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

Optimal. Leaf size=96 \[ -\frac{f^2 F^{a+b c+b d x}}{b^2 d^2 \log ^2(F)}+e^2 F^{a+b c} \text{Ei}(b d x \log (F))+\frac{2 e f F^{a+b c+b d x}}{b d \log (F)}+\frac{f^2 x F^{a+b c+b d x}}{b d \log (F)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.257749, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2199, 2194, 2178, 2176} \[ -\frac{f^2 F^{a+b c+b d x}}{b^2 d^2 \log ^2(F)}+e^2 F^{a+b c} \text{Ei}(b d x \log (F))+\frac{2 e f F^{a+b c+b d x}}{b d \log (F)}+\frac{f^2 x F^{a+b c+b d x}}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rubi steps

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

Mathematica [A]  time = 0.125699, size = 54, normalized size = 0.56 \[ F^{a+b c} \left (\frac{f F^{b d x} (b d \log (F) (2 e+f x)-f)}{b^2 d^2 \log ^2(F)}+e^2 \text{Ei}(b d x \log (F))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.045, size = 126, normalized size = 1.3 \begin{align*} -{e}^{2}{F}^{bc}{F}^{a}{\it Ei} \left ( 1,bc\ln \left ( F \right ) +\ln \left ( F \right ) a-bdx\ln \left ( F \right ) - \left ( bc+a \right ) \ln \left ( F \right ) \right ) +{\frac{{f}^{2}{F}^{bdx}{F}^{bc+a}x}{bd\ln \left ( F \right ) }}-{\frac{{f}^{2}{F}^{bdx}{F}^{bc+a}}{{b}^{2}{d}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}}}+2\,{\frac{fe{F}^{bdx}{F}^{bc+a}}{bd\ln \left ( F \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.15666, size = 117, normalized size = 1.22 \begin{align*} F^{b c + a} e^{2}{\rm Ei}\left (b d x \log \left (F\right )\right ) + \frac{2 \, F^{b d x + b c + a} e f}{b d \log \left (F\right )} + \frac{{\left (F^{b c + a} b d x \log \left (F\right ) - F^{b c + a}\right )} F^{b d x} f^{2}}{b^{2} d^{2} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.52764, size = 180, normalized size = 1.88 \begin{align*} \frac{F^{b c + a} b^{2} d^{2} e^{2}{\rm Ei}\left (b d x \log \left (F\right )\right ) \log \left (F\right )^{2} -{\left (f^{2} -{\left (b d f^{2} x + 2 \, b d e f\right )} \log \left (F\right )\right )} F^{b d x + b c + a}}{b^{2} d^{2} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(F**(a + b*(c + d*x))*(e + f*x)**2/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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