3.73 \(\int \frac{F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx\)

Optimal. Leaf size=321 \[ \frac{1}{24} b^4 d^4 e^2 \log ^4(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{24 x^2}-\frac{b^3 d^3 e^2 \log ^3(F) F^{a+b c+b d x}}{24 x}+\frac{1}{3} b^3 d^3 e f \log ^3(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{b^2 d^2 e f \log ^2(F) F^{a+b c+b d x}}{3 x}+\frac{1}{2} b^2 d^2 f^2 \log ^2(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{e^2 F^{a+b c+b d x}}{4 x^4}-\frac{b d e^2 \log (F) F^{a+b c+b d x}}{12 x^3}-\frac{2 e f F^{a+b c+b d x}}{3 x^3}-\frac{b d e f \log (F) F^{a+b c+b d x}}{3 x^2}-\frac{f^2 F^{a+b c+b d x}}{2 x^2}-\frac{b d f^2 \log (F) F^{a+b c+b d x}}{2 x} \]

[Out]

-(e^2*F^(a + b*c + b*d*x))/(4*x^4) - (2*e*f*F^(a + b*c + b*d*x))/(3*x^3) - (f^2*F^(a + b*c + b*d*x))/(2*x^2) -
(b*d*e^2*F^(a + b*c + b*d*x)*Log[F])/(12*x^3) - (b*d*e*f*F^(a + b*c + b*d*x)*Log[F])/(3*x^2) - (b*d*f^2*F^(a
+ b*c + b*d*x)*Log[F])/(2*x) - (b^2*d^2*e^2*F^(a + b*c + b*d*x)*Log[F]^2)/(24*x^2) - (b^2*d^2*e*f*F^(a + b*c +
b*d*x)*Log[F]^2)/(3*x) + (b^2*d^2*f^2*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^2)/2 - (b^3*d^3*e^2*F^(a
+ b*c + b*d*x)*Log[F]^3)/(24*x) + (b^3*d^3*e*f*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^3)/3 + (b^4*d^4
*e^2*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^4)/24

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Rubi [A]  time = 0.577845, antiderivative size = 321, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2199, 2177, 2178} \[ \frac{1}{24} b^4 d^4 e^2 \log ^4(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{24 x^2}-\frac{b^3 d^3 e^2 \log ^3(F) F^{a+b c+b d x}}{24 x}+\frac{1}{3} b^3 d^3 e f \log ^3(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{b^2 d^2 e f \log ^2(F) F^{a+b c+b d x}}{3 x}+\frac{1}{2} b^2 d^2 f^2 \log ^2(F) F^{a+b c} \text{Ei}(b d x \log (F))-\frac{e^2 F^{a+b c+b d x}}{4 x^4}-\frac{b d e^2 \log (F) F^{a+b c+b d x}}{12 x^3}-\frac{2 e f F^{a+b c+b d x}}{3 x^3}-\frac{b d e f \log (F) F^{a+b c+b d x}}{3 x^2}-\frac{f^2 F^{a+b c+b d x}}{2 x^2}-\frac{b d f^2 \log (F) F^{a+b c+b d x}}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^5,x]

[Out]

-(e^2*F^(a + b*c + b*d*x))/(4*x^4) - (2*e*f*F^(a + b*c + b*d*x))/(3*x^3) - (f^2*F^(a + b*c + b*d*x))/(2*x^2) -
(b*d*e^2*F^(a + b*c + b*d*x)*Log[F])/(12*x^3) - (b*d*e*f*F^(a + b*c + b*d*x)*Log[F])/(3*x^2) - (b*d*f^2*F^(a
+ b*c + b*d*x)*Log[F])/(2*x) - (b^2*d^2*e^2*F^(a + b*c + b*d*x)*Log[F]^2)/(24*x^2) - (b^2*d^2*e*f*F^(a + b*c +
b*d*x)*Log[F]^2)/(3*x) + (b^2*d^2*f^2*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^2)/2 - (b^3*d^3*e^2*F^(a
+ b*c + b*d*x)*Log[F]^3)/(24*x) + (b^3*d^3*e*f*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^3)/3 + (b^4*d^4
*e^2*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^4)/24

