3.10 \(\int \frac {(A+B x^2) (b x^2+c x^4)}{x^7} \, dx\)

Optimal. Leaf size=29 \[ -\frac {A c+b B}{2 x^2}-\frac {A b}{4 x^4}+B c \log (x) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 29, 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, 446, 76} \[ -\frac {A c+b B}{2 x^2}-\frac {A b}{4 x^4}+B c \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4))/x^7,x]

[Out]

-(A*b)/(4*x^4) - (b*B + A*c)/(2*x^2) + B*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 )}{x^7} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )}{x^5} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) (b+c x)}{x^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {A b}{x^3}+\frac {b B+A c}{x^2}+\frac {B c}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac {A b}{4 x^4}-\frac {b B+A c}{2 x^2}+B c \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 31, normalized size = 1.07 \[ \frac {-A c-b B}{2 x^2}-\frac {A b}{4 x^4}+B c \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4))/x^7,x]

[Out]

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

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fricas [A]  time = 0.82, size = 31, normalized size = 1.07 \[ \frac {4 \, B c x^{4} \log \relax (x) - 2 \, {\left (B b + A c\right )} x^{2} - A b}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^7,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 39, normalized size = 1.34 \[ \frac {1}{2} \, B c \log \left (x^{2}\right ) - \frac {3 \, B c x^{4} + 2 \, B b x^{2} + 2 \, A c x^{2} + A b}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^7,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 28, normalized size = 0.97 \[ B c \ln \relax (x )-\frac {A c}{2 x^{2}}-\frac {B b}{2 x^{2}}-\frac {A b}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)/x^7,x)

[Out]

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

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maxima [A]  time = 1.40, size = 30, normalized size = 1.03 \[ \frac {1}{2} \, B c \log \left (x^{2}\right ) - \frac {2 \, {\left (B b + A c\right )} x^{2} + A b}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^7,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 29, normalized size = 1.00 \[ B\,c\,\ln \relax (x)-\frac {\left (\frac {A\,c}{2}+\frac {B\,b}{2}\right )\,x^2+\frac {A\,b}{4}}{x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4))/x^7,x)

[Out]

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

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sympy [A]  time = 0.37, size = 29, normalized size = 1.00 \[ B c \log {\relax (x )} + \frac {- A b + x^{2} \left (- 2 A c - 2 B b\right )}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)/x**7,x)

[Out]

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

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