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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.053257, antiderivative size = 51, normalized size of antiderivative = 1., 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} \[ -\frac{A b^2}{2 x^2}+\frac{1}{2} c x^2 (A c+2 b B)+b \log (x) (2 A c+b B)+\frac{1}{4} B c^2 x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0279669, size = 49, normalized size = 0.96 \[ \frac{1}{4} \left (-\frac{2 A b^2}{x^2}+2 c x^2 (A c+2 b B)+4 b \log (x) (2 A c+b B)+B c^2 x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 50, normalized size = 1. \begin{align*}{\frac{B{c}^{2}{x}^{4}}{4}}+{\frac{A{x}^{2}{c}^{2}}{2}}+B{x}^{2}bc+2\,A\ln \left ( x \right ) bc+B\ln \left ( x \right ){b}^{2}-{\frac{A{b}^{2}}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.08898, size = 70, normalized size = 1.37 \begin{align*} \frac{1}{4} \, B c^{2} x^{4} + \frac{1}{2} \,{\left (2 \, B b c + A c^{2}\right )} x^{2} + \frac{1}{2} \,{\left (B b^{2} + 2 \, A b c\right )} \log \left (x^{2}\right ) - \frac{A b^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.506127, size = 122, normalized size = 2.39 \begin{align*} \frac{B c^{2} x^{6} + 2 \,{\left (2 \, B b c + A c^{2}\right )} x^{4} + 4 \,{\left (B b^{2} + 2 \, A b c\right )} x^{2} \log \left (x\right ) - 2 \, A b^{2}}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.371721, size = 48, normalized size = 0.94 \begin{align*} - \frac{A b^{2}}{2 x^{2}} + \frac{B c^{2} x^{4}}{4} + b \left (2 A c + B b\right ) \log{\left (x \right )} + x^{2} \left (\frac{A c^{2}}{2} + B b c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.2133, size = 95, normalized size = 1.86 \begin{align*} \frac{1}{4} \, B c^{2} x^{4} + B b c x^{2} + \frac{1}{2} \, A c^{2} x^{2} + \frac{1}{2} \,{\left (B b^{2} + 2 \, A b c\right )} \log \left (x^{2}\right ) - \frac{B b^{2} x^{2} + 2 \, A b c x^{2} + A b^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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