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^5} \, dx &=\int \left (\frac{e^2 F^{a+b c+b d x}}{x^5}+\frac{2 e f F^{a+b c+b d x}}{x^4}+\frac{f^2 F^{a+b c+b d x}}{x^3}\right ) \, dx\\ &=e^2 \int \frac{F^{a+b c+b d x}}{x^5} \, dx+(2 e f) \int \frac{F^{a+b c+b d x}}{x^4} \, dx+f^2 \int \frac{F^{a+b c+b d x}}{x^3} \, dx\\ &=-\frac{e^2 F^{a+b c+b d x}}{4 x^4}-\frac{2 e f F^{a+b c+b d x}}{3 x^3}-\frac{f^2 F^{a+b c+b d x}}{2 x^2}+\frac{1}{4} \left (b d e^2 \log (F)\right ) \int \frac{F^{a+b c+b d x}}{x^4} \, dx+\frac{1}{3} (2 b d e f \log (F)) \int \frac{F^{a+b c+b d x}}{x^3} \, dx+\frac{1}{2} \left (b d f^2 \log (F)\right ) \int \frac{F^{a+b c+b d x}}{x^2} \, dx\\ &=-\frac{e^2 F^{a+b c+b d x}}{4 x^4}-\frac{2 e f F^{a+b c+b d x}}{3 x^3}-\frac{f^2 F^{a+b c+b d x}}{2 x^2}-\frac{b d e^2 F^{a+b c+b d x} \log (F)}{12 x^3}-\frac{b d e f F^{a+b c+b d x} \log (F)}{3 x^2}-\frac{b d f^2 F^{a+b c+b d x} \log (F)}{2 x}+\frac{1}{12} \left (b^2 d^2 e^2 \log ^2(F)\right ) \int \frac{F^{a+b c+b d x}}{x^3} \, dx+\frac{1}{3} \left (b^2 d^2 e f \log ^2(F)\right ) \int \frac{F^{a+b c+b d x}}{x^2} \, dx+\frac{1}{2} \left (b^2 d^2 f^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}}{4 x^4}-\frac{2 e f F^{a+b c+b d x}}{3 x^3}-\frac{f^2 F^{a+b c+b d x}}{2 x^2}-\frac{b d e^2 F^{a+b c+b d x} \log (F)}{12 x^3}-\frac{b d e f F^{a+b c+b d x} \log (F)}{3 x^2}-\frac{b d f^2 F^{a+b c+b d x} \log (F)}{2 x}-\frac{b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{24 x^2}-\frac{b^2 d^2 e f F^{a+b c+b d x} \log ^2(F)}{3 x}+\frac{1}{2} b^2 d^2 f^2 F^{a+b c} \text{Ei}(b d x \log (F)) \log ^2(F)+\frac{1}{24} \left (b^3 d^3 e^2 \log ^3(F)\right ) \int \frac{F^{a+b c+b d x}}{x^2} \, dx+\frac{1}{3} \left (b^3 d^3 e f \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}}{4 x^4}-\frac{2 e f F^{a+b c+b d x}}{3 x^3}-\frac{f^2 F^{a+b c+b d x}}{2 x^2}-\frac{b d e^2 F^{a+b c+b d x} \log (F)}{12 x^3}-\frac{b d e f F^{a+b c+b d x} \log (F)}{3 x^2}-\frac{b d f^2 F^{a+b c+b d x} \log (F)}{2 x}-\frac{b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{24 x^2}-\frac{b^2 d^2 e f F^{a+b c+b d x} \log ^2(F)}{3 x}+\frac{1}{2} b^2 d^2 f^2 F^{a+b c} \text{Ei}(b d x \log (F)) \log ^2(F)-\frac{b^3 d^3 e^2 F^{a+b c+b d x} \log ^3(F)}{24 x}+\frac{1}{3} b^3 d^3 e f F^{a+b c} \text{Ei}(b d x \log (F)) \log ^3(F)+\frac{1}{24} \left (b^4 d^4 e^2 \log ^4(F)\right ) \int \frac{F^{a+b c+b d x}}{x} \, dx\\ &=-\frac{e^2 F^{a+b c+b d x}}{4 x^4}-\frac{2 e f F^{a+b c+b d x}}{3 x^3}-\frac{f^2 F^{a+b c+b d x}}{2 x^2}-\frac{b d e^2 F^{a+b c+b d x} \log (F)}{12 x^3}-\frac{b d e f F^{a+b c+b d x} \log (F)}{3 x^2}-\frac{b d f^2 F^{a+b c+b d x} \log (F)}{2 x}-\frac{b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{24 x^2}-\frac{b^2 d^2 e f F^{a+b c+b d x} \log ^2(F)}{3 x}+\frac{1}{2} b^2 d^2 f^2 F^{a+b c} \text{Ei}(b d x \log (F)) \log ^2(F)-\frac{b^3 d^3 e^2 F^{a+b c+b d x} \log ^3(F)}{24 x}+\frac{1}{3} b^3 d^3 e f F^{a+b c} \text{Ei}(b d x \log (F)) \log ^3(F)+\frac{1}{24} b^4 d^4 e^2 F^{a+b c} \text{Ei}(b d x \log (F)) \log ^4(F)\\ \end{align*}

