\(\int (c x)^m (b x^2+c x^4) \, dx\) [405]

Optimal result

Integrand size = 17, antiderivative size = 34 \[ \int (c x)^m \left (b x^2+c x^4\right ) \, dx=\frac {b (c x)^{3+m}}{c^3 (3+m)}+\frac {(c x)^{5+m}}{c^4 (5+m)} \]

[Out]

b*(c*x)^(3+m)/c^3/(3+m)+(c*x)^(5+m)/c^4/(5+m)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14} \[ \int (c x)^m \left (b x^2+c x^4\right ) \, dx=\frac {b (c x)^{m+3}}{c^3 (m+3)}+\frac {(c x)^{m+5}}{c^4 (m+5)} \]

[In]

Int[(c*x)^m*(b*x^2 + c*x^4),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int (c x)^m \left (b x^2+c x^4\right ) \, dx=x^3 (c x)^m \left (\frac {b}{3+m}+\frac {c x^2}{5+m}\right ) \]

[In]

Integrate[(c*x)^m*(b*x^2 + c*x^4),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06

method	result	size
norman	\(\frac {b \,x^{3} {\mathrm e}^{m \ln \left (c x \right )}}{3+m}+\frac {c \,x^{5} {\mathrm e}^{m \ln \left (c x \right )}}{5+m}\)	\(36\)
gosper	\(\frac {\left (c x \right )^{m} \left (c m \,x^{2}+3 c \,x^{2}+b m +5 b \right ) x^{3}}{\left (5+m \right ) \left (3+m \right )}\)	\(39\)
risch	\(\frac {\left (c x \right )^{m} \left (c m \,x^{2}+3 c \,x^{2}+b m +5 b \right ) x^{3}}{\left (5+m \right ) \left (3+m \right )}\)	\(39\)
parallelrisch	\(\frac {x^{5} \left (c x \right )^{m} c m +3 x^{5} \left (c x \right )^{m} c +x^{3} \left (c x \right )^{m} b m +5 x^{3} \left (c x \right )^{m} b}{\left (5+m \right ) \left (3+m \right )}\)	\(57\)

[In]

int((c*x)^m*(c*x^4+b*x^2),x,method=_RETURNVERBOSE)

[Out]

b/(3+m)*x^3*exp(m*ln(c*x))+c/(5+m)*x^5*exp(m*ln(c*x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int (c x)^m \left (b x^2+c x^4\right ) \, dx=\frac {{\left ({\left (c m + 3 \, c\right )} x^{5} + {\left (b m + 5 \, b\right )} x^{3}\right )} \left (c x\right )^{m}}{m^{2} + 8 \, m + 15} \]

[In]

integrate((c*x)^m*(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

((c*m + 3*c)*x^5 + (b*m + 5*b)*x^3)*(c*x)^m/(m^2 + 8*m + 15)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (27) = 54\).

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 3.29 \[ \int (c x)^m \left (b x^2+c x^4\right ) \, dx=\begin {cases} \frac {- \frac {b}{2 x^{2}} + c \log {\left (x \right )}}{c^{5}} & \text {for}\: m = -5 \\\frac {b \log {\left (x \right )} + \frac {c x^{2}}{2}}{c^{3}} & \text {for}\: m = -3 \\\frac {b m x^{3} \left (c x\right )^{m}}{m^{2} + 8 m + 15} + \frac {5 b x^{3} \left (c x\right )^{m}}{m^{2} + 8 m + 15} + \frac {c m x^{5} \left (c x\right )^{m}}{m^{2} + 8 m + 15} + \frac {3 c x^{5} \left (c x\right )^{m}}{m^{2} + 8 m + 15} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x)**m*(c*x**4+b*x**2),x)

[Out]

Piecewise(((-b/(2*x**2) + c*log(x))/c**5, Eq(m, -5)), ((b*log(x) + c*x**2/2)/c**3, Eq(m, -3)), (b*m*x**3*(c*x)
**m/(m**2 + 8*m + 15) + 5*b*x**3*(c*x)**m/(m**2 + 8*m + 15) + c*m*x**5*(c*x)**m/(m**2 + 8*m + 15) + 3*c*x**5*(
c*x)**m/(m**2 + 8*m + 15), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int (c x)^m \left (b x^2+c x^4\right ) \, dx=\frac {c^{m + 1} x^{5} x^{m}}{m + 5} + \frac {b c^{m} x^{3} x^{m}}{m + 3} \]

[In]

integrate((c*x)^m*(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

c^(m + 1)*x^5*x^m/(m + 5) + b*c^m*x^3*x^m/(m + 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.65 \[ \int (c x)^m \left (b x^2+c x^4\right ) \, dx=\frac {\left (c x\right )^{m} c m x^{5} + 3 \, \left (c x\right )^{m} c x^{5} + \left (c x\right )^{m} b m x^{3} + 5 \, \left (c x\right )^{m} b x^{3}}{m^{2} + 8 \, m + 15} \]

[In]

integrate((c*x)^m*(c*x^4+b*x^2),x, algorithm="giac")

[Out]

((c*x)^m*c*m*x^5 + 3*(c*x)^m*c*x^5 + (c*x)^m*b*m*x^3 + 5*(c*x)^m*b*x^3)/(m^2 + 8*m + 15)

Mupad [B] (verification not implemented)

Time = 13.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int (c x)^m \left (b x^2+c x^4\right ) \, dx=\frac {x^3\,{\left (c\,x\right )}^m\,\left (5\,b+b\,m+3\,c\,x^2+c\,m\,x^2\right )}{m^2+8\,m+15} \]

[In]

int((c*x)^m*(b*x^2 + c*x^4),x)

[Out]

(x^3*(c*x)^m*(5*b + b*m + 3*c*x^2 + c*m*x^2))/(8*m + m^2 + 15)