3.455 \(\int \frac{f^{a+b x+c x^2}}{(b+2 c x)^3} \, dx\)

Optimal. Leaf size=69 \[ \frac{\log (f) f^{a-\frac{b^2}{4 c}} \text{Ei}\left (\frac{(b+2 c x)^2 \log (f)}{4 c}\right )}{16 c^2}-\frac{f^{a+b x+c x^2}}{4 c (b+2 c x)^2} \]

[Out]

-f^(a + b*x + c*x^2)/(4*c*(b + 2*c*x)^2) + (f^(a - b^2/(4*c))*ExpIntegralEi[((b + 2*c*x)^2*Log[f])/(4*c)]*Log[
f])/(16*c^2)

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Rubi [A]  time = 0.0706892, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2239, 2238} \[ \frac{\log (f) f^{a-\frac{b^2}{4 c}} \text{Ei}\left (\frac{(b+2 c x)^2 \log (f)}{4 c}\right )}{16 c^2}-\frac{f^{a+b x+c x^2}}{4 c (b+2 c x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-f^(a + b*x + c*x^2)/(4*c*(b + 2*c*x)^2) + (f^(a - b^2/(4*c))*ExpIntegralEi[((b + 2*c*x)^2*Log[f])/(4*c)]*Log[
f])/(16*c^2)

Rule 2239

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*F
^(a + b*x + c*x^2))/(e*(m + 1)), x] - Dist[(2*c*Log[F])/(e^2*(m + 1)), Int[(d + e*x)^(m + 2)*F^(a + b*x + c*x^
2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0] && LtQ[m, -1]

Rule 2238

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(1*F^(a - b^2/(4*c))*ExpI
ntegralEi[((b + 2*c*x)^2*Log[F])/(4*c)])/(2*e), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rubi steps

\begin{align*} \int \frac{f^{a+b x+c x^2}}{(b+2 c x)^3} \, dx &=-\frac{f^{a+b x+c x^2}}{4 c (b+2 c x)^2}+\frac{\log (f) \int \frac{f^{a+b x+c x^2}}{b+2 c x} \, dx}{4 c}\\ &=-\frac{f^{a+b x+c x^2}}{4 c (b+2 c x)^2}+\frac{f^{a-\frac{b^2}{4 c}} \text{Ei}\left (\frac{(b+2 c x)^2 \log (f)}{4 c}\right ) \log (f)}{16 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0877809, size = 79, normalized size = 1.14 \[ \frac{f^{a-\frac{b^2}{4 c}} \left (\log (f) (b+2 c x)^2 \text{Ei}\left (\frac{(b+2 c x)^2 \log (f)}{4 c}\right )-4 c f^{\frac{(b+2 c x)^2}{4 c}}\right )}{16 c^2 (b+2 c x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x + c*x^2)/(b + 2*c*x)^3,x]

[Out]

(f^(a - b^2/(4*c))*(-4*c*f^((b + 2*c*x)^2/(4*c)) + (b + 2*c*x)^2*ExpIntegralEi[((b + 2*c*x)^2*Log[f])/(4*c)]*L
og[f]))/(16*c^2*(b + 2*c*x)^2)

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Maple [A]  time = 0.031, size = 88, normalized size = 1.3 \begin{align*} -{\frac{1}{4\,c \left ( 2\,cx+b \right ) ^{2}}{f}^{{\frac{ \left ( 2\,cx+b \right ) ^{2}}{4\,c}}}{f}^{{\frac{4\,ac-{b}^{2}}{4\,c}}}}-{\frac{\ln \left ( f \right ) }{16\,{c}^{2}}{f}^{{\frac{4\,ac-{b}^{2}}{4\,c}}}{\it Ei} \left ( 1,-{\frac{ \left ( 2\,cx+b \right ) ^{2}\ln \left ( f \right ) }{4\,c}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/c/(2*c*x+b)^2*f^(1/4*(2*c*x+b)^2/c)*f^(1/4*(4*a*c-b^2)/c)-1/16/c^2*ln(f)*f^(1/4*(4*a*c-b^2)/c)*Ei(1,-1/4*
(2*c*x+b)^2*ln(f)/c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{c x^{2} + b x + a}}{{\left (2 \, c x + b\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(c*x^2+b*x+a)/(2*c*x+b)^3,x, algorithm="maxima")

[Out]

integrate(f^(c*x^2 + b*x + a)/(2*c*x + b)^3, x)

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Fricas [A]  time = 1.56674, size = 234, normalized size = 3.39 \begin{align*} -\frac{4 \, c f^{c x^{2} + b x + a} - \frac{{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )}{\rm Ei}\left (\frac{{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (f\right )}{4 \, c}\right ) \log \left (f\right )}{f^{\frac{b^{2} - 4 \, a c}{4 \, c}}}}{16 \,{\left (4 \, c^{4} x^{2} + 4 \, b c^{3} x + b^{2} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(c*x^2+b*x+a)/(2*c*x+b)^3,x, algorithm="fricas")

[Out]

-1/16*(4*c*f^(c*x^2 + b*x + a) - (4*c^2*x^2 + 4*b*c*x + b^2)*Ei(1/4*(4*c^2*x^2 + 4*b*c*x + b^2)*log(f)/c)*log(
f)/f^(1/4*(b^2 - 4*a*c)/c))/(4*c^4*x^2 + 4*b*c^3*x + b^2*c^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{a + b x + c x^{2}}}{\left (b + 2 c x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(c*x**2+b*x+a)/(2*c*x+b)**3,x)

[Out]

Integral(f**(a + b*x + c*x**2)/(b + 2*c*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{c x^{2} + b x + a}}{{\left (2 \, c x + b\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(c*x^2+b*x+a)/(2*c*x+b)^3,x, algorithm="giac")

[Out]

integrate(f^(c*x^2 + b*x + a)/(2*c*x + b)^3, x)