Mathematica [A]  time = 0.296526, size = 156, normalized size = 0.49 \[ \frac{F^{a+b c} \left (b^2 d^2 x^4 \log ^2(F) \left (b^2 d^2 e^2 \log ^2(F)+8 b d e f \log (F)+12 f^2\right ) \text{Ei}(b d x \log (F))-F^{b d x} \left (b^3 d^3 e^2 x^3 \log ^3(F)+b^2 d^2 e x^2 \log ^2(F) (e+8 f x)+2 b d x \log (F) \left (e^2+4 e f x+6 f^2 x^2\right )+2 \left (3 e^2+8 e f x+6 f^2 x^2\right )\right )\right )}{24 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^5,x]

[Out]

(F^(a + b*c)*(b^2*d^2*x^4*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^2*(12*f^2 + 8*b*d*e*f*Log[F] + b^2*d^2*e^2*Log[F]
^2) - F^(b*d*x)*(2*(3*e^2 + 8*e*f*x + 6*f^2*x^2) + 2*b*d*x*(e^2 + 4*e*f*x + 6*f^2*x^2)*Log[F] + b^2*d^2*e*x^2*
(e + 8*f*x)*Log[F]^2 + b^3*d^3*e^2*x^3*Log[F]^3)))/(24*x^4)

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Maple [A]  time = 0.066, size = 382, normalized size = 1.2 \begin{align*} -{\frac{ \left ( \ln \left ( F \right ) \right ) ^{4}{b}^{4}{d}^{4}{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 ) }{24}}-{\frac{2\,fe{F}^{bdx}{F}^{bc+a}}{3\,{x}^{3}}}-{\frac{fe\ln \left ( F \right ) bd{F}^{bdx}{F}^{bc+a}}{3\,{x}^{2}}}-{\frac{{b}^{2}{d}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}fe{F}^{bdx}{F}^{bc+a}}{3\,x}}-{\frac{ \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}{d}^{3}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 ) }{3}}-{\frac{bd\ln \left ( F \right ){f}^{2}{F}^{bdx}{F}^{bc+a}}{2\,x}}-{\frac{ \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{d}^{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{{f}^{2}{F}^{bdx}{F}^{bc+a}}{2\,{x}^{2}}}-{\frac{{e}^{2}{F}^{bdx}{F}^{bc+a}}{4\,{x}^{4}}}-{\frac{\ln \left ( F \right ) bd{e}^{2}{F}^{bdx}{F}^{bc+a}}{12\,{x}^{3}}}-{\frac{ \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}{d}^{2}{e}^{2}{F}^{bdx}{F}^{bc+a}}{24\,{x}^{2}}}-{\frac{ \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}{d}^{3}{e}^{2}{F}^{bdx}{F}^{bc+a}}{24\,x}} \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^5,x)

[Out]

