3.2.57 \(\int \frac {(a+b x^2)^2 (c+d x^2)^2}{x^3} \, dx\) [157]

Optimal. Leaf size=84 \[ -\frac {a^2 c^2}{2 x^2}+\frac {1}{2} \left (b^2 c^2+4 a b c d+a^2 d^2\right ) x^2+\frac {1}{2} b d (b c+a d) x^4+\frac {1}{6} b^2 d^2 x^6+2 a c (b c+a d) \log (x) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 90} \begin {gather*} \frac {1}{2} x^2 \left (a^2 d^2+4 a b c d+b^2 c^2\right )-\frac {a^2 c^2}{2 x^2}+\frac {1}{2} b d x^4 (a d+b c)+2 a c \log (x) (a d+b c)+\frac {1}{6} b^2 d^2 x^6 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^2)/x^3,x]

[Out]

-1/2*(a^2*c^2)/x^2 + ((b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^2)/2 + (b*d*(b*c + a*d)*x^4)/2 + (b^2*d^2*x^6)/6 + 2*a
*c*(b*c + a*d)*Log[x]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

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Mathematica [A]
time = 0.03, size = 83, normalized size = 0.99 \begin {gather*} \frac {1}{6} \left (3 a b d x^2 \left (4 c+d x^2\right )+\frac {3 a^2 \left (-c^2+d^2 x^4\right )}{x^2}+b^2 x^2 \left (3 c^2+3 c d x^2+d^2 x^4\right )+12 a c (b c+a d) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^2)/x^3,x]

[Out]

(3*a*b*d*x^2*(4*c + d*x^2) + (3*a^2*(-c^2 + d^2*x^4))/x^2 + b^2*x^2*(3*c^2 + 3*c*d*x^2 + d^2*x^4) + 12*a*c*(b*
c + a*d)*Log[x])/6

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Maple [A]
time = 0.07, size = 88, normalized size = 1.05

method	result	size
default	\(\frac {b^{2} d^{2} x^{6}}{6}+\frac {a b \,d^{2} x^{4}}{2}+\frac {b^{2} c d \,x^{4}}{2}+\frac {a^{2} d^{2} x^{2}}{2}+2 a b c d \,x^{2}+\frac {b^{2} c^{2} x^{2}}{2}-\frac {a^{2} c^{2}}{2 x^{2}}+2 a c \left (a d +b c \right ) \ln \left (x \right )\)	\(88\)
norman	\(\frac {\left (\frac {1}{2} a b \,d^{2}+\frac {1}{2} b^{2} c d \right ) x^{6}+\left (\frac {1}{2} a^{2} d^{2}+2 a b c d +\frac {1}{2} b^{2} c^{2}\right ) x^{4}-\frac {a^{2} c^{2}}{2}+\frac {b^{2} d^{2} x^{8}}{6}}{x^{2}}+\left (2 a^{2} c d +2 a b \,c^{2}\right ) \ln \left (x \right )\)	\(90\)
risch	\(\frac {b^{2} d^{2} x^{6}}{6}+\frac {a b \,d^{2} x^{4}}{2}+\frac {b^{2} c d \,x^{4}}{2}+\frac {a^{2} d^{2} x^{2}}{2}+2 a b c d \,x^{2}+\frac {b^{2} c^{2} x^{2}}{2}-\frac {a^{2} c^{2}}{2 x^{2}}+2 \ln \left (x \right ) a^{2} c d +2 \ln \left (x \right ) a b \,c^{2}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.35, size = 85, normalized size = 1.01 \begin {gather*} \frac {1}{6} \, b^{2} d^{2} x^{6} + \frac {1}{2} \, {\left (b^{2} c d + a b d^{2}\right )} x^{4} + \frac {1}{2} \, {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{2} - \frac {a^{2} c^{2}}{2 \, x^{2}} + {\left (a b c^{2} + a^{2} c d\right )} \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)^2/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.75, size = 88, normalized size = 1.05 \begin {gather*} \frac {b^{2} d^{2} x^{8} + 3 \, {\left (b^{2} c d + a b d^{2}\right )} x^{6} + 3 \, {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{4} - 3 \, a^{2} c^{2} + 12 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} \log \left (x\right )}{6 \, 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)^2/x^3,x, algorithm="fricas")

[Out]

1/6*(b^2*d^2*x^8 + 3*(b^2*c*d + a*b*d^2)*x^6 + 3*(b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4 - 3*a^2*c^2 + 12*(a*b*c^2
+ a^2*c*d)*x^2*log(x))/x^2

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Sympy [A]
time = 0.16, size = 87, normalized size = 1.04 \begin {gather*} - \frac {a^{2} c^{2}}{2 x^{2}} + 2 a c \left (a d + b c\right ) \log {\left (x \right )} + \frac {b^{2} d^{2} x^{6}}{6} + x^{4} \left (\frac {a b d^{2}}{2} + \frac {b^{2} c d}{2}\right ) + x^{2} \left (\frac {a^{2} d^{2}}{2} + 2 a b c d + \frac {b^{2} c^{2}}{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)**2/x**3,x)

[Out]

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

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Giac [A]
time = 1.40, size = 114, normalized size = 1.36 \begin {gather*} \frac {1}{6} \, b^{2} d^{2} x^{6} + \frac {1}{2} \, b^{2} c d x^{4} + \frac {1}{2} \, a b d^{2} x^{4} + \frac {1}{2} \, b^{2} c^{2} x^{2} + 2 \, a b c d x^{2} + \frac {1}{2} \, a^{2} d^{2} x^{2} + {\left (a b c^{2} + a^{2} c d\right )} \log \left (x^{2}\right ) - \frac {2 \, a b c^{2} x^{2} + 2 \, a^{2} c d x^{2} + a^{2} c^{2}}{2 \, 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)^2/x^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.02, size = 82, normalized size = 0.98 \begin {gather*} x^2\,\left (\frac {a^2\,d^2}{2}+2\,a\,b\,c\,d+\frac {b^2\,c^2}{2}\right )+\ln \left (x\right )\,\left (2\,d\,a^2\,c+2\,b\,a\,c^2\right )-\frac {a^2\,c^2}{2\,x^2}+\frac {b^2\,d^2\,x^6}{6}+\frac {b\,d\,x^4\,\left (a\,d+b\,c\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^2)/x^3,x)

[Out]

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

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