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

Optimal. Leaf size=229 \[ -\frac{a^2 c \log (f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{b^3}+\frac{a^2 (a+b x) f^{\frac{c}{a+b x}}}{b^3}-\frac{c^3 \log ^3(f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{6 b^3}+\frac{a c^2 \log ^2(f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{b^3}+\frac{c^2 \log ^2(f) (a+b x) f^{\frac{c}{a+b x}}}{6 b^3}+\frac{(a+b x)^3 f^{\frac{c}{a+b x}}}{3 b^3}-\frac{a (a+b x)^2 f^{\frac{c}{a+b x}}}{b^3}+\frac{c \log (f) (a+b x)^2 f^{\frac{c}{a+b x}}}{6 b^3}-\frac{a c \log (f) (a+b x) f^{\frac{c}{a+b x}}}{b^3} \]

[Out]

(a^2*f^(c/(a + b*x))*(a + b*x))/b^3 - (a*f^(c/(a + b*x))*(a + b*x)^2)/b^3 + (f^(c/(a + b*x))*(a + b*x)^3)/(3*b
^3) - (a*c*f^(c/(a + b*x))*(a + b*x)*Log[f])/b^3 + (c*f^(c/(a + b*x))*(a + b*x)^2*Log[f])/(6*b^3) - (a^2*c*Exp
IntegralEi[(c*Log[f])/(a + b*x)]*Log[f])/b^3 + (c^2*f^(c/(a + b*x))*(a + b*x)*Log[f]^2)/(6*b^3) + (a*c^2*ExpIn
tegralEi[(c*Log[f])/(a + b*x)]*Log[f]^2)/b^3 - (c^3*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f]^3)/(6*b^3)

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Rubi [A]  time = 0.22341, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2226, 2206, 2210, 2214} \[ -\frac{a^2 c \log (f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{b^3}+\frac{a^2 (a+b x) f^{\frac{c}{a+b x}}}{b^3}-\frac{c^3 \log ^3(f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{6 b^3}+\frac{a c^2 \log ^2(f) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{b^3}+\frac{c^2 \log ^2(f) (a+b x) f^{\frac{c}{a+b x}}}{6 b^3}+\frac{(a+b x)^3 f^{\frac{c}{a+b x}}}{3 b^3}-\frac{a (a+b x)^2 f^{\frac{c}{a+b x}}}{b^3}+\frac{c \log (f) (a+b x)^2 f^{\frac{c}{a+b x}}}{6 b^3}-\frac{a c \log (f) (a+b x) f^{\frac{c}{a+b x}}}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[f^(c/(a + b*x))*x^2,x]

[Out]

(a^2*f^(c/(a + b*x))*(a + b*x))/b^3 - (a*f^(c/(a + b*x))*(a + b*x)^2)/b^3 + (f^(c/(a + b*x))*(a + b*x)^3)/(3*b
^3) - (a*c*f^(c/(a + b*x))*(a + b*x)*Log[f])/b^3 + (c*f^(c/(a + b*x))*(a + b*x)^2*Log[f])/(6*b^3) - (a^2*c*Exp
IntegralEi[(c*Log[f])/(a + b*x)]*Log[f])/b^3 + (c^2*f^(c/(a + b*x))*(a + b*x)*Log[f]^2)/(6*b^3) + (a*c^2*ExpIn
tegralEi[(c*Log[f])/(a + b*x)]*Log[f]^2)/b^3 - (c^3*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f]^3)/(6*b^3)

