3.37 \(\int \frac{f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx\)

Optimal. Leaf size=127 \[ \frac{c^{3/2} f^{a-\frac{5 e}{2}} \tan ^{-1}\left (\frac{\sqrt{d} f^{\frac{1}{2} (2 b x+e)}}{\sqrt{c}}\right )}{b d^{5/2} \log (f)}-\frac{c f^{\frac{1}{2} (2 a-5 e)+\frac{1}{2} (2 b x+e)}}{b d^2 \log (f)}+\frac{f^{\frac{1}{2} (2 a-5 e)+\frac{3}{2} (2 b x+e)}}{3 b d \log (f)} \]

[Out]

-((c*f^((2*a - 5*e)/2 + (e + 2*b*x)/2))/(b*d^2*Log[f])) + f^((2*a - 5*e)/2 + (3*(e + 2*b*x))/2)/(3*b*d*Log[f])
+ (c^(3/2)*f^(a - (5*e)/2)*ArcTan[(Sqrt[d]*f^((e + 2*b*x)/2))/Sqrt[c]])/(b*d^(5/2)*Log[f])

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Rubi [A]  time = 0.079692, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2248, 302, 205} \[ \frac{c^{3/2} f^{a-\frac{5 e}{2}} \tan ^{-1}\left (\frac{\sqrt{d} f^{\frac{1}{2} (2 b x+e)}}{\sqrt{c}}\right )}{b d^{5/2} \log (f)}-\frac{c f^{\frac{1}{2} (2 a-5 e)+\frac{1}{2} (2 b x+e)}}{b d^2 \log (f)}+\frac{f^{\frac{1}{2} (2 a-5 e)+\frac{3}{2} (2 b x+e)}}{3 b d \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + 5*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

-((c*f^((2*a - 5*e)/2 + (e + 2*b*x)/2))/(b*d^2*Log[f])) + f^((2*a - 5*e)/2 + (3*(e + 2*b*x))/2)/(3*b*d*Log[f])
+ (c^(3/2)*f^(a - (5*e)/2)*ArcTan[(Sqrt[d]*f^((e + 2*b*x)/2))/Sqrt[c]])/(b*d^(5/2)*Log[f])

Rule 2248

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [A]  time = 0.0622489, size = 86, normalized size = 0.68 \[ \frac{3 c^{3/2} f^{a-\frac{5 e}{2}} \tan ^{-1}\left (\frac{\sqrt{d} f^{b x+\frac{e}{2}}}{\sqrt{c}}\right )+\sqrt{d} f^{a+b x-2 e} \left (d f^{2 b x+e}-3 c\right )}{3 b d^{5/2} \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + 5*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

(Sqrt[d]*f^(a - 2*e + b*x)*(-3*c + d*f^(e + 2*b*x)) + 3*c^(3/2)*f^(a - (5*e)/2)*ArcTan[(Sqrt[d]*f^(e/2 + b*x))
/Sqrt[c]])/(3*b*d^(5/2)*Log[f])

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Maple [B]  time = 0.066, size = 212, normalized size = 1.7 \begin{align*}{\frac{1}{3\,\ln \left ( f \right ) bd} \left ({f}^{bx+{\frac{a}{5}}} \right ) ^{3} \left ({f}^{{\frac{e}{2}}} \right ) ^{-2} \left ({f}^{-{\frac{a}{5}}} \right ) ^{-2}}-{\frac{c}{{d}^{2}\ln \left ( f \right ) b}{f}^{bx+{\frac{a}{5}}} \left ({f}^{{\frac{e}{2}}} \right ) ^{-4} \left ({f}^{-{\frac{a}{5}}} \right ) ^{-4}}+{\frac{c}{2\,b{d}^{3}\ln \left ( f \right ) }\sqrt{-cd}\ln \left ({f}^{bx+{\frac{a}{5}}}+{\frac{1}{d}\sqrt{-cd} \left ({f}^{{\frac{e}{2}}} \right ) ^{-1} \left ({f}^{-{\frac{a}{5}}} \right ) ^{-1}} \right ) \left ({f}^{-{\frac{a}{5}}} \right ) ^{-5} \left ({f}^{{\frac{e}{2}}} \right ) ^{-5}}-{\frac{c}{2\,b{d}^{3}\ln \left ( f \right ) }\sqrt{-cd}\ln \left ({f}^{bx+{\frac{a}{5}}}-{\frac{1}{d}\sqrt{-cd} \left ({f}^{{\frac{e}{2}}} \right ) ^{-1} \left ({f}^{-{\frac{a}{5}}} \right ) ^{-1}} \right ) \left ({f}^{-{\frac{a}{5}}} \right ) ^{-5} \left ({f}^{{\frac{e}{2}}} \right ) ^{-5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(5*b*x+a)/(c+d*f^(2*b*x+e)),x)

