3.633 \(\int \frac{e^{a+b x+c x^2} (b+2 c x)}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=85 \[ \frac{4}{3} \sqrt{\pi } \text{Erfi}\left (\sqrt{a+b x+c x^2}\right )-\frac{4 e^{a+b x+c x^2}}{3 \sqrt{a+b x+c x^2}}-\frac{2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.348027, antiderivative size = 85, 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, 2177, 2180, 2204} \[ \frac{4}{3} \sqrt{\pi } \text{Erfi}\left (\sqrt{a+b x+c x^2}\right )-\frac{4 e^{a+b x+c x^2}}{3 \sqrt{a+b x+c x^2}}-\frac{2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^(a + b*x + c*x^2)*(b + 2*c*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && 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 \frac{e^{a+b x+c x^2} (b+2 c x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=\operatorname{Subst}\left (\int \frac{e^x}{x^{5/2}} \, dx,x,a+b x+c x^2\right )\\ &=-\frac{2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac{2}{3} \operatorname{Subst}\left (\int \frac{e^x}{x^{3/2}} \, dx,x,a+b x+c x^2\right )\\ &=-\frac{2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 e^{a+b x+c x^2}}{3 \sqrt{a+b x+c x^2}}+\frac{4}{3} \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{x}} \, dx,x,a+b x+c x^2\right )\\ &=-\frac{2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 e^{a+b x+c x^2}}{3 \sqrt{a+b x+c x^2}}+\frac{8}{3} \operatorname{Subst}\left (\int e^{x^2} \, dx,x,\sqrt{a+b x+c x^2}\right )\\ &=-\frac{2 e^{a+b x+c x^2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{4 e^{a+b x+c x^2}}{3 \sqrt{a+b x+c x^2}}+\frac{4}{3} \sqrt{\pi } \text{erfi}\left (\sqrt{a+b x+c x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.127352, size = 77, normalized size = 0.91 \[ -\frac{2 \left (2 (-a-x (b+c x))^{3/2} \text{Gamma}\left (\frac{1}{2},-a-x (b+c x)\right )+e^{a+x (b+c x)} (2 (a+x (b+c x))+1)\right )}{3 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(a + b*x + c*x^2)*(b + 2*c*x))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(E^(a + x*(b + c*x))*(1 + 2*(a + x*(b + c*x))) + 2*(-a - x*(b + c*x))^(3/2)*Gamma[1/2, -a - x*(b + c*x)]))
/(3*(a + x*(b + c*x))^(3/2))

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Maple [A]  time = 0.041, size = 70, normalized size = 0.8 \begin{align*} -{\frac{2\,{{\rm e}^{c{x}^{2}+bx+a}}}{3} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}+{\frac{4\,\sqrt{\pi }}{3}{\it erfi} \left ( \sqrt{c{x}^{2}+bx+a} \right ) }-{\frac{4\,{{\rm e}^{c{x}^{2}+bx+a}}}{3}{\frac{1}{\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)^(5/2),x)

[Out]

-2/3*exp(c*x^2+b*x+a)/(c*x^2+b*x+a)^(3/2)+4/3*erfi((c*x^2+b*x+a)^(1/2))*Pi^(1/2)-4/3*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 \frac{{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*x + b)*e^(c*x^2 + b*x + a)/(c*x^2 + b*x + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \,{\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x +{\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \,{\left (a b^{2} + a^{2} c\right )} x^{2}}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x + a)*(2*c*x + b)*e^(c*x^2 + b*x + a)/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^2)*x^4
+ 3*a^2*b*x + (b^3 + 6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x^2), 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)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="giac")

[Out]

integrate((2*c*x + b)*e^(c*x^2 + b*x + a)/(c*x^2 + b*x + a)^(5/2), x)