3.72 \(\int \frac{F^{a+b (c+d x)} (e+f x)^2}{x^4} \, dx\)

Optimal. Leaf size=217 \[ \frac{1}{6} b^3 d^3 e^2 \log ^3(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{6 x}+b^2 d^2 e f \log ^2(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{e^2 F^{a+b c+b d x}}{3 x^3}-\frac{b d e^2 \log (F) F^{a+b c+b d x}}{6 x^2}-\frac{e f F^{a+b c+b d x}}{x^2}-\frac{b d e f \log (F) F^{a+b c+b d x}}{x}+b d f^2 \log (F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{f^2 F^{a+b c+b d x}}{x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.458988, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2199, 2177, 2178} \[ \frac{1}{6} b^3 d^3 e^2 \log ^3(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{6 x}+b^2 d^2 e f \log ^2(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{e^2 F^{a+b c+b d x}}{3 x^3}-\frac{b d e^2 \log (F) F^{a+b c+b d x}}{6 x^2}-\frac{e f F^{a+b c+b d x}}{x^2}-\frac{b d e f \log (F) F^{a+b c+b d x}}{x}+b d f^2 \log (F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{f^2 F^{a+b c+b d x}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^4,x]

[Out]

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

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^4} \, dx &=\int \left (\frac{e^2 F^{a+b c+b d x}}{x^4}+\frac{2 e f F^{a+b c+b d x}}{x^3}+\frac{f^2 F^{a+b c+b d x}}{x^2}\right ) \, dx\\ &=e^2 \int \frac{F^{a+b c+b d x}}{x^4} \, dx+(2 e f) \int \frac{F^{a+b c+b d x}}{x^3} \, dx+f^2 \int \frac{F^{a+b c+b d x}}{x^2} \, dx\\ &=-\frac{e^2 F^{a+b c+b d x}}{3 x^3}-\frac{e f F^{a+b c+b d x}}{x^2}-\frac{f^2 F^{a+b c+b d x}}{x}+\frac{1}{3} \left (b d e^2 \log (F)\right ) \int \frac{F^{a+b c+b d x}}{x^3} \, dx+(b d e f \log (F)) \int \frac{F^{a+b c+b d x}}{x^2} \, dx+\left (b d f^2 \log (F)\right ) \int \frac{F^{a+b c+b d x}}{x} \, dx\\ &=-\frac{e^2 F^{a+b c+b d x}}{3 x^3}-\frac{e f F^{a+b c+b d x}}{x^2}-\frac{f^2 F^{a+b c+b d x}}{x}-\frac{b d e^2 F^{a+b c+b d x} \log (F)}{6 x^2}-\frac{b d e f F^{a+b c+b d x} \log (F)}{x}+b d f^2 F^{a+b c} \text{Ei}(b d x \log (F)) \log (F)+\frac{1}{6} \left (b^2 d^2 e^2 \log ^2(F)\right ) \int \frac{F^{a+b c+b d x}}{x^2} \, dx+\left (b^2 d^2 e f \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}}{3 x^3}-\frac{e f F^{a+b c+b d x}}{x^2}-\frac{f^2 F^{a+b c+b d x}}{x}-\frac{b d e^2 F^{a+b c+b d x} \log (F)}{6 x^2}-\frac{b d e f F^{a+b c+b d x} \log (F)}{x}+b d f^2 F^{a+b c} \text{Ei}(b d x \log (F)) \log (F)-\frac{b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{6 x}+b^2 d^2 e f F^{a+b c} \text{Ei}(b d x \log (F)) \log ^2(F)+\frac{1}{6} \left (b^3 d^3 e^2 \log ^3(F)\right ) \int \frac{F^{a+b c+b d x}}{x} \, dx\\ &=-\frac{e^2 F^{a+b c+b d x}}{3 x^3}-\frac{e f F^{a+b c+b d x}}{x^2}-\frac{f^2 F^{a+b c+b d x}}{x}-\frac{b d e^2 F^{a+b c+b d x} \log (F)}{6 x^2}-\frac{b d e f F^{a+b c+b d x} \log (F)}{x}+b d f^2 F^{a+b c} \text{Ei}(b d x \log (F)) \log (F)-\frac{b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{6 x}+b^2 d^2 e f F^{a+b c} \text{Ei}(b d x \log (F)) \log ^2(F)+\frac{1}{6} b^3 d^3 e^2 F^{a+b c} \text{Ei}(b d x \log (F)) \log ^3(F)\\ \end{align*}