[Out]

1/3/(f^(1/2*e))^2/(f^(-1/5*a))^2/d/ln(f)/b*(f^(b*x+1/5*a))^3-c/(f^(1/2*e))^4/(f^(-1/5*a))^4/d^2/ln(f)/b*f^(b*x
+1/5*a)+1/2/d^3*(-c*d)^(1/2)*c/b/(f^(-1/5*a))^5/(f^(1/2*e))^5/ln(f)*ln(f^(b*x+1/5*a)+1/d*(-c*d)^(1/2)/(f^(-1/5
*a))/(f^(1/2*e)))-1/2/d^3*(-c*d)^(1/2)*c/b/(f^(-1/5*a))^5/(f^(1/2*e))^5/ln(f)*ln(f^(b*x+1/5*a)-1/d*(-c*d)^(1/2
)/(f^(-1/5*a))/(f^(1/2*e)))

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Maxima [A]  time = 1.73309, size = 211, normalized size = 1.66 \begin{align*} \frac{c^{2} f^{a - 2 \, e} \log \left (\frac{d{\left (f^{5 \, b x + a}\right )}^{\frac{1}{5}} f^{e} - \sqrt{-c d f^{e}} f^{\frac{1}{5} \, a}}{d{\left (f^{5 \, b x + a}\right )}^{\frac{1}{5}} f^{e} + \sqrt{-c d f^{e}} f^{\frac{1}{5} \, a}}\right )}{2 \, \sqrt{-c d f^{e}} b d^{2} \log \left (f\right )} + \frac{d{\left (f^{5 \, b x + a}\right )}^{\frac{3}{5}} f^{\frac{2}{5} \, a + e} - 3 \, c{\left (f^{5 \, b x + a}\right )}^{\frac{1}{5}} f^{\frac{4}{5} \, a}}{3 \, b d^{2} f^{2 \, e} \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(5*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="maxima")

[Out]

1/2*c^2*f^(a - 2*e)*log((d*(f^(5*b*x + a))^(1/5)*f^e - sqrt(-c*d*f^e)*f^(1/5*a))/(d*(f^(5*b*x + a))^(1/5)*f^e
+ sqrt(-c*d*f^e)*f^(1/5*a)))/(sqrt(-c*d*f^e)*b*d^2*log(f)) + 1/3*(d*(f^(5*b*x + a))^(3/5)*f^(2/5*a + e) - 3*c*
(f^(5*b*x + a))^(1/5)*f^(4/5*a))/(b*d^2*f^(2*e)*log(f))

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Fricas [A]  time = 1.62651, size = 501, normalized size = 3.94 \begin{align*} \left [\frac{3 \, c f^{a - \frac{5}{2} \, e} \sqrt{-\frac{c}{d}} \log \left (\frac{2 \, d f^{b x + \frac{1}{2} \, e} \sqrt{-\frac{c}{d}} + d f^{2 \, b x + e} - c}{d f^{2 \, b x + e} + c}\right ) + 2 \, d f^{3 \, b x + \frac{3}{2} \, e} f^{a - \frac{5}{2} \, e} - 6 \, c f^{b x + \frac{1}{2} \, e} f^{a - \frac{5}{2} \, e}}{6 \, b d^{2} \log \left (f\right )}, \frac{3 \, c f^{a - \frac{5}{2} \, e} \sqrt{\frac{c}{d}} \arctan \left (\frac{d f^{b x + \frac{1}{2} \, e} \sqrt{\frac{c}{d}}}{c}\right ) + d f^{3 \, b x + \frac{3}{2} \, e} f^{a - \frac{5}{2} \, e} - 3 \, c f^{b x + \frac{1}{2} \, e} f^{a - \frac{5}{2} \, e}}{3 \, b d^{2} \log \left (f\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(5*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="fricas")

[Out]

[1/6*(3*c*f^(a - 5/2*e)*sqrt(-c/d)*log((2*d*f^(b*x + 1/2*e)*sqrt(-c/d) + d*f^(2*b*x + e) - c)/(d*f^(2*b*x + e)
+ c)) + 2*d*f^(3*b*x + 3/2*e)*f^(a - 5/2*e) - 6*c*f^(b*x + 1/2*e)*f^(a - 5/2*e))/(b*d^2*log(f)), 1/3*(3*c*f^(
a - 5/2*e)*sqrt(c/d)*arctan(d*f^(b*x + 1/2*e)*sqrt(c/d)/c) + d*f^(3*b*x + 3/2*e)*f^(a - 5/2*e) - 3*c*f^(b*x +
1/2*e)*f^(a - 5/2*e))/(b*d^2*log(f))]

