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

Optimal. Leaf size=154 \[ -\frac{a^2 (a+b x) \left (-c \log (f) (a+b x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},-c \log (f) (a+b x)^n\right )}{b^3 n}-\frac{(a+b x)^3 \left (-c \log (f) (a+b x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},-c \log (f) (a+b x)^n\right )}{b^3 n}+\frac{2 a (a+b x)^2 \left (-c \log (f) (a+b x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},-c \log (f) (a+b x)^n\right )}{b^3 n} \]

[Out]

-(((a + b*x)^3*Gamma[3/n, -(c*(a + b*x)^n*Log[f])])/(b^3*n*(-(c*(a + b*x)^n*Log[f]))^(3/n))) + (2*a*(a + b*x)^
2*Gamma[2/n, -(c*(a + b*x)^n*Log[f])])/(b^3*n*(-(c*(a + b*x)^n*Log[f]))^(2/n)) - (a^2*(a + b*x)*Gamma[n^(-1),
-(c*(a + b*x)^n*Log[f])])/(b^3*n*(-(c*(a + b*x)^n*Log[f]))^n^(-1))

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Rubi [A]  time = 0.108436, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2226, 2208, 2218} \[ -\frac{a^2 (a+b x) \left (-c \log (f) (a+b x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},-c \log (f) (a+b x)^n\right )}{b^3 n}-\frac{(a+b x)^3 \left (-c \log (f) (a+b x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},-c \log (f) (a+b x)^n\right )}{b^3 n}+\frac{2 a (a+b x)^2 \left (-c \log (f) (a+b x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},-c \log (f) (a+b x)^n\right )}{b^3 n} \]

Antiderivative was successfully verified.

[In]

Int[f^(c*(a + b*x)^n)*x^2,x]

[Out]

-(((a + b*x)^3*Gamma[3/n, -(c*(a + b*x)^n*Log[f])])/(b^3*n*(-(c*(a + b*x)^n*Log[f]))^(3/n))) + (2*a*(a + b*x)^
2*Gamma[2/n, -(c*(a + b*x)^n*Log[f])])/(b^3*n*(-(c*(a + b*x)^n*Log[f]))^(2/n)) - (a^2*(a + b*x)*Gamma[n^(-1),
-(c*(a + b*x)^n*Log[f])])/(b^3*n*(-(c*(a + b*x)^n*Log[f]))^n^(-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]

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin{align*} \int f^{c (a+b x)^n} x^2 \, dx &=\int \left (\frac{a^2 f^{c (a+b x)^n}}{b^2}-\frac{2 a f^{c (a+b x)^n} (a+b x)}{b^2}+\frac{f^{c (a+b x)^n} (a+b x)^2}{b^2}\right ) \, dx\\ &=\frac{\int f^{c (a+b x)^n} (a+b x)^2 \, dx}{b^2}-\frac{(2 a) \int f^{c (a+b x)^n} (a+b x) \, dx}{b^2}+\frac{a^2 \int f^{c (a+b x)^n} \, dx}{b^2}\\ &=-\frac{(a+b x)^3 \Gamma \left (\frac{3}{n},-c (a+b x)^n \log (f)\right ) \left (-c (a+b x)^n \log (f)\right )^{-3/n}}{b^3 n}+\frac{2 a (a+b x)^2 \Gamma \left (\frac{2}{n},-c (a+b x)^n \log (f)\right ) \left (-c (a+b x)^n \log (f)\right )^{-2/n}}{b^3 n}-\frac{a^2 (a+b x) \Gamma \left (\frac{1}{n},-c (a+b x)^n \log (f)\right ) \left (-c (a+b x)^n \log (f)\right )^{-1/n}}{b^3 n}\\ \end{align*}

Mathematica [A]  time = 0.0689383, size = 136, normalized size = 0.88 \[ -\frac{(a+b x) \left (-c \log (f) (a+b x)^n\right )^{-3/n} \left (a \left (-c \log (f) (a+b x)^n\right )^{\frac{1}{n}} \left (a \left (-c \log (f) (a+b x)^n\right )^{\frac{1}{n}} \text{Gamma}\left (\frac{1}{n},-c \log (f) (a+b x)^n\right )-2 (a+b x) \text{Gamma}\left (\frac{2}{n},-c \log (f) (a+b x)^n\right )\right )+(a+b x)^2 \text{Gamma}\left (\frac{3}{n},-c \log (f) (a+b x)^n\right )\right )}{b^3 n} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(c*(a + b*x)^n)*x^2,x]

[Out]

-(((a + b*x)*((a + b*x)^2*Gamma[3/n, -(c*(a + b*x)^n*Log[f])] + a*(-(c*(a + b*x)^n*Log[f]))^n^(-1)*(-2*(a + b*
x)*Gamma[2/n, -(c*(a + b*x)^n*Log[f])] + a*Gamma[n^(-1), -(c*(a + b*x)^n*Log[f])]*(-(c*(a + b*x)^n*Log[f]))^n^
(-1))))/(b^3*n*(-(c*(a + b*x)^n*Log[f]))^(3/n)))

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{f}^{c \left ( bx+a \right ) ^{n}}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c*(b*x+a)^n)*x^2,x)

[Out]

int(f^(c*(b*x+a)^n)*x^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^((b*x + a)^n*c)*x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (f^{{\left (b x + a\right )}^{n} c} x^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(f^((b*x + a)^n*c)*x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(c*(a + b*x)**n)*x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^((b*x + a)^n*c)*x^2, x)