3.3.22 \(\int x^m (A+B x^2) (b x^2+c x^4) \, dx\)

Optimal. Leaf size=45 \[ \frac {x^{m+5} (A c+b B)}{m+5}+\frac {A b x^{m+3}}{m+3}+\frac {B c x^{m+7}}{m+7} \]

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Rubi [A]  time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1584, 448} \begin {gather*} \frac {x^{m+5} (A c+b B)}{m+5}+\frac {A b x^{m+3}}{m+3}+\frac {B c x^{m+7}}{m+7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m*(A + B*x^2)*(b*x^2 + c*x^4),x]

[Out]

(A*b*x^(3 + m))/(3 + m) + ((b*B + A*c)*x^(5 + m))/(5 + m) + (B*c*x^(7 + m))/(7 + 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]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int x^m \left (A+B x^2\right ) \left (b x^2+c x^4\right ) \, dx &=\int x^{2+m} \left (A+B x^2\right ) \left (b+c x^2\right ) \, dx\\ &=\int \left (A b x^{2+m}+(b B+A c) x^{4+m}+B c x^{6+m}\right ) \, dx\\ &=\frac {A b x^{3+m}}{3+m}+\frac {(b B+A c) x^{5+m}}{5+m}+\frac {B c x^{7+m}}{7+m}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 42, normalized size = 0.93 \begin {gather*} x^{m+3} \left (\frac {x^2 (A c+b B)}{m+5}+\frac {A b}{m+3}+\frac {B c x^4}{m+7}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^m*(A + B*x^2)*(b*x^2 + c*x^4),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.05, size = 0, normalized size = 0.00 \begin {gather*} \int x^m \left (A+B x^2\right ) \left (b x^2+c x^4\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^m*(A + B*x^2)*(b*x^2 + c*x^4),x]

[Out]

Defer[IntegrateAlgebraic][x^m*(A + B*x^2)*(b*x^2 + c*x^4), x]

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fricas [B]  time = 0.42, size = 94, normalized size = 2.09 \begin {gather*} \frac {{\left ({\left (B c m^{2} + 8 \, B c m + 15 \, B c\right )} x^{7} + {\left ({\left (B b + A c\right )} m^{2} + 21 \, B b + 21 \, A c + 10 \, {\left (B b + A c\right )} m\right )} x^{5} + {\left (A b m^{2} + 12 \, A b m + 35 \, A b\right )} x^{3}\right )} x^{m}}{m^{3} + 15 \, m^{2} + 71 \, m + 105} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*c*m^2 + 8*B*c*m + 15*B*c)*x^7 + ((B*b + A*c)*m^2 + 21*B*b + 21*A*c + 10*(B*b + A*c)*m)*x^5 + (A*b*m^2 + 12
*A*b*m + 35*A*b)*x^3)*x^m/(m^3 + 15*m^2 + 71*m + 105)

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giac [B]  time = 0.16, size = 149, normalized size = 3.31 \begin {gather*} \frac {B c m^{2} x^{7} x^{m} + 8 \, B c m x^{7} x^{m} + B b m^{2} x^{5} x^{m} + A c m^{2} x^{5} x^{m} + 15 \, B c x^{7} x^{m} + 10 \, B b m x^{5} x^{m} + 10 \, A c m x^{5} x^{m} + A b m^{2} x^{3} x^{m} + 21 \, B b x^{5} x^{m} + 21 \, A c x^{5} x^{m} + 12 \, A b m x^{3} x^{m} + 35 \, A b x^{3} x^{m}}{m^{3} + 15 \, m^{2} + 71 \, m + 105} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*c*m^2*x^7*x^m + 8*B*c*m*x^7*x^m + B*b*m^2*x^5*x^m + A*c*m^2*x^5*x^m + 15*B*c*x^7*x^m + 10*B*b*m*x^5*x^m + 1
0*A*c*m*x^5*x^m + A*b*m^2*x^3*x^m + 21*B*b*x^5*x^m + 21*A*c*x^5*x^m + 12*A*b*m*x^3*x^m + 35*A*b*x^3*x^m)/(m^3
+ 15*m^2 + 71*m + 105)

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maple [B]  time = 0.05, size = 110, normalized size = 2.44 \begin {gather*} \frac {\left (B c \,m^{2} x^{4}+8 B c m \,x^{4}+A c \,m^{2} x^{2}+B b \,m^{2} x^{2}+15 B c \,x^{4}+10 A c m \,x^{2}+10 B b m \,x^{2}+A b \,m^{2}+21 A c \,x^{2}+21 B b \,x^{2}+12 A b m +35 A b \right ) x^{m +3}}{\left (m +7\right ) \left (m +5\right ) \left (m +3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(B*x^2+A)*(c*x^4+b*x^2),x)

