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

Optimal. Leaf size=55 \[ \frac {1}{4} A b^2 x^4+\frac {1}{8} c x^8 (A c+2 b B)+\frac {1}{6} b x^6 (2 A c+b B)+\frac {1}{10} B c^2 x^{10} \]

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Rubi [A]  time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1584, 446, 76} \begin {gather*} \frac {1}{4} A b^2 x^4+\frac {1}{8} c x^8 (A c+2 b B)+\frac {1}{6} b x^6 (2 A c+b B)+\frac {1}{10} B c^2 x^{10} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b^2*x^4)/4 + (b*(b*B + 2*A*c)*x^6)/6 + (c*(2*b*B + A*c)*x^8)/8 + (B*c^2*x^10)/10

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])

Rule 446

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 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 )^2}{x} \, dx &=\int x^3 \left (A+B x^2\right ) \left (b+c x^2\right )^2 \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int x (A+B x) (b+c x)^2 \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (A b^2 x+b (b B+2 A c) x^2+c (2 b B+A c) x^3+B c^2 x^4\right ) \, dx,x,x^2\right )\\ &=\frac {1}{4} A b^2 x^4+\frac {1}{6} b (b B+2 A c) x^6+\frac {1}{8} c (2 b B+A c) x^8+\frac {1}{10} B c^2 x^{10}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 55, normalized size = 1.00 \begin {gather*} \frac {1}{4} A b^2 x^4+\frac {1}{8} c x^8 (A c+2 b B)+\frac {1}{6} b x^6 (2 A c+b B)+\frac {1}{10} B c^2 x^{10} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b^2*x^4)/4 + (b*(b*B + 2*A*c)*x^6)/6 + (c*(2*b*B + A*c)*x^8)/8 + (B*c^2*x^10)/10

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IntegrateAlgebraic [A]  time = 0.03, size = 57, normalized size = 1.04 \begin {gather*} \frac {1}{120} x^4 \left (30 A b^2+40 A b c x^2+15 A c^2 x^4+20 b^2 B x^2+30 b B c x^4+12 B c^2 x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(30*A*b^2 + 20*b^2*B*x^2 + 40*A*b*c*x^2 + 30*b*B*c*x^4 + 15*A*c^2*x^4 + 12*B*c^2*x^6))/120

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fricas [A]  time = 0.38, size = 51, normalized size = 0.93 \begin {gather*} \frac {1}{10} \, B c^{2} x^{10} + \frac {1}{8} \, {\left (2 \, B b c + A c^{2}\right )} x^{8} + \frac {1}{4} \, A b^{2} x^{4} + \frac {1}{6} \, {\left (B b^{2} + 2 \, A b c\right )} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*B*c^2*x^10 + 1/8*(2*B*b*c + A*c^2)*x^8 + 1/4*A*b^2*x^4 + 1/6*(B*b^2 + 2*A*b*c)*x^6

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giac [A]  time = 0.15, size = 53, normalized size = 0.96 \begin {gather*} \frac {1}{10} \, B c^{2} x^{10} + \frac {1}{4} \, B b c x^{8} + \frac {1}{8} \, A c^{2} x^{8} + \frac {1}{6} \, B b^{2} x^{6} + \frac {1}{3} \, A b c x^{6} + \frac {1}{4} \, A b^{2} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*B*c^2*x^10 + 1/4*B*b*c*x^8 + 1/8*A*c^2*x^8 + 1/6*B*b^2*x^6 + 1/3*A*b*c*x^6 + 1/4*A*b^2*x^4

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maple [A]  time = 0.04, size = 52, normalized size = 0.95 \begin {gather*} \frac {B \,c^{2} x^{10}}{10}+\frac {\left (A \,c^{2}+2 b B c \right ) x^{8}}{8}+\frac {A \,b^{2} x^{4}}{4}+\frac {\left (2 A b c +B \,b^{2}\right ) x^{6}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*B*c^2*x^10+1/8*(A*c^2+2*B*b*c)*x^8+1/6*(2*A*b*c+B*b^2)*x^6+1/4*A*b^2*x^4

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maxima [A]  time = 1.37, size = 51, normalized size = 0.93 \begin {gather*} \frac {1}{10} \, B c^{2} x^{10} + \frac {1}{8} \, {\left (2 \, B b c + A c^{2}\right )} x^{8} + \frac {1}{4} \, A b^{2} x^{4} + \frac {1}{6} \, {\left (B b^{2} + 2 \, A b c\right )} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*B*c^2*x^10 + 1/8*(2*B*b*c + A*c^2)*x^8 + 1/4*A*b^2*x^4 + 1/6*(B*b^2 + 2*A*b*c)*x^6

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mupad [B]  time = 0.05, size = 51, normalized size = 0.93 \begin {gather*} x^6\,\left (\frac {B\,b^2}{6}+\frac {A\,c\,b}{3}\right )+x^8\,\left (\frac {A\,c^2}{8}+\frac {B\,b\,c}{4}\right )+\frac {A\,b^2\,x^4}{4}+\frac {B\,c^2\,x^{10}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^6*((B*b^2)/6 + (A*b*c)/3) + x^8*((A*c^2)/8 + (B*b*c)/4) + (A*b^2*x^4)/4 + (B*c^2*x^10)/10

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sympy [A]  time = 0.08, size = 53, normalized size = 0.96 \begin {gather*} \frac {A b^{2} x^{4}}{4} + \frac {B c^{2} x^{10}}{10} + x^{8} \left (\frac {A c^{2}}{8} + \frac {B b c}{4}\right ) + x^{6} \left (\frac {A b c}{3} + \frac {B b^{2}}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*b**2*x**4/4 + B*c**2*x**10/10 + x**8*(A*c**2/8 + B*b*c/4) + x**6*(A*b*c/3 + B*b**2/6)

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