3.71 \(\int \frac{F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx\)

Optimal. Leaf size=136 \[ \frac{1}{2} b^2 d^2 e^2 \log ^2(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{e^2 F^{a+b c+b d x}}{2 x^2}-\frac{b d e^2 \log (F) F^{a+b c+b d x}}{2 x}+2 b d e f \log (F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{2 e f F^{a+b c+b d x}}{x}+f^2 F^{a+b c} \text{Ei}(b d x \log (F)) \]

[Out]

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

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Rubi [A]  time = 0.362979, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2199, 2177, 2178} \[ \frac{1}{2} b^2 d^2 e^2 \log ^2(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{e^2 F^{a+b c+b d x}}{2 x^2}-\frac{b d e^2 \log (F) F^{a+b c+b d x}}{2 x}+2 b d e f \log (F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{2 e f F^{a+b c+b d x}}{x}+f^2 F^{a+b c} \text{Ei}(b d x \log (F)) \]

Antiderivative was successfully verified.

[In]

Int[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^3,x]

[Out]

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

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 2177

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

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

Rubi steps

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

Mathematica [A]  time = 0.14826, size = 76, normalized size = 0.56 \[ \frac{F^{a+b c} \left (x^2 \left (b^2 d^2 e^2 \log ^2(F)+4 b d e f \log (F)+2 f^2\right ) \text{Ei}(b d x \log (F))-e F^{b d x} (b d e x \log (F)+e+4 f x)\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^3,x]

[Out]

(F^(a + b*c)*(-(e*F^(b*d*x)*(e + 4*f*x + b*d*e*x*Log[F])) + x^2*ExpIntegralEi[b*d*x*Log[F]]*(2*f^2 + 4*b*d*e*f
*Log[F] + b^2*d^2*e^2*Log[F]^2)))/(2*x^2)

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Maple [A]  time = 0.059, size = 204, normalized size = 1.5 \begin{align*} -{\frac{{b}^{2}{d}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{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 ) }{2}}-{f}^{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 ) -2\,{\frac{fe{F}^{bdx}{F}^{bc+a}}{x}}-2\,bd\ln \left ( F \right ) fe{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{\ln \left ( F \right ) bd{e}^{2}{F}^{bdx}{F}^{bc+a}}{2\,x}}-{\frac{{e}^{2}{F}^{bdx}{F}^{bc+a}}{2\,{x}^{2}}} \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^3,x)

[Out]

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

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Maxima [A]  time = 1.32388, size = 100, normalized size = 0.74 \begin{align*} -F^{b c + a} b^{2} d^{2} e^{2} \Gamma \left (-2, -b d x \log \left (F\right )\right ) \log \left (F\right )^{2} + 2 \, F^{b c + a} b d e f \Gamma \left (-1, -b d x \log \left (F\right )\right ) \log \left (F\right ) + F^{b c + a} f^{2}{\rm Ei}\left (b d x \log \left (F\right )\right ) \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^3,x, algorithm="maxima")

[Out]

-F^(b*c + a)*b^2*d^2*e^2*gamma(-2, -b*d*x*log(F))*log(F)^2 + 2*F^(b*c + a)*b*d*e*f*gamma(-1, -b*d*x*log(F))*lo
g(F) + F^(b*c + a)*f^2*Ei(b*d*x*log(F))

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Fricas [A]  time = 1.50513, size = 215, normalized size = 1.58 \begin{align*} \frac{{\left (b^{2} d^{2} e^{2} x^{2} \log \left (F\right )^{2} + 4 \, b d e f x^{2} \log \left (F\right ) + 2 \, f^{2} x^{2}\right )} F^{b c + a}{\rm Ei}\left (b d x \log \left (F\right )\right ) -{\left (b d e^{2} x \log \left (F\right ) + 4 \, e f x + e^{2}\right )} F^{b d x + b c + a}}{2 \, x^{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^3,x, algorithm="fricas")

[Out]

1/2*((b^2*d^2*e^2*x^2*log(F)^2 + 4*b*d*e*f*x^2*log(F) + 2*f^2*x^2)*F^(b*c + a)*Ei(b*d*x*log(F)) - (b*d*e^2*x*l
og(F) + 4*e*f*x + e^2)*F^(b*d*x + b*c + a))/x^2

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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^{3}}\, 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**3,x)

[Out]

Integral(F**(a + b*(c + d*x))*(e + f*x)**2/x**3, x)

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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^{3}}\,{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^3,x, algorithm="giac")

[Out]

integrate((f*x + e)^2*F^((d*x + c)*b + a)/x^3, x)