[Out]

x^(m+3)*(B*c*m^2*x^4+8*B*c*m*x^4+A*c*m^2*x^2+B*b*m^2*x^2+15*B*c*x^4+10*A*c*m*x^2+10*B*b*m*x^2+A*b*m^2+21*A*c*x
^2+21*B*b*x^2+12*A*b*m+35*A*b)/(m+7)/(m+5)/(m+3)

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maxima [A]  time = 1.30, size = 53, normalized size = 1.18 \begin {gather*} \frac {B c x^{m + 7}}{m + 7} + \frac {B b x^{m + 5}}{m + 5} + \frac {A c x^{m + 5}}{m + 5} + \frac {A b x^{m + 3}}{m + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*c*x^(m + 7)/(m + 7) + B*b*x^(m + 5)/(m + 5) + A*c*x^(m + 5)/(m + 5) + A*b*x^(m + 3)/(m + 3)

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mupad [B]  time = 0.25, size = 97, normalized size = 2.16 \begin {gather*} x^m\,\left (\frac {x^5\,\left (A\,c+B\,b\right )\,\left (m^2+10\,m+21\right )}{m^3+15\,m^2+71\,m+105}+\frac {A\,b\,x^3\,\left (m^2+12\,m+35\right )}{m^3+15\,m^2+71\,m+105}+\frac {B\,c\,x^7\,\left (m^2+8\,m+15\right )}{m^3+15\,m^2+71\,m+105}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(A + B*x^2)*(b*x^2 + c*x^4),x)

[Out]

x^m*((x^5*(A*c + B*b)*(10*m + m^2 + 21))/(71*m + 15*m^2 + m^3 + 105) + (A*b*x^3*(12*m + m^2 + 35))/(71*m + 15*
m^2 + m^3 + 105) + (B*c*x^7*(8*m + m^2 + 15))/(71*m + 15*m^2 + m^3 + 105))

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sympy [A]  time = 1.62, size = 415, normalized size = 9.22 \begin {gather*} \begin {cases} - \frac {A b}{4 x^{4}} - \frac {A c}{2 x^{2}} - \frac {B b}{2 x^{2}} + B c \log {\relax (x )} & \text {for}\: m = -7 \\- \frac {A b}{2 x^{2}} + A c \log {\relax (x )} + B b \log {\relax (x )} + \frac {B c x^{2}}{2} & \text {for}\: m = -5 \\A b \log {\relax (x )} + \frac {A c x^{2}}{2} + \frac {B b x^{2}}{2} + \frac {B c x^{4}}{4} & \text {for}\: m = -3 \\\frac {A b m^{2} x^{3} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {12 A b m x^{3} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {35 A b x^{3} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {A c m^{2} x^{5} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {10 A c m x^{5} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {21 A c x^{5} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {B b m^{2} x^{5} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {10 B b m x^{5} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {21 B b x^{5} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {B c m^{2} x^{7} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {8 B c m x^{7} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} + \frac {15 B c x^{7} x^{m}}{m^{3} + 15 m^{2} + 71 m + 105} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-A*b/(4*x**4) - A*c/(2*x**2) - B*b/(2*x**2) + B*c*log(x), Eq(m, -7)), (-A*b/(2*x**2) + A*c*log(x) +
B*b*log(x) + B*c*x**2/2, Eq(m, -5)), (A*b*log(x) + A*c*x**2/2 + B*b*x**2/2 + B*c*x**4/4, Eq(m, -3)), (A*b*m**
2*x**3*x**m/(m**3 + 15*m**2 + 71*m + 105) + 12*A*b*m*x**3*x**m/(m**3 + 15*m**2 + 71*m + 105) + 35*A*b*x**3*x**
m/(m**3 + 15*m**2 + 71*m + 105) + A*c*m**2*x**5*x**m/(m**3 + 15*m**2 + 71*m + 105) + 10*A*c*m*x**5*x**m/(m**3
+ 15*m**2 + 71*m + 105) + 21*A*c*x**5*x**m/(m**3 + 15*m**2 + 71*m + 105) + B*b*m**2*x**5*x**m/(m**3 + 15*m**2
+ 71*m + 105) + 10*B*b*m*x**5*x**m/(m**3 + 15*m**2 + 71*m + 105) + 21*B*b*x**5*x**m/(m**3 + 15*m**2 + 71*m + 1
05) + B*c*m**2*x**7*x**m/(m**3 + 15*m**2 + 71*m + 105) + 8*B*c*m*x**7*x**m/(m**3 + 15*m**2 + 71*m + 105) + 15*
B*c*x**7*x**m/(m**3 + 15*m**2 + 71*m + 105), True))

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