3.202 \(\int f^{c (a+b x)^3} x^2 \, dx\)

Optimal. Leaf size=120 \[ -\frac{a^2 (a+b x) \text{Gamma}\left (\frac{1}{3},-c \log (f) (a+b x)^3\right )}{3 b^3 \sqrt [3]{-c \log (f) (a+b x)^3}}+\frac{2 a (a+b x)^2 \text{Gamma}\left (\frac{2}{3},-c \log (f) (a+b x)^3\right )}{3 b^3 \left (-c \log (f) (a+b x)^3\right )^{2/3}}+\frac{f^{c (a+b x)^3}}{3 b^3 c \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.0889143, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2226, 2208, 2218, 2209} \[ -\frac{a^2 (a+b x) \text{Gamma}\left (\frac{1}{3},-c \log (f) (a+b x)^3\right )}{3 b^3 \sqrt [3]{-c \log (f) (a+b x)^3}}+\frac{2 a (a+b x)^2 \text{Gamma}\left (\frac{2}{3},-c \log (f) (a+b x)^3\right )}{3 b^3 \left (-c \log (f) (a+b x)^3\right )^{2/3}}+\frac{f^{c (a+b x)^3}}{3 b^3 c \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(c*(a + b*x)^3)*x^2,x]

[Out]

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

Rule 2209

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

Rubi steps

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

Mathematica [A]  time = 0.199875, size = 111, normalized size = 0.92 \[ \frac{-\frac{a^2 (a+b x) \text{Gamma}\left (\frac{1}{3},-c \log (f) (a+b x)^3\right )}{\sqrt [3]{-c \log (f) (a+b x)^3}}+\frac{2 a (a+b x)^2 \text{Gamma}\left (\frac{2}{3},-c \log (f) (a+b x)^3\right )}{\left (-c \log (f) (a+b x)^3\right )^{2/3}}+\frac{f^{c (a+b x)^3}}{c \log (f)}}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(c*(a + b*x)^3)*x^2,x]

[Out]

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

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{f}^{c \left ( bx+a \right ) ^{3}}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c*(b*x+a)^3)*x^2,x)

[Out]

int(f^(c*(b*x+a)^3)*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 )}^{3} c} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(c*(b*x+a)^3)*x^2,x, algorithm="maxima")

[Out]

integrate(f^((b*x + a)^3*c)*x^2, x)

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Fricas [A]  time = 1.55489, size = 373, normalized size = 3.11 \begin{align*} \frac{\left (-b^{3} c \log \left (f\right )\right )^{\frac{2}{3}} a^{2} \Gamma \left (\frac{1}{3}, -{\left (b^{3} c x^{3} + 3 \, a b^{2} c x^{2} + 3 \, a^{2} b c x + a^{3} c\right )} \log \left (f\right )\right ) - 2 \, \left (-b^{3} c \log \left (f\right )\right )^{\frac{1}{3}} a b \Gamma \left (\frac{2}{3}, -{\left (b^{3} c x^{3} + 3 \, a b^{2} c x^{2} + 3 \, a^{2} b c x + a^{3} c\right )} \log \left (f\right )\right ) + b^{2} f^{b^{3} c x^{3} + 3 \, a b^{2} c x^{2} + 3 \, a^{2} b c x + a^{3} c}}{3 \, b^{5} c \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(c*(b*x+a)^3)*x^2,x, algorithm="fricas")

[Out]

1/3*((-b^3*c*log(f))^(2/3)*a^2*gamma(1/3, -(b^3*c*x^3 + 3*a*b^2*c*x^2 + 3*a^2*b*c*x + a^3*c)*log(f)) - 2*(-b^3
*c*log(f))^(1/3)*a*b*gamma(2/3, -(b^3*c*x^3 + 3*a*b^2*c*x^2 + 3*a^2*b*c*x + a^3*c)*log(f)) + b^2*f^(b^3*c*x^3
+ 3*a*b^2*c*x^2 + 3*a^2*b*c*x + a^3*c))/(b^5*c*log(f))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(c*(b*x+a)**3)*x**2,x)

[Out]

Integral(f**(c*(a + b*x)**3)*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 )}^{3} c} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(c*(b*x+a)^3)*x^2,x, algorithm="giac")

[Out]

integrate(f^((b*x + a)^3*c)*x^2, x)