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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 120, normalized size of antiderivative = 1.00, 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 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]))

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]

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 = 0.07, 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]

-1/2*(a*f^(c/(a + b*x))*(a - c*Log[f]))/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)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 71, normalized size = 0.59 \[ \frac {{\left (b^{2} x^{2} - a^{2} + {\left (b c x + a c\right )} \log \relax (f)\right )} f^{\frac {c}{b x + a}} - {\left (c^{2} \log \relax (f)^{2} - 2 \, a c \log \relax (f)\right )} {\rm Ei}\left (\frac {c \log \relax (f)}{b x + a}\right )}{2 \, b^{2}} \]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{\frac {c}{b x + a}} x\,{d x} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.11, size = 126, normalized size = 1.05 \[ \frac {c x \,f^{\frac {c}{b x +a}} \ln \relax (f )}{2 b}+\frac {c^{2} \Ei \left (1, -\frac {c \ln \relax (f )}{b x +a}\right ) \ln \relax (f )^{2}}{2 b^{2}}+\frac {x^{2} f^{\frac {c}{b x +a}}}{2}+\frac {a c \,f^{\frac {c}{b x +a}} \ln \relax (f )}{2 b^{2}}-\frac {a c \Ei \left (1, -\frac {c \ln \relax (f )}{b x +a}\right ) \ln \relax (f )}{b^{2}}-\frac {a^{2} f^{\frac {c}{b x +a}}}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(1/(b*x+a)*c)*x,x)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (b x^{2} + c x \log \relax (f)\right )} f^{\frac {c}{b x + a}}}{2 \, b} - \int \frac {{\left (a^{2} c \log \relax (f) - {\left (b c^{2} \log \relax (f)^{2} - 2 \, a b c \log \relax (f)\right )} x\right )} f^{\frac {c}{b x + a}}}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [B]  time = 3.65, size = 136, normalized size = 1.13 \[ \frac {\frac {b\,f^{\frac {c}{a+b\,x}}\,x^3}{2}+f^{\frac {c}{a+b\,x}}\,x^2\,\left (\frac {a}{2}+\frac {c\,\ln \relax (f)}{2}\right )-\frac {a^2\,f^{\frac {c}{a+b\,x}}\,\left (a-c\,\ln \relax (f)\right )}{2\,b^2}-\frac {f^{\frac {c}{a+b\,x}}\,x\,\left (a^2-2\,a\,c\,\ln \relax (f)\right )}{2\,b}}{a+b\,x}-\frac {\mathrm {ei}\left (\frac {c\,\ln \relax (f)}{a+b\,x}\right )\,\left (c^2\,{\ln \relax (f)}^2-2\,a\,c\,\ln \relax (f)\right )}{2\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c/(a + b*x))*x,x)

[Out]

((b*f^(c/(a + b*x))*x^3)/2 + f^(c/(a + b*x))*x^2*(a/2 + (c*log(f))/2) - (a^2*f^(c/(a + b*x))*(a - c*log(f)))/(
2*b^2) - (f^(c/(a + b*x))*x*(a^2 - 2*a*c*log(f)))/(2*b))/(a + b*x) - (ei((c*log(f))/(a + b*x))*(c^2*log(f)^2 -
2*a*c*log(f)))/(2*b^2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{\frac {c}{a + b x}} x\, dx \]

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)

________________________________________________________________________________________