\(\int F^{a+b (c+d x)^2} (e+f x)^2 \, dx\) [385]

Optimal result

Integrand size = 21, antiderivative size = 170 \[ \int F^{a+b (c+d x)^2} (e+f x)^2 \, dx=-\frac {f^2 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{4 b^{3/2} d^3 \log ^{\frac {3}{2}}(F)}+\frac {f (d e-c f) F^{a+b (c+d x)^2}}{b d^3 \log (F)}+\frac {f^2 F^{a+b (c+d x)^2} (c+d x)}{2 b d^3 \log (F)}+\frac {(d e-c f)^2 F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{2 \sqrt {b} d^3 \sqrt {\log (F)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2258, 2235, 2240, 2243} \[ \int F^{a+b (c+d x)^2} (e+f x)^2 \, dx=-\frac {\sqrt {\pi } f^2 F^a \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{4 b^{3/2} d^3 \log ^{\frac {3}{2}}(F)}+\frac {\sqrt {\pi } F^a (d e-c f)^2 \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{2 \sqrt {b} d^3 \sqrt {\log (F)}}+\frac {f (d e-c f) F^{a+b (c+d x)^2}}{b d^3 \log (F)}+\frac {f^2 (c+d x) F^{a+b (c+d x)^2}}{2 b d^3 \log (F)} \]

[In]

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

[Out]

-1/4*(f^2*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(b^(3/2)*d^3*Log[F]^(3/2)) + (f*(d*e - c*f)*F^(a
+ b*(c + d*x)^2))/(b*d^3*Log[F]) + (f^2*F^(a + b*(c + d*x)^2)*(c + d*x))/(2*b*d^3*Log[F]) + ((d*e - c*f)^2*F^a
*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(2*Sqrt[b]*d^3*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 2240

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

Rule 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2258

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

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

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.62 \[ \int F^{a+b (c+d x)^2} (e+f x)^2 \, dx=\frac {F^a \left (2 \sqrt {b} f F^{b (c+d x)^2} (2 d e-c f+d f x) \sqrt {\log (F)}+\sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \left (-f^2+2 b (d e-c f)^2 \log (F)\right )\right )}{4 b^{3/2} d^3 \log ^{\frac {3}{2}}(F)} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(383\) vs. \(2(144)=288\).

Time = 0.33 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.26

method	result	size
risch	\(-\frac {F^{b \,c^{2}} F^{a} e^{2} \sqrt {\pi }\, F^{-b \,c^{2}} \operatorname {erf}\left (-d \sqrt {-b \ln \left (F \right )}\, x +\frac {b c \ln \left (F \right )}{\sqrt {-b \ln \left (F \right )}}\right )}{2 d \sqrt {-b \ln \left (F \right )}}+\frac {f^{2} F^{a} F^{b \,c^{2}} x \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b \,d^{2}}-\frac {f^{2} F^{a} F^{b \,c^{2}} c \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 d^{3} \ln \left (F \right ) b}-\frac {f^{2} F^{a} F^{b \,c^{2}} c^{2} \sqrt {\pi }\, F^{-b \,c^{2}} \operatorname {erf}\left (-d \sqrt {-b \ln \left (F \right )}\, x +\frac {b c \ln \left (F \right )}{\sqrt {-b \ln \left (F \right )}}\right )}{2 d^{3} \sqrt {-b \ln \left (F \right )}}+\frac {f^{2} F^{a} F^{b \,c^{2}} \sqrt {\pi }\, F^{-b \,c^{2}} \operatorname {erf}\left (-d \sqrt {-b \ln \left (F \right )}\, x +\frac {b c \ln \left (F \right )}{\sqrt {-b \ln \left (F \right )}}\right )}{4 \ln \left (F \right ) b \,d^{3} \sqrt {-b \ln \left (F \right )}}+\frac {F^{b \,c^{2}} F^{a} e f \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{\ln \left (F \right ) b \,d^{2}}+\frac {F^{b \,c^{2}} F^{a} e f c \sqrt {\pi }\, F^{-b \,c^{2}} \operatorname {erf}\left (-d \sqrt {-b \ln \left (F \right )}\, x +\frac {b c \ln \left (F \right )}{\sqrt {-b \ln \left (F \right )}}\right )}{d^{2} \sqrt {-b \ln \left (F \right )}}\)	\(384\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.79 \[ \int F^{a+b (c+d x)^2} (e+f x)^2 \, dx=\frac {\sqrt {\pi } \sqrt {-b d^{2} \log \left (F\right )} {\left (f^{2} - 2 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (F\right )\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) + 2 \, {\left (b d^{2} f^{2} x + 2 \, b d^{2} e f - b c d f^{2}\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a} \log \left (F\right )}{4 \, b^{2} d^{4} \log \left (F\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (144) = 288\).

