3.298 \(\int \frac{F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx\)

Optimal. Leaf size=49 \[ -\frac{F^a \left (-b \log (F) (c+d x)^3\right )^{4/3} \text{Gamma}\left (-\frac{4}{3},-b \log (F) (c+d x)^3\right )}{3 d (c+d x)^4} \]

[Out]

-(F^a*Gamma[-4/3, -(b*(c + d*x)^3*Log[F])]*(-(b*(c + d*x)^3*Log[F]))^(4/3))/(3*d*(c + d*x)^4)

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Rubi [A]  time = 0.0626988, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {2218} \[ -\frac{F^a \left (-b \log (F) (c+d x)^3\right )^{4/3} \text{Gamma}\left (-\frac{4}{3},-b \log (F) (c+d x)^3\right )}{3 d (c+d x)^4} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^3)/(c + d*x)^5,x]

[Out]

-(F^a*Gamma[-4/3, -(b*(c + d*x)^3*Log[F])]*(-(b*(c + d*x)^3*Log[F]))^(4/3))/(3*d*(c + d*x)^4)

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 \frac{F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx &=-\frac{F^a \Gamma \left (-\frac{4}{3},-b (c+d x)^3 \log (F)\right ) \left (-b (c+d x)^3 \log (F)\right )^{4/3}}{3 d (c+d x)^4}\\ \end{align*}

Mathematica [A]  time = 0.0231699, size = 49, normalized size = 1. \[ -\frac{F^a \left (-b \log (F) (c+d x)^3\right )^{4/3} \text{Gamma}\left (-\frac{4}{3},-b \log (F) (c+d x)^3\right )}{3 d (c+d x)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^3)/(c + d*x)^5,x]

[Out]

-(F^a*Gamma[-4/3, -(b*(c + d*x)^3*Log[F])]*(-(b*(c + d*x)^3*Log[F]))^(4/3))/(3*d*(c + d*x)^4)

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Maple [F]  time = 0.054, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{a+b \left ( dx+c \right ) ^{3}}}{ \left ( dx+c \right ) ^{5}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x)

[Out]

int(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^3*b + a)/(d*x + c)^5, x)

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Fricas [B]  time = 1.618, size = 501, normalized size = 10.22 \begin{align*} \frac{3 \,{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4}\right )} \left (-b d^{3} \log \left (F\right )\right )^{\frac{1}{3}} F^{a} \Gamma \left (\frac{2}{3}, -{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right ) \log \left (F\right ) -{\left (3 \,{\left (b d^{4} x^{3} + 3 \, b c d^{3} x^{2} + 3 \, b c^{2} d^{2} x + b c^{3} d\right )} \log \left (F\right ) + d\right )} F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{4 \,{\left (d^{6} x^{4} + 4 \, c d^{5} x^{3} + 6 \, c^{2} d^{4} x^{2} + 4 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x, algorithm="fricas")

[Out]

1/4*(3*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4)*(-b*d^3*log(F))^(1/3)*F^a*gamma(2/3
, -(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*log(F))*log(F) - (3*(b*d^4*x^3 + 3*b*c*d^3*x^2 + 3*b*c^2*
d^2*x + b*c^3*d)*log(F) + d)*F^(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a))/(d^6*x^4 + 4*c*d^5*x^3 +
6*c^2*d^4*x^2 + 4*c^3*d^3*x + c^4*d^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**3)/(d*x+c)**5,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^5,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^3*b + a)/(d*x + c)^5, x)