3.1.2 \(\int x (A+B x^2) (b x^2+c x^4) \, dx\) [2]

Optimal. Leaf size=33 \[ \frac {1}{4} A b x^4+\frac {1}{6} (b B+A c) x^6+\frac {1}{8} B c x^8 \]

[Out]

1/4*A*b*x^4+1/6*(A*c+B*b)*x^6+1/8*B*c*x^8

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Rubi [A]
time = 0.03, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1598, 457, 77} \begin {gather*} \frac {1}{6} x^6 (A c+b B)+\frac {1}{4} A b x^4+\frac {1}{8} B c x^8 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x^2)*(b*x^2 + c*x^4),x]

[Out]

(A*b*x^4)/4 + ((b*B + A*c)*x^6)/6 + (B*c*x^8)/8

Rule 77

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])

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1598

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 \left (A+B x^2\right ) \left (b x^2+c x^4\right ) \, dx &=\int x^3 \left (A+B x^2\right ) \left (b+c x^2\right ) \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int x (A+B x) (b+c x) \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (A b x+(b B+A c) x^2+B c x^3\right ) \, dx,x,x^2\right )\\ &=\frac {1}{4} A b x^4+\frac {1}{6} (b B+A c) x^6+\frac {1}{8} B c x^8\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 33, normalized size = 1.00 \begin {gather*} \frac {1}{4} A b x^4+\frac {1}{6} (b B+A c) x^6+\frac {1}{8} B c x^8 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b*x^4)/4 + ((b*B + A*c)*x^6)/6 + (B*c*x^8)/8

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Maple [A]
time = 0.21, size = 28, normalized size = 0.85

method	result	size
default	\(\frac {A b \,x^{4}}{4}+\frac {\left (A c +B b \right ) x^{6}}{6}+\frac {B c \,x^{8}}{8}\)	\(28\)
norman	\(\frac {B c \,x^{8}}{8}+\left (\frac {A c}{6}+\frac {B b}{6}\right ) x^{6}+\frac {A b \,x^{4}}{4}\)	\(29\)
risch	\(\frac {1}{4} A b \,x^{4}+\frac {1}{6} x^{6} A c +\frac {1}{6} b B \,x^{6}+\frac {1}{8} B c \,x^{8}\)	\(30\)
gosper	\(\frac {x^{4} \left (3 B c \,x^{4}+4 A c \,x^{2}+4 b B \,x^{2}+6 A b \right )}{24}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x^2+A)*(c*x^4+b*x^2),x,method=_RETURNVERBOSE)

[Out]

1/4*A*b*x^4+1/6*(A*c+B*b)*x^6+1/8*B*c*x^8

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Maxima [A]
time = 0.29, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{8} \, B c x^{8} + \frac {1}{6} \, {\left (B b + A c\right )} x^{6} + \frac {1}{4} \, A b x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*c*x^8 + 1/6*(B*b + A*c)*x^6 + 1/4*A*b*x^4

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Fricas [A]
time = 3.16, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{8} \, B c x^{8} + \frac {1}{6} \, {\left (B b + A c\right )} x^{6} + \frac {1}{4} \, A b x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*c*x^8 + 1/6*(B*b + A*c)*x^6 + 1/4*A*b*x^4

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Sympy [A]
time = 0.01, size = 29, normalized size = 0.88 \begin {gather*} \frac {A b x^{4}}{4} + \frac {B c x^{8}}{8} + x^{6} \left (\frac {A c}{6} + \frac {B b}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*b*x**4/4 + B*c*x**8/8 + x**6*(A*c/6 + B*b/6)

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Giac [A]
time = 1.66, size = 29, normalized size = 0.88 \begin {gather*} \frac {1}{8} \, B c x^{8} + \frac {1}{6} \, B b x^{6} + \frac {1}{6} \, A c x^{6} + \frac {1}{4} \, A b x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*c*x^8 + 1/6*B*b*x^6 + 1/6*A*c*x^6 + 1/4*A*b*x^4

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Mupad [B]
time = 0.06, size = 28, normalized size = 0.85 \begin {gather*} \frac {B\,c\,x^8}{8}+\left (\frac {A\,c}{6}+\frac {B\,b}{6}\right )\,x^6+\frac {A\,b\,x^4}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(A + B*x^2)*(b*x^2 + c*x^4),x)

[Out]

x^6*((A*c)/6 + (B*b)/6) + (A*b*x^4)/4 + (B*c*x^8)/8

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