3.619 \(\int e^{a+b x+c x^2} (b+2 c x) (a+b x+c x^2)^m \, dx\)

Optimal. Leaf size=49 \[ \left (-a-b x-c x^2\right )^{-m} \left (a+b x+c x^2\right )^m \text{Gamma}\left (m+1,-a-b x-c x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.199597, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {6707, 2181} \[ \left (-a-b x-c x^2\right )^{-m} \left (a+b x+c x^2\right )^m \text{Gamma}\left (m+1,-a-b x-c x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int e^{a+b x+c x^2} (b+2 c x) \left (a+b x+c x^2\right )^m \, dx &=\operatorname{Subst}\left (\int e^x x^m \, dx,x,a+b x+c x^2\right )\\ &=\left (-a-b x-c x^2\right )^{-m} \left (a+b x+c x^2\right )^m \Gamma \left (1+m,-a-b x-c x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0523048, size = 44, normalized size = 0.9 \[ (-a-x (b+c x))^{-m} (a+x (b+c x))^m \text{Gamma}(m+1,-a-x (b+c x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.413, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{c{x}^{2}+bx+a}} \left ( 2\,cx+b \right ) \left ( c{x}^{2}+bx+a \right ) ^{m}\, dx \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)^m,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (2 \, c x + b\right )}{\left (c x^{2} + b x + a\right )}^{m} 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)^m,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.732206, size = 57, normalized size = 1.16 \begin{align*} \cos \left (\pi m\right ) \Gamma \left (m + 1, -c x^{2} - b x - a\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)^m,x, algorithm="fricas")

[Out]

cos(pi*m)*gamma(m + 1, -c*x^2 - b*x - a)

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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)**m,x)

[Out]

Timed out

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

[Out]

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