\(\int (d x)^m (b x+c x^2)^2 \, dx\) [113]

Optimal result

Integrand size = 17, antiderivative size = 58 \[ \int (d x)^m \left (b x+c x^2\right )^2 \, dx=\frac {b^2 (d x)^{3+m}}{d^3 (3+m)}+\frac {2 b c (d x)^{4+m}}{d^4 (4+m)}+\frac {c^2 (d x)^{5+m}}{d^5 (5+m)} \]

[Out]

b^2*(d*x)^(3+m)/d^3/(3+m)+2*b*c*(d*x)^(4+m)/d^4/(4+m)+c^2*(d*x)^(5+m)/d^5/(5+m)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {661, 45} \[ \int (d x)^m \left (b x+c x^2\right )^2 \, dx=\frac {b^2 (d x)^{m+3}}{d^3 (m+3)}+\frac {2 b c (d x)^{m+4}}{d^4 (m+4)}+\frac {c^2 (d x)^{m+5}}{d^5 (m+5)} \]

[In]

Int[(d*x)^m*(b*x + c*x^2)^2,x]

[Out]

(b^2*(d*x)^(3 + m))/(d^3*(3 + m)) + (2*b*c*(d*x)^(4 + m))/(d^4*(4 + m)) + (c^2*(d*x)^(5 + m))/(d^5*(5 + m))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 661

Int[((e_.)*(x_))^(m_.)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e*x)^(m + p)*(b + c*x)
^p, x], x] /; FreeQ[{b, c, e, m}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (d x)^{2+m} (b+c x)^2 \, dx}{d^2} \\ & = \frac {\int \left (b^2 (d x)^{2+m}+\frac {2 b c (d x)^{3+m}}{d}+\frac {c^2 (d x)^{4+m}}{d^2}\right ) \, dx}{d^2} \\ & = \frac {b^2 (d x)^{3+m}}{d^3 (3+m)}+\frac {2 b c (d x)^{4+m}}{d^4 (4+m)}+\frac {c^2 (d x)^{5+m}}{d^5 (5+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.71 \[ \int (d x)^m \left (b x+c x^2\right )^2 \, dx=x^3 (d x)^m \left (\frac {b^2}{3+m}+\frac {2 b c x}{4+m}+\frac {c^2 x^2}{5+m}\right ) \]

[In]

Integrate[(d*x)^m*(b*x + c*x^2)^2,x]

[Out]

x^3*(d*x)^m*(b^2/(3 + m) + (2*b*c*x)/(4 + m) + (c^2*x^2)/(5 + m))

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02

method	result	size
norman	\(\frac {b^{2} x^{3} {\mathrm e}^{m \ln \left (d x \right )}}{3+m}+\frac {c^{2} x^{5} {\mathrm e}^{m \ln \left (d x \right )}}{5+m}+\frac {2 b c \,x^{4} {\mathrm e}^{m \ln \left (d x \right )}}{4+m}\)	\(59\)
gosper	\(\frac {\left (d x \right )^{m} \left (c^{2} m^{2} x^{2}+2 b c \,m^{2} x +7 m \,x^{2} c^{2}+b^{2} m^{2}+16 b c m x +12 c^{2} x^{2}+9 b^{2} m +30 b c x +20 b^{2}\right ) x^{3}}{\left (5+m \right ) \left (4+m \right ) \left (3+m \right )}\)	\(90\)
risch	\(\frac {\left (d x \right )^{m} \left (c^{2} m^{2} x^{2}+2 b c \,m^{2} x +7 m \,x^{2} c^{2}+b^{2} m^{2}+16 b c m x +12 c^{2} x^{2}+9 b^{2} m +30 b c x +20 b^{2}\right ) x^{3}}{\left (5+m \right ) \left (4+m \right ) \left (3+m \right )}\)	\(90\)
parallelrisch	\(\frac {x^{5} \left (d x \right )^{m} c^{2} m^{2}+7 x^{5} \left (d x \right )^{m} c^{2} m +2 x^{4} \left (d x \right )^{m} b c \,m^{2}+12 x^{5} \left (d x \right )^{m} c^{2}+16 x^{4} \left (d x \right )^{m} b c m +x^{3} \left (d x \right )^{m} b^{2} m^{2}+30 x^{4} \left (d x \right )^{m} b c +9 x^{3} \left (d x \right )^{m} b^{2} m +20 x^{3} \left (d x \right )^{m} b^{2}}{\left (5+m \right ) \left (4+m \right ) \left (3+m \right )}\)	\(142\)

[In]

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

[Out]

b^2/(3+m)*x^3*exp(m*ln(d*x))+c^2/(5+m)*x^5*exp(m*ln(d*x))+2*b*c/(4+m)*x^4*exp(m*ln(d*x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.53 \[ \int (d x)^m \left (b x+c x^2\right )^2 \, dx=\frac {{\left ({\left (c^{2} m^{2} + 7 \, c^{2} m + 12 \, c^{2}\right )} x^{5} + 2 \, {\left (b c m^{2} + 8 \, b c m + 15 \, b c\right )} x^{4} + {\left (b^{2} m^{2} + 9 \, b^{2} m + 20 \, b^{2}\right )} x^{3}\right )} \left (d x\right )^{m}}{m^{3} + 12 \, m^{2} + 47 \, m + 60} \]

[In]

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

[Out]

((c^2*m^2 + 7*c^2*m + 12*c^2)*x^5 + 2*(b*c*m^2 + 8*b*c*m + 15*b*c)*x^4 + (b^2*m^2 + 9*b^2*m + 20*b^2)*x^3)*(d*
x)^m/(m^3 + 12*m^2 + 47*m + 60)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (51) = 102\).