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Sympy [B]  time = 2.28615, size = 366, normalized size = 2.88 \begin{align*} \begin{cases} \frac{\left (- 3 b c d e^{\frac{4 a \log{\left (f \right )}}{5}} e^{e \log{\left (f \right )}} e^{\frac{\left (a + 5 b x\right ) \log{\left (f \right )}}{5}} \log{\left (f \right )} + b d^{2} e^{\frac{2 a \log{\left (f \right )}}{5}} e^{2 e \log{\left (f \right )}} e^{\frac{3 \left (a + 5 b x\right ) \log{\left (f \right )}}{5}} \log{\left (f \right )}\right ) e^{- 3 e \log{\left (f \right )}}}{3 b^{2} d^{3} \log{\left (f \right )}^{2}} & \text{for}\: 3 b^{2} d^{3} e^{3 e \log{\left (f \right )}} \log{\left (f \right )}^{2} \neq 0 \\- \frac{x \left (c^{3} e^{\frac{16 a \log{\left (f \right )}}{5}} + c^{2} d e^{\frac{14 a \log{\left (f \right )}}{5}} e^{e \log{\left (f \right )}} - c d^{2} e^{\frac{12 a \log{\left (f \right )}}{5}} e^{2 e \log{\left (f \right )}} - d^{3} e^{2 a \log{\left (f \right )}} e^{3 e \log{\left (f \right )}}\right )}{c^{2} d^{2} e^{\frac{12 a \log{\left (f \right )}}{5}} e^{2 e \log{\left (f \right )}} + 2 c d^{3} e^{2 a \log{\left (f \right )}} e^{3 e \log{\left (f \right )}} + d^{4} e^{\frac{8 a \log{\left (f \right )}}{5}} e^{4 e \log{\left (f \right )}}} & \text{otherwise} \end{cases} + \operatorname{RootSum}{\left (4 z^{2} b^{2} d^{5} e^{5 e \log{\left (f \right )}} \log{\left (f \right )}^{2} + c^{3} e^{2 a \log{\left (f \right )}}, \left ( i \mapsto i \log{\left (\frac{2 i b d^{2} e^{- \frac{4 a \log{\left (f \right )}}{5}} e^{2 e \log{\left (f \right )}} \log{\left (f \right )}}{c} + e^{\frac{\left (a + 5 b x\right ) \log{\left (f \right )}}{5}} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(5*b*x+a)/(c+d*f**(2*b*x+e)),x)

[Out]

Piecewise(((-3*b*c*d*exp(4*a*log(f)/5)*exp(e*log(f))*exp((a + 5*b*x)*log(f)/5)*log(f) + b*d**2*exp(2*a*log(f)/
5)*exp(2*e*log(f))*exp(3*(a + 5*b*x)*log(f)/5)*log(f))*exp(-3*e*log(f))/(3*b**2*d**3*log(f)**2), Ne(3*b**2*d**
3*exp(3*e*log(f))*log(f)**2, 0)), (-x*(c**3*exp(16*a*log(f)/5) + c**2*d*exp(14*a*log(f)/5)*exp(e*log(f)) - c*d
**2*exp(12*a*log(f)/5)*exp(2*e*log(f)) - d**3*exp(2*a*log(f))*exp(3*e*log(f)))/(c**2*d**2*exp(12*a*log(f)/5)*e
xp(2*e*log(f)) + 2*c*d**3*exp(2*a*log(f))*exp(3*e*log(f)) + d**4*exp(8*a*log(f)/5)*exp(4*e*log(f))), True)) +
RootSum(4*_z**2*b**2*d**5*exp(5*e*log(f))*log(f)**2 + c**3*exp(2*a*log(f)), Lambda(_i, _i*log(2*_i*b*d**2*exp(
-4*a*log(f)/5)*exp(2*e*log(f))*log(f)/c + exp((a + 5*b*x)*log(f)/5))))

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Giac [A]  time = 1.21433, size = 165, normalized size = 1.3 \begin{align*} \frac{1}{3} \, f^{a}{\left (\frac{3 \, c^{2} \arctan \left (\frac{d f^{b x} f^{e}}{\sqrt{c d f^{e}}}\right )}{\sqrt{c d f^{e}} b d^{2} f^{2 \, e} \log \left (f\right )} + \frac{b^{2} d^{2} f^{3 \, b x} f^{2 \, e} \log \left (f\right )^{2} - 3 \, b^{2} c d f^{b x} f^{e} \log \left (f\right )^{2}}{b^{3} d^{3} f^{3 \, e} \log \left (f\right )^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(5*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="giac")

[Out]

1/3*f^a*(3*c^2*arctan(d*f^(b*x)*f^e/sqrt(c*d*f^e))/(sqrt(c*d*f^e)*b*d^2*f^(2*e)*log(f)) + (b^2*d^2*f^(3*b*x)*f
^(2*e)*log(f)^2 - 3*b^2*c*d*f^(b*x)*f^e*log(f)^2)/(b^3*d^3*f^(3*e)*log(f)^3))