\(\int (b+2 c x+3 d x^2) (b x+c x^2+d x^3)^n \, dx\) [181]

Optimal result

Integrand size = 29, antiderivative size = 24 \[ \int \left (b+2 c x+3 d x^2\right ) \left (b x+c x^2+d x^3\right )^n \, dx=\frac {\left (b x+c x^2+d x^3\right )^{1+n}}{1+n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {1602} \[ \int \left (b+2 c x+3 d x^2\right ) \left (b x+c x^2+d x^3\right )^n \, dx=\frac {\left (b x+c x^2+d x^3\right )^{n+1}}{n+1} \]

[In]

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

[Out]

(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 (b x+c x^2+d x^3\right )^{1+n}}{1+n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \left (b+2 c x+3 d x^2\right ) \left (b x+c x^2+d x^3\right )^n \, dx=\frac {(x (b+x (c+d x)))^{1+n}}{1+n} \]

[In]

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

[Out]

(x*(b + x*(c + d*x)))^(1 + n)/(1 + n)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {\left (x^{3} d +c \,x^{2}+b x \right )^{1+n}}{1+n}\)	\(25\)
default	\(\frac {\left (x^{3} d +c \,x^{2}+b x \right )^{1+n}}{1+n}\)	\(25\)
risch	\(\frac {x \left (d \,x^{2}+c x +b \right ) {\left (x \left (d \,x^{2}+c x +b \right )\right )}^{n}}{1+n}\)	\(32\)
gosper	\(\frac {x \left (d \,x^{2}+c x +b \right ) \left (x^{3} d +c \,x^{2}+b x \right )^{n}}{1+n}\)	\(34\)
parallelrisch	\(\frac {x^{3} {\left (x \left (d \,x^{2}+c x +b \right )\right )}^{n} d^{2}+x^{2} {\left (x \left (d \,x^{2}+c x +b \right )\right )}^{n} c d +x {\left (x \left (d \,x^{2}+c x +b \right )\right )}^{n} b d}{d \left (1+n \right )}\)	\(70\)
norman	\(\frac {b x \,{\mathrm e}^{n \ln \left (x^{3} d +c \,x^{2}+b x \right )}}{1+n}+\frac {c \,x^{2} {\mathrm e}^{n \ln \left (x^{3} d +c \,x^{2}+b x \right )}}{1+n}+\frac {d \,x^{3} {\mathrm e}^{n \ln \left (x^{3} d +c \,x^{2}+b x \right )}}{1+n}\)	\(84\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \left (b+2 c x+3 d x^2\right ) \left (b x+c x^2+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + c x^{2} + b x\right )} {\left (d x^{3} + c x^{2} + b x\right )}^{n}}{n + 1} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \left (b+2 c x+3 d x^2\right ) \left (b x+c x^2+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + c x^{2} + b x\right )}^{n + 1}}{n + 1} \]

[In]

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

[Out]

(d*x^3 + c*x^2 + b*x)^(n + 1)/(n + 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \left (b+2 c x+3 d x^2\right ) \left (b x+c x^2+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + c x^{2} + b x\right )}^{n + 1}}{n + 1} \]

[In]

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

[Out]

(d*x^3 + c*x^2 + b*x)^(n + 1)/(n + 1)

Mupad [B] (verification not implemented)

Time = 9.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \left (b+2 c x+3 d x^2\right ) \left (b x+c x^2+d x^3\right )^n \, dx=\left (\frac {b\,x}{n+1}+\frac {c\,x^2}{n+1}+\frac {d\,x^3}{n+1}\right )\,{\left (d\,x^3+c\,x^2+b\,x\right )}^n \]

[In]

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

[Out]

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