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

Optimal. Leaf size=81 \[ \frac{\sqrt{\pi } F^a (d e-c f) \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{2 \sqrt{b} d^2 \sqrt{\log (F)}}+\frac{f F^{a+b (c+d x)^2}}{2 b d^2 \log (F)} \]

[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]])

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Rubi [A]  time = 0.146504, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2226, 2204, 2209} \[ \frac{\sqrt{\pi } F^a (d e-c f) \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{2 \sqrt{b} d^2 \sqrt{\log (F)}}+\frac{f F^{a+b (c+d x)^2}}{2 b d^2 \log (F)} \]

Antiderivative was successfully verified.

[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 2226

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]

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 2209

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]

Rubi steps

\begin{align*} \int F^{a+b (c+d x)^2} (e+f x) \, dx &=\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]  time = 0.0671947, size = 74, normalized size = 0.91 \[ \frac{F^a \left (\sqrt{\pi } \sqrt{b} \sqrt{\log (F)} (d e-c f) \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )+f F^{b (c+d x)^2}\right )}{2 b d^2 \log (F)} \]

Antiderivative was successfully verified.

[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])

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Maple [A]  time = 0.035, size = 132, normalized size = 1.6 \begin{align*} -{\frac{e\sqrt{\pi }{F}^{a}}{2\,d}{\it Erf} \left ( -d\sqrt{-b\ln \left ( F \right ) }x+{bc\ln \left ( F \right ){\frac{1}{\sqrt{-b\ln \left ( F \right ) }}}} \right ){\frac{1}{\sqrt{-b\ln \left ( F \right ) }}}}+{\frac{f{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{2\,\ln \left ( F \right ) b{d}^{2}}}+{\frac{cf\sqrt{\pi }{F}^{a}}{2\,{d}^{2}}{\it Erf} \left ( -d\sqrt{-b\ln \left ( F \right ) }x+{bc\ln \left ( F \right ){\frac{1}{\sqrt{-b\ln \left ( F \right ) }}}} \right ){\frac{1}{\sqrt{-b\ln \left ( F \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.217, size = 269, normalized size = 3.32 \begin{align*} -\frac{{\left (\frac{\sqrt{\pi }{\left (b d^{2} x + b c d\right )} b c d{\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 d^{2} \log \left (F\right )\right )^{\frac{3}{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 d^{2} \log \left (F\right )}{\left (b d^{2} \log \left (F\right )\right )^{\frac{3}{2}}}\right )} F^{a} f}{2 \, \sqrt{b d^{2} \log \left (F\right )}} + \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Fricas [A]  time = 1.54248, size = 201, normalized size = 2.48 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.27421, size = 171, normalized size = 2.11 \begin{align*} -\frac{\sqrt{\pi } \operatorname{erf}\left (-\sqrt{-b \log \left (F\right )} d{\left (x + \frac{c}{d}\right )}\right ) e^{\left (a \log \left (F\right ) + 1\right )}}{2 \, \sqrt{-b \log \left (F\right )} d} + \frac{\frac{\sqrt{\pi } F^{a} c f \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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