-1/24*ln(F)^4*b^4*d^4*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))-2/3*f*e*F^(b*d*x)*F^(b
*c+a)/x^3-1/3*ln(F)*b*d*f*e*F^(b*d*x)*F^(b*c+a)/x^2-1/3*ln(F)^2*b^2*d^2*f*e*F^(b*d*x)*F^(b*c+a)/x-1/3*ln(F)^3*
b^3*d^3*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*ln(F)*b*d*f^2*F^(b*d*x)*F^(b*c+a
)/x-1/2*ln(F)^2*b^2*d^2*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))-1/2*f^2*F^(b*d*x)*F^
(b*c+a)/x^2-1/4*e^2*F^(b*d*x)*F^(b*c+a)/x^4-1/12*ln(F)*b*d*e^2*F^(b*d*x)*F^(b*c+a)/x^3-1/24*ln(F)^2*b^2*d^2*e^
2*F^(b*d*x)*F^(b*c+a)/x^2-1/24*ln(F)^3*b^3*d^3*e^2*F^(b*d*x)*F^(b*c+a)/x

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Maxima [A]  time = 1.24327, size = 126, normalized size = 0.39 \begin{align*} -F^{b c + a} b^{4} d^{4} e^{2} \Gamma \left (-4, -b d x \log \left (F\right )\right ) \log \left (F\right )^{4} + 2 \, F^{b c + a} b^{3} d^{3} e f \Gamma \left (-3, -b d x \log \left (F\right )\right ) \log \left (F\right )^{3} - F^{b c + a} b^{2} d^{2} f^{2} \Gamma \left (-2, -b d x \log \left (F\right )\right ) \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^5,x, algorithm="maxima")

[Out]

-F^(b*c + a)*b^4*d^4*e^2*gamma(-4, -b*d*x*log(F))*log(F)^4 + 2*F^(b*c + a)*b^3*d^3*e*f*gamma(-3, -b*d*x*log(F)
)*log(F)^3 - F^(b*c + a)*b^2*d^2*f^2*gamma(-2, -b*d*x*log(F))*log(F)^2

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Fricas [A]  time = 1.50121, size = 421, normalized size = 1.31 \begin{align*} \frac{{\left (b^{4} d^{4} e^{2} x^{4} \log \left (F\right )^{4} + 8 \, b^{3} d^{3} e f x^{4} \log \left (F\right )^{3} + 12 \, b^{2} d^{2} f^{2} x^{4} \log \left (F\right )^{2}\right )} F^{b c + a}{\rm Ei}\left (b d x \log \left (F\right )\right ) -{\left (b^{3} d^{3} e^{2} x^{3} \log \left (F\right )^{3} + 12 \, f^{2} x^{2} + 16 \, e f x +{\left (8 \, b^{2} d^{2} e f x^{3} + b^{2} d^{2} e^{2} x^{2}\right )} \log \left (F\right )^{2} + 6 \, e^{2} + 2 \,{\left (6 \, b d f^{2} x^{3} + 4 \, b d e f x^{2} + b d e^{2} x\right )} \log \left (F\right )\right )} F^{b d x + b c + a}}{24 \, x^{4}} \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^5,x, algorithm="fricas")

[Out]

1/24*((b^4*d^4*e^2*x^4*log(F)^4 + 8*b^3*d^3*e*f*x^4*log(F)^3 + 12*b^2*d^2*f^2*x^4*log(F)^2)*F^(b*c + a)*Ei(b*d
*x*log(F)) - (b^3*d^3*e^2*x^3*log(F)^3 + 12*f^2*x^2 + 16*e*f*x + (8*b^2*d^2*e*f*x^3 + b^2*d^2*e^2*x^2)*log(F)^
2 + 6*e^2 + 2*(6*b*d*f^2*x^3 + 4*b*d*e*f*x^2 + b*d*e^2*x)*log(F))*F^(b*d*x + b*c + a))/x^4

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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^{5}}\, 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**5,x)

[Out]

Integral(F**(a + b*(c + d*x))*(e + f*x)**2/x**5, 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^{5}}\,{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^5,x, algorithm="giac")

[Out]

integrate((f*x + e)^2*F^((d*x + c)*b + a)/x^5, x)