\(\int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx\) [71]

Optimal result

Integrand size = 22, antiderivative size = 136 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=-\frac {e^2 F^{a+b c+b d x}}{2 x^2}-\frac {2 e f F^{a+b c+b d x}}{x}+f^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {b d e^2 F^{a+b c+b d x} \log (F)}{2 x}+2 b d e f F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log (F)+\frac {1}{2} b^2 d^2 e^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log ^2(F) \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2230, 2208, 2209} \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\frac {1}{2} b^2 d^2 e^2 \log ^2(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{2 x}+2 b d e f \log (F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {2 e f F^{a+b c+b d x}}{x}+f^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \]

[In]

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

[Out]

-1/2*(e^2*F^(a + b*c + b*d*x))/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 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 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 (\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] (verified)

Time = 0.33 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.56 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\frac {F^{a+b c} \left (-e F^{b d x} (e+4 f x+b d e x \log (F))+x^2 \operatorname {ExpIntegralEi}(b d x \log (F)) \left (2 f^2+4 b d e f \log (F)+b^2 d^2 e^2 \log ^2(F)\right )\right )}{2 x^2} \]

[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)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.53

method	result	size
risch	\(-\frac {\ln \left (F \right )^{2} F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) b^{2} d^{2} e^{2} x^{2}+4 \ln \left (F \right ) F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) b d e f \,x^{2}+\ln \left (F \right ) F^{b d x} F^{c b +a} b d \,e^{2} x +2 F^{c b} F^{a} \operatorname {Ei}_{1}\left (c b \ln \left (F \right )+a \ln \left (F \right )-b d x \ln \left (F \right )-\left (c b +a \right ) \ln \left (F \right )\right ) f^{2} x^{2}+4 F^{b d x} F^{c b +a} e f x +F^{b d x} F^{c b +a} e^{2}}{2 x^{2}}\)	\(208\)
meijerg	\(F^{c b +a} f^{2} \left (\ln \left (x \right )+\ln \left (-b d \right )+\ln \left (\ln \left (F \right )\right )-\ln \left (-b d x \ln \left (F \right )\right )-\operatorname {Ei}_{1}\left (-b d x \ln \left (F \right )\right )\right )-2 b d \ln \left (F \right ) F^{c b +a} f e \left (\frac {1}{b d x \ln \left (F \right )}+1-\ln \left (x \right )-\ln \left (-b d \right )-\ln \left (\ln \left (F \right )\right )-\frac {2+2 b d x \ln \left (F \right )}{2 b d x \ln \left (F \right )}+\frac {{\mathrm e}^{b d x \ln \left (F \right )}}{b d x \ln \left (F \right )}+\ln \left (-b d x \ln \left (F \right )\right )+\operatorname {Ei}_{1}\left (-b d x \ln \left (F \right )\right )\right )+F^{c b +a} e^{2} \ln \left (F \right )^{2} b^{2} d^{2} \left (-\frac {1}{2 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {1}{b d x \ln \left (F \right )}-\frac {3}{4}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-b d \right )}{2}+\frac {\ln \left (\ln \left (F \right )\right )}{2}+\frac {9 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+12 b d x \ln \left (F \right )+6}{12 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\left (3 b d x \ln \left (F \right )+3\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{6 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\ln \left (-b d x \ln \left (F \right )\right )}{2}-\frac {\operatorname {Ei}_{1}\left (-b d x \ln \left (F \right )\right )}{2}\right )\)	\(314\)

[In]

int(F^(a+b*(d*x+c))*(f*x+e)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\frac {{\left (b^{2} d^{2} e^{2} x^{2} \log \left (F\right )^{2} + 4 \, b d e f x^{2} \log \left (F\right ) + 2 \, f^{2} x^{2}\right )} F^{b c + a} {\rm Ei}\left (b d x \log \left (F\right )\right ) - {\left (b d e^{2} x \log \left (F\right ) + 4 \, e f x + e^{2}\right )} F^{b d x + b c + a}}{2 \, x^{2}} \]

[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

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\int \frac {F^{a + b \left (c + d x\right )} \left (e + f x\right )^{2}}{x^{3}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=-F^{b c + a} b^{2} d^{2} e^{2} \Gamma \left (-2, -b d x \log \left (F\right )\right ) \log \left (F\right )^{2} + 2 \, F^{b c + a} b d e f \Gamma \left (-1, -b d x \log \left (F\right )\right ) \log \left (F\right ) + F^{b c + a} f^{2} {\rm Ei}\left (b d x \log \left (F\right )\right ) \]

[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))

Giac [F]

\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=\int { \frac {{\left (f x + e\right )}^{2} F^{{\left (d x + c\right )} b + a}}{x^{3}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.98 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^3} \, dx=F^{a+b\,c}\,f^2\,\mathrm {ei}\left (b\,d\,x\,\ln \left (F\right )\right )-\frac {2\,F^{b\,d\,x}\,F^{a+b\,c}\,e\,f}{x}-F^{a+b\,c}\,b^2\,d^2\,e^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{2}+F^{b\,d\,x}\,\left (\frac {1}{2\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{2\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}\right )\right )-2\,F^{a+b\,c}\,b\,d\,e\,f\,\ln \left (F\right )\,\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right ) \]

[In]

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

[Out]

F^(a + b*c)*f^2*ei(b*d*x*log(F)) - (2*F^(b*d*x)*F^(a + b*c)*e*f)/x - F^(a + b*c)*b^2*d^2*e^2*log(F)^2*(expint(
-b*d*x*log(F))/2 + F^(b*d*x)*(1/(2*b*d*x*log(F)) + 1/(2*b^2*d^2*x^2*log(F)^2))) - 2*F^(a + b*c)*b*d*e*f*log(F)
*expint(-b*d*x*log(F))