3.3.35 \(\int f^{\frac {c}{(a+b x)^3}} x \, dx\) [235]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 92, normalized size of antiderivative = 1.00, 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 = {2258, 2239, 2250} \begin {gather*} \frac {(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^2}-\frac {a (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^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[f^(c/(a + b*x)^3)*x,x]

[Out]

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

Rule 2239

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 2250

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

Rule 2258

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

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Mathematica [A]
time = 0.05, size = 86, normalized size = 0.93 \begin {gather*} \frac {(a+b x) \left (-a \Gamma \left (-\frac {1}{3},-\frac {c \log (f)}{(a+b x)^3}\right ) \sqrt [3]{-\frac {c \log (f)}{(a+b x)^3}}+(a+b x) \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}\right )}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[f^(c/(a + b*x)^3)*x,x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int f^{\frac {c}{\left (b x +a \right )^{3}}} x\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.10, size = 154, normalized size = 1.67 \begin {gather*} -\frac {b^{2} \left (-\frac {c \log \left (f\right )}{b^{3}}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -\frac {c \log \left (f\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}\right ) - 2 \, a b \left (-\frac {c \log \left (f\right )}{b^{3}}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -\frac {c \log \left (f\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}\right ) - {\left (b^{2} x^{2} - a^{2}\right )} f^{\frac {c}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(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)) - 2*a*b*(-c*lo
g(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)) - (b^2*x^2 - a^2)*f^(c/(b^3*x^
3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)))/b^2

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

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^(c/(b*x + a)^3)*x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c/(a + b*x)^3)*x,x)

[Out]

int(f^(c/(a + b*x)^3)*x, x)

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