3.304 \(\int F^{a+\frac{b}{c+d x}} (c+d x)^2 \, dx\)

Optimal. Leaf size=119 \[ -\frac{b^3 F^a \log ^3(F) \text{Ei}\left (\frac{b \log (F)}{c+d x}\right )}{6 d}+\frac{b^2 \log ^2(F) (c+d x) F^{a+\frac{b}{c+d x}}}{6 d}+\frac{(c+d x)^3 F^{a+\frac{b}{c+d x}}}{3 d}+\frac{b \log (F) (c+d x)^2 F^{a+\frac{b}{c+d x}}}{6 d} \]

[Out]

(F^(a + b/(c + d*x))*(c + d*x)^3)/(3*d) + (b*F^(a + b/(c + d*x))*(c + d*x)^2*Log[F])/(6*d) + (b^2*F^(a + b/(c
+ d*x))*(c + d*x)*Log[F]^2)/(6*d) - (b^3*F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)]*Log[F]^3)/(6*d)

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Rubi [A]  time = 0.132549, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2214, 2206, 2210} \[ -\frac{b^3 F^a \log ^3(F) \text{Ei}\left (\frac{b \log (F)}{c+d x}\right )}{6 d}+\frac{b^2 \log ^2(F) (c+d x) F^{a+\frac{b}{c+d x}}}{6 d}+\frac{(c+d x)^3 F^{a+\frac{b}{c+d x}}}{3 d}+\frac{b \log (F) (c+d x)^2 F^{a+\frac{b}{c+d x}}}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x))*(c + d*x)^2,x]

[Out]

(F^(a + b/(c + d*x))*(c + d*x)^3)/(3*d) + (b*F^(a + b/(c + d*x))*(c + d*x)^2*Log[F])/(6*d) + (b^2*F^(a + b/(c
+ d*x))*(c + d*x)*Log[F]^2)/(6*d) - (b^3*F^a*ExpIntegralEi[(b*Log[F])/(c + d*x)]*Log[F]^3)/(6*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 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

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 F^{a+\frac{b}{c+d x}} (c+d x)^2 \, dx &=\frac{F^{a+\frac{b}{c+d x}} (c+d x)^3}{3 d}+\frac{1}{3} (b \log (F)) \int F^{a+\frac{b}{c+d x}} (c+d x) \, dx\\ &=\frac{F^{a+\frac{b}{c+d x}} (c+d x)^3}{3 d}+\frac{b F^{a+\frac{b}{c+d x}} (c+d x)^2 \log (F)}{6 d}+\frac{1}{6} \left (b^2 \log ^2(F)\right ) \int F^{a+\frac{b}{c+d x}} \, dx\\ &=\frac{F^{a+\frac{b}{c+d x}} (c+d x)^3}{3 d}+\frac{b F^{a+\frac{b}{c+d x}} (c+d x)^2 \log (F)}{6 d}+\frac{b^2 F^{a+\frac{b}{c+d x}} (c+d x) \log ^2(F)}{6 d}+\frac{1}{6} \left (b^3 \log ^3(F)\right ) \int \frac{F^{a+\frac{b}{c+d x}}}{c+d x} \, dx\\ &=\frac{F^{a+\frac{b}{c+d x}} (c+d x)^3}{3 d}+\frac{b F^{a+\frac{b}{c+d x}} (c+d x)^2 \log (F)}{6 d}+\frac{b^2 F^{a+\frac{b}{c+d x}} (c+d x) \log ^2(F)}{6 d}-\frac{b^3 F^a \text{Ei}\left (\frac{b \log (F)}{c+d x}\right ) \log ^3(F)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.0683396, size = 76, normalized size = 0.64 \[ \frac{F^a \left ((c+d x) F^{\frac{b}{c+d x}} \left (b^2 \log ^2(F)+b \log (F) (c+d x)+2 (c+d x)^2\right )-b^3 \log ^3(F) \text{Ei}\left (\frac{b \log (F)}{c+d x}\right )\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x))*(c + d*x)^2,x]

