\(\int (a+b x) (1+(a x+\frac {b x^2}{2})^n) \, dx\) [208]

Optimal result

Integrand size = 22, antiderivative size = 34 \[ \int (a+b x) \left (1+\left (a x+\frac {b x^2}{2}\right )^n\right ) \, dx=a x+\frac {b x^2}{2}+\frac {\left (a x+\frac {b x^2}{2}\right )^{1+n}}{1+n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {1605} \[ \int (a+b x) \left (1+\left (a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {\left (a x+\frac {b x^2}{2}\right )^{n+1}}{n+1}+a x+\frac {b x^2}{2} \]

[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 1605

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*} \text {integral}& = \text {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] (verified)

Time = 0.34 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int (a+b x) \left (1+\left (a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {x (2 a+b x) \left (1+n+\left (a x+\frac {b x^2}{2}\right )^n\right )}{2 (1+n)} \]

[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))

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(a x +\frac {b \,x^{2}}{2}+\frac {\left (a x +\frac {1}{2} b \,x^{2}\right )^{1+n}}{1+n}\)	\(31\)
default	\(a x +\frac {b \,x^{2}}{2}+\frac {\left (a x +\frac {1}{2} b \,x^{2}\right )^{1+n}}{1+n}\)	\(31\)
risch	\(a x +\frac {b \,x^{2}}{2}+\frac {x \left (b x +2 a \right ) \left (x \left (b x +2 a \right )\right )^{n} \left (\frac {1}{2}\right )^{n}}{2+2 n}\)	\(40\)
norman	\(a x +\frac {a x \,{\mathrm e}^{n \ln \left (a x +\frac {1}{2} b \,x^{2}\right )}}{1+n}+\frac {b \,x^{2}}{2}+\frac {b \,x^{2} {\mathrm e}^{n \ln \left (a x +\frac {1}{2} b \,x^{2}\right )}}{2+2 n}\)	\(58\)
parallelrisch	\(\frac {x^{2} \left (\frac {x \left (b x +2 a \right )}{2}\right )^{n} b^{2}+x^{2} b^{2} n +b^{2} x^{2}+2 x \left (\frac {x \left (b x +2 a \right )}{2}\right )^{n} a b +2 a b n x +2 a b x -4 a^{2} n -4 a^{2}}{2 b \left (1+n \right )}\)	\(85\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int (a+b x) \left (1+\left (a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {{\left (b n + b\right )} x^{2} + {\left (b x^{2} + 2 \, a x\right )} {\left (\frac {1}{2} \, b x^{2} + a x\right )}^{n} + 2 \, {\left (a n + a\right )} x}{2 \, {\left (n + 1\right )}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (26) = 52\).

Time = 18.29 (sec) , antiderivative size = 228, normalized size of antiderivative = 6.71 \[ \int (a+b x) \left (1+\left (a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\begin {cases} a \left (x + \frac {\log {\left (x \right )}}{a}\right ) & \text {for}\: b = 0 \wedge n = -1 \\a \left (\frac {n x}{n + 1} + \frac {x \left (a x\right )^{n}}{n + 1} + \frac {x}{n + 1}\right ) & \text {for}\: b = 0 \\a x + \frac {b x^{2}}{2} + \log {\left (x \right )} + \log {\left (\frac {2 a}{b} + x \right )} & \text {for}\: n = -1 \\\frac {2 \cdot 2^{n} a b n x}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac {2 \cdot 2^{n} a b x}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac {2^{n} b^{2} n x^{2}}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac {2^{n} b^{2} x^{2}}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac {2 a b x \left (2 a x + b x^{2}\right )^{n}}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} + \frac {b^{2} x^{2} \left (2 a x + b x^{2}\right )^{n}}{2 \cdot 2^{n} b n + 2 \cdot 2^{n} b} & \text {otherwise} \end {cases} \]

[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*(n*x/(n + 1) + x*(a*x)**n/(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))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int (a+b x) \left (1+\left (a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {1}{2} \, b x^{2} + a x + \frac {{\left (b x^{2} + 2 \, a x\right )} e^{\left (n \log \left (b x + 2 \, a\right ) + n \log \left (x\right )\right )}}{2^{n + 1} n + 2^{n + 1}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int (a+b x) \left (1+\left (a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {1}{2} \, b x^{2} + a x + \frac {{\left (\frac {1}{2} \, b x^{2} + a x\right )}^{n + 1}}{n + 1} \]

[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)

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int (a+b x) \left (1+\left (a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {x\,\left (2\,a+b\,x\right )\,\left (n+{\left (\frac {b\,x^2}{2}+a\,x\right )}^n+1\right )}{2\,\left (n+1\right )} \]

[In]

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

[Out]

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