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

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

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 85, normalized size of antiderivative = 1.00, 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, 2177, 2178} \[ b d e^2 \log (F) F^{a+b c} \text {Ei}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{x}+2 e f F^{a+b c} \text {Ei}(b d x \log (F))+\frac {f^2 F^{a+b c+b d x}}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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 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

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 58, normalized size = 0.68 \[ F^{a+b c} \left (F^{b d x} \left (\frac {f^2}{b d \log (F)}-\frac {e^2}{x}\right )+e (b d e \log (F)+2 f) \text {Ei}(b d x \log (F))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 83, normalized size = 0.98 \[ \frac {{\left (b^{2} d^{2} e^{2} x \log \relax (F)^{2} + 2 \, b d e f x \log \relax (F)\right )} F^{b c + a} {\rm Ei}\left (b d x \log \relax (F)\right ) - {\left (b d e^{2} \log \relax (F) - f^{2} x\right )} F^{b d x + b c + a}}{b d x \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{2} F^{{\left (d x + c\right )} b + a}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 135, normalized size = 1.59 \[ -b d \,e^{2} F^{a} F^{b c} \Ei \left (1, -b d x \ln \relax (F )+b c \ln \relax (F )+a \ln \relax (F )-\left (b c +a \right ) \ln \relax (F )\right ) \ln \relax (F )-2 e f \,F^{a} F^{b c} \Ei \left (1, -b d x \ln \relax (F )+b c \ln \relax (F )+a \ln \relax (F )-\left (b c +a \right ) \ln \relax (F )\right )-\frac {e^{2} F^{b d x} F^{b c +a}}{x}+\frac {f^{2} F^{b d x} F^{b c +a}}{b d \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.92, size = 68, normalized size = 0.80 \[ F^{b c + a} b d e^{2} \Gamma \left (-1, -b d x \log \relax (F)\right ) \log \relax (F) + 2 \, F^{b c + a} e f {\rm Ei}\left (b d x \log \relax (F)\right ) + \frac {F^{b d x + b c + a} f^{2}}{b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.59, size = 89, normalized size = 1.05 \[ 2\,F^{a+b\,c}\,e\,f\,\mathrm {ei}\left (b\,d\,x\,\ln \relax (F)\right )-\frac {F^{b\,d\,x}\,F^{a+b\,c}\,e^2}{x}+\frac {F^{a+b\,c+b\,d\,x}\,f^2}{b\,d\,\ln \relax (F)}-F^{a+b\,c}\,b\,d\,e^2\,\ln \relax (F)\,\mathrm {expint}\left (-b\,d\,x\,\ln \relax (F)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + b \left (c + d x\right )} \left (e + f x\right )^{2}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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