Mathematica [A]  time = 0.223677, size = 116, normalized size = 0.53 \[ \frac{F^{a+b c} \left (b d x^3 \log (F) \left (b^2 d^2 e^2 \log ^2(F)+6 b d e f \log (F)+6 f^2\right ) \text{Ei}(b d x \log (F))-F^{b d x} \left (b^2 d^2 e^2 x^2 \log ^2(F)+b d e x \log (F) (e+6 f x)+2 \left (e^2+3 e f x+3 f^2 x^2\right )\right )\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^4,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.065, size = 290, normalized size = 1.3 \begin{align*} -{\frac{ \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}{d}^{3}{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 ) }{6}}-{\frac{fe{F}^{bdx}{F}^{bc+a}}{{x}^{2}}}-{\frac{fe\ln \left ( F \right ) bd{F}^{bdx}{F}^{bc+a}}{x}}- \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{d}^{2}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{{e}^{2}{F}^{bdx}{F}^{bc+a}}{3\,{x}^{3}}}-{\frac{\ln \left ( F \right ) bd{e}^{2}{F}^{bdx}{F}^{bc+a}}{6\,{x}^{2}}}-{\frac{ \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{d}^{2}{e}^{2}{F}^{bdx}{F}^{bc+a}}{6\,x}}-{\frac{{f}^{2}{F}^{bdx}{F}^{bc+a}}{x}}-\ln \left ( F \right ) bd{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 ) \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^4,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.21015, size = 115, normalized size = 0.53 \begin{align*} F^{b c + a} b^{3} d^{3} e^{2} \Gamma \left (-3, -b d x \log \left (F\right )\right ) \log \left (F\right )^{3} - 2 \, F^{b c + a} b^{2} d^{2} e f \Gamma \left (-2, -b d x \log \left (F\right )\right ) \log \left (F\right )^{2} + F^{b c + a} b d f^{2} \Gamma \left (-1, -b d x \log \left (F\right )\right ) \log \left (F\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^4,x, algorithm="maxima")

[Out]

F^(b*c + a)*b^3*d^3*e^2*gamma(-3, -b*d*x*log(F))*log(F)^3 - 2*F^(b*c + a)*b^2*d^2*e*f*gamma(-2, -b*d*x*log(F))
*log(F)^2 + F^(b*c + a)*b*d*f^2*gamma(-1, -b*d*x*log(F))*log(F)

________________________________________________________________________________________

Fricas [A]  time = 1.53015, size = 317, normalized size = 1.46 \begin{align*} \frac{{\left (b^{3} d^{3} e^{2} x^{3} \log \left (F\right )^{3} + 6 \, b^{2} d^{2} e f x^{3} \log \left (F\right )^{2} + 6 \, b d f^{2} x^{3} \log \left (F\right )\right )} F^{b c + a}{\rm Ei}\left (b d x \log \left (F\right )\right ) -{\left (b^{2} d^{2} e^{2} x^{2} \log \left (F\right )^{2} + 6 \, f^{2} x^{2} + 6 \, e f x + 2 \, e^{2} +{\left (6 \, b d e f x^{2} + b d e^{2} x\right )} \log \left (F\right )\right )} F^{b d x + b c + a}}{6 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

Integral(F**(a + b*(c + d*x))*(e + f*x)**2/x**4, 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^{4}}\,{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^4,x, algorithm="giac")

[Out]

integrate((f*x + e)^2*F^((d*x + c)*b + a)/x^4, x)