3.5.45 \(\int f^{a+b x+c x^2} (d+e x)^2 \, dx\) [445]

Optimal. Leaf size=189 \[ -\frac {e^2 f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 c^{3/2} \log ^{\frac {3}{2}}(f)}+\frac {e (2 c d-b e) f^{a+b x+c x^2}}{4 c^2 \log (f)}+\frac {e f^{a+b x+c x^2} (d+e x)}{2 c \log (f)}+\frac {(2 c d-b e)^2 f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{8 c^{5/2} \sqrt {\log (f)}} \]

[Out]

1/4*e*(-b*e+2*c*d)*f^(c*x^2+b*x+a)/c^2/ln(f)+1/2*e*f^(c*x^2+b*x+a)*(e*x+d)/c/ln(f)-1/4*e^2*f^(a-1/4*b^2/c)*erf
i(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(1/2))*Pi^(1/2)/c^(3/2)/ln(f)^(3/2)+1/8*(-b*e+2*c*d)^2*f^(a-1/4*b^2/c)*erfi(1/2*
(2*c*x+b)*ln(f)^(1/2)/c^(1/2))*Pi^(1/2)/c^(5/2)/ln(f)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2273, 2272, 2266, 2235} \begin {gather*} \frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} (2 c d-b e)^2 \text {Erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{8 c^{5/2} \sqrt {\log (f)}}-\frac {\sqrt {\pi } e^2 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} \log ^{\frac {3}{2}}(f)}+\frac {e (2 c d-b e) f^{a+b x+c x^2}}{4 c^2 \log (f)}+\frac {e (d+e x) f^{a+b x+c x^2}}{2 c \log (f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(e^2*f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(c^(3/2)*Log[f]^(3/2)) + (e
*(2*c*d - b*e)*f^(a + b*x + c*x^2))/(4*c^2*Log[f]) + (e*f^(a + b*x + c*x^2)*(d + e*x))/(2*c*Log[f]) + ((2*c*d
- b*e)^2*f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(8*c^(5/2)*Sqrt[Log[f]])

Rule 2235

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 2266

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 2272

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

Rule 2273

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[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], 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] && NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

Rubi steps

\begin {align*} \int f^{a+b x+c x^2} (d+e x)^2 \, dx &=\frac {e f^{a+b x+c x^2} (d+e x)}{2 c \log (f)}-\frac {(-2 c d+b e) \int f^{a+b x+c x^2} (d+e x) \, dx}{2 c}-\frac {e^2 \int f^{a+b x+c x^2} \, dx}{2 c \log (f)}\\ &=\frac {e (2 c d-b e) f^{a+b x+c x^2}}{4 c^2 \log (f)}+\frac {e f^{a+b x+c x^2} (d+e x)}{2 c \log (f)}+\frac {(2 c d-b e)^2 \int f^{a+b x+c x^2} \, dx}{4 c^2}-\frac {\left (e^2 f^{a-\frac {b^2}{4 c}}\right ) \int f^{\frac {(b+2 c x)^2}{4 c}} \, dx}{2 c \log (f)}\\ &=-\frac {e^2 f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 c^{3/2} \log ^{\frac {3}{2}}(f)}+\frac {e (2 c d-b e) f^{a+b x+c x^2}}{4 c^2 \log (f)}+\frac {e f^{a+b x+c x^2} (d+e x)}{2 c \log (f)}+\frac {\left ((2 c d-b e)^2 f^{a-\frac {b^2}{4 c}}\right ) \int f^{\frac {(b+2 c x)^2}{4 c}} \, dx}{4 c^2}\\ &=-\frac {e^2 f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 c^{3/2} \log ^{\frac {3}{2}}(f)}+\frac {e (2 c d-b e) f^{a+b x+c x^2}}{4 c^2 \log (f)}+\frac {e f^{a+b x+c x^2} (d+e x)}{2 c \log (f)}+\frac {(2 c d-b e)^2 f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{8 c^{5/2} \sqrt {\log (f)}}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 123, normalized size = 0.65 \begin {gather*} \frac {f^{a-\frac {b^2}{4 c}} \left (2 \sqrt {c} e f^{\frac {(b+2 c x)^2}{4 c}} (4 c d-b e+2 c e x) \sqrt {\log (f)}+\sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \left (-2 c e^2+(-2 c d+b e)^2 \log (f)\right )\right )}{8 c^{5/2} \log ^{\frac {3}{2}}(f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(f^(a - b^2/(4*c))*(2*Sqrt[c]*e*f^((b + 2*c*x)^2/(4*c))*(4*c*d - b*e + 2*c*e*x)*Sqrt[Log[f]] + Sqrt[Pi]*Erfi[(
(b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])]*(-2*c*e^2 + (-2*c*d + b*e)^2*Log[f])))/(8*c^(5/2)*Log[f]^(3/2))