Time = 0.32 (sec) , antiderivative size = 330, normalized size of antiderivative = 5.69 \[ \int (d x)^m \left (b x+c x^2\right )^2 \, dx=\begin {cases} \frac {- \frac {b^{2}}{2 x^{2}} - \frac {2 b c}{x} + c^{2} \log {\left (x \right )}}{d^{5}} & \text {for}\: m = -5 \\\frac {- \frac {b^{2}}{x} + 2 b c \log {\left (x \right )} + c^{2} x}{d^{4}} & \text {for}\: m = -4 \\\frac {b^{2} \log {\left (x \right )} + 2 b c x + \frac {c^{2} x^{2}}{2}}{d^{3}} & \text {for}\: m = -3 \\\frac {b^{2} m^{2} x^{3} \left (d x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {9 b^{2} m x^{3} \left (d x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {20 b^{2} x^{3} \left (d x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {2 b c m^{2} x^{4} \left (d x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {16 b c m x^{4} \left (d x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {30 b c x^{4} \left (d x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {c^{2} m^{2} x^{5} \left (d x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {7 c^{2} m x^{5} \left (d x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} + \frac {12 c^{2} x^{5} \left (d x\right )^{m}}{m^{3} + 12 m^{2} + 47 m + 60} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((-b**2/(2*x**2) - 2*b*c/x + c**2*log(x))/d**5, Eq(m, -5)), ((-b**2/x + 2*b*c*log(x) + c**2*x)/d**4,
Eq(m, -4)), ((b**2*log(x) + 2*b*c*x + c**2*x**2/2)/d**3, Eq(m, -3)), (b**2*m**2*x**3*(d*x)**m/(m**3 + 12*m**2
+ 47*m + 60) + 9*b**2*m*x**3*(d*x)**m/(m**3 + 12*m**2 + 47*m + 60) + 20*b**2*x**3*(d*x)**m/(m**3 + 12*m**2 +
47*m + 60) + 2*b*c*m**2*x**4*(d*x)**m/(m**3 + 12*m**2 + 47*m + 60) + 16*b*c*m*x**4*(d*x)**m/(m**3 + 12*m**2 +
47*m + 60) + 30*b*c*x**4*(d*x)**m/(m**3 + 12*m**2 + 47*m + 60) + c**2*m**2*x**5*(d*x)**m/(m**3 + 12*m**2 + 47*
m + 60) + 7*c**2*m*x**5*(d*x)**m/(m**3 + 12*m**2 + 47*m + 60) + 12*c**2*x**5*(d*x)**m/(m**3 + 12*m**2 + 47*m +
60), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int (d x)^m \left (b x+c x^2\right )^2 \, dx=\frac {c^{2} d^{m} x^{5} x^{m}}{m + 5} + \frac {2 \, b c d^{m} x^{4} x^{m}}{m + 4} + \frac {b^{2} d^{m} x^{3} x^{m}}{m + 3} \]

[In]

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

[Out]

c^2*d^m*x^5*x^m/(m + 5) + 2*b*c*d^m*x^4*x^m/(m + 4) + b^2*d^m*x^3*x^m/(m + 3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (58) = 116\).

Time = 0.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.43 \[ \int (d x)^m \left (b x+c x^2\right )^2 \, dx=\frac {\left (d x\right )^{m} c^{2} m^{2} x^{5} + 2 \, \left (d x\right )^{m} b c m^{2} x^{4} + 7 \, \left (d x\right )^{m} c^{2} m x^{5} + \left (d x\right )^{m} b^{2} m^{2} x^{3} + 16 \, \left (d x\right )^{m} b c m x^{4} + 12 \, \left (d x\right )^{m} c^{2} x^{5} + 9 \, \left (d x\right )^{m} b^{2} m x^{3} + 30 \, \left (d x\right )^{m} b c x^{4} + 20 \, \left (d x\right )^{m} b^{2} x^{3}}{m^{3} + 12 \, m^{2} + 47 \, m + 60} \]

[In]

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

[Out]

((d*x)^m*c^2*m^2*x^5 + 2*(d*x)^m*b*c*m^2*x^4 + 7*(d*x)^m*c^2*m*x^5 + (d*x)^m*b^2*m^2*x^3 + 16*(d*x)^m*b*c*m*x^
4 + 12*(d*x)^m*c^2*x^5 + 9*(d*x)^m*b^2*m*x^3 + 30*(d*x)^m*b*c*x^4 + 20*(d*x)^m*b^2*x^3)/(m^3 + 12*m^2 + 47*m +
60)

Mupad [B] (verification not implemented)

Time = 9.41 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.67 \[ \int (d x)^m \left (b x+c x^2\right )^2 \, dx={\left (d\,x\right )}^m\,\left (\frac {b^2\,x^3\,\left (m^2+9\,m+20\right )}{m^3+12\,m^2+47\,m+60}+\frac {c^2\,x^5\,\left (m^2+7\,m+12\right )}{m^3+12\,m^2+47\,m+60}+\frac {2\,b\,c\,x^4\,\left (m^2+8\,m+15\right )}{m^3+12\,m^2+47\,m+60}\right ) \]

[In]

int((b*x + c*x^2)^2*(d*x)^m,x)

[Out]

(d*x)^m*((b^2*x^3*(9*m + m^2 + 20))/(47*m + 12*m^2 + m^3 + 60) + (c^2*x^5*(7*m + m^2 + 12))/(47*m + 12*m^2 + m
^3 + 60) + (2*b*c*x^4*(8*m + m^2 + 15))/(47*m + 12*m^2 + m^3 + 60))