Time = 0.43 (sec) , antiderivative size = 422, normalized size of antiderivative = 2.48 \[ \int F^{a+b (c+d x)^2} (e+f x)^2 \, dx=-\frac {{\left (\frac {\sqrt {\pi } {\left (b d^{2} x + b c d\right )} b c {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b d^{2} x + b c d\right )}^{2} \log \left (F\right )}{b d^{2}}}\right ) - 1\right )} \log \left (F\right )^{2}}{\left (b \log \left (F\right )\right )^{\frac {3}{2}} d^{2} \sqrt {-\frac {{\left (b d^{2} x + b c d\right )}^{2} \log \left (F\right )}{b d^{2}}}} - \frac {F^{\frac {{\left (b d^{2} x + b c d\right )}^{2}}{b d^{2}}} b \log \left (F\right )}{\left (b \log \left (F\right )\right )^{\frac {3}{2}} d}\right )} F^{a} e f}{\sqrt {b \log \left (F\right )} d} + \frac {{\left (\frac {\sqrt {\pi } {\left (b d^{2} x + b c d\right )} b^{2} c^{2} {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b d^{2} x + b c d\right )}^{2} \log \left (F\right )}{b d^{2}}}\right ) - 1\right )} \log \left (F\right )^{3}}{\left (b \log \left (F\right )\right )^{\frac {5}{2}} d^{3} \sqrt {-\frac {{\left (b d^{2} x + b c d\right )}^{2} \log \left (F\right )}{b d^{2}}}} - \frac {2 \, F^{\frac {{\left (b d^{2} x + b c d\right )}^{2}}{b d^{2}}} b^{2} c \log \left (F\right )^{2}}{\left (b \log \left (F\right )\right )^{\frac {5}{2}} d^{2}} - \frac {{\left (b d^{2} x + b c d\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {{\left (b d^{2} x + b c d\right )}^{2} \log \left (F\right )}{b d^{2}}\right ) \log \left (F\right )^{3}}{\left (b \log \left (F\right )\right )^{\frac {5}{2}} d^{5} \left (-\frac {{\left (b d^{2} x + b c d\right )}^{2} \log \left (F\right )}{b d^{2}}\right )^{\frac {3}{2}}}\right )} F^{a} f^{2}}{2 \, \sqrt {b \log \left (F\right )} d} + \frac {\sqrt {\pi } F^{b c^{2} + a} e^{2} \operatorname {erf}\left (\sqrt {-b \log \left (F\right )} d x - \frac {b c \log \left (F\right )}{\sqrt {-b \log \left (F\right )}}\right )}{2 \, \sqrt {-b \log \left (F\right )} F^{b c^{2}} d} \]

[In]

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

[Out]

-(sqrt(pi)*(b*d^2*x + b*c*d)*b*c*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^2/((b*log(F))^(3/
2)*d^2*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - F^((b*d^2*x + b*c*d)^2/(b*d^2))*b*log(F)/((b*log(F))^(3/2)
*d))*F^a*e*f/(sqrt(b*log(F))*d) + 1/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^2*c^2*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F
)/(b*d^2))) - 1)*log(F)^3/((b*log(F))^(5/2)*d^3*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 2*F^((b*d^2*x + b
*c*d)^2/(b*d^2))*b^2*c*log(F)^2/((b*log(F))^(5/2)*d^2) - (b*d^2*x + b*c*d)^3*gamma(3/2, -(b*d^2*x + b*c*d)^2*l
og(F)/(b*d^2))*log(F)^3/((b*log(F))^(5/2)*d^5*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)))*F^a*f^2/(sqrt(b*lo
g(F))*d) + 1/2*sqrt(pi)*F^(b*c^2 + a)*e^2*erf(sqrt(-b*log(F))*d*x - b*c*log(F)/sqrt(-b*log(F)))/(sqrt(-b*log(F
))*F^(b*c^2)*d)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.89 \[ \int F^{a+b (c+d x)^2} (e+f x)^2 \, dx=-\frac {\frac {\sqrt {\pi } {\left (2 \, b d^{2} e^{2} \log \left (F\right ) - 4 \, b c d e f \log \left (F\right ) + 2 \, b c^{2} f^{2} \log \left (F\right ) - f^{2}\right )} F^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (F\right )} d {\left (x + \frac {c}{d}\right )}\right )}{\sqrt {-b \log \left (F\right )} b d \log \left (F\right )} - \frac {2 \, {\left (d f^{2} {\left (x + \frac {c}{d}\right )} + 2 \, d e f - 2 \, c f^{2}\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{b d \log \left (F\right )}}{4 \, d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.14 \[ \int F^{a+b (c+d x)^2} (e+f x)^2 \, dx=\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,f^2\,x}{2\,b\,d^2\,\ln \left (F\right )}-\frac {F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (F\right )\,d^2+b\,c\,\ln \left (F\right )\,d}{\sqrt {b\,d^2\,\ln \left (F\right )}}\right )\,\left (-2\,b\,\ln \left (F\right )\,c^2\,f^2+4\,b\,\ln \left (F\right )\,c\,d\,e\,f-2\,b\,\ln \left (F\right )\,d^2\,e^2+f^2\right )}{4\,b\,d^2\,\ln \left (F\right )\,\sqrt {b\,d^2\,\ln \left (F\right )}}-F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,\left (\frac {c\,f^2}{2\,b\,d^3\,\ln \left (F\right )}-\frac {e\,f}{b\,d^2\,\ln \left (F\right )}\right ) \]

[In]

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

[Out]

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