 \, a + e}} \arctan \left (\frac{\sqrt{c d f^{-2 \, a + e}}}{d f^{b x + a} f^{-2 \, a + e}}\right )}{b c d f^{-2 \, a + e} \log \left (f\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-c*d*f^(-2*a + e))*log((d*f^(2*b*x + 2*a)*f^(-2*a + e) - 2*sqrt(-c*d*f^(-2*a + e))*f^(b*x + a) - c)
/(d*f^(2*b*x + 2*a)*f^(-2*a + e) + c))/(b*c*d*f^(-2*a + e)*log(f)), -sqrt(c*d*f^(-2*a + e))*arctan(sqrt(c*d*f^
(-2*a + e))/(d*f^(b*x + a)*f^(-2*a + e)))/(b*c*d*f^(-2*a + e)*log(f))]

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Sympy [A]  time = 1.60283, size = 51, normalized size = 1.02 \begin{align*} \operatorname{RootSum}{\left (4 z^{2} b^{2} c d e^{e \log{\left (f \right )}} \log{\left (f \right )}^{2} + e^{2 a \log{\left (f \right )}}, \left ( i \mapsto i \log{\left (2 i b c \log{\left (f \right )} + f^{a + b x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x+a)/(c+d*f**(2*b*x+e)),x)

[Out]

RootSum(4*_z**2*b**2*c*d*exp(e*log(f))*log(f)**2 + exp(2*a*log(f)), Lambda(_i, _i*log(2*_i*b*c*log(f) + f**(a
+ b*x))))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{b x + a}}{d f^{2 \, b x + e} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(b*x + a)/(d*f^(2*b*x + e) + c), x)