3.21 \(\int \frac {(A+B x^2) (b x^2+c x^4)^2}{x^9} \, dx\)

Optimal. Leaf size=51 \[ -\frac {A b^2}{4 x^4}-\frac {b (2 A c+b B)}{2 x^2}+c \log (x) (A c+2 b B)+\frac {1}{2} B c^2 x^2 \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 51, 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} \[ -\frac {A b^2}{4 x^4}-\frac {b (2 A c+b B)}{2 x^2}+c \log (x) (A c+2 b B)+\frac {1}{2} B c^2 x^2 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^9,x]

[Out]

-(A*b^2)/(4*x^4) - (b*(b*B + 2*A*c))/(2*x^2) + (B*c^2*x^2)/2 + c*(2*b*B + A*c)*Log[x]

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^9} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^2}{x^5} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) (b+c x)^2}{x^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (B c^2+\frac {A b^2}{x^3}+\frac {b (b B+2 A c)}{x^2}+\frac {c (2 b B+A c)}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac {A b^2}{4 x^4}-\frac {b (b B+2 A c)}{2 x^2}+\frac {1}{2} B c^2 x^2+c (2 b B+A c) \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 50, normalized size = 0.98 \[ c \log (x) (A c+2 b B)-\frac {A b \left (b+4 c x^2\right )+2 B x^2 \left (b^2-c^2 x^4\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^9,x]

[Out]

-1/4*(A*b*(b + 4*c*x^2) + 2*B*x^2*(b^2 - c^2*x^4))/x^4 + c*(2*b*B + A*c)*Log[x]

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fricas [A]  time = 0.85, size = 55, normalized size = 1.08 \[ \frac {2 \, B c^{2} x^{6} + 4 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} \log \relax (x) - A b^{2} - 2 \, {\left (B b^{2} + 2 \, A b c\right )} x^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^2/x^9,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 72, normalized size = 1.41 \[ \frac {1}{2} \, B c^{2} x^{2} + \frac {1}{2} \, {\left (2 \, B b c + A c^{2}\right )} \log \left (x^{2}\right ) - \frac {6 \, B b c x^{4} + 3 \, A c^{2} x^{4} + 2 \, B b^{2} x^{2} + 4 \, A b c x^{2} + A b^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^2/x^9,x, algorithm="giac")

[Out]

1/2*B*c^2*x^2 + 1/2*(2*B*b*c + A*c^2)*log(x^2) - 1/4*(6*B*b*c*x^4 + 3*A*c^2*x^4 + 2*B*b^2*x^2 + 4*A*b*c*x^2 +
A*b^2)/x^4

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maple [A]  time = 0.06, size = 51, normalized size = 1.00 \[ \frac {B \,c^{2} x^{2}}{2}+A \,c^{2} \ln \relax (x )+2 B b c \ln \relax (x )-\frac {A b c}{x^{2}}-\frac {B \,b^{2}}{2 x^{2}}-\frac {A \,b^{2}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^2/x^9,x)

[Out]

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

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maxima [A]  time = 1.37, size = 54, normalized size = 1.06 \[ \frac {1}{2} \, B c^{2} x^{2} + \frac {1}{2} \, {\left (2 \, B b c + A c^{2}\right )} \log \left (x^{2}\right ) - \frac {A b^{2} + 2 \, {\left (B b^{2} + 2 \, A b c\right )} x^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^2/x^9,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.08, size = 51, normalized size = 1.00 \[ \ln \relax (x)\,\left (A\,c^2+2\,B\,b\,c\right )-\frac {x^2\,\left (\frac {B\,b^2}{2}+A\,c\,b\right )+\frac {A\,b^2}{4}}{x^4}+\frac {B\,c^2\,x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^9,x)

[Out]

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

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sympy [A]  time = 0.55, size = 51, normalized size = 1.00 \[ \frac {B c^{2} x^{2}}{2} + c \left (A c + 2 B b\right ) \log {\relax (x )} + \frac {- A b^{2} + x^{2} \left (- 4 A b c - 2 B b^{2}\right )}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**2/x**9,x)

[Out]

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

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