3.321 \(\int x^m (a+b x^2) (A+B x^2) \, dx\)

Optimal. Leaf size=45 \[ \frac {x^{m+3} (a B+A b)}{m+3}+\frac {a A x^{m+1}}{m+1}+\frac {b B x^{m+5}}{m+5} \]

[Out]

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

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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 = {448} \[ \frac {x^{m+3} (a B+A b)}{m+3}+\frac {a A x^{m+1}}{m+1}+\frac {b B x^{m+5}}{m+5} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x^2)*(A + B*x^2),x]

[Out]

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

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin {align*} \int x^m \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx &=\int \left (a A x^m+(A b+a B) x^{2+m}+b B x^{4+m}\right ) \, dx\\ &=\frac {a A x^{1+m}}{1+m}+\frac {(A b+a B) x^{3+m}}{3+m}+\frac {b B x^{5+m}}{5+m}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.93 \[ x^{m+1} \left (\frac {x^2 (a B+A b)}{m+3}+\frac {a A}{m+1}+\frac {b B x^4}{m+5}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.50, size = 92, normalized size = 2.04 \[ \frac {{\left ({\left (B b m^{2} + 4 \, B b m + 3 \, B b\right )} x^{5} + {\left ({\left (B a + A b\right )} m^{2} + 5 \, B a + 5 \, A b + 6 \, {\left (B a + A b\right )} m\right )} x^{3} + {\left (A a m^{2} + 8 \, A a m + 15 \, A a\right )} x\right )} x^{m}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*b*m^2 + 4*B*b*m + 3*B*b)*x^5 + ((B*a + A*b)*m^2 + 5*B*a + 5*A*b + 6*(B*a + A*b)*m)*x^3 + (A*a*m^2 + 8*A*a*
m + 15*A*a)*x)*x^m/(m^3 + 9*m^2 + 23*m + 15)

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giac [B]  time = 0.36, size = 143, normalized size = 3.18 \[ \frac {B b m^{2} x^{5} x^{m} + 4 \, B b m x^{5} x^{m} + B a m^{2} x^{3} x^{m} + A b m^{2} x^{3} x^{m} + 3 \, B b x^{5} x^{m} + 6 \, B a m x^{3} x^{m} + 6 \, A b m x^{3} x^{m} + A a m^{2} x x^{m} + 5 \, B a x^{3} x^{m} + 5 \, A b x^{3} x^{m} + 8 \, A a m x x^{m} + 15 \, A a x x^{m}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b*m^2*x^5*x^m + 4*B*b*m*x^5*x^m + B*a*m^2*x^3*x^m + A*b*m^2*x^3*x^m + 3*B*b*x^5*x^m + 6*B*a*m*x^3*x^m + 6*A
*b*m*x^3*x^m + A*a*m^2*x*x^m + 5*B*a*x^3*x^m + 5*A*b*x^3*x^m + 8*A*a*m*x*x^m + 15*A*a*x*x^m)/(m^3 + 9*m^2 + 23
*m + 15)

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maple [B]  time = 0.00, size = 110, normalized size = 2.44 \[ \frac {\left (B b \,m^{2} x^{4}+4 B b m \,x^{4}+A b \,m^{2} x^{2}+B a \,m^{2} x^{2}+3 B b \,x^{4}+6 A b m \,x^{2}+6 B a m \,x^{2}+A a \,m^{2}+5 A b \,x^{2}+5 B a \,x^{2}+8 A a m +15 A a \right ) x^{m +1}}{\left (m +5\right ) \left (m +3\right ) \left (m +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(m+1)*(B*b*m^2*x^4+4*B*b*m*x^4+A*b*m^2*x^2+B*a*m^2*x^2+3*B*b*x^4+6*A*b*m*x^2+6*B*a*m*x^2+A*a*m^2+5*A*b*x^2+5
*B*a*x^2+8*A*a*m+15*A*a)/(m+5)/(m+3)/(m+1)

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maxima [A]  time = 1.06, size = 53, normalized size = 1.18 \[ \frac {B b x^{m + 5}}{m + 5} + \frac {B a x^{m + 3}}{m + 3} + \frac {A b x^{m + 3}}{m + 3} + \frac {A a x^{m + 1}}{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.30, size = 95, normalized size = 2.11 \[ x^m\,\left (\frac {x^3\,\left (A\,b+B\,a\right )\,\left (m^2+6\,m+5\right )}{m^3+9\,m^2+23\,m+15}+\frac {B\,b\,x^5\,\left (m^2+4\,m+3\right )}{m^3+9\,m^2+23\,m+15}+\frac {A\,a\,x\,\left (m^2+8\,m+15\right )}{m^3+9\,m^2+23\,m+15}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(A + B*x^2)*(a + b*x^2),x)

[Out]

x^m*((x^3*(A*b + B*a)*(6*m + m^2 + 5))/(23*m + 9*m^2 + m^3 + 15) + (B*b*x^5*(4*m + m^2 + 3))/(23*m + 9*m^2 + m
^3 + 15) + (A*a*x*(8*m + m^2 + 15))/(23*m + 9*m^2 + m^3 + 15))

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sympy [A]  time = 0.94, size = 410, normalized size = 9.11 \[ \begin {cases} - \frac {A a}{4 x^{4}} - \frac {A b}{2 x^{2}} - \frac {B a}{2 x^{2}} + B b \log {\relax (x )} & \text {for}\: m = -5 \\- \frac {A a}{2 x^{2}} + A b \log {\relax (x )} + B a \log {\relax (x )} + \frac {B b x^{2}}{2} & \text {for}\: m = -3 \\A a \log {\relax (x )} + \frac {A b x^{2}}{2} + \frac {B a x^{2}}{2} + \frac {B b x^{4}}{4} & \text {for}\: m = -1 \\\frac {A a m^{2} x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {8 A a m x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {15 A a x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {A b m^{2} x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {6 A b m x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {5 A b x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {B a m^{2} x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {6 B a m x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {5 B a x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {B b m^{2} x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {4 B b m x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {3 B b x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-A*a/(4*x**4) - A*b/(2*x**2) - B*a/(2*x**2) + B*b*log(x), Eq(m, -5)), (-A*a/(2*x**2) + A*b*log(x) +
B*a*log(x) + B*b*x**2/2, Eq(m, -3)), (A*a*log(x) + A*b*x**2/2 + B*a*x**2/2 + B*b*x**4/4, Eq(m, -1)), (A*a*m**
2*x*x**m/(m**3 + 9*m**2 + 23*m + 15) + 8*A*a*m*x*x**m/(m**3 + 9*m**2 + 23*m + 15) + 15*A*a*x*x**m/(m**3 + 9*m*
*2 + 23*m + 15) + A*b*m**2*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + 6*A*b*m*x**3*x**m/(m**3 + 9*m**2 + 23*m + 1
5) + 5*A*b*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + B*a*m**2*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + 6*B*a*m*x*
*3*x**m/(m**3 + 9*m**2 + 23*m + 15) + 5*B*a*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + B*b*m**2*x**5*x**m/(m**3 +
9*m**2 + 23*m + 15) + 4*B*b*m*x**5*x**m/(m**3 + 9*m**2 + 23*m + 15) + 3*B*b*x**5*x**m/(m**3 + 9*m**2 + 23*m +
15), True))

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