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Maple [A]
time = 0.09, size = 307, normalized size = 1.62

method	result	size
risch	\(-\frac {d^{2} \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \erf \left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{2 \sqrt {-c \ln \left (f \right )}}+\frac {e^{2} x \,f^{c \,x^{2}} f^{b x} f^{a}}{2 c \ln \left (f \right )}-\frac {e^{2} b \,f^{c \,x^{2}} f^{b x} f^{a}}{4 c^{2} \ln \left (f \right )}-\frac {e^{2} b^{2} \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \erf \left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{8 c^{2} \sqrt {-c \ln \left (f \right )}}+\frac {e^{2} \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \erf \left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 c \ln \left (f \right ) \sqrt {-c \ln \left (f \right )}}+\frac {e d \,f^{c \,x^{2}} f^{b x} f^{a}}{c \ln \left (f \right )}+\frac {e d b \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \erf \left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{2 c \sqrt {-c \ln \left (f \right )}}\)	\(307\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*d^2*Pi^(1/2)*f^a*f^(-1/4*b^2/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*b*ln(f)/(-c*ln(f))^(1/2))+1/
2*e^2/c/ln(f)*x*f^(c*x^2)*f^(b*x)*f^a-1/4*e^2/c^2*b/ln(f)*f^(c*x^2)*f^(b*x)*f^a-1/8*e^2/c^2*b^2*Pi^(1/2)*f^a*f
^(-1/4*b^2/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*b*ln(f)/(-c*ln(f))^(1/2))+1/4*e^2/c/ln(f)*Pi^(1/2)*
f^a*f^(-1/4*b^2/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*b*ln(f)/(-c*ln(f))^(1/2))+e*d/c/ln(f)*f^(c*x^2
)*f^(b*x)*f^a+1/2*e*d*b/c*Pi^(1/2)*f^a*f^(-1/4*b^2/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*b*ln(f)/(-c
*ln(f))^(1/2))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (153) = 306\).
time = 0.39, size = 332, normalized size = 1.76 \begin {gather*} -\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 \left (f\right )}{c}}\right ) - 1\right )} \log \left (f\right )^{2}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}} \left (c \log \left (f\right )\right )^{\frac {3}{2}}} - \frac {2 \, c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \left (f\right )}{\left (c \log \left (f\right )\right )^{\frac {3}{2}}}\right )} d e f^{a - \frac {b^{2}}{4 \, c}}}{2 \, \sqrt {c \log \left (f\right )}} + \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 \left (f\right )}{c}}\right ) - 1\right )} \log \left (f\right )^{3}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}} \left (c \log \left (f\right )\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 \left (f\right )}{4 \, c}\right ) \log \left (f\right )^{3}}{\left (-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}\right )^{\frac {3}{2}} \left (c \log \left (f\right )\right )^{\frac {5}{2}}} - \frac {4 \, b c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \left (f\right )^{2}}{\left (c \log \left (f\right )\right )^{\frac {5}{2}}}\right )} e^{2} f^{a - \frac {b^{2}}{4 \, c}}}{8 \, \sqrt {c \log \left (f\right )}} + \frac {\sqrt {\pi } d^{2} f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right )}}\right )}{2 \, \sqrt {-c \log \left (f\right )} f^{\frac {b^{2}}{4 \, c}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(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))*d*e*f^(a - 1/4*b^2/c)/sqrt(c*log(
f)) + 1/8*(sqrt(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))*e^2*f^(a - 1/4
*b^2/c)/sqrt(c*log(f)) + 1/2*sqrt(pi)*d^2*f^a*erf(sqrt(-c*log(f))*x - 1/2*b*log(f)/sqrt(-c*log(f)))/(sqrt(-c*l
og(f))*f^(1/4*b^2/c))

