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

Optimal. Leaf size=84 \[ \frac {\sqrt {\pi } \sqrt {\log (f)} f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2}}-\frac {f^{a+b x+c x^2}}{2 c (b+2 c x)} \]

[Out]

-1/2*f^(c*x^2+b*x+a)/c/(2*c*x+b)+1/4*f^(a-1/4/c*b^2)*erfi(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(1/2))*Pi^(1/2)*ln(f)^(1
/2)/c^(3/2)

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Rubi [A]  time = 0.06, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2239, 2234, 2204} \[ \frac {\sqrt {\pi } \sqrt {\log (f)} f^{a-\frac {b^2}{4 c}} \text {Erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2}}-\frac {f^{a+b x+c x^2}}{2 c (b+2 c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-f^(a + b*x + c*x^2)/(2*c*(b + 2*c*x)) + (f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c
])]*Sqrt[Log[f]])/(4*c^(3/2))

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

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

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Mathematica [A]  time = 0.11, size = 96, normalized size = 1.14 \[ \frac {f^{a-\frac {b^2}{4 c}} \left (\sqrt {\pi } \sqrt {\log (f)} (b+2 c x) \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )-2 \sqrt {c} f^{\frac {(b+2 c x)^2}{4 c}}\right )}{4 c^{3/2} (b+2 c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^(a - b^2/(4*c))*(-2*Sqrt[c]*f^((b + 2*c*x)^2/(4*c)) + Sqrt[Pi]*(b + 2*c*x)*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/
(2*Sqrt[c])]*Sqrt[Log[f]]))/(4*c^(3/2)*(b + 2*c*x))

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fricas [A]  time = 0.44, size = 85, normalized size = 1.01 \[ -\frac {2 \, c f^{c x^{2} + b x + a} + \frac {\sqrt {\pi } {\left (2 \, c x + b\right )} \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \relax (f)}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{4 \, {\left (2 \, c^{3} x + b c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*c*f^(c*x^2 + b*x + a) + sqrt(pi)*(2*c*x + b)*sqrt(-c*log(f))*erf(1/2*(2*c*x + b)*sqrt(-c*log(f))/c)/f^
(1/4*(b^2 - 4*a*c)/c))/(2*c^3*x + b*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 101, normalized size = 1.20 \[ -\frac {f^{\frac {4 a c -b^{2}}{4 c}} f^{\frac {\left (2 c x +b \right )^{2}}{4 c}}}{2 \left (2 c x +b \right ) c}+\frac {\sqrt {\pi }\, f^{\frac {4 a c -b^{2}}{4 c}} \erf \left (\frac {\sqrt {-\frac {\ln \relax (f )}{c}}\, \left (2 c x +b \right )}{2}\right ) \ln \relax (f )}{4 \sqrt {-\frac {\ln \relax (f )}{c}}\, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.14, size = 76, normalized size = 0.90 \[ \frac {f^{a-\frac {b^2}{4\,c}}\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\frac {b\,\ln \relax (f)}{2}+c\,x\,\ln \relax (f)}{\sqrt {c\,\ln \relax (f)}}\right )\,\ln \relax (f)}{4\,c\,\sqrt {c\,\ln \relax (f)}}-\frac {f^a\,f^{c\,x^2}\,f^{b\,x}}{2\,c\,\left (b+2\,c\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(f^(a - b^2/(4*c))*pi^(1/2)*erfi(((b*log(f))/2 + c*x*log(f))/(c*log(f))^(1/2))*log(f))/(4*c*(c*log(f))^(1/2))
- (f^a*f^(c*x^2)*f^(b*x))/(2*c*(b + 2*c*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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