\(\int x^m (a x+b x^3+c x^5) \, dx\) [65]

Optimal result

Integrand size = 18, antiderivative size = 37 \[ \int x^m \left (a x+b x^3+c x^5\right ) \, dx=\frac {a x^{2+m}}{2+m}+\frac {b x^{4+m}}{4+m}+\frac {c x^{6+m}}{6+m} \]

[Out]

a*x^(2+m)/(2+m)+b*x^(4+m)/(4+m)+c*x^(6+m)/(6+m)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14} \[ \int x^m \left (a x+b x^3+c x^5\right ) \, dx=\frac {a x^{m+2}}{m+2}+\frac {b x^{m+4}}{m+4}+\frac {c x^{m+6}}{m+6} \]

[In]

Int[x^m*(a*x + b*x^3 + c*x^5),x]

[Out]

(a*x^(2 + m))/(2 + m) + (b*x^(4 + m))/(4 + m) + (c*x^(6 + m))/(6 + 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 (a x^{1+m}+b x^{3+m}+c x^{5+m}\right ) \, dx \\ & = \frac {a x^{2+m}}{2+m}+\frac {b x^{4+m}}{4+m}+\frac {c x^{6+m}}{6+m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int x^m \left (a x+b x^3+c x^5\right ) \, dx=x^{2+m} \left (\frac {a}{2+m}+\frac {b x^2}{4+m}+\frac {c x^4}{6+m}\right ) \]

[In]

Integrate[x^m*(a*x + b*x^3 + c*x^5),x]

[Out]

x^(2 + m)*(a/(2 + m) + (b*x^2)/(4 + m) + (c*x^4)/(6 + m))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27

method	result	size
norman	\(\frac {a \,x^{2} {\mathrm e}^{m \ln \left (x \right )}}{2+m}+\frac {b \,x^{4} {\mathrm e}^{m \ln \left (x \right )}}{4+m}+\frac {c \,x^{6} {\mathrm e}^{m \ln \left (x \right )}}{6+m}\)	\(47\)
gosper	\(\frac {x^{2+m} \left (c \,m^{2} x^{4}+6 c m \,x^{4}+b \,m^{2} x^{2}+8 c \,x^{4}+8 b m \,x^{2}+a \,m^{2}+12 b \,x^{2}+10 a m +24 a \right )}{\left (2+m \right ) \left (4+m \right ) \left (6+m \right )}\)	\(77\)
risch	\(\frac {x^{m} \left (c \,m^{2} x^{4}+6 c m \,x^{4}+b \,m^{2} x^{2}+8 c \,x^{4}+8 b m \,x^{2}+a \,m^{2}+12 b \,x^{2}+10 a m +24 a \right ) x^{2}}{\left (6+m \right ) \left (4+m \right ) \left (2+m \right )}\)	\(78\)
parallelrisch	\(\frac {x^{6} x^{m} c \,m^{2}+6 x^{6} x^{m} c m +8 x^{6} x^{m} c +x^{4} x^{m} b \,m^{2}+8 x^{4} x^{m} b m +12 x^{4} x^{m} b +x^{2} x^{m} a \,m^{2}+10 x^{2} x^{m} a m +24 x^{2} x^{m} a}{\left (6+m \right ) \left (4+m \right ) \left (2+m \right )}\)	\(108\)

[In]

int(x^m*(c*x^5+b*x^3+a*x),x,method=_RETURNVERBOSE)

[Out]

a/(2+m)*x^2*exp(m*ln(x))+b/(4+m)*x^4*exp(m*ln(x))+c/(6+m)*x^6*exp(m*ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.92 \[ \int x^m \left (a x+b x^3+c x^5\right ) \, dx=\frac {{\left ({\left (c m^{2} + 6 \, c m + 8 \, c\right )} x^{6} + {\left (b m^{2} + 8 \, b m + 12 \, b\right )} x^{4} + {\left (a m^{2} + 10 \, a m + 24 \, a\right )} x^{2}\right )} x^{m}}{m^{3} + 12 \, m^{2} + 44 \, m + 48} \]

[In]

integrate(x^m*(c*x^5+b*x^3+a*x),x, algorithm="fricas")

[Out]

((c*m^2 + 6*c*m + 8*c)*x^6 + (b*m^2 + 8*b*m + 12*b)*x^4 + (a*m^2 + 10*a*m + 24*a)*x^2)*x^m/(m^3 + 12*m^2 + 44*
m + 48)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (29) = 58\).

