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

Optimal. Leaf size=34 \[ \frac{\left (a x+\frac{b x^2}{2}\right )^{n+1}}{n+1}+a x+\frac{b x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.0084399, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {1591} \[ \frac{\left (a x+\frac{b x^2}{2}\right )^{n+1}}{n+1}+a x+\frac{b x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0547979, size = 34, normalized size = 1. \[ \frac{x (2 a+b x) \left (\left (a x+\frac{b x^2}{2}\right )^n+n+1\right )}{2 (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 31, normalized size = 0.9 \begin{align*} ax+{\frac{b{x}^{2}}{2}}+{\frac{1}{1+n} \left ( 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+(a*x+1/2*b*x^2)^n),x)

[Out]

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

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Maxima [A]  time = 1.69603, size = 70, normalized size = 2.06 \begin{align*} \frac{1}{2} \, b x^{2} + a x + \frac{{\left (b x^{2} + 2 \, a x\right )} e^{\left (n \log \left (b x + 2 \, a\right ) + n \log \left (x\right )\right )}}{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+(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)*e^(n*log(b*x + 2*a) + n*log(x))/(2^(n + 1)*n + 2^(n + 1))

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Fricas [A]  time = 1.39302, size = 112, normalized size = 3.29 \begin{align*} \frac{{\left (b n + b\right )} x^{2} +{\left (b x^{2} + 2 \, a x\right )}{\left (\frac{1}{2} \, b x^{2} + a x\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+(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)*(1/2*b*x^2 + a*x)^n + 2*(a*n + a)*x)/(n + 1)

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Sympy [A]  time = 87.1078, size = 230, normalized size = 6.76 \begin{align*} \begin{cases} a \left (x + \frac{\log{\left (x \right )}}{a}\right ) & \text{for}\: b = 0 \wedge n = -1 \\a \left (\frac{a^{n} x x^{n}}{n + 1} + \frac{n x}{n + 1} + \frac{x}{n + 1}\right ) & \text{for}\: b = 0 \\a x + \frac{b x^{2}}{2} + \log{\left (x \right )} + \log{\left (\frac{2 a}{b} + x \right )} & \text{for}\: n = -1 \\\frac{2 \cdot 2^{n} a b n x}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac{2 \cdot 2^{n} a b x}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac{2^{n} b^{2} n x^{2}}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac{2^{n} b^{2} x^{2}}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac{2 a b x \left (2 a x + b x^{2}\right )^{n}}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac{b^{2} x^{2} \left (2 a x + b x^{2}\right )^{n}}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*(x + log(x)/a), Eq(b, 0) & Eq(n, -1)), (a*(a**n*x*x**n/(n + 1) + n*x/(n + 1) + x/(n + 1)), Eq(b,
0)), (a*x + b*x**2/2 + log(x) + log(2*a/b + x), Eq(n, -1)), (2*2**n*a*b*n*x/(2*2**n*b*n + 2*2**n*b) + 2*2**n*a
*b*x/(2*2**n*b*n + 2*2**n*b) + 2**n*b**2*n*x**2/(2*2**n*b*n + 2*2**n*b) + 2**n*b**2*x**2/(2*2**n*b*n + 2*2**n*
b) + 2*a*b*x*(2*a*x + b*x**2)**n/(2*2**n*b*n + 2*2**n*b) + b**2*x**2*(2*a*x + b*x**2)**n/(2*2**n*b*n + 2*2**n*
b), True))

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Giac [A]  time = 1.12755, size = 41, normalized size = 1.21 \begin{align*} \frac{1}{2} \, b x^{2} + a x + \frac{{\left (\frac{1}{2} \, b x^{2} + a x\right )}^{n + 1}}{n + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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