3.209 \(\int (a+b x) (1+(c+a x+\frac{b x^2}{2})^n) \, dx\)

Optimal. Leaf size=35 \[ \frac{\left (a x+\frac{b x^2}{2}+c\right )^{n+1}}{n+1}+a x+\frac{b x^2}{2} \]

[Out]

a*x + (b*x^2)/2 + (c + a*x + (b*x^2)/2)^(1 + n)/(1 + n)

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Rubi [A]  time = 0.0090755, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {1591} \[ \frac{\left (a x+\frac{b x^2}{2}+c\right )^{n+1}}{n+1}+a x+\frac{b x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*(1 + (c + a*x + (b*x^2)/2)^n),x]

[Out]

a*x + (b*x^2)/2 + (c + a*x + (b*x^2)/2)^(1 + n)/(1 + n)

Rule 1591

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

Rubi steps

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

Mathematica [A]  time = 0.0540426, size = 35, normalized size = 1. \[ \frac{\left (a x+\frac{b x^2}{2}+c\right )^{n+1}}{n+1}+a x+\frac{b x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*(1 + (c + a*x + (b*x^2)/2)^n),x]

[Out]

a*x + (b*x^2)/2 + (c + a*x + (b*x^2)/2)^(1 + n)/(1 + n)

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Maple [A]  time = 0.003, size = 33, normalized size = 0.9 \begin{align*} c+ax+{\frac{b{x}^{2}}{2}}+{\frac{1}{1+n} \left ( c+ax+{\frac{b{x}^{2}}{2}} \right ) ^{1+n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c+a*x+1/2*b*x^2+(c+a*x+1/2*b*x^2)^(1+n)/(1+n)

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Maxima [A]  time = 1.66477, size = 73, normalized size = 2.09 \begin{align*} \frac{1}{2} \, b x^{2} + a x + \frac{{\left (b x^{2} + 2 \, a x + 2 \, c\right )}{\left (b x^{2} + 2 \, a x + 2 \, c\right )}^{n}}{2^{n + 1} n + 2^{n + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*x^2 + a*x + (b*x^2 + 2*a*x + 2*c)*(b*x^2 + 2*a*x + 2*c)^n/(2^(n + 1)*n + 2^(n + 1))

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Fricas [A]  time = 1.37815, size = 126, normalized size = 3.6 \begin{align*} \frac{{\left (b n + b\right )} x^{2} +{\left (b x^{2} + 2 \, a x + 2 \, c\right )}{\left (\frac{1}{2} \, b x^{2} + a x + c\right )}^{n} + 2 \,{\left (a n + a\right )} x}{2 \,{\left (n + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*n + b)*x^2 + (b*x^2 + 2*a*x + 2*c)*(1/2*b*x^2 + a*x + c)^n + 2*(a*n + a)*x)/(n + 1)

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

[Out]

Timed out

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Giac [B]  time = 1.21129, size = 107, normalized size = 3.06 \begin{align*} \frac{{\left (\frac{1}{2} \, b x^{2} + a x + c\right )}^{n} b x^{2} + b n x^{2} + 2 \,{\left (\frac{1}{2} \, b x^{2} + a x + c\right )}^{n} a x + 2 \, a n x + b x^{2} + 2 \,{\left (\frac{1}{2} \, b x^{2} + a x + c\right )}^{n} c + 2 \, a x}{2 \,{\left (n + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((1/2*b*x^2 + a*x + c)^n*b*x^2 + b*n*x^2 + 2*(1/2*b*x^2 + a*x + c)^n*a*x + 2*a*n*x + b*x^2 + 2*(1/2*b*x^2
+ a*x + c)^n*c + 2*a*x)/(n + 1)