3.629 \(\int e^{a+b x+c x^2} (b+2 c x) (a+b x+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=82 \[ \frac{3}{4} \sqrt{\pi } \text{Erfi}\left (\sqrt{a+b x+c x^2}\right )+e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{3/2}-\frac{3}{2} e^{a+b x+c x^2} \sqrt{a+b x+c x^2} \]

[Out]

(-3*E^(a + b*x + c*x^2)*Sqrt[a + b*x + c*x^2])/2 + E^(a + b*x + c*x^2)*(a + b*x + c*x^2)^(3/2) + (3*Sqrt[Pi]*E
rfi[Sqrt[a + b*x + c*x^2]])/4

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Rubi [A]  time = 0.362943, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {6707, 2176, 2180, 2204} \[ \frac{3}{4} \sqrt{\pi } \text{Erfi}\left (\sqrt{a+b x+c x^2}\right )+e^{a+b x+c x^2} \left (a+b x+c x^2\right )^{3/2}-\frac{3}{2} e^{a+b x+c x^2} \sqrt{a+b x+c x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x + c*x^2)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2),x]

[Out]

(-3*E^(a + b*x + c*x^2)*Sqrt[a + b*x + c*x^2])/2 + E^(a + b*x + c*x^2)*(a + b*x + c*x^2)^(3/2) + (3*Sqrt[Pi]*E
rfi[Sqrt[a + b*x + c*x^2]])/4

Rule 6707

Int[(F_)^(v_)*(u_)*(w_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Dist[q, Subst[Int[x^m*F^x,
x], x, v], x] /;  !FalseQ[q]] /; FreeQ[{F, m}, x] && EqQ[w, v]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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]

Rubi steps

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

Mathematica [A]  time = 0.109704, size = 47, normalized size = 0.57 \[ -\frac{\sqrt{a+x (b+c x)} \text{Gamma}\left (\frac{5}{2},-a-x (b+c x)\right )}{\sqrt{-a-x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x + c*x^2)*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2),x]

[Out]

-((Sqrt[a + x*(b + c*x)]*Gamma[5/2, -a - x*(b + c*x)])/Sqrt[-a - x*(b + c*x)])

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Maple [A]  time = 0.041, size = 69, normalized size = 0.8 \begin{align*}{{\rm e}^{c{x}^{2}+bx+a}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}+{\frac{3\,\sqrt{\pi }}{4}{\it erfi} \left ( \sqrt{c{x}^{2}+bx+a} \right ) }-{\frac{3\,{{\rm e}^{c{x}^{2}+bx+a}}}{2}\sqrt{c{x}^{2}+bx+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*x^2+b*x+a)*(2*c*x+b)*(c*x^2+b*x+a)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)*(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)*(2*c*x + b)*e^(c*x^2 + b*x + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (2 \, c^{2} x^{3} + 3 \, b c x^{2} + a b +{\left (b^{2} + 2 \, a c\right )} x\right )} \sqrt{c x^{2} + b x + a} e^{\left (c x^{2} + b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)*(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

integral((2*c^2*x^3 + 3*b*c*x^2 + a*b + (b^2 + 2*a*c)*x)*sqrt(c*x^2 + b*x + a)*e^(c*x^2 + b*x + a), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*x**2+b*x+a)*(2*c*x+b)*(c*x**2+b*x+a)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)*(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)*(2*c*x + b)*e^(c*x^2 + b*x + a), x)