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

Optimal. Leaf size=142 \[ \frac {a^2 (a+b x) \sqrt [3]{-\frac {c \log (f)}{(a+b x)^3}} \Gamma \left (-\frac {1}{3},-\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^3}-\frac {c \log (f) \text {Ei}\left (\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^3}+\frac {(a+b x)^3 f^{\frac {c}{(a+b x)^3}}}{3 b^3}-\frac {2 a (a+b x)^2 \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{2/3} \Gamma \left (-\frac {2}{3},-\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^3} \]

[Out]

1/3*f^(c/(b*x+a)^3)*(b*x+a)^3/b^3-1/3*c*Ei(c*ln(f)/(b*x+a)^3)*ln(f)/b^3+1/3*a^2*(b*x+a)*GAMMA(-1/3,-c*ln(f)/(b
*x+a)^3)*(-c*ln(f)/(b*x+a)^3)^(1/3)/b^3-2/3*a*(b*x+a)^2*GAMMA(-2/3,-c*ln(f)/(b*x+a)^3)*(-c*ln(f)/(b*x+a)^3)^(2
/3)/b^3

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Rubi [A]  time = 0.12, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2226, 2208, 2218, 2214, 2210} \[ \frac {a^2 (a+b x) \sqrt [3]{-\frac {c \log (f)}{(a+b x)^3}} \text {Gamma}\left (-\frac {1}{3},-\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^3}-\frac {2 a (a+b x)^2 \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{2/3} \text {Gamma}\left (-\frac {2}{3},-\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^3}-\frac {c \log (f) \text {Ei}\left (\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^3}+\frac {(a+b x)^3 f^{\frac {c}{(a+b x)^3}}}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Int[f^(c/(a + b*x)^3)*x^2,x]

[Out]

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

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 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]))

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 127, normalized size = 0.89 \[ \frac {a^2 (a+b x) \sqrt [3]{-\frac {c \log (f)}{(a+b x)^3}} \Gamma \left (-\frac {1}{3},-\frac {c \log (f)}{(a+b x)^3}\right )-c \log (f) \text {Ei}\left (\frac {c \log (f)}{(a+b x)^3}\right )+(a+b x)^3 f^{\frac {c}{(a+b x)^3}}-2 a (a+b x)^2 \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{2/3} \Gamma \left (-\frac {2}{3},-\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(c/(a + b*x)^3)*x^2,x]

[Out]

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

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fricas [A]  time = 0.47, size = 194, normalized size = 1.37 \[ \frac {3 \, a b^{2} \left (-\frac {c \log \relax (f)}{b^{3}}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -\frac {c \log \relax (f)}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}\right ) - 3 \, a^{2} b \left (-\frac {c \log \relax (f)}{b^{3}}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -\frac {c \log \relax (f)}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}\right ) - c {\rm Ei}\left (\frac {c \log \relax (f)}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}\right ) \log \relax (f) + {\left (b^{3} x^{3} + a^{3}\right )} f^{\frac {c}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}}}{3 \, b^{3}} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{\frac {c}{{\left (b x + a\right )}^{3}}} x^{2}\,{d x} \]

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^(c/(b*x + a)^3)*x^2, x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int x^{2} f^{\frac {c}{\left (b x +a \right )^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(1/(b*x+a)^3*c)*x^2,x)

[Out]

int(f^(1/(b*x+a)^3*c)*x^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, f^{\frac {c}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}} x^{3} + b c \int \frac {f^{\frac {c}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}} x^{3}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}}\,{d x} \log \relax (f) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int f^{\frac {c}{{\left (a+b\,x\right )}^3}}\,x^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c/(a + b*x)^3)*x^2,x)

[Out]

int(f^(c/(a + b*x)^3)*x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{\frac {c}{\left (a + b x\right )^{3}}} x^{2}\, dx \]

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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