\(\int \frac {f^{a+b x+c x^2}}{(b+2 c x)^3} \, dx\) [455]

Optimal result

Integrand size = 21, antiderivative size = 69 \[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^3} \, dx=-\frac {f^{a+b x+c x^2}}{4 c (b+2 c x)^2}+\frac {f^{a-\frac {b^2}{4 c}} \operatorname {ExpIntegralEi}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right ) \log (f)}{16 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2271, 2270} \[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^3} \, dx=\frac {\log (f) f^{a-\frac {b^2}{4 c}} \operatorname {ExpIntegralEi}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right )}{16 c^2}-\frac {f^{a+b x+c x^2}}{4 c (b+2 c x)^2} \]

[In]

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

[Out]

-1/4*f^(a + b*x + c*x^2)/(c*(b + 2*c*x)^2) + (f^(a - b^2/(4*c))*ExpIntegralEi[((b + 2*c*x)^2*Log[f])/(4*c)]*Lo
g[f])/(16*c^2)

Rule 2270

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

Rule 2271

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*} \text {integral}& = -\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] (verified)

Time = 0.41 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^3} \, dx=\frac {f^{a-\frac {b^2}{4 c}} \left (-4 c f^{\frac {(b+2 c x)^2}{4 c}}+(b+2 c x)^2 \operatorname {ExpIntegralEi}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right ) \log (f)\right )}{16 c^2 (b+2 c x)^2} \]

[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)

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.28

method	result	size
risch	\(-\frac {f^{\frac {\left (2 x c +b \right )^{2}}{4 c}} f^{\frac {4 c a -b^{2}}{4 c}}}{4 c \left (2 x c +b \right )^{2}}-\frac {\ln \left (f \right ) f^{\frac {4 c a -b^{2}}{4 c}} \operatorname {Ei}_{1}\left (-\frac {\left (2 x c +b \right )^{2} \ln \left (f \right )}{4 c}\right )}{16 c^{2}}\)	\(88\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.54 \[ \int \frac {f^{a+b x+c x^2}}{(b+2 c x)^3} \, dx=-\frac {4 \, c f^{c x^{2} + b x + a} - \frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} {\rm Ei}\left (\frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (f\right )}{4 \, c}\right ) \log \left (f\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{16 \, {\left (4 \, c^{4} x^{2} + 4 \, b c^{3} x + b^{2} c^{2}\right )}} \]

[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)

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[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)