3.219 \(\int f^{\frac{c}{a+b x}} x \, dx\)

Optimal. Leaf size=120 \[ -\frac{c^2 \log ^2(f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{2 b^2}+\frac{a c \log (f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{b^2}+\frac{(a+b x)^2 f^{\frac{c}{a+b x}}}{2 b^2}-\frac{a (a+b x) f^{\frac{c}{a+b x}}}{b^2}+\frac{c \log (f) (a+b x) f^{\frac{c}{a+b x}}}{2 b^2} \]

[Out]

-((a*f^(c/(a + b*x))*(a + b*x))/b^2) + (f^(c/(a + b*x))*(a + b*x)^2)/(2*b^2) + (c*f^(c/(a + b*x))*(a + b*x)*Lo
g[f])/(2*b^2) + (a*c*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f])/b^2 - (c^2*ExpIntegralEi[(c*Log[f])/(a + b*x)
]*Log[f]^2)/(2*b^2)

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Rubi [A]  time = 0.116233, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2226, 2206, 2210, 2214} \[ -\frac{c^2 \log ^2(f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{2 b^2}+\frac{a c \log (f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{b^2}+\frac{(a+b x)^2 f^{\frac{c}{a+b x}}}{2 b^2}-\frac{a (a+b x) f^{\frac{c}{a+b x}}}{b^2}+\frac{c \log (f) (a+b x) f^{\frac{c}{a+b x}}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[f^(c/(a + b*x))*x,x]

[Out]

-((a*f^(c/(a + b*x))*(a + b*x))/b^2) + (f^(c/(a + b*x))*(a + b*x)^2)/(2*b^2) + (c*f^(c/(a + b*x))*(a + b*x)*Lo
g[f])/(2*b^2) + (a*c*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f])/b^2 - (c^2*ExpIntegralEi[(c*Log[f])/(a + b*x)
]*Log[f]^2)/(2*b^2)

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rubi steps

\begin{align*} \int f^{\frac{c}{a+b x}} x \, dx &=\int \left (-\frac{a f^{\frac{c}{a+b x}}}{b}+\frac{f^{\frac{c}{a+b x}} (a+b x)}{b}\right ) \, dx\\ &=\frac{\int f^{\frac{c}{a+b x}} (a+b x) \, dx}{b}-\frac{a \int f^{\frac{c}{a+b x}} \, dx}{b}\\ &=-\frac{a f^{\frac{c}{a+b x}} (a+b x)}{b^2}+\frac{f^{\frac{c}{a+b x}} (a+b x)^2}{2 b^2}+\frac{(c \log (f)) \int f^{\frac{c}{a+b x}} \, dx}{2 b}-\frac{(a c \log (f)) \int \frac{f^{\frac{c}{a+b x}}}{a+b x} \, dx}{b}\\ &=-\frac{a f^{\frac{c}{a+b x}} (a+b x)}{b^2}+\frac{f^{\frac{c}{a+b x}} (a+b x)^2}{2 b^2}+\frac{c f^{\frac{c}{a+b x}} (a+b x) \log (f)}{2 b^2}+\frac{a c \text{Ei}\left (\frac{c \log (f)}{a+b x}\right ) \log (f)}{b^2}+\frac{\left (c^2 \log ^2(f)\right ) \int \frac{f^{\frac{c}{a+b x}}}{a+b x} \, dx}{2 b}\\ &=-\frac{a f^{\frac{c}{a+b x}} (a+b x)}{b^2}+\frac{f^{\frac{c}{a+b x}} (a+b x)^2}{2 b^2}+\frac{c f^{\frac{c}{a+b x}} (a+b x) \log (f)}{2 b^2}+\frac{a c \text{Ei}\left (\frac{c \log (f)}{a+b x}\right ) \log (f)}{b^2}-\frac{c^2 \text{Ei}\left (\frac{c \log (f)}{a+b x}\right ) \log ^2(f)}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0655748, size = 82, normalized size = 0.68 \[ \frac{c \log (f) (2 a-c \log (f)) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )+b x f^{\frac{c}{a+b x}} (b x+c \log (f))}{2 b^2}-\frac{a (a-c \log (f)) f^{\frac{c}{a+b x}}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(c/(a + b*x))*x,x]

[Out]

-(a*f^(c/(a + b*x))*(a - c*Log[f]))/(2*b^2) + (c*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f]*(2*a - c*Log[f]) +
b*f^(c/(a + b*x))*x*(b*x + c*Log[f]))/(2*b^2)

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Maple [A]  time = 0.068, size = 126, normalized size = 1.1 \begin{align*}{\frac{{x}^{2}}{2}{f}^{{\frac{c}{bx+a}}}}-{\frac{{a}^{2}}{2\,{b}^{2}}{f}^{{\frac{c}{bx+a}}}}+{\frac{c\ln \left ( f \right ) x}{2\,b}{f}^{{\frac{c}{bx+a}}}}+{\frac{ac\ln \left ( f \right ) }{2\,{b}^{2}}{f}^{{\frac{c}{bx+a}}}}+{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{c}^{2}}{2\,{b}^{2}}{\it Ei} \left ( 1,-{\frac{c\ln \left ( f \right ) }{bx+a}} \right ) }-{\frac{ac\ln \left ( f \right ) }{{b}^{2}}{\it Ei} \left ( 1,-{\frac{c\ln \left ( f \right ) }{bx+a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f^(c/(b*x+a))*x^2-1/2/b^2*f^(c/(b*x+a))*a^2+1/2/b*ln(f)*c*f^(c/(b*x+a))*x+1/2/b^2*ln(f)*c*f^(c/(b*x+a))*a+
1/2/b^2*ln(f)^2*c^2*Ei(1,-c*ln(f)/(b*x+a))-1/b^2*ln(f)*c*a*Ei(1,-c*ln(f)/(b*x+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (b x^{2} + c x \log \left (f\right )\right )} f^{\frac{c}{b x + a}}}{2 \, b} - \int \frac{{\left (a^{2} c \log \left (f\right ) -{\left (b c^{2} \log \left (f\right )^{2} - 2 \, a b c \log \left (f\right )\right )} x\right )} f^{\frac{c}{b x + a}}}{2 \,{\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2 + c*x*log(f))*f^(c/(b*x + a))/b - integrate(1/2*(a^2*c*log(f) - (b*c^2*log(f)^2 - 2*a*b*c*log(f))*x
)*f^(c/(b*x + a))/(b^3*x^2 + 2*a*b^2*x + a^2*b), x)

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Fricas [A]  time = 1.55185, size = 163, normalized size = 1.36 \begin{align*} \frac{{\left (b^{2} x^{2} - a^{2} +{\left (b c x + a c\right )} \log \left (f\right )\right )} f^{\frac{c}{b x + a}} -{\left (c^{2} \log \left (f\right )^{2} - 2 \, a c \log \left (f\right )\right )}{\rm Ei}\left (\frac{c \log \left (f\right )}{b x + a}\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b^2*x^2 - a^2 + (b*c*x + a*c)*log(f))*f^(c/(b*x + a)) - (c^2*log(f)^2 - 2*a*c*log(f))*Ei(c*log(f)/(b*x +
a)))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(c/(a + b*x))*x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(c/(b*x + a))*x, x)