3.14 \(\int \frac {(a+b x^2)^2 (A+B x^2)}{x} \, dx\)

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

[Out]

a*A*b*x^2+1/4*A*b^2*x^4+1/6*B*(b*x^2+a)^3/b+a^2*A*ln(x)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {446, 80, 43} \[ a^2 A \log (x)+a A b x^2+\frac {B \left (a+b x^2\right )^3}{6 b}+\frac {1}{4} A b^2 x^4 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(A + B*x^2))/x,x]

[Out]

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

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])

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 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]]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 51, normalized size = 1.19 \[ a^2 A \log (x)+\frac {1}{4} b x^4 (2 a B+A b)+\frac {1}{2} a x^2 (a B+2 A b)+\frac {1}{6} b^2 B x^6 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 49, normalized size = 1.14 \[ \frac {1}{6} \, B b^{2} x^{6} + \frac {1}{4} \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + A a^{2} \log \relax (x) + \frac {1}{2} \, {\left (B a^{2} + 2 \, A a b\right )} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.31, size = 53, normalized size = 1.23 \[ \frac {1}{6} \, B b^{2} x^{6} + \frac {1}{2} \, B a b x^{4} + \frac {1}{4} \, A b^{2} x^{4} + \frac {1}{2} \, B a^{2} x^{2} + A a b x^{2} + \frac {1}{2} \, A a^{2} \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 51, normalized size = 1.19 \[ \frac {B \,b^{2} x^{6}}{6}+\frac {A \,b^{2} x^{4}}{4}+\frac {B a b \,x^{4}}{2}+A a b \,x^{2}+\frac {B \,a^{2} x^{2}}{2}+A \,a^{2} \ln \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.30, size = 52, normalized size = 1.21 \[ \frac {1}{6} \, B b^{2} x^{6} + \frac {1}{4} \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + \frac {1}{2} \, A a^{2} \log \left (x^{2}\right ) + \frac {1}{2} \, {\left (B a^{2} + 2 \, A a b\right )} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 48, normalized size = 1.12 \[ x^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )+x^4\,\left (\frac {A\,b^2}{4}+\frac {B\,a\,b}{2}\right )+\frac {B\,b^2\,x^6}{6}+A\,a^2\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^2)/x,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 49, normalized size = 1.14 \[ A a^{2} \log {\relax (x )} + \frac {B b^{2} x^{6}}{6} + x^{4} \left (\frac {A b^{2}}{4} + \frac {B a b}{2}\right ) + x^{2} \left (A a b + \frac {B a^{2}}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________