3.457 \(\int f^{b x+c x^2} (b+2 c x)^2 \, dx\)

Optimal. Leaf size=75 \[ \frac {(b+2 c x) f^{b x+c x^2}}{\log (f)}-\frac {\sqrt {\pi } \sqrt {c} f^{-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{\log ^{\frac {3}{2}}(f)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2237, 2234, 2204} \[ \frac {(b+2 c x) f^{b x+c x^2}}{\log (f)}-\frac {\sqrt {\pi } \sqrt {c} f^{-\frac {b^2}{4 c}} \text {Erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{\log ^{\frac {3}{2}}(f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(b*x + c*x^2)*(b + 2*c*x)^2,x]

[Out]

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

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 2237

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 84, normalized size = 1.12 \[ \frac {f^{-\frac {b^2}{4 c}} \left (\sqrt {\log (f)} (b+2 c x) f^{\frac {(b+2 c x)^2}{4 c}}-\sqrt {\pi } \sqrt {c} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )\right )}{\log ^{\frac {3}{2}}(f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 68, normalized size = 0.91 \[ \frac {{\left (2 \, c x + b\right )} f^{c x^{2} + b x} \log \relax (f) + \frac {\sqrt {\pi } \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \relax (f)}}{2 \, c}\right )}{f^{\frac {b^{2}}{4 \, c}}}}{\log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.46, size = 77, normalized size = 1.03 \[ \frac {c {\left (2 \, x + \frac {b}{c}\right )} e^{\left (c x^{2} \log \relax (f) + b x \log \relax (f)\right )}}{\log \relax (f)} + \frac {\sqrt {\pi } c \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \relax (f)} {\left (2 \, x + \frac {b}{c}\right )}\right )}{\sqrt {-c \log \relax (f)} f^{\frac {b^{2}}{4 \, c}} \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 90, normalized size = 1.20 \[ \frac {2 c x \,f^{b x} f^{c \,x^{2}}}{\ln \relax (f )}+\frac {b \,f^{b x} f^{c \,x^{2}}}{\ln \relax (f )}+\frac {\sqrt {\pi }\, c \,f^{-\frac {b^{2}}{4 c}} \erf \left (\frac {b \ln \relax (f )}{2 \sqrt {-c \ln \relax (f )}}-\sqrt {-c \ln \relax (f )}\, x \right )}{\sqrt {-c \ln \relax (f )}\, \ln \relax (f )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.04, size = 329, normalized size = 4.39 \[ \frac {\sqrt {\pi } b^{2} \operatorname {erf}\left (\sqrt {-c \log \relax (f)} x - \frac {b \log \relax (f)}{2 \, \sqrt {-c \log \relax (f)}}\right )}{2 \, \sqrt {-c \log \relax (f)} f^{\frac {b^{2}}{4 \, c}}} - \frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}}\right ) - 1\right )} \log \relax (f)^{2}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}} \left (c \log \relax (f)\right )^{\frac {3}{2}}} - \frac {2 \, c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \relax (f)}{\left (c \log \relax (f)\right )^{\frac {3}{2}}}\right )} b c}{\sqrt {c \log \relax (f)} f^{\frac {b^{2}}{4 \, c}}} + \frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b^{2} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}}\right ) - 1\right )} \log \relax (f)^{3}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}} \left (c \log \relax (f)\right )^{\frac {5}{2}}} - \frac {4 \, {\left (2 \, c x + b\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{4 \, c}\right ) \log \relax (f)^{3}}{\left (-\frac {{\left (2 \, c x + b\right )}^{2} \log \relax (f)}{c}\right )^{\frac {3}{2}} \left (c \log \relax (f)\right )^{\frac {5}{2}}} - \frac {4 \, b c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \relax (f)^{2}}{\left (c \log \relax (f)\right )^{\frac {5}{2}}}\right )} c^{2}}{2 \, \sqrt {c \log \relax (f)} f^{\frac {b^{2}}{4 \, c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(pi)*b^2*erf(sqrt(-c*log(f))*x - 1/2*b*log(f)/sqrt(-c*log(f)))/(sqrt(-c*log(f))*f^(1/4*b^2/c)) - (sqrt
(pi)*(2*c*x + b)*b*(erf(1/2*sqrt(-(2*c*x + b)^2*log(f)/c)) - 1)*log(f)^2/(sqrt(-(2*c*x + b)^2*log(f)/c)*(c*log
(f))^(3/2)) - 2*c*f^(1/4*(2*c*x + b)^2/c)*log(f)/(c*log(f))^(3/2))*b*c/(sqrt(c*log(f))*f^(1/4*b^2/c)) + 1/2*(s
qrt(pi)*(2*c*x + b)*b^2*(erf(1/2*sqrt(-(2*c*x + b)^2*log(f)/c)) - 1)*log(f)^3/(sqrt(-(2*c*x + b)^2*log(f)/c)*(
c*log(f))^(5/2)) - 4*(2*c*x + b)^3*gamma(3/2, -1/4*(2*c*x + b)^2*log(f)/c)*log(f)^3/((-(2*c*x + b)^2*log(f)/c)
^(3/2)*(c*log(f))^(5/2)) - 4*b*c*f^(1/4*(2*c*x + b)^2/c)*log(f)^2/(c*log(f))^(5/2))*c^2/(sqrt(c*log(f))*f^(1/4
*b^2/c))

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mupad [B]  time = 3.65, size = 86, normalized size = 1.15 \[ \frac {b\,f^{c\,x^2}\,f^{b\,x}}{\ln \relax (f)}+\frac {2\,c\,f^{c\,x^2}\,f^{b\,x}\,x}{\ln \relax (f)}-\frac {c\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\frac {b\,\ln \relax (f)}{2}+c\,x\,\ln \relax (f)}{\sqrt {c\,\ln \relax (f)}}\right )}{f^{\frac {b^2}{4\,c}}\,\ln \relax (f)\,\sqrt {c\,\ln \relax (f)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x + c*x^2)*(b + 2*c*x)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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