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^2 \, dx &=\int \left (\frac{a^2 f^{\frac{c}{a+b x}}}{b^2}-\frac{2 a f^{\frac{c}{a+b x}} (a+b x)}{b^2}+\frac{f^{\frac{c}{a+b x}} (a+b x)^2}{b^2}\right ) \, dx\\ &=\frac{\int f^{\frac{c}{a+b x}} (a+b x)^2 \, dx}{b^2}-\frac{(2 a) \int f^{\frac{c}{a+b x}} (a+b x) \, dx}{b^2}+\frac{a^2 \int f^{\frac{c}{a+b x}} \, dx}{b^2}\\ &=\frac{a^2 f^{\frac{c}{a+b x}} (a+b x)}{b^3}-\frac{a f^{\frac{c}{a+b x}} (a+b x)^2}{b^3}+\frac{f^{\frac{c}{a+b x}} (a+b x)^3}{3 b^3}+\frac{(c \log (f)) \int f^{\frac{c}{a+b x}} (a+b x) \, dx}{3 b^2}-\frac{(a c \log (f)) \int f^{\frac{c}{a+b x}} \, dx}{b^2}+\frac{\left (a^2 c \log (f)\right ) \int \frac{f^{\frac{c}{a+b x}}}{a+b x} \, dx}{b^2}\\ &=\frac{a^2 f^{\frac{c}{a+b x}} (a+b x)}{b^3}-\frac{a f^{\frac{c}{a+b x}} (a+b x)^2}{b^3}+\frac{f^{\frac{c}{a+b x}} (a+b x)^3}{3 b^3}-\frac{a c f^{\frac{c}{a+b x}} (a+b x) \log (f)}{b^3}+\frac{c f^{\frac{c}{a+b x}} (a+b x)^2 \log (f)}{6 b^3}-\frac{a^2 c \text{Ei}\left (\frac{c \log (f)}{a+b x}\right ) \log (f)}{b^3}+\frac{\left (c^2 \log ^2(f)\right ) \int f^{\frac{c}{a+b x}} \, dx}{6 b^2}-\frac{\left (a c^2 \log ^2(f)\right ) \int \frac{f^{\frac{c}{a+b x}}}{a+b x} \, dx}{b^2}\\ &=\frac{a^2 f^{\frac{c}{a+b x}} (a+b x)}{b^3}-\frac{a f^{\frac{c}{a+b x}} (a+b x)^2}{b^3}+\frac{f^{\frac{c}{a+b x}} (a+b x)^3}{3 b^3}-\frac{a c f^{\frac{c}{a+b x}} (a+b x) \log (f)}{b^3}+\frac{c f^{\frac{c}{a+b x}} (a+b x)^2 \log (f)}{6 b^3}-\frac{a^2 c \text{Ei}\left (\frac{c \log (f)}{a+b x}\right ) \log (f)}{b^3}+\frac{c^2 f^{\frac{c}{a+b x}} (a+b x) \log ^2(f)}{6 b^3}+\frac{a c^2 \text{Ei}\left (\frac{c \log (f)}{a+b x}\right ) \log ^2(f)}{b^3}+\frac{\left (c^3 \log ^3(f)\right ) \int \frac{f^{\frac{c}{a+b x}}}{a+b x} \, dx}{6 b^2}\\ &=\frac{a^2 f^{\frac{c}{a+b x}} (a+b x)}{b^3}-\frac{a f^{\frac{c}{a+b x}} (a+b x)^2}{b^3}+\frac{f^{\frac{c}{a+b x}} (a+b x)^3}{3 b^3}-\frac{a c f^{\frac{c}{a+b x}} (a+b x) \log (f)}{b^3}+\frac{c f^{\frac{c}{a+b x}} (a+b x)^2 \log (f)}{6 b^3}-\frac{a^2 c \text{Ei}\left (\frac{c \log (f)}{a+b x}\right ) \log (f)}{b^3}+\frac{c^2 f^{\frac{c}{a+b x}} (a+b x) \log ^2(f)}{6 b^3}+\frac{a c^2 \text{Ei}\left (\frac{c \log (f)}{a+b x}\right ) \log ^2(f)}{b^3}-\frac{c^3 \text{Ei}\left (\frac{c \log (f)}{a+b x}\right ) \log ^3(f)}{6 b^3}\\ \end{align*}

Mathematica [A]  time = 0.113264, size = 128, normalized size = 0.56 \[ \frac{b x f^{\frac{c}{a+b x}} \left (\log (f) (b c x-4 a c)+2 b^2 x^2+c^2 \log ^2(f)\right )-c \log (f) \left (6 a^2-6 a c \log (f)+c^2 \log ^2(f)\right ) \text{Ei}\left (\frac{c \log (f)}{a+b x}\right )}{6 b^3}+\frac{a \left (2 a^2-5 a c \log (f)+c^2 \log ^2(f)\right ) f^{\frac{c}{a+b x}}}{6 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(c/(a + b*x))*x^2,x]

[Out]

(a*f^(c/(a + b*x))*(2*a^2 - 5*a*c*Log[f] + c^2*Log[f]^2))/(6*b^3) + (-(c*ExpIntegralEi[(c*Log[f])/(a + b*x)]*L
og[f]*(6*a^2 - 6*a*c*Log[f] + c^2*Log[f]^2)) + b*f^(c/(a + b*x))*x*(2*b^2*x^2 + (-4*a*c + b*c*x)*Log[f] + c^2*
Log[f]^2))/(6*b^3)

