\(\int e^{a+b x+c x^2} (b+2 c x) \sqrt {a+b x+c x^2} \, dx\) [630]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [A] (verification not implemented)
   Maxima [F]
   Giac [C] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 33, antiderivative size = 52 \[ \int e^{a+b x+c x^2} (b+2 c x) \sqrt {a+b x+c x^2} \, dx=e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\frac {1}{2} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6839, 2207, 2211, 2235} \[ \int e^{a+b x+c x^2} (b+2 c x) \sqrt {a+b x+c x^2} \, dx=e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\frac {1}{2} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right ) \]

[In]

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

[Out]

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

Rule 2207

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] &&  !TrueQ[$UseGamma]

Rule 2211

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] &&  !TrueQ[$UseGamma]

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 6839

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*} \text {integral}& = \text {Subst}\left (\int e^x \sqrt {x} \, dx,x,a+b x+c x^2\right ) \\ & = e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\frac {1}{2} \text {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,a+b x+c x^2\right ) \\ & = e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\text {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {a+b x+c x^2}\right ) \\ & = e^{a+b x+c x^2} \sqrt {a+b x+c x^2}-\frac {1}{2} \sqrt {\pi } \text {erfi}\left (\sqrt {a+b x+c x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 1.68 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88 \[ \int e^{a+b x+c x^2} (b+2 c x) \sqrt {a+b x+c x^2} \, dx=\frac {\sqrt {a+x (b+c x)} \Gamma \left (\frac {3}{2},-a-x (b+c x)\right )}{\sqrt {-a-x (b+c x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85

method result size
derivativedivides \(-\frac {\operatorname {erfi}\left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{2}+{\mathrm e}^{c \,x^{2}+b x +a} \sqrt {c \,x^{2}+b x +a}\) \(44\)
default \(-\frac {\operatorname {erfi}\left (\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {\pi }}{2}+{\mathrm e}^{c \,x^{2}+b x +a} \sqrt {c \,x^{2}+b x +a}\) \(44\)

[In]

int(exp(c*x^2+b*x+a)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

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

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.85 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.50 \[ \int e^{a+b x+c x^2} (b+2 c x) \sqrt {a+b x+c x^2} \, dx=\frac {\left (\sqrt {- a - b x - c x^{2}} e^{a + b x + c x^{2}} + \frac {\sqrt {\pi } \operatorname {erfc}{\left (\sqrt {- a - b x - c x^{2}} \right )}}{2}\right ) \sqrt {a + b x + c x^{2}}}{\sqrt {- a - b x - c x^{2}}} \]

[In]

integrate(exp(c*x**2+b*x+a)*(2*c*x+b)*(c*x**2+b*x+a)**(1/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int e^{a+b x+c x^2} (b+2 c x) \sqrt {a+b x+c x^2} \, dx=-\frac {1}{2} i \, \sqrt {\pi } \operatorname {erf}\left (-i \, \sqrt {c x^{2} + b x + a}\right ) + \sqrt {c x^{2} + b x + a} e^{\left (c x^{2} + b x + a\right )} \]

[In]

integrate(exp(c*x^2+b*x+a)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/2*I*sqrt(pi)*erf(-I*sqrt(c*x^2 + b*x + a)) + sqrt(c*x^2 + b*x + a)*e^(c*x^2 + b*x + a)

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.46 \[ \int e^{a+b x+c x^2} (b+2 c x) \sqrt {a+b x+c x^2} \, dx=\frac {\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-c\,x^2-b\,x-a}\right )\,\sqrt {c\,x^2+b\,x+a}}{2\,\sqrt {-c\,x^2-b\,x-a}}+{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,{\mathrm {e}}^{c\,x^2}\,\sqrt {c\,x^2+b\,x+a} \]

[In]

int(exp(a + b*x + c*x^2)*(b + 2*c*x)*(a + b*x + c*x^2)^(1/2),x)

[Out]

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