3.1.1 \(\int x^m (A+B x) (b x+c x^2) \, dx\)

Optimal. Leaf size=45 \[ \frac {x^{m+3} (A c+b B)}{m+3}+\frac {A b x^{m+2}}{m+2}+\frac {B c x^{m+4}}{m+4} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {765} \begin {gather*} \frac {x^{m+3} (A c+b B)}{m+3}+\frac {A b x^{m+2}}{m+2}+\frac {B c x^{m+4}}{m+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m*(A + B*x)*(b*x + c*x^2),x]

[Out]

(A*b*x^(2 + m))/(2 + m) + ((b*B + A*c)*x^(3 + m))/(3 + m) + (B*c*x^(4 + m))/(4 + m)

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int x^m (A+B x) \left (b x+c x^2\right ) \, dx &=\int \left (A b x^{1+m}+(b B+A c) x^{2+m}+B c x^{3+m}\right ) \, dx\\ &=\frac {A b x^{2+m}}{2+m}+\frac {(b B+A c) x^{3+m}}{3+m}+\frac {B c x^{4+m}}{4+m}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 57, normalized size = 1.27 \begin {gather*} \frac {x^{m+2} (A (m+4) (b (m+3)+c (m+2) x)+B (m+2) x (b (m+4)+c (m+3) x))}{(m+2) (m+3) (m+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^m*(A + B*x)*(b*x + c*x^2),x]

[Out]

(x^(2 + m)*(A*(4 + m)*(b*(3 + m) + c*(2 + m)*x) + B*(2 + m)*x*(b*(4 + m) + c*(3 + m)*x)))/((2 + m)*(3 + m)*(4
+ m))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.05, size = 0, normalized size = 0.00 \begin {gather*} \int x^m (A+B x) \left (b x+c x^2\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^m*(A + B*x)*(b*x + c*x^2),x]

[Out]

Defer[IntegrateAlgebraic][x^m*(A + B*x)*(b*x + c*x^2), x]

________________________________________________________________________________________

fricas [B]  time = 0.42, size = 94, normalized size = 2.09 \begin {gather*} \frac {{\left ({\left (B c m^{2} + 5 \, B c m + 6 \, B c\right )} x^{4} + {\left ({\left (B b + A c\right )} m^{2} + 8 \, B b + 8 \, A c + 6 \, {\left (B b + A c\right )} m\right )} x^{3} + {\left (A b m^{2} + 7 \, A b m + 12 \, A b\right )} x^{2}\right )} x^{m}}{m^{3} + 9 \, m^{2} + 26 \, m + 24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*c*m^2 + 5*B*c*m + 6*B*c)*x^4 + ((B*b + A*c)*m^2 + 8*B*b + 8*A*c + 6*(B*b + A*c)*m)*x^3 + (A*b*m^2 + 7*A*b*
m + 12*A*b)*x^2)*x^m/(m^3 + 9*m^2 + 26*m + 24)

________________________________________________________________________________________

giac [B]  time = 0.23, size = 149, normalized size = 3.31 \begin {gather*} \frac {B c m^{2} x^{4} x^{m} + B b m^{2} x^{3} x^{m} + A c m^{2} x^{3} x^{m} + 5 \, B c m x^{4} x^{m} + A b m^{2} x^{2} x^{m} + 6 \, B b m x^{3} x^{m} + 6 \, A c m x^{3} x^{m} + 6 \, B c x^{4} x^{m} + 7 \, A b m x^{2} x^{m} + 8 \, B b x^{3} x^{m} + 8 \, A c x^{3} x^{m} + 12 \, A b x^{2} x^{m}}{m^{3} + 9 \, m^{2} + 26 \, m + 24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*c*m^2*x^4*x^m + B*b*m^2*x^3*x^m + A*c*m^2*x^3*x^m + 5*B*c*m*x^4*x^m + A*b*m^2*x^2*x^m + 6*B*b*m*x^3*x^m + 6
*A*c*m*x^3*x^m + 6*B*c*x^4*x^m + 7*A*b*m*x^2*x^m + 8*B*b*x^3*x^m + 8*A*c*x^3*x^m + 12*A*b*x^2*x^m)/(m^3 + 9*m^
2 + 26*m + 24)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 98, normalized size = 2.18 \begin {gather*} \frac {\left (B c \,m^{2} x^{2}+A c \,m^{2} x +B b \,m^{2} x +5 B c m \,x^{2}+A b \,m^{2}+6 A c m x +6 B b m x +6 B c \,x^{2}+7 A b m +8 A c x +8 B b x +12 A b \right ) x^{m +2}}{\left (m +4\right ) \left (m +3\right ) \left (m +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(B*x+A)*(c*x^2+b*x),x)

