\(\int F^{a+b (c+d x)^2} (e+f x) \, dx\) [386]

Optimal result

Integrand size = 19, antiderivative size = 81 \[ \int F^{a+b (c+d x)^2} (e+f x) \, dx=\frac {f F^{a+b (c+d x)^2}}{2 b d^2 \log (F)}+\frac {(d e-c f) F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{2 \sqrt {b} d^2 \sqrt {\log (F)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2258, 2235, 2240} \[ \int F^{a+b (c+d x)^2} (e+f x) \, dx=\frac {\sqrt {\pi } F^a (d e-c f) \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{2 \sqrt {b} d^2 \sqrt {\log (F)}}+\frac {f F^{a+b (c+d x)^2}}{2 b d^2 \log (F)} \]

[In]

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

[Out]

(f*F^(a + b*(c + d*x)^2))/(2*b*d^2*Log[F]) + ((d*e - c*f)*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(
2*Sqrt[b]*d^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 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 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) F^{a+b (c+d x)^2}}{d}+\frac {f F^{a+b (c+d x)^2} (c+d x)}{d}\right ) \, dx \\ & = \frac {f \int F^{a+b (c+d x)^2} (c+d x) \, dx}{d}+\frac {(d e-c f) \int F^{a+b (c+d x)^2} \, dx}{d} \\ & = \frac {f F^{a+b (c+d x)^2}}{2 b d^2 \log (F)}+\frac {(d e-c f) F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right )}{2 \sqrt {b} d^2 \sqrt {\log (F)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int F^{a+b (c+d x)^2} (e+f x) \, dx=\frac {F^a \left (f F^{b (c+d x)^2}+\sqrt {b} (d e-c f) \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \sqrt {\log (F)}\right )}{2 b d^2 \log (F)} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(161\) vs. \(2(67)=134\).

Time = 0.03 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.00

method	result	size
risch	\(-\frac {F^{b \,c^{2}} F^{a} e \sqrt {\pi }\, F^{-b \,c^{2}} \operatorname {erf}\left (-d \sqrt {-b \ln \left (F \right )}\, x +\frac {b c \ln \left (F \right )}{\sqrt {-b \ln \left (F \right )}}\right )}{2 d \sqrt {-b \ln \left (F \right )}}+\frac {F^{b \,c^{2}} F^{a} f \,F^{b \,d^{2} x^{2}} F^{2 b c d x}}{2 \ln \left (F \right ) b \,d^{2}}+\frac {F^{b \,c^{2}} F^{a} f c \sqrt {\pi }\, F^{-b \,c^{2}} \operatorname {erf}\left (-d \sqrt {-b \ln \left (F \right )}\, x +\frac {b c \ln \left (F \right )}{\sqrt {-b \ln \left (F \right )}}\right )}{2 d^{2} \sqrt {-b \ln \left (F \right )}}\)	\(162\)

[In]

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

[Out]

-1/2*F^(b*c^2)*F^a*e*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^(b*c^2)*F^a*f/ln(F)/b/d^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)+1/2*F^(b*c^2)*F^a*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 = 85, normalized size of antiderivative = 1.05 \[ \int F^{a+b (c+d x)^2} (e+f x) \, dx=-\frac {\sqrt {\pi } \sqrt {-b d^{2} \log \left (F\right )} {\left (d e - c f\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) - F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a} d f}{2 \, b d^{3} \log \left (F\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (67) = 134\).

Time = 0.34 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.41 \[ \int F^{a+b (c+d x)^2} (e+f x) \, dx=-\frac {{\left (\frac {\sqrt {\pi } {\left (b d^{2} x + b c d\right )} b c {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b d^{2} x + b c d\right )}^{2} \log \left (F\right )}{b d^{2}}}\right ) - 1\right )} \log \left (F\right )^{2}}{\left (b \log \left (F\right )\right )^{\frac {3}{2}} d^{2} \sqrt {-\frac {{\left (b d^{2} x + b c d\right )}^{2} \log \left (F\right )}{b d^{2}}}} - \frac {F^{\frac {{\left (b d^{2} x + b c d\right )}^{2}}{b d^{2}}} b \log \left (F\right )}{\left (b \log \left (F\right )\right )^{\frac {3}{2}} d}\right )} F^{a} f}{2 \, \sqrt {b \log \left (F\right )} d} + \frac {\sqrt {\pi } F^{b c^{2} + a} e \operatorname {erf}\left (\sqrt {-b \log \left (F\right )} d x - \frac {b c \log \left (F\right )}{\sqrt {-b \log \left (F\right )}}\right )}{2 \, \sqrt {-b \log \left (F\right )} F^{b c^{2}} d} \]

[In]

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

[Out]

-1/2*(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*f/(sqrt(b*log(F))*d) + 1/2*sqrt(pi)*F^(b*c^2 + a)*e*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 = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15 \[ \int F^{a+b (c+d x)^2} (e+f x) \, dx=-\frac {\frac {\sqrt {\pi } {\left (d e - c f\right )} F^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (F\right )} d {\left (x + \frac {c}{d}\right )}\right )}{\sqrt {-b \log \left (F\right )} d} - \frac {f e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{b d \log \left (F\right )}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19 \[ \int F^{a+b (c+d x)^2} (e+f x) \, dx=\frac {F^{b\,d^2\,x^2}\,F^a\,F^{b\,c^2}\,F^{2\,b\,c\,d\,x}\,f}{2\,b\,d^2\,\ln \left (F\right )}-\frac {F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (F\right )\,d^2+b\,c\,\ln \left (F\right )\,d}{\sqrt {b\,d^2\,\ln \left (F\right )}}\right )\,\left (c\,f-d\,e\right )}{2\,d\,\sqrt {b\,d^2\,\ln \left (F\right )}} \]

[In]

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

[Out]

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