\(\int (a+b x) (1+(c+a x+\frac {b x^2}{2})^n) \, dx\) [209]

Optimal result

Integrand size = 23, antiderivative size = 35 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^n\right ) \, dx=a x+\frac {b x^2}{2}+\frac {\left (c+a x+\frac {b x^2}{2}\right )^{1+n}}{1+n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1605} \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {\left (a x+\frac {b x^2}{2}+c\right )^{n+1}}{n+1}+a x+\frac {b x^2}{2} \]

[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 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,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 [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(73\) vs. \(2(35)=70\).

Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.09 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {2 c \left (c+a x+\frac {b x^2}{2}\right )^n+2 a x \left (1+n+\left (c+a x+\frac {b x^2}{2}\right )^n\right )+b x^2 \left (1+n+\left (c+a x+\frac {b x^2}{2}\right )^n\right )}{2 (1+n)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(c +a x +\frac {b \,x^{2}}{2}+\frac {\left (c +a x +\frac {1}{2} b \,x^{2}\right )^{1+n}}{1+n}\)	\(33\)
default	\(c +a x +\frac {b \,x^{2}}{2}+\frac {\left (c +a x +\frac {1}{2} b \,x^{2}\right )^{1+n}}{1+n}\)	\(33\)
risch	\(a x +\frac {b \,x^{2}}{2}+\frac {\left (b \,x^{2}+2 a x +2 c \right ) \left (b \,x^{2}+2 a x +2 c \right )^{n} \left (\frac {1}{2}\right )^{n}}{2+2 n}\)	\(49\)
norman	\(a x +\frac {c \,{\mathrm e}^{n \ln \left (c +a x +\frac {1}{2} b \,x^{2}\right )}}{1+n}+\frac {a x \,{\mathrm e}^{n \ln \left (c +a x +\frac {1}{2} b \,x^{2}\right )}}{1+n}+\frac {b \,x^{2}}{2}+\frac {b \,x^{2} {\mathrm e}^{n \ln \left (c +a x +\frac {1}{2} b \,x^{2}\right )}}{2+2 n}\)	\(82\)
parallelrisch	\(\frac {\left (c +a x +\frac {1}{2} b \,x^{2}\right )^{n} b^{2} x^{2}+x^{2} b^{2} n +b^{2} x^{2}+2 \left (c +a x +\frac {1}{2} b \,x^{2}\right )^{n} a b x +2 a b n x +2 a b x +2 \left (c +a x +\frac {1}{2} b \,x^{2}\right )^{n} b c -4 a^{2} n -2 b c n -4 a^{2}-2 b c}{2 b \left (1+n \right )}\)	\(113\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {{\left (b n + b\right )} x^{2} + {\left (b x^{2} + 2 \, a x + 2 \, c\right )} {\left (\frac {1}{2} \, b x^{2} + a x + c\right )}^{n} + 2 \, {\left (a n + a\right )} x}{2 \, {\left (n + 1\right )}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.54 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {1}{2} \, b x^{2} + a x + \frac {{\left (b x^{2} + 2 \, a x + 2 \, c\right )} {\left (b x^{2} + 2 \, a x + 2 \, c\right )}^{n}}{2^{n + 1} n + 2^{n + 1}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^n\right ) \, dx=\frac {1}{2} \, b x^{2} + a x + c + \frac {{\left (\frac {1}{2} \, b x^{2} + a x + c\right )}^{n + 1}}{n + 1} \]

[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)

Mupad [B] (verification not implemented)

Time = 9.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.66 \[ \int (a+b x) \left (1+\left (c+a x+\frac {b x^2}{2}\right )^n\right ) \, dx=a\,x+{\left (\frac {b\,x^2}{2}+a\,x+c\right )}^n\,\left (\frac {2\,c}{2\,n+2}+\frac {b\,x^2}{2\,n+2}+\frac {2\,a\,x}{2\,n+2}\right )+\frac {b\,x^2}{2} \]

[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