3.2.46 \(\int \frac {(a+b x^2)^2 (c+d x^2)}{x} \, dx\) [146]

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

[Out]

a*b*c*x^2+1/4*b^2*c*x^4+1/6*d*(b*x^2+a)^3/b+a^2*c*ln(x)

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Rubi [A]
time = 0.02, 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 = {457, 81, 45} \begin {gather*} a^2 c \log (x)+a b c x^2+\frac {d \left (a+b x^2\right )^3}{6 b}+\frac {1}{4} b^2 c x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2))/x,x]

[Out]

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

Rule 45

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 81

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 457

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

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Mathematica [A]
time = 0.01, size = 51, normalized size = 1.19 \begin {gather*} \frac {1}{2} a (2 b c+a d) x^2+\frac {1}{4} b (b c+2 a d) x^4+\frac {1}{6} b^2 d x^6+a^2 c \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2))/x,x]

[Out]

(a*(2*b*c + a*d)*x^2)/2 + (b*(b*c + 2*a*d)*x^4)/4 + (b^2*d*x^6)/6 + a^2*c*Log[x]

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Maple [A]
time = 0.07, size = 51, normalized size = 1.19

method	result	size
norman	\(\left (\frac {1}{2} a^{2} d +a b c \right ) x^{2}+\left (\frac {1}{2} a b d +\frac {1}{4} b^{2} c \right ) x^{4}+\frac {b^{2} d \,x^{6}}{6}+a^{2} c \ln \left (x \right )\)	\(49\)
default	\(\frac {b^{2} d \,x^{6}}{6}+\frac {a b d \,x^{4}}{2}+\frac {b^{2} c \,x^{4}}{4}+\frac {a^{2} d \,x^{2}}{2}+a b c \,x^{2}+a^{2} c \ln \left (x \right )\)	\(51\)
risch	\(\frac {b^{2} d \,x^{6}}{6}+\frac {a b d \,x^{4}}{2}+\frac {b^{2} c \,x^{4}}{4}+\frac {a^{2} d \,x^{2}}{2}+a b c \,x^{2}+a^{2} c \ln \left (x \right )\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)/x,x,method=_RETURNVERBOSE)

[Out]

1/6*b^2*d*x^6+1/2*a*b*d*x^4+1/4*b^2*c*x^4+1/2*a^2*d*x^2+a*b*c*x^2+a^2*c*ln(x)

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Maxima [A]
time = 0.37, size = 52, normalized size = 1.21 \begin {gather*} \frac {1}{6} \, b^{2} d x^{6} + \frac {1}{4} \, {\left (b^{2} c + 2 \, a b d\right )} x^{4} + \frac {1}{2} \, a^{2} c \log \left (x^{2}\right ) + \frac {1}{2} \, {\left (2 \, a b c + a^{2} d\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^2*d*x^6 + 1/4*(b^2*c + 2*a*b*d)*x^4 + 1/2*a^2*c*log(x^2) + 1/2*(2*a*b*c + a^2*d)*x^2

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Fricas [A]
time = 0.97, size = 49, normalized size = 1.14 \begin {gather*} \frac {1}{6} \, b^{2} d x^{6} + \frac {1}{4} \, {\left (b^{2} c + 2 \, a b d\right )} x^{4} + a^{2} c \log \left (x\right ) + \frac {1}{2} \, {\left (2 \, a b c + a^{2} d\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^2*d*x^6 + 1/4*(b^2*c + 2*a*b*d)*x^4 + a^2*c*log(x) + 1/2*(2*a*b*c + a^2*d)*x^2

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Sympy [A]
time = 0.05, size = 49, normalized size = 1.14 \begin {gather*} a^{2} c \log {\left (x \right )} + \frac {b^{2} d x^{6}}{6} + x^{4} \left (\frac {a b d}{2} + \frac {b^{2} c}{4}\right ) + x^{2} \left (\frac {a^{2} d}{2} + a b c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)/x,x)

[Out]

a**2*c*log(x) + b**2*d*x**6/6 + x**4*(a*b*d/2 + b**2*c/4) + x**2*(a**2*d/2 + a*b*c)

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Giac [A]
time = 0.92, size = 53, normalized size = 1.23 \begin {gather*} \frac {1}{6} \, b^{2} d x^{6} + \frac {1}{4} \, b^{2} c x^{4} + \frac {1}{2} \, a b d x^{4} + a b c x^{2} + \frac {1}{2} \, a^{2} d x^{2} + \frac {1}{2} \, a^{2} c \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^2*d*x^6 + 1/4*b^2*c*x^4 + 1/2*a*b*d*x^4 + a*b*c*x^2 + 1/2*a^2*d*x^2 + 1/2*a^2*c*log(x^2)

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Mupad [B]
time = 0.02, size = 48, normalized size = 1.12 \begin {gather*} x^2\,\left (\frac {d\,a^2}{2}+b\,c\,a\right )+x^4\,\left (\frac {c\,b^2}{4}+\frac {a\,d\,b}{2}\right )+\frac {b^2\,d\,x^6}{6}+a^2\,c\,\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2))/x,x)

[Out]

x^2*((a^2*d)/2 + a*b*c) + x^4*((b^2*c)/4 + (a*b*d)/2) + (b^2*d*x^6)/6 + a^2*c*log(x)

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