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]

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

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Rubi [A]  time = 0.07, antiderivative size = 69, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 0.41, size = 106, normalized size = 1.54 \[ -\frac {4 \, c f^{c x^{2} + b x + a} - \frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} {\rm Ei}\left (\frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \relax (f)}{4 \, c}\right ) \log \relax (f)}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{16 \, {\left (4 \, c^{4} x^{2} + 4 \, b c^{3} x + b^{2} c^{2}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{c x^{2} + b x + a}}{{\left (2 \, c x + b\right )}^{3}}\,{d x} \]

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)

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maple [A]  time = 0.05, size = 88, normalized size = 1.28 \[ -\frac {f^{\frac {4 a c -b^{2}}{4 c}} f^{\frac {\left (2 c x +b \right )^{2}}{4 c}}}{4 \left (2 c x +b \right )^{2} c}-\frac {f^{\frac {4 a c -b^{2}}{4 c}} \Ei \left (1, -\frac {\left (2 c x +b \right )^{2} \ln \relax (f )}{4 c}\right ) \ln \relax (f )}{16 c^{2}} \]

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/c*ln(f))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{c x^{2} + b x + a}}{{\left (2 \, c x + b\right )}^{3}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {f^{c\,x^2+b\,x+a}}{{\left (b+2\,c\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + b x + c x^{2}}}{\left (b + 2 c x\right )^{3}}\, dx \]

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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