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

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.06, size = 35, normalized size = 1.00 \[ \frac {\left (a x+\frac {b x^2}{2}+c\right )^{n+1}}{n+1}+a x+\frac {b x^2}{2} \]

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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fricas [A]  time = 0.95, size = 52, normalized size = 1.49 \[ \frac {{\left (b n + b\right )} x^{2} + {\left (b x^{2} + 2 \, a x + 2 \, c\right )} {\left (\frac {1}{2} \, b x^{2} + a x + c\right )}^{n} + 2 \, {\left (a n + a\right )} x}{2 \, {\left (n + 1\right )}} \]

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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giac [A]  time = 0.25, size = 32, normalized size = 0.91 \[ \frac {1}{2} \, b x^{2} + a x + c + \frac {{\left (\frac {1}{2} \, b x^{2} + a x + c\right )}^{n + 1}}{n + 1} \]

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

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maple [A]  time = 0.00, size = 33, normalized size = 0.94 \[ \frac {b \,x^{2}}{2}+a x +c +\frac {\left (\frac {1}{2} b \,x^{2}+a x +c \right )^{n +1}}{n +1} \]

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)^(n+1)/(n+1)

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maxima [A]  time = 1.16, size = 54, normalized size = 1.54 \[ \frac {1}{2} \, b x^{2} + a x + \frac {{\left (b x^{2} + 2 \, a x + 2 \, c\right )} {\left (b x^{2} + 2 \, a x + 2 \, c\right )}^{n}}{2^{n + 1} n + 2^{n + 1}} \]

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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mupad [B]  time = 2.11, size = 58, normalized size = 1.66 \[ a\,x+{\left (\frac {b\,x^2}{2}+a\,x+c\right )}^n\,\left (\frac {2\,c}{2\,n+2}+\frac {b\,x^2}{2\,n+2}+\frac {2\,a\,x}{2\,n+2}\right )+\frac {b\,x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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