\(\int (d x)^m (b x+c x^2) \, dx\) [114]

Optimal result

Integrand size = 15, antiderivative size = 35 \[ \int (d x)^m \left (b x+c x^2\right ) \, dx=\frac {b (d x)^{2+m}}{d^2 (2+m)}+\frac {c (d x)^{3+m}}{d^3 (3+m)} \]

[Out]

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

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14} \[ \int (d x)^m \left (b x+c x^2\right ) \, dx=\frac {b (d x)^{m+2}}{d^2 (m+2)}+\frac {c (d x)^{m+3}}{d^3 (m+3)} \]

[In]

Int[(d*x)^m*(b*x + c*x^2),x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int (d x)^m \left (b x+c x^2\right ) \, dx=x^2 (d x)^m \left (\frac {b}{2+m}+\frac {c x}{3+m}\right ) \]

[In]

Integrate[(d*x)^m*(b*x + c*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00

method	result	size
gosper	\(\frac {\left (d x \right )^{m} \left (c m x +b m +2 c x +3 b \right ) x^{2}}{\left (3+m \right ) \left (2+m \right )}\)	\(35\)
risch	\(\frac {\left (d x \right )^{m} \left (c m x +b m +2 c x +3 b \right ) x^{2}}{\left (3+m \right ) \left (2+m \right )}\)	\(35\)
norman	\(\frac {b \,x^{2} {\mathrm e}^{m \ln \left (d x \right )}}{2+m}+\frac {c \,x^{3} {\mathrm e}^{m \ln \left (d x \right )}}{3+m}\)	\(36\)
parallelrisch	\(\frac {x^{3} \left (d x \right )^{m} c m +2 x^{3} \left (d x \right )^{m} c +\left (d x \right )^{m} b \,x^{2} m +3 \left (d x \right )^{m} b \,x^{2}}{\left (3+m \right ) \left (2+m \right )}\)	\(57\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int (d x)^m \left (b x+c x^2\right ) \, dx=\frac {{\left ({\left (c m + 2 \, c\right )} x^{3} + {\left (b m + 3 \, b\right )} x^{2}\right )} \left (d x\right )^{m}}{m^{2} + 5 \, m + 6} \]

[In]

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

[Out]

((c*m + 2*c)*x^3 + (b*m + 3*b)*x^2)*(d*x)^m/(m^2 + 5*m + 6)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (29) = 58\).

Time = 0.22 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.00 \[ \int (d x)^m \left (b x+c x^2\right ) \, dx=\begin {cases} \frac {- \frac {b}{x} + c \log {\left (x \right )}}{d^{3}} & \text {for}\: m = -3 \\\frac {b \log {\left (x \right )} + c x}{d^{2}} & \text {for}\: m = -2 \\\frac {b m x^{2} \left (d x\right )^{m}}{m^{2} + 5 m + 6} + \frac {3 b x^{2} \left (d x\right )^{m}}{m^{2} + 5 m + 6} + \frac {c m x^{3} \left (d x\right )^{m}}{m^{2} + 5 m + 6} + \frac {2 c x^{3} \left (d x\right )^{m}}{m^{2} + 5 m + 6} & \text {otherwise} \end {cases} \]

[In]

integrate((d*x)**m*(c*x**2+b*x),x)

[Out]

Piecewise(((-b/x + c*log(x))/d**3, Eq(m, -3)), ((b*log(x) + c*x)/d**2, Eq(m, -2)), (b*m*x**2*(d*x)**m/(m**2 +
5*m + 6) + 3*b*x**2*(d*x)**m/(m**2 + 5*m + 6) + c*m*x**3*(d*x)**m/(m**2 + 5*m + 6) + 2*c*x**3*(d*x)**m/(m**2 +
5*m + 6), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int (d x)^m \left (b x+c x^2\right ) \, dx=\frac {c d^{m} x^{3} x^{m}}{m + 3} + \frac {b d^{m} x^{2} x^{m}}{m + 2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.60 \[ \int (d x)^m \left (b x+c x^2\right ) \, dx=\frac {\left (d x\right )^{m} c m x^{3} + \left (d x\right )^{m} b m x^{2} + 2 \, \left (d x\right )^{m} c x^{3} + 3 \, \left (d x\right )^{m} b x^{2}}{m^{2} + 5 \, m + 6} \]

[In]

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

[Out]

((d*x)^m*c*m*x^3 + (d*x)^m*b*m*x^2 + 2*(d*x)^m*c*x^3 + 3*(d*x)^m*b*x^2)/(m^2 + 5*m + 6)

Mupad [B] (verification not implemented)

Time = 9.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int (d x)^m \left (b x+c x^2\right ) \, dx=\frac {x^2\,{\left (d\,x\right )}^m\,\left (3\,b+b\,m+2\,c\,x+c\,m\,x\right )}{m^2+5\,m+6} \]

[In]

int((b*x + c*x^2)*(d*x)^m,x)

[Out]

(x^2*(d*x)^m*(3*b + b*m + 2*c*x + c*m*x))/(5*m + m^2 + 6)