3.17 \(\int \frac{(A+B x^2) (b x^2+c x^4)^2}{x^5} \, dx\)

Optimal. Leaf size=43 \[ A b^2 \log (x)+A b c x^2+\frac{1}{4} A c^2 x^4+\frac{B \left (b+c x^2\right )^3}{6 c} \]

[Out]

A*b*c*x^2 + (A*c^2*x^4)/4 + (B*(b + c*x^2)^3)/(6*c) + A*b^2*Log[x]

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Rubi [A]  time = 0.0402506, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1584, 446, 80, 43} \[ A b^2 \log (x)+A b c x^2+\frac{1}{4} A c^2 x^4+\frac{B \left (b+c x^2\right )^3}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^5,x]

[Out]

A*b*c*x^2 + (A*c^2*x^4)/4 + (B*(b + c*x^2)^3)/(6*c) + A*b^2*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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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])

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^2}{x^5} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )^2}{x} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) (b+c x)^2}{x} \, dx,x,x^2\right )\\ &=\frac{B \left (b+c x^2\right )^3}{6 c}+\frac{1}{2} A \operatorname{Subst}\left (\int \frac{(b+c x)^2}{x} \, dx,x,x^2\right )\\ &=\frac{B \left (b+c x^2\right )^3}{6 c}+\frac{1}{2} A \operatorname{Subst}\left (\int \left (2 b c+\frac{b^2}{x}+c^2 x\right ) \, dx,x,x^2\right )\\ &=A b c x^2+\frac{1}{4} A c^2 x^4+\frac{B \left (b+c x^2\right )^3}{6 c}+A b^2 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0163226, size = 51, normalized size = 1.19 \[ A b^2 \log (x)+\frac{1}{4} c x^4 (A c+2 b B)+\frac{1}{2} b x^2 (2 A c+b B)+\frac{1}{6} B c^2 x^6 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^5,x]

[Out]

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

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Maple [A]  time = 0.002, size = 51, normalized size = 1.2 \begin{align*}{\frac{B{c}^{2}{x}^{6}}{6}}+{\frac{A{c}^{2}{x}^{4}}{4}}+{\frac{B{x}^{4}bc}{2}}+Abc{x}^{2}+{\frac{B{x}^{2}{b}^{2}}{2}}+A{b}^{2}\ln \left ( x \right ) \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^5,x)

[Out]

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

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Maxima [A]  time = 1.1075, size = 70, normalized size = 1.63 \begin{align*} \frac{1}{6} \, B c^{2} x^{6} + \frac{1}{4} \,{\left (2 \, B b c + A c^{2}\right )} x^{4} + \frac{1}{2} \, A b^{2} \log \left (x^{2}\right ) + \frac{1}{2} \,{\left (B b^{2} + 2 \, A b c\right )} 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^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.450696, size = 116, normalized size = 2.7 \begin{align*} \frac{1}{6} \, B c^{2} x^{6} + \frac{1}{4} \,{\left (2 \, B b c + A c^{2}\right )} x^{4} + A b^{2} \log \left (x\right ) + \frac{1}{2} \,{\left (B b^{2} + 2 \, A b c\right )} 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^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.289352, size = 49, normalized size = 1.14 \begin{align*} A b^{2} \log{\left (x \right )} + \frac{B c^{2} x^{6}}{6} + x^{4} \left (\frac{A c^{2}}{4} + \frac{B b c}{2}\right ) + x^{2} \left (A b c + \frac{B b^{2}}{2}\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**5,x)

[Out]

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

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Giac [A]  time = 1.27175, size = 72, normalized size = 1.67 \begin{align*} \frac{1}{6} \, B c^{2} x^{6} + \frac{1}{2} \, B b c x^{4} + \frac{1}{4} \, A c^{2} x^{4} + \frac{1}{2} \, B b^{2} x^{2} + A b c x^{2} + \frac{1}{2} \, A b^{2} \log \left (x^{2}\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^5,x, algorithm="giac")

[Out]

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