Time = 0.34 (sec) , antiderivative size = 280, normalized size of antiderivative = 7.57 \[ \int x^m \left (a x+b x^3+c x^5\right ) \, dx=\begin {cases} - \frac {a}{4 x^{4}} - \frac {b}{2 x^{2}} + c \log {\left (x \right )} & \text {for}\: m = -6 \\- \frac {a}{2 x^{2}} + b \log {\left (x \right )} + \frac {c x^{2}}{2} & \text {for}\: m = -4 \\a \log {\left (x \right )} + \frac {b x^{2}}{2} + \frac {c x^{4}}{4} & \text {for}\: m = -2 \\\frac {a m^{2} x^{2} x^{m}}{m^{3} + 12 m^{2} + 44 m + 48} + \frac {10 a m x^{2} x^{m}}{m^{3} + 12 m^{2} + 44 m + 48} + \frac {24 a x^{2} x^{m}}{m^{3} + 12 m^{2} + 44 m + 48} + \frac {b m^{2} x^{4} x^{m}}{m^{3} + 12 m^{2} + 44 m + 48} + \frac {8 b m x^{4} x^{m}}{m^{3} + 12 m^{2} + 44 m + 48} + \frac {12 b x^{4} x^{m}}{m^{3} + 12 m^{2} + 44 m + 48} + \frac {c m^{2} x^{6} x^{m}}{m^{3} + 12 m^{2} + 44 m + 48} + \frac {6 c m x^{6} x^{m}}{m^{3} + 12 m^{2} + 44 m + 48} + \frac {8 c x^{6} x^{m}}{m^{3} + 12 m^{2} + 44 m + 48} & \text {otherwise} \end {cases} \]

[In]

integrate(x**m*(c*x**5+b*x**3+a*x),x)

[Out]

Piecewise((-a/(4*x**4) - b/(2*x**2) + c*log(x), Eq(m, -6)), (-a/(2*x**2) + b*log(x) + c*x**2/2, Eq(m, -4)), (a
*log(x) + b*x**2/2 + c*x**4/4, Eq(m, -2)), (a*m**2*x**2*x**m/(m**3 + 12*m**2 + 44*m + 48) + 10*a*m*x**2*x**m/(
m**3 + 12*m**2 + 44*m + 48) + 24*a*x**2*x**m/(m**3 + 12*m**2 + 44*m + 48) + b*m**2*x**4*x**m/(m**3 + 12*m**2 +
44*m + 48) + 8*b*m*x**4*x**m/(m**3 + 12*m**2 + 44*m + 48) + 12*b*x**4*x**m/(m**3 + 12*m**2 + 44*m + 48) + c*m
**2*x**6*x**m/(m**3 + 12*m**2 + 44*m + 48) + 6*c*m*x**6*x**m/(m**3 + 12*m**2 + 44*m + 48) + 8*c*x**6*x**m/(m**
3 + 12*m**2 + 44*m + 48), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int x^m \left (a x+b x^3+c x^5\right ) \, dx=\frac {c x^{m + 6}}{m + 6} + \frac {b x^{m + 4}}{m + 4} + \frac {a x^{m + 2}}{m + 2} \]

[In]

integrate(x^m*(c*x^5+b*x^3+a*x),x, algorithm="maxima")

[Out]

c*x^(m + 6)/(m + 6) + b*x^(m + 4)/(m + 4) + a*x^(m + 2)/(m + 2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (37) = 74\).

Time = 0.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.89 \[ \int x^m \left (a x+b x^3+c x^5\right ) \, dx=\frac {c m^{2} x^{6} x^{m} + 6 \, c m x^{6} x^{m} + b m^{2} x^{4} x^{m} + 8 \, c x^{6} x^{m} + 8 \, b m x^{4} x^{m} + a m^{2} x^{2} x^{m} + 12 \, b x^{4} x^{m} + 10 \, a m x^{2} x^{m} + 24 \, a x^{2} x^{m}}{m^{3} + 12 \, m^{2} + 44 \, m + 48} \]

[In]

integrate(x^m*(c*x^5+b*x^3+a*x),x, algorithm="giac")

[Out]

(c*m^2*x^6*x^m + 6*c*m*x^6*x^m + b*m^2*x^4*x^m + 8*c*x^6*x^m + 8*b*m*x^4*x^m + a*m^2*x^2*x^m + 12*b*x^4*x^m +
10*a*m*x^2*x^m + 24*a*x^2*x^m)/(m^3 + 12*m^2 + 44*m + 48)

Mupad [B] (verification not implemented)

Time = 8.55 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.41 \[ \int x^m \left (a x+b x^3+c x^5\right ) \, dx=x^m\,\left (\frac {a\,x^2\,\left (m^2+10\,m+24\right )}{m^3+12\,m^2+44\,m+48}+\frac {b\,x^4\,\left (m^2+8\,m+12\right )}{m^3+12\,m^2+44\,m+48}+\frac {c\,x^6\,\left (m^2+6\,m+8\right )}{m^3+12\,m^2+44\,m+48}\right ) \]

[In]

int(x^m*(a*x + b*x^3 + c*x^5),x)

[Out]

x^m*((a*x^2*(10*m + m^2 + 24))/(44*m + 12*m^2 + m^3 + 48) + (b*x^4*(8*m + m^2 + 12))/(44*m + 12*m^2 + m^3 + 48
) + (c*x^6*(6*m + m^2 + 8))/(44*m + 12*m^2 + m^3 + 48))