3.288 \(\int \frac{F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx\)

Optimal. Leaf size=53 \[ \frac{b F^a \log (F) \text{Ei}\left (b (c+d x)^3 \log (F)\right )}{3 d}-\frac{F^{a+b (c+d x)^3}}{3 d (c+d x)^3} \]

[Out]

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

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Rubi [A]  time = 0.125406, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2214, 2210} \[ \frac{b F^a \log (F) \text{Ei}\left (b (c+d x)^3 \log (F)\right )}{3 d}-\frac{F^{a+b (c+d x)^3}}{3 d (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

\begin{align*} \int \frac{F^{a+b (c+d x)^3}}{(c+d x)^4} \, dx &=-\frac{F^{a+b (c+d x)^3}}{3 d (c+d x)^3}+(b \log (F)) \int \frac{F^{a+b (c+d x)^3}}{c+d x} \, dx\\ &=-\frac{F^{a+b (c+d x)^3}}{3 d (c+d x)^3}+\frac{b F^a \text{Ei}\left (b (c+d x)^3 \log (F)\right ) \log (F)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.0316843, size = 47, normalized size = 0.89 \[ \frac{F^a \left (b \log (F) \text{Ei}\left (b (c+d x)^3 \log (F)\right )-\frac{F^{b (c+d x)^3}}{(c+d x)^3}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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 ) ^{4}}}\, 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)^4,x)

[Out]

int(F^(a+b*(d*x+c)^3)/(d*x+c)^4,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 )}^{4}}\,{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)^4,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^3*b + a)/(d*x + c)^4, x)

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Fricas [B]  time = 1.49807, size = 315, normalized size = 5.94 \begin{align*} \frac{{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} F^{a}{\rm Ei}\left ({\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 ) - F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{3 \,{\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\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)^4,x, algorithm="fricas")

[Out]

1/3*((b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*F^a*Ei((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) - F^(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a))/(d^4*x^3 + 3*c*d^3*x^2 + 3*c^2*d^2
*x + c^3*d)

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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)**4,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 )}^{4}}\,{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)^4,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^3*b + a)/(d*x + c)^4, x)