3.374 \(\int x^m (a+b x) (A+B x) \, dx\)

Optimal. Leaf size=45 \[ \frac{x^{m+2} (a B+A b)}{m+2}+\frac{a A x^{m+1}}{m+1}+\frac{b B x^{m+3}}{m+3} \]

[Out]

(a*A*x^(1 + m))/(1 + m) + ((A*b + a*B)*x^(2 + m))/(2 + m) + (b*B*x^(3 + m))/(3 + m)

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Rubi [A]  time = 0.0181551, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {76} \[ \frac{x^{m+2} (a B+A b)}{m+2}+\frac{a A x^{m+1}}{m+1}+\frac{b B x^{m+3}}{m+3} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x)*(A + B*x),x]

[Out]

(a*A*x^(1 + m))/(1 + m) + ((A*b + a*B)*x^(2 + m))/(2 + m) + (b*B*x^(3 + m))/(3 + m)

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

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

Mathematica [A]  time = 0.0326054, size = 57, normalized size = 1.27 \[ \frac{x^{m+1} (a (m+3) (A (m+2)+B (m+1) x)+b (m+1) x (A (m+3)+B (m+2) x))}{(m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x)*(A + B*x),x]

[Out]

(x^(1 + m)*(a*(3 + m)*(A*(2 + m) + B*(1 + m)*x) + b*(1 + m)*x*(A*(3 + m) + B*(2 + m)*x)))/((1 + m)*(2 + m)*(3
+ m))

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Maple [B]  time = 0.003, size = 98, normalized size = 2.2 \begin{align*}{\frac{{x}^{1+m} \left ( Bb{m}^{2}{x}^{2}+Ab{m}^{2}x+Ba{m}^{2}x+3\,Bbm{x}^{2}+Aa{m}^{2}+4\,Abmx+4\,Bamx+2\,bB{x}^{2}+5\,Aam+3\,Abx+3\,Bax+6\,Aa \right ) }{ \left ( 3+m \right ) \left ( 2+m \right ) \left ( 1+m \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1+m)*(B*b*m^2*x^2+A*b*m^2*x+B*a*m^2*x+3*B*b*m*x^2+A*a*m^2+4*A*b*m*x+4*B*a*m*x+2*B*b*x^2+5*A*a*m+3*A*b*x+3*B
*a*x+6*A*a)/(3+m)/(2+m)/(1+m)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.28358, size = 208, normalized size = 4.62 \begin{align*} \frac{{\left ({\left (B b m^{2} + 3 \, B b m + 2 \, B b\right )} x^{3} +{\left ({\left (B a + A b\right )} m^{2} + 3 \, B a + 3 \, A b + 4 \,{\left (B a + A b\right )} m\right )} x^{2} +{\left (A a m^{2} + 5 \, A a m + 6 \, A a\right )} x\right )} x^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*b*m^2 + 3*B*b*m + 2*B*b)*x^3 + ((B*a + A*b)*m^2 + 3*B*a + 3*A*b + 4*(B*a + A*b)*m)*x^2 + (A*a*m^2 + 5*A*a*
m + 6*A*a)*x)*x^m/(m^3 + 6*m^2 + 11*m + 6)

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Sympy [A]  time = 0.665872, size = 389, normalized size = 8.64 \begin{align*} \begin{cases} - \frac{A a}{2 x^{2}} - \frac{A b}{x} - \frac{B a}{x} + B b \log{\left (x \right )} & \text{for}\: m = -3 \\- \frac{A a}{x} + A b \log{\left (x \right )} + B a \log{\left (x \right )} + B b x & \text{for}\: m = -2 \\A a \log{\left (x \right )} + A b x + B a x + \frac{B b x^{2}}{2} & \text{for}\: m = -1 \\\frac{A a m^{2} x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{5 A a m x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{6 A a x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{A b m^{2} x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{4 A b m x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{3 A b x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{B a m^{2} x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{4 B a m x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{3 B a x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{B b m^{2} x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{3 B b m x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{2 B b x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(b*x+a)*(B*x+A),x)

[Out]

Piecewise((-A*a/(2*x**2) - A*b/x - B*a/x + B*b*log(x), Eq(m, -3)), (-A*a/x + A*b*log(x) + B*a*log(x) + B*b*x,
Eq(m, -2)), (A*a*log(x) + A*b*x + B*a*x + B*b*x**2/2, Eq(m, -1)), (A*a*m**2*x*x**m/(m**3 + 6*m**2 + 11*m + 6)
+ 5*A*a*m*x*x**m/(m**3 + 6*m**2 + 11*m + 6) + 6*A*a*x*x**m/(m**3 + 6*m**2 + 11*m + 6) + A*b*m**2*x**2*x**m/(m*
*3 + 6*m**2 + 11*m + 6) + 4*A*b*m*x**2*x**m/(m**3 + 6*m**2 + 11*m + 6) + 3*A*b*x**2*x**m/(m**3 + 6*m**2 + 11*m
+ 6) + B*a*m**2*x**2*x**m/(m**3 + 6*m**2 + 11*m + 6) + 4*B*a*m*x**2*x**m/(m**3 + 6*m**2 + 11*m + 6) + 3*B*a*x
**2*x**m/(m**3 + 6*m**2 + 11*m + 6) + B*b*m**2*x**3*x**m/(m**3 + 6*m**2 + 11*m + 6) + 3*B*b*m*x**3*x**m/(m**3
+ 6*m**2 + 11*m + 6) + 2*B*b*x**3*x**m/(m**3 + 6*m**2 + 11*m + 6), True))

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Giac [B]  time = 1.175, size = 193, normalized size = 4.29 \begin{align*} \frac{B b m^{2} x^{3} x^{m} + B a m^{2} x^{2} x^{m} + A b m^{2} x^{2} x^{m} + 3 \, B b m x^{3} x^{m} + A a m^{2} x x^{m} + 4 \, B a m x^{2} x^{m} + 4 \, A b m x^{2} x^{m} + 2 \, B b x^{3} x^{m} + 5 \, A a m x x^{m} + 3 \, B a x^{2} x^{m} + 3 \, A b x^{2} x^{m} + 6 \, A a x x^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b*m^2*x^3*x^m + B*a*m^2*x^2*x^m + A*b*m^2*x^2*x^m + 3*B*b*m*x^3*x^m + A*a*m^2*x*x^m + 4*B*a*m*x^2*x^m + 4*A
*b*m*x^2*x^m + 2*B*b*x^3*x^m + 5*A*a*m*x*x^m + 3*B*a*x^2*x^m + 3*A*b*x^2*x^m + 6*A*a*x*x^m)/(m^3 + 6*m^2 + 11*
m + 6)