\(\int (b+2 c x+3 d x^2) (a+b x+c x^2+d x^3)^n \, dx\) [180]

Optimal result

Integrand size = 30, antiderivative size = 25 \[ \int \left (b+2 c x+3 d x^2\right ) \left (a+b x+c x^2+d x^3\right )^n \, dx=\frac {\left (a+b x+c x^2+d x^3\right )^{1+n}}{1+n} \]

[Out]

(d*x^3+c*x^2+b*x+a)^(1+n)/(1+n)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {1602} \[ \int \left (b+2 c x+3 d x^2\right ) \left (a+b x+c x^2+d x^3\right )^n \, dx=\frac {\left (a+b x+c x^2+d x^3\right )^{n+1}}{n+1} \]

[In]

Int[(b + 2*c*x + 3*d*x^2)*(a + b*x + c*x^2 + d*x^3)^n,x]

[Out]

(a + b*x + c*x^2 + d*x^3)^(1 + n)/(1 + n)

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+b x+c x^2+d x^3\right )^{1+n}}{1+n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \left (b+2 c x+3 d x^2\right ) \left (a+b x+c x^2+d x^3\right )^n \, dx=\frac {(a+x (b+x (c+d x)))^{1+n}}{1+n} \]

[In]

Integrate[(b + 2*c*x + 3*d*x^2)*(a + b*x + c*x^2 + d*x^3)^n,x]

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04

method	result	size
gosper	\(\frac {\left (x^{3} d +c \,x^{2}+b x +a \right )^{1+n}}{1+n}\)	\(26\)
derivativedivides	\(\frac {\left (x^{3} d +c \,x^{2}+b x +a \right )^{1+n}}{1+n}\)	\(26\)
default	\(\frac {\left (x^{3} d +c \,x^{2}+b x +a \right )^{1+n}}{1+n}\)	\(26\)
risch	\(\frac {\left (x^{3} d +c \,x^{2}+b x +a \right ) \left (x^{3} d +c \,x^{2}+b x +a \right )^{n}}{1+n}\)	\(39\)
parallelrisch	\(\frac {x^{3} \left (x^{3} d +c \,x^{2}+b x +a \right )^{n} c d +x^{2} \left (x^{3} d +c \,x^{2}+b x +a \right )^{n} c^{2}+x \left (x^{3} d +c \,x^{2}+b x +a \right )^{n} b c +\left (x^{3} d +c \,x^{2}+b x +a \right )^{n} a c}{c \left (1+n \right )}\)	\(99\)
norman	\(\frac {a \,{\mathrm e}^{n \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}}{1+n}+\frac {b x \,{\mathrm e}^{n \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}}{1+n}+\frac {c \,x^{2} {\mathrm e}^{n \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}}{1+n}+\frac {d \,x^{3} {\mathrm e}^{n \ln \left (x^{3} d +c \,x^{2}+b x +a \right )}}{1+n}\)	\(113\)

[In]

int((3*d*x^2+2*c*x+b)*(d*x^3+c*x^2+b*x+a)^n,x,method=_RETURNVERBOSE)

[Out]

(d*x^3+c*x^2+b*x+a)^(1+n)/(1+n)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \left (b+2 c x+3 d x^2\right ) \left (a+b x+c x^2+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + c x^{2} + b x + a\right )} {\left (d x^{3} + c x^{2} + b x + a\right )}^{n}}{n + 1} \]

[In]

integrate((3*d*x^2+2*c*x+b)*(d*x^3+c*x^2+b*x+a)^n,x, algorithm="fricas")

[Out]

(d*x^3 + c*x^2 + b*x + a)*(d*x^3 + c*x^2 + b*x + a)^n/(n + 1)

Sympy [F(-1)]

Timed out. \[ \int \left (b+2 c x+3 d x^2\right ) \left (a+b x+c x^2+d x^3\right )^n \, dx=\text {Timed out} \]

[In]

integrate((3*d*x**2+2*c*x+b)*(d*x**3+c*x**2+b*x+a)**n,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \left (b+2 c x+3 d x^2\right ) \left (a+b x+c x^2+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + c x^{2} + b x + a\right )}^{n + 1}}{n + 1} \]

[In]

integrate((3*d*x^2+2*c*x+b)*(d*x^3+c*x^2+b*x+a)^n,x, algorithm="maxima")

[Out]

(d*x^3 + c*x^2 + b*x + a)^(n + 1)/(n + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \left (b+2 c x+3 d x^2\right ) \left (a+b x+c x^2+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + c x^{2} + b x + a\right )}^{n + 1}}{n + 1} \]

[In]

integrate((3*d*x^2+2*c*x+b)*(d*x^3+c*x^2+b*x+a)^n,x, algorithm="giac")

[Out]

(d*x^3 + c*x^2 + b*x + a)^(n + 1)/(n + 1)

Mupad [B] (verification not implemented)

Time = 9.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \left (b+2 c x+3 d x^2\right ) \left (a+b x+c x^2+d x^3\right )^n \, dx={\left (d\,x^3+c\,x^2+b\,x+a\right )}^n\,\left (\frac {a}{n+1}+\frac {b\,x}{n+1}+\frac {c\,x^2}{n+1}+\frac {d\,x^3}{n+1}\right ) \]

[In]

int((b + 2*c*x + 3*d*x^2)*(a + b*x + c*x^2 + d*x^3)^n,x)

[Out]

(a + b*x + c*x^2 + d*x^3)^n*(a/(n + 1) + (b*x)/(n + 1) + (c*x^2)/(n + 1) + (d*x^3)/(n + 1))