3.5.61 \(\int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx\) [461]

Optimal. Leaf size=66 \[ -\frac {f^{b x+c x^2}}{4 c (b+2 c x)^2}+\frac {f^{-\frac {b^2}{4 c}} \text {Ei}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right ) \log (f)}{16 c^2} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2271, 2270} \begin {gather*} \frac {\log (f) f^{-\frac {b^2}{4 c}} \text {Ei}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right )}{16 c^2}-\frac {f^{b x+c x^2}}{4 c (b+2 c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[f^(b*x + c*x^2)/(b + 2*c*x)^3,x]

[Out]

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

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*} \int \frac {f^{b x+c x^2}}{(b+2 c x)^3} \, dx &=-\frac {f^{b x+c x^2}}{4 c (b+2 c x)^2}+\frac {\log (f) \int \frac {f^{b x+c x^2}}{b+2 c x} \, dx}{4 c}\\ &=-\frac {f^{b x+c x^2}}{4 c (b+2 c x)^2}+\frac {f^{-\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 = 0.21, size = 77, normalized size = 1.17 \begin {gather*} \frac {f^{-\frac {b^2}{4 c}} \left (-4 c f^{\frac {(b+2 c x)^2}{4 c}}+(b+2 c x)^2 \text {Ei}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right ) \log (f)\right )}{16 c^2 (b+2 c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 74, normalized size = 1.12

method	result	size
risch	\(-\frac {f^{\frac {\left (2 c x +b \right )^{2}}{4 c}} f^{-\frac {b^{2}}{4 c}}}{4 c \left (2 c x +b \right )^{2}}-\frac {\ln \left (f \right ) f^{-\frac {b^{2}}{4 c}} \expIntegral \left (1, -\frac {\left (2 c x +b \right )^{2} \ln \left (f \right )}{4 c}\right )}{16 c^{2}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 100, normalized size = 1.52 \begin {gather*} -\frac {4 \, c f^{c x^{2} + b x} - \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 \, c}}}}{16 \, {\left (4 \, c^{4} x^{2} + 4 \, b c^{3} x + b^{2} c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {f^{c\,x^2+b\,x}}{{\left (b+2\,c\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x + c*x^2)/(b + 2*c*x)^3,x)

[Out]

int(f^(b*x + c*x^2)/(b + 2*c*x)^3, x)

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