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Fricas [A]
time = 0.38, size = 128, normalized size = 0.68 \begin {gather*} \frac {2 \, {\left (4 \, c^{2} d e + {\left (2 \, c^{2} x - b c\right )} e^{2}\right )} f^{c x^{2} + b x + a} \log \left (f\right ) + \frac {\sqrt {\pi } {\left (2 \, c e^{2} - {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \log \left (f\right )\right )} \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{8 \, c^{3} \log \left (f\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(4*c^2*d*e + (2*c^2*x - b*c)*e^2)*f^(c*x^2 + b*x + a)*log(f) + sqrt(pi)*(2*c*e^2 - (4*c^2*d^2 - 4*b*c*d
*e + b^2*e^2)*log(f))*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))/(c^3*log
(f)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.79, size = 252, normalized size = 1.33 \begin {gather*} -\frac {\sqrt {\pi } d^{2} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right )}{4 \, c}\right )}}{2 \, \sqrt {-c \log \left (f\right )}} + \frac {\frac {\sqrt {\pi } b d \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right ) - 4 \, c}{4 \, c}\right )}}{\sqrt {-c \log \left (f\right )}} + \frac {2 \, d e^{\left (c x^{2} \log \left (f\right ) + b x \log \left (f\right ) + a \log \left (f\right ) + 1\right )}}{\log \left (f\right )}}{2 \, c} - \frac {\frac {\sqrt {\pi } {\left (b^{2} \log \left (f\right ) - 2 \, c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right ) - 8 \, c}{4 \, c}\right )}}{\sqrt {-c \log \left (f\right )} \log \left (f\right )} - \frac {2 \, {\left (c {\left (2 \, x + \frac {b}{c}\right )} - 2 \, b\right )} e^{\left (c x^{2} \log \left (f\right ) + b x \log \left (f\right ) + a \log \left (f\right ) + 2\right )}}{\log \left (f\right )}}{8 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(pi)*d^2*erf(-1/2*sqrt(-c*log(f))*(2*x + b/c))*e^(-1/4*(b^2*log(f) - 4*a*c*log(f))/c)/sqrt(-c*log(f))
+ 1/2*(sqrt(pi)*b*d*erf(-1/2*sqrt(-c*log(f))*(2*x + b/c))*e^(-1/4*(b^2*log(f) - 4*a*c*log(f) - 4*c)/c)/sqrt(-
c*log(f)) + 2*d*e^(c*x^2*log(f) + b*x*log(f) + a*log(f) + 1)/log(f))/c - 1/8*(sqrt(pi)*(b^2*log(f) - 2*c)*erf(
-1/2*sqrt(-c*log(f))*(2*x + b/c))*e^(-1/4*(b^2*log(f) - 4*a*c*log(f) - 8*c)/c)/(sqrt(-c*log(f))*log(f)) - 2*(c
*(2*x + b/c) - 2*b)*e^(c*x^2*log(f) + b*x*log(f) + a*log(f) + 2)/log(f))/c^2

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Mupad [B]
time = 3.85, size = 153, normalized size = 0.81 \begin {gather*} f^a\,f^{c\,x^2}\,f^{b\,x}\,\left (\frac {d\,e}{c\,\ln \left (f\right )}-\frac {b\,e^2}{4\,c^2\,\ln \left (f\right )}\right )+\frac {e^2\,f^a\,f^{c\,x^2}\,f^{b\,x}\,x}{2\,c\,\ln \left (f\right )}-\frac {f^{a-\frac {b^2}{4\,c}}\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\frac {b\,\ln \left (f\right )}{2}+c\,x\,\ln \left (f\right )}{\sqrt {c\,\ln \left (f\right )}}\right )\,\left (-\frac {\ln \left (f\right )\,b^2\,e^2}{8}+\frac {\ln \left (f\right )\,b\,c\,d\,e}{2}-\frac {\ln \left (f\right )\,c^2\,d^2}{2}+\frac {c\,e^2}{4}\right )}{c^2\,\ln \left (f\right )\,\sqrt {c\,\ln \left (f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f^a*f^(c*x^2)*f^(b*x)*((d*e)/(c*log(f)) - (b*e^2)/(4*c^2*log(f))) + (e^2*f^a*f^(c*x^2)*f^(b*x)*x)/(2*c*log(f))
- (f^(a - b^2/(4*c))*pi^(1/2)*erfi(((b*log(f))/2 + c*x*log(f))/(c*log(f))^(1/2))*((c*e^2)/4 - (b^2*e^2*log(f)
)/8 - (c^2*d^2*log(f))/2 + (b*c*d*e*log(f))/2))/(c^2*log(f)*(c*log(f))^(1/2))

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