\(\int (2 c x+3 d x^2) (c x^2+d x^3)^n \, dx\) [187]

Optimal result

Integrand size = 25, antiderivative size = 21 \[ \int \left (2 c x+3 d x^2\right ) \left (c x^2+d x^3\right )^n \, dx=\frac {\left (c x^2+d x^3\right )^{1+n}}{1+n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {1602} \[ \int \left (2 c x+3 d x^2\right ) \left (c x^2+d x^3\right )^n \, dx=\frac {\left (c x^2+d x^3\right )^{n+1}}{n+1} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \left (2 c x+3 d x^2\right ) \left (c x^2+d x^3\right )^n \, dx=\frac {\left (x^2 (c+d x)\right )^{1+n}}{1+n} \]

[In]

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

[Out]

(x^2*(c + d*x))^(1 + n)/(1 + n)

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\left (x^{3} d +c \,x^{2}\right )^{1+n}}{1+n}\)	\(22\)
default	\(\frac {\left (x^{3} d +c \,x^{2}\right )^{1+n}}{1+n}\)	\(22\)
risch	\(\frac {x^{2} \left (d x +c \right ) \left (x^{2} \left (d x +c \right )\right )^{n}}{1+n}\)	\(26\)
gosper	\(\frac {\left (x^{3} d +c \,x^{2}\right )^{n} x^{2} \left (d x +c \right )}{1+n}\)	\(28\)
parallelrisch	\(\frac {x^{3} \left (x^{2} \left (d x +c \right )\right )^{n} c d +x^{2} \left (x^{2} \left (d x +c \right )\right )^{n} c^{2}}{c \left (1+n \right )}\)	\(46\)
norman	\(\frac {c \,x^{2} {\mathrm e}^{n \ln \left (x^{3} d +c \,x^{2}\right )}}{1+n}+\frac {d \,x^{3} {\mathrm e}^{n \ln \left (x^{3} d +c \,x^{2}\right )}}{1+n}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \left (2 c x+3 d x^2\right ) \left (c x^2+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + c x^{2}\right )} {\left (d x^{3} + c x^{2}\right )}^{n}}{n + 1} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (15) = 30\).

Time = 0.43 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.52 \[ \int \left (2 c x+3 d x^2\right ) \left (c x^2+d x^3\right )^n \, dx=\begin {cases} \frac {c x^{2} \left (c x^{2} + d x^{3}\right )^{n}}{n + 1} + \frac {d x^{3} \left (c x^{2} + d x^{3}\right )^{n}}{n + 1} & \text {for}\: n \neq -1 \\2 \log {\left (x \right )} + \log {\left (\frac {c}{d} + x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((c*x**2*(c*x**2 + d*x**3)**n/(n + 1) + d*x**3*(c*x**2 + d*x**3)**n/(n + 1), Ne(n, -1)), (2*log(x) +
log(c/d + x), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \left (2 c x+3 d x^2\right ) \left (c x^2+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + c x^{2}\right )}^{n + 1}}{n + 1} \]

[In]

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

[Out]

(d*x^3 + c*x^2)^(n + 1)/(n + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \left (2 c x+3 d x^2\right ) \left (c x^2+d x^3\right )^n \, dx=\frac {{\left (d x^{3} + c x^{2}\right )}^{n + 1}}{n + 1} \]

[In]

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

[Out]

(d*x^3 + c*x^2)^(n + 1)/(n + 1)

Mupad [B] (verification not implemented)

Time = 9.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \left (2 c x+3 d x^2\right ) \left (c x^2+d x^3\right )^n \, dx=\frac {x^2\,{\left (d\,x^3+c\,x^2\right )}^n\,\left (c+d\,x\right )}{n+1} \]

[In]

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

[Out]

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