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

Optimal. Leaf size=92 \[ \frac{a (a+b x) \text{Gamma}\left (\frac{1}{3},-c \log (f) (a+b x)^3\right )}{3 b^2 \sqrt [3]{-c \log (f) (a+b x)^3}}-\frac{(a+b x)^2 \text{Gamma}\left (\frac{2}{3},-c \log (f) (a+b x)^3\right )}{3 b^2 \left (-c \log (f) (a+b x)^3\right )^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.0494645, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2226, 2208, 2218} \[ \frac{a (a+b x) \text{Gamma}\left (\frac{1}{3},-c \log (f) (a+b x)^3\right )}{3 b^2 \sqrt [3]{-c \log (f) (a+b x)^3}}-\frac{(a+b x)^2 \text{Gamma}\left (\frac{2}{3},-c \log (f) (a+b x)^3\right )}{3 b^2 \left (-c \log (f) (a+b x)^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[f^(c*(a + b*x)^3)*x,x]

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0471472, size = 86, normalized size = 0.93 \[ -\frac{(a+b x) \left ((a+b x) \text{Gamma}\left (\frac{2}{3},-c \log (f) (a+b x)^3\right )-a \sqrt [3]{-c \log (f) (a+b x)^3} \text{Gamma}\left (\frac{1}{3},-c \log (f) (a+b x)^3\right )\right )}{3 b^2 \left (-c \log (f) (a+b x)^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(c*(a + b*x)^3)*x,x]

[Out]

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

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Maple [F]  time = 0.018, size = 0, normalized size = 0. \begin{align*} \int{f}^{c \left ( bx+a \right ) ^{3}}x\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^((b*x + a)^3*c)*x, x)

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Fricas [A]  time = 1.56721, size = 288, normalized size = 3.13 \begin{align*} -\frac{\left (-b^{3} c \log \left (f\right )\right )^{\frac{2}{3}} a \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 ) - \left (-b^{3} c \log \left (f\right )\right )^{\frac{1}{3}} 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 )}{3 \, b^{4} 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,x, algorithm="fricas")

[Out]

-1/3*((-b^3*c*log(f))^(2/3)*a*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)) - (-b^3*c*
log(f))^(1/3)*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^4*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\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^((b*x + a)^3*c)*x, x)