\(\int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx\) [70]

Optimal result

Integrand size = 22, antiderivative size = 85 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=-\frac {e^2 F^{a+b c+b d x}}{x}+2 e f F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))+\frac {f^2 F^{a+b c+b d x}}{b d \log (F)}+b d e^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log (F) \]

[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)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2230, 2225, 2208, 2209} \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=b d e^2 \log (F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{x}+2 e f F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))+\frac {f^2 F^{a+b c+b d x}}{b d \log (F)} \]

[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 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

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

Rule 2225

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 2230

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] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=F^{a+b c} \left (F^{b d x} \left (-\frac {e^2}{x}+\frac {f^2}{b d \log (F)}\right )+e \operatorname {ExpIntegralEi}(b d x \log (F)) (2 f+b d e \log (F))\right ) \]

[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]))

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.80

method	result	size
risch	\(-\frac {\ln \left (F \right )^{2} F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) b^{2} d^{2} e^{2} x +2 \ln \left (F \right ) F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) b d e f x +\ln \left (F \right ) F^{b d x} F^{c b +a} b d \,e^{2}-F^{b d x} F^{c b +a} f^{2} x}{\ln \left (F \right ) b d x}\)	\(153\)
meijerg	\(-\frac {F^{c b +a} f^{2} \left (1-{\mathrm e}^{b d x \ln \left (F \right )}\right )}{b d \ln \left (F \right )}+2 F^{c b +a} f e \left (\ln \left (x \right )+\ln \left (-b d \right )+\ln \left (\ln \left (F \right )\right )-\ln \left (-b d x \ln \left (F \right )\right )-\operatorname {Ei}_{1}\left (-b d x \ln \left (F \right )\right )\right )-F^{c b +a} e^{2} b d \ln \left (F \right ) \left (\frac {1}{b d x \ln \left (F \right )}+1-\ln \left (x \right )-\ln \left (-b d \right )-\ln \left (\ln \left (F \right )\right )-\frac {2+2 b d x \ln \left (F \right )}{2 b d x \ln \left (F \right )}+\frac {{\mathrm e}^{b d x \ln \left (F \right )}}{b d x \ln \left (F \right )}+\ln \left (-b d x \ln \left (F \right )\right )+\operatorname {Ei}_{1}\left (-b d x \ln \left (F \right )\right )\right )\)	\(188\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=\frac {{\left (b^{2} d^{2} e^{2} x \log \left (F\right )^{2} + 2 \, b d e f x \log \left (F\right )\right )} F^{b c + a} {\rm Ei}\left (b d x \log \left (F\right )\right ) - {\left (b d e^{2} \log \left (F\right ) - f^{2} x\right )} F^{b d x + b c + a}}{b d x \log \left (F\right )} \]

[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))

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=F^{b c + a} b d e^{2} \Gamma \left (-1, -b d x \log \left (F\right )\right ) \log \left (F\right ) + 2 \, F^{b c + a} e f {\rm Ei}\left (b d x \log \left (F\right )\right ) + \frac {F^{b d x + b c + a} f^{2}}{b d \log \left (F\right )} \]

[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))

Giac [F]

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

[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)

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^2} \, dx=2\,F^{a+b\,c}\,e\,f\,\mathrm {ei}\left (b\,d\,x\,\ln \left (F\right )\right )-\frac {F^{b\,d\,x}\,F^{a+b\,c}\,e^2}{x}+\frac {F^{a+b\,c+b\,d\,x}\,f^2}{b\,d\,\ln \left (F\right )}-F^{a+b\,c}\,b\,d\,e^2\,\ln \left (F\right )\,\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right ) \]

[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))