3.4 \(\int \frac {(A+B x^2) (b x^2+c x^4)}{x} \, dx\)

Optimal. Leaf size=33 \[ \frac {1}{4} x^4 (A c+b B)+\frac {1}{2} A b x^2+\frac {1}{6} B c x^6 \]

[Out]

1/2*A*b*x^2+1/4*(A*c+B*b)*x^4+1/6*B*c*x^6

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Rubi [A]  time = 0.04, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1584, 444, 43} \[ \frac {1}{4} x^4 (A c+b B)+\frac {1}{2} A b x^2+\frac {1}{6} B c x^6 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b*x^2)/2 + ((b*B + A*c)*x^4)/4 + (B*c*x^6)/6

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 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 \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )}{x} \, dx &=\int x \left (A+B x^2\right ) \left (b+c x^2\right ) \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int (A+B x) (b+c x) \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (A b+(b B+A c) x+B c x^2\right ) \, dx,x,x^2\right )\\ &=\frac {1}{2} A b x^2+\frac {1}{4} (b B+A c) x^4+\frac {1}{6} B c x^6\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.00 \[ \frac {1}{4} x^4 (A c+b B)+\frac {1}{2} A b x^2+\frac {1}{6} B c x^6 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b*x^2)/2 + ((b*B + A*c)*x^4)/4 + (B*c*x^6)/6

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fricas [A]  time = 0.56, size = 27, normalized size = 0.82 \[ \frac {1}{6} \, B c x^{6} + \frac {1}{4} \, {\left (B b + A c\right )} x^{4} + \frac {1}{2} \, A b x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*c*x^6 + 1/4*(B*b + A*c)*x^4 + 1/2*A*b*x^2

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giac [A]  time = 0.15, size = 29, normalized size = 0.88 \[ \frac {1}{6} \, B c x^{6} + \frac {1}{4} \, B b x^{4} + \frac {1}{4} \, A c x^{4} + \frac {1}{2} \, A b x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*c*x^6 + 1/4*B*b*x^4 + 1/4*A*c*x^4 + 1/2*A*b*x^2

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maple [A]  time = 0.04, size = 28, normalized size = 0.85 \[ \frac {B c \,x^{6}}{6}+\frac {A b \,x^{2}}{2}+\frac {\left (A c +b B \right ) x^{4}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*A*b*x^2+1/4*(A*c+B*b)*x^4+1/6*B*c*x^6

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maxima [A]  time = 1.31, size = 27, normalized size = 0.82 \[ \frac {1}{6} \, B c x^{6} + \frac {1}{4} \, {\left (B b + A c\right )} x^{4} + \frac {1}{2} \, A b x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*c*x^6 + 1/4*(B*b + A*c)*x^4 + 1/2*A*b*x^2

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mupad [B]  time = 0.06, size = 28, normalized size = 0.85 \[ \frac {B\,c\,x^6}{6}+\left (\frac {A\,c}{4}+\frac {B\,b}{4}\right )\,x^4+\frac {A\,b\,x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^4*((A*c)/4 + (B*b)/4) + (A*b*x^2)/2 + (B*c*x^6)/6

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sympy [A]  time = 0.07, size = 29, normalized size = 0.88 \[ \frac {A b x^{2}}{2} + \frac {B c x^{6}}{6} + x^{4} \left (\frac {A c}{4} + \frac {B b}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*b*x**2/2 + B*c*x**6/6 + x**4*(A*c/4 + B*b/4)

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