[Out]

x^(m+2)*(B*c*m^2*x^2+A*c*m^2*x+B*b*m^2*x+5*B*c*m*x^2+A*b*m^2+6*A*c*m*x+6*B*b*m*x+6*B*c*x^2+7*A*b*m+8*A*c*x+8*B
*b*x+12*A*b)/(m+4)/(m+3)/(m+2)

________________________________________________________________________________________

maxima [A]  time = 0.86, size = 53, normalized size = 1.18 \begin {gather*} \frac {B c x^{m + 4}}{m + 4} + \frac {B b x^{m + 3}}{m + 3} + \frac {A c x^{m + 3}}{m + 3} + \frac {A b x^{m + 2}}{m + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*c*x^(m + 4)/(m + 4) + B*b*x^(m + 3)/(m + 3) + A*c*x^(m + 3)/(m + 3) + A*b*x^(m + 2)/(m + 2)

________________________________________________________________________________________

mupad [B]  time = 1.13, size = 97, normalized size = 2.16 \begin {gather*} x^m\,\left (\frac {x^3\,\left (A\,c+B\,b\right )\,\left (m^2+6\,m+8\right )}{m^3+9\,m^2+26\,m+24}+\frac {A\,b\,x^2\,\left (m^2+7\,m+12\right )}{m^3+9\,m^2+26\,m+24}+\frac {B\,c\,x^4\,\left (m^2+5\,m+6\right )}{m^3+9\,m^2+26\,m+24}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(b*x + c*x^2)*(A + B*x),x)

[Out]

x^m*((x^3*(A*c + B*b)*(6*m + m^2 + 8))/(26*m + 9*m^2 + m^3 + 24) + (A*b*x^2*(7*m + m^2 + 12))/(26*m + 9*m^2 +
m^3 + 24) + (B*c*x^4*(5*m + m^2 + 6))/(26*m + 9*m^2 + m^3 + 24))

________________________________________________________________________________________

sympy [A]  time = 0.81, size = 394, normalized size = 8.76 \begin {gather*} \begin {cases} - \frac {A b}{2 x^{2}} - \frac {A c}{x} - \frac {B b}{x} + B c \log {\relax (x )} & \text {for}\: m = -4 \\- \frac {A b}{x} + A c \log {\relax (x )} + B b \log {\relax (x )} + B c x & \text {for}\: m = -3 \\A b \log {\relax (x )} + A c x + B b x + \frac {B c x^{2}}{2} & \text {for}\: m = -2 \\\frac {A b m^{2} x^{2} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {7 A b m x^{2} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {12 A b x^{2} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {A c m^{2} x^{3} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {6 A c m x^{3} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {8 A c x^{3} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {B b m^{2} x^{3} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {6 B b m x^{3} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {8 B b x^{3} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {B c m^{2} x^{4} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {5 B c m x^{4} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} + \frac {6 B c x^{4} x^{m}}{m^{3} + 9 m^{2} + 26 m + 24} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(B*x+A)*(c*x**2+b*x),x)

[Out]

Piecewise((-A*b/(2*x**2) - A*c/x - B*b/x + B*c*log(x), Eq(m, -4)), (-A*b/x + A*c*log(x) + B*b*log(x) + B*c*x,
Eq(m, -3)), (A*b*log(x) + A*c*x + B*b*x + B*c*x**2/2, Eq(m, -2)), (A*b*m**2*x**2*x**m/(m**3 + 9*m**2 + 26*m +
24) + 7*A*b*m*x**2*x**m/(m**3 + 9*m**2 + 26*m + 24) + 12*A*b*x**2*x**m/(m**3 + 9*m**2 + 26*m + 24) + A*c*m**2*
x**3*x**m/(m**3 + 9*m**2 + 26*m + 24) + 6*A*c*m*x**3*x**m/(m**3 + 9*m**2 + 26*m + 24) + 8*A*c*x**3*x**m/(m**3
+ 9*m**2 + 26*m + 24) + B*b*m**2*x**3*x**m/(m**3 + 9*m**2 + 26*m + 24) + 6*B*b*m*x**3*x**m/(m**3 + 9*m**2 + 26
*m + 24) + 8*B*b*x**3*x**m/(m**3 + 9*m**2 + 26*m + 24) + B*c*m**2*x**4*x**m/(m**3 + 9*m**2 + 26*m + 24) + 5*B*
c*m*x**4*x**m/(m**3 + 9*m**2 + 26*m + 24) + 6*B*c*x**4*x**m/(m**3 + 9*m**2 + 26*m + 24), True))

________________________________________________________________________________________