3.3.19 \(\int f^{\frac {c}{a+b x}} x \, dx\) [219]

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

[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

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Rubi [A]
time = 0.08, 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 = {2258, 2237, 2241, 2245} \begin {gather*} -\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} \end {gather*}

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 2237

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 2241

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 2245

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 2258

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*}

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Mathematica [A]
time = 0.05, size = 82, normalized size = 0.68 \begin {gather*} -\frac {a f^{\frac {c}{a+b x}} (a-c \log (f))}{2 b^2}+\frac {c \text {Ei}\left (\frac {c \log (f)}{a+b x}\right ) \log (f) (2 a-c \log (f))+b f^{\frac {c}{a+b x}} x (b x+c \log (f))}{2 b^2} \end {gather*}

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)

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Maple [A]
time = 0.08, size = 126, normalized size = 1.05

method	result	size
risch	\(\frac {f^{\frac {c}{b x +a}} x^{2}}{2}-\frac {f^{\frac {c}{b x +a}} a^{2}}{2 b^{2}}+\frac {c \ln \left (f \right ) f^{\frac {c}{b x +a}} x}{2 b}+\frac {c \ln \left (f \right ) f^{\frac {c}{b x +a}} a}{2 b^{2}}+\frac {c^{2} \ln \left (f \right )^{2} \expIntegral \left (1, -\frac {c \ln \left (f \right )}{b x +a}\right )}{2 b^{2}}-\frac {c \ln \left (f \right ) a \expIntegral \left (1, -\frac {c \ln \left (f \right )}{b x +a}\right )}{b^{2}}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.40, size = 71, normalized size = 0.59 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int f^{\frac {c}{a + b x}} x\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 3.65, size = 136, normalized size = 1.13 \begin {gather*} \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 \left (f\right )}{2}\right )-\frac {a^2\,f^{\frac {c}{a+b\,x}}\,\left (a-c\,\ln \left (f\right )\right )}{2\,b^2}-\frac {f^{\frac {c}{a+b\,x}}\,x\,\left (a^2-2\,a\,c\,\ln \left (f\right )\right )}{2\,b}}{a+b\,x}-\frac {\mathrm {ei}\left (\frac {c\,\ln \left (f\right )}{a+b\,x}\right )\,\left (c^2\,{\ln \left (f\right )}^2-2\,a\,c\,\ln \left (f\right )\right )}{2\,b^2} \end {gather*}

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)

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