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

Optimal. Leaf size=51 \[ 2 \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right )-\frac {2 e^{a+b x+c x^2}}{\sqrt {a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6707, 2177, 2180, 2204} \[ 2 \sqrt {\pi } \text {Erfi}\left (\sqrt {a+b x+c x^2}\right )-\frac {2 e^{a+b x+c x^2}}{\sqrt {a+b x+c x^2}} \]

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]

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

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]

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]

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

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Mathematica [A]  time = 0.10, size = 62, normalized size = 1.22 \[ \frac {2 \sqrt {-a-x (b+c x)} \Gamma \left (\frac {1}{2},-a-x (b+c x)\right )-2 e^{a+x (b+c x)}}{\sqrt {a+x (b+c x)}} \]

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]

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

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

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

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maple [A]  time = 0.04, size = 45, normalized size = 0.88 \[ 2 \sqrt {\pi }\, \erfi \left (\sqrt {c \,x^{2}+b x +a}\right )-\frac {2 \,{\mathrm e}^{c \,x^{2}+b x +a}}{\sqrt {c \,x^{2}+b x +a}} \]

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]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c x + b\right )} e^{\left (c x^{2} + b x + a\right )}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

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

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mupad [B]  time = 4.00, size = 79, normalized size = 1.55 \[ -\frac {{\mathrm {e}}^{c\,x^2+b\,x+a}\,\left (2\,c\,x^2+2\,b\,x+2\,a\right )+2\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-c\,x^2-b\,x-a}\right )\,{\left (-c\,x^2-b\,x-a\right )}^{3/2}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(exp(a + b*x + c*x^2)*(2*a + 2*b*x + 2*c*x^2) + 2*pi^(1/2)*erfc((- a - b*x - c*x^2)^(1/2))*(- a - b*x - c*x^2
)^(3/2))/(a + b*x + c*x^2)^(3/2)

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sympy [A]  time = 6.49, size = 80, normalized size = 1.57 \[ \frac {\left (- 2 \sqrt {\pi } \operatorname {erfc}{\left (\sqrt {- a - b x - c x^{2}} \right )} + \frac {2 e^{a + b x + c x^{2}}}{\sqrt {- a - b x - c x^{2}}}\right ) \left (- a - b x - c x^{2}\right )^{\frac {3}{2}}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}} \]

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]

(-2*sqrt(pi)*erfc(sqrt(-a - b*x - c*x**2)) + 2*exp(a + b*x + c*x**2)/sqrt(-a - b*x - c*x**2))*(-a - b*x - c*x*
*2)**(3/2)/(a + b*x + c*x**2)**(3/2)

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