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

Optimal. Leaf size=85 \[ -\frac{b^2 F^a \log ^2(F) \text{Ei}\left (\frac{b \log (F)}{c+d x}\right )}{2 d}+\frac{(c+d x)^2 F^{a+\frac{b}{c+d x}}}{2 d}+\frac{b \log (F) (c+d x) F^{a+\frac{b}{c+d x}}}{2 d} \]

[Out]

(F^(a + b/(c + d*x))*(c + d*x)^2)/(2*d) + (b*F^(a + b/(c + d*x))*(c + d*x)*Log[F])/(2*d) - (b^2*F^a*ExpIntegra
lEi[(b*Log[F])/(c + d*x)]*Log[F]^2)/(2*d)

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Rubi [A]  time = 0.0798134, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2214, 2206, 2210} \[ -\frac{b^2 F^a \log ^2(F) \text{Ei}\left (\frac{b \log (F)}{c+d x}\right )}{2 d}+\frac{(c+d x)^2 F^{a+\frac{b}{c+d x}}}{2 d}+\frac{b \log (F) (c+d x) F^{a+\frac{b}{c+d x}}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x))*(c + d*x),x]

[Out]

(F^(a + b/(c + d*x))*(c + d*x)^2)/(2*d) + (b*F^(a + b/(c + d*x))*(c + d*x)*Log[F])/(2*d) - (b^2*F^a*ExpIntegra
lEi[(b*Log[F])/(c + d*x)]*Log[F]^2)/(2*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) \, dx &=\frac{F^{a+\frac{b}{c+d x}} (c+d x)^2}{2 d}+\frac{1}{2} (b \log (F)) \int F^{a+\frac{b}{c+d x}} \, dx\\ &=\frac{F^{a+\frac{b}{c+d x}} (c+d x)^2}{2 d}+\frac{b F^{a+\frac{b}{c+d x}} (c+d x) \log (F)}{2 d}+\frac{1}{2} \left (b^2 \log ^2(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)^2}{2 d}+\frac{b F^{a+\frac{b}{c+d x}} (c+d x) \log (F)}{2 d}-\frac{b^2 F^a \text{Ei}\left (\frac{b \log (F)}{c+d x}\right ) \log ^2(F)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0456404, size = 58, normalized size = 0.68 \[ \frac{F^a \left ((c+d x) F^{\frac{b}{c+d x}} (b \log (F)+c+d x)-b^2 \log ^2(F) \text{Ei}\left (\frac{b \log (F)}{c+d x}\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.083, size = 133, normalized size = 1.6 \begin{align*}{\frac{d{F}^{a}{x}^{2}}{2}{F}^{{\frac{b}{dx+c}}}}+{F}^{a}{F}^{{\frac{b}{dx+c}}}cx+{\frac{{F}^{a}{c}^{2}}{2\,d}{F}^{{\frac{b}{dx+c}}}}+{\frac{b\ln \left ( F \right ){F}^{a}x}{2}{F}^{{\frac{b}{dx+c}}}}+{\frac{b\ln \left ( F \right ){F}^{a}c}{2\,d}{F}^{{\frac{b}{dx+c}}}}+{\frac{{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{F}^{a}}{2\,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),x)

[Out]

1/2*d*F^a*F^(b/(d*x+c))*x^2+F^a*F^(b/(d*x+c))*c*x+1/2/d*F^a*F^(b/(d*x+c))*c^2+1/2*b*ln(F)*F^a*F^(b/(d*x+c))*x+
1/2/d*b*ln(F)*F^a*F^(b/(d*x+c))*c+1/2/d*b^2*ln(F)^2*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}{2} \,{\left (F^{a} d x^{2} +{\left (F^{a} b \log \left (F\right ) + 2 \, F^{a} c\right )} x\right )} F^{\frac{b}{d x + c}} + \int \frac{{\left (F^{a} b^{2} d x \log \left (F\right )^{2} - F^{a} b c^{2} \log \left (F\right )\right )} F^{\frac{b}{d x + c}}}{2 \,{\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),x, algorithm="maxima")

[Out]

1/2*(F^a*d*x^2 + (F^a*b*log(F) + 2*F^a*c)*x)*F^(b/(d*x + c)) + integrate(1/2*(F^a*b^2*d*x*log(F)^2 - F^a*b*c^2
*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 = 1.59668, size = 180, normalized size = 2.12 \begin{align*} -\frac{F^{a} b^{2}{\rm Ei}\left (\frac{b \log \left (F\right )}{d x + c}\right ) \log \left (F\right )^{2} -{\left (d^{2} x^{2} + 2 \, c d x + c^{2} +{\left (b d x + b c\right )} \log \left (F\right )\right )} F^{\frac{a d x + a c + b}{d x + c}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(F**(a + b/(c + d*x))*(c + d*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )} 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),x, algorithm="giac")

[Out]

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