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Maple [A]  time = 0.072, size = 227, normalized size = 1. \begin{align*}{\frac{{a}^{3}}{3\,{b}^{3}}{f}^{{\frac{c}{bx+a}}}}+{\frac{{a}^{2}c\ln \left ( f \right ) }{{b}^{3}}{\it Ei} \left ( 1,-{\frac{c\ln \left ( f \right ) }{bx+a}} \right ) }+{\frac{{x}^{3}}{3}{f}^{{\frac{c}{bx+a}}}}+{\frac{c\ln \left ( f \right ){x}^{2}}{6\,b}{f}^{{\frac{c}{bx+a}}}}-{\frac{2\,ac\ln \left ( f \right ) x}{3\,{b}^{2}}{f}^{{\frac{c}{bx+a}}}}-{\frac{5\,{a}^{2}c\ln \left ( f \right ) }{6\,{b}^{3}}{f}^{{\frac{c}{bx+a}}}}+{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{c}^{2}x}{6\,{b}^{2}}{f}^{{\frac{c}{bx+a}}}}+{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}a{c}^{2}}{6\,{b}^{3}}{f}^{{\frac{c}{bx+a}}}}+{\frac{ \left ( \ln \left ( f \right ) \right ) ^{3}{c}^{3}}{6\,{b}^{3}}{\it Ei} \left ( 1,-{\frac{c\ln \left ( f \right ) }{bx+a}} \right ) }-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}a{c}^{2}}{{b}^{3}}{\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^2,x)

[Out]

1/3/b^3*a^3*f^(c/(b*x+a))+1/b^3*ln(f)*c*a^2*Ei(1,-c*ln(f)/(b*x+a))+1/3*f^(c/(b*x+a))*x^3+1/6/b*ln(f)*c*f^(c/(b
*x+a))*x^2-2/3/b^2*ln(f)*c*f^(c/(b*x+a))*a*x-5/6/b^3*ln(f)*c*f^(c/(b*x+a))*a^2+1/6/b^2*ln(f)^2*c^2*f^(c/(b*x+a
))*x+1/6/b^3*ln(f)^2*c^2*f^(c/(b*x+a))*a+1/6/b^3*ln(f)^3*c^3*Ei(1,-c*ln(f)/(b*x+a))-1/b^3*ln(f)^2*c^2*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 (2 \, b^{2} x^{3} + b c x^{2} \log \left (f\right ) +{\left (c^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )\right )} x\right )} f^{\frac{c}{b x + a}}}{6 \, b^{2}} + \int -\frac{{\left (a^{2} c^{2} \log \left (f\right )^{2} - 4 \, a^{3} c \log \left (f\right ) -{\left (b c^{3} \log \left (f\right )^{3} - 6 \, a b c^{2} \log \left (f\right )^{2} + 6 \, a^{2} b c \log \left (f\right )\right )} x\right )} f^{\frac{c}{b x + a}}}{6 \,{\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^2*x^3 + b*c*x^2*log(f) + (c^2*log(f)^2 - 4*a*c*log(f))*x)*f^(c/(b*x + a))/b^2 + integrate(-1/6*(a^2*c
^2*log(f)^2 - 4*a^3*c*log(f) - (b*c^3*log(f)^3 - 6*a*b*c^2*log(f)^2 + 6*a^2*b*c*log(f))*x)*f^(c/(b*x + a))/(b^
4*x^2 + 2*a*b^3*x + a^2*b^2), x)

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Fricas [A]  time = 1.546, size = 263, normalized size = 1.15 \begin{align*} \frac{{\left (2 \, b^{3} x^{3} + 2 \, a^{3} +{\left (b c^{2} x + a c^{2}\right )} \log \left (f\right )^{2} +{\left (b^{2} c x^{2} - 4 \, a b c x - 5 \, a^{2} c\right )} \log \left (f\right )\right )} f^{\frac{c}{b x + a}} -{\left (c^{3} \log \left (f\right )^{3} - 6 \, a c^{2} \log \left (f\right )^{2} + 6 \, a^{2} c \log \left (f\right )\right )}{\rm Ei}\left (\frac{c \log \left (f\right )}{b x + a}\right )}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((2*b^3*x^3 + 2*a^3 + (b*c^2*x + a*c^2)*log(f)^2 + (b^2*c*x^2 - 4*a*b*c*x - 5*a^2*c)*log(f))*f^(c/(b*x + a
)) - (c^3*log(f)^3 - 6*a*c^2*log(f)^2 + 6*a^2*c*log(f))*Ei(c*log(f)/(b*x + a)))/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(c/(a + b*x))*x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(c/(b*x + a))*x^2, x)