[Out]

(F^a*(-(b^3*ExpIntegralEi[(b*Log[F])/(c + d*x)]*Log[F]^3) + F^(b/(c + d*x))*(c + d*x)*(2*(c + d*x)^2 + b*(c +
d*x)*Log[F] + b^2*Log[F]^2)))/(6*d)

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Maple [B]  time = 0.085, size = 234, normalized size = 2. \begin{align*}{\frac{{d}^{2}{F}^{a}{x}^{3}}{3}{F}^{{\frac{b}{dx+c}}}}+d{F}^{a}{F}^{{\frac{b}{dx+c}}}c{x}^{2}+{F}^{a}{F}^{{\frac{b}{dx+c}}}{c}^{2}x+{\frac{{F}^{a}{c}^{3}}{3\,d}{F}^{{\frac{b}{dx+c}}}}+{\frac{\ln \left ( F \right ) bd{F}^{a}{x}^{2}}{6}{F}^{{\frac{b}{dx+c}}}}+{\frac{b\ln \left ( F \right ){F}^{a}cx}{3}{F}^{{\frac{b}{dx+c}}}}+{\frac{b\ln \left ( F \right ){F}^{a}{c}^{2}}{6\,d}{F}^{{\frac{b}{dx+c}}}}+{\frac{{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{F}^{a}x}{6}{F}^{{\frac{b}{dx+c}}}}+{\frac{{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{F}^{a}c}{6\,d}{F}^{{\frac{b}{dx+c}}}}+{\frac{{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}{F}^{a}}{6\,d}{\it Ei} \left ( 1,-{\frac{b\ln \left ( F \right ) }{dx+c}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b/(d*x+c))*(d*x+c)^2,x)

[Out]

1/3*d^2*F^a*F^(b/(d*x+c))*x^3+d*F^a*F^(b/(d*x+c))*c*x^2+F^a*F^(b/(d*x+c))*c^2*x+1/3/d*F^a*F^(b/(d*x+c))*c^3+1/
6*d*b*ln(F)*F^a*F^(b/(d*x+c))*x^2+1/3*b*ln(F)*F^a*F^(b/(d*x+c))*c*x+1/6/d*b*ln(F)*F^a*F^(b/(d*x+c))*c^2+1/6*b^
2*ln(F)^2*F^a*F^(b/(d*x+c))*x+1/6/d*b^2*ln(F)^2*F^a*F^(b/(d*x+c))*c+1/6/d*b^3*ln(F)^3*F^a*Ei(1,-b*ln(F)/(d*x+c
))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \,{\left (2 \, F^{a} d^{2} x^{3} +{\left (F^{a} b d \log \left (F\right ) + 6 \, F^{a} c d\right )} x^{2} +{\left (F^{a} b^{2} \log \left (F\right )^{2} + 2 \, F^{a} b c \log \left (F\right ) + 6 \, F^{a} c^{2}\right )} x\right )} F^{\frac{b}{d x + c}} + \int \frac{{\left (F^{a} b^{3} d x \log \left (F\right )^{3} - F^{a} b^{2} c^{2} \log \left (F\right )^{2} - 2 \, F^{a} b c^{3} \log \left (F\right )\right )} F^{\frac{b}{d x + c}}}{6 \,{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.56683, size = 270, normalized size = 2.27 \begin{align*} -\frac{F^{a} b^{3}{\rm Ei}\left (\frac{b \log \left (F\right )}{d x + c}\right ) \log \left (F\right )^{3} -{\left (2 \, d^{3} x^{3} + 6 \, c d^{2} x^{2} + 6 \, c^{2} d x + 2 \, c^{3} +{\left (b^{2} d x + b^{2} c\right )} \log \left (F\right )^{2} +{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right )} F^{\frac{a d x + a c + b}{d x + c}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c))*(d*x+c)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2} F^{a + \frac{b}{d x + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c))*(d*x+c)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*F^(a + b/(d*x + c)), x)