3.2.48 \(\int \frac {(a+b x^2)^2 (c+d x^2)}{x^3} \, dx\) [148]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {457, 77} \begin {gather*} -\frac {a^2 c}{2 x^2}+\frac {1}{2} b x^2 (2 a d+b c)+a \log (x) (a d+2 b c)+\frac {1}{4} b^2 d x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

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])

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

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Mathematica [A]
time = 0.02, size = 49, normalized size = 0.96 \begin {gather*} \frac {1}{4} \left (-\frac {2 a^2 c}{x^2}+2 b (b c+2 a d) x^2+b^2 d x^4+4 a (2 b c+a d) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 48, normalized size = 0.94

method	result	size
default	\(\frac {b^{2} d \,x^{4}}{4}+a b d \,x^{2}+\frac {b^{2} c \,x^{2}}{2}-\frac {a^{2} c}{2 x^{2}}+a \left (a d +2 b c \right ) \ln \left (x \right )\)	\(48\)
norman	\(\frac {\left (a b d +\frac {1}{2} b^{2} c \right ) x^{4}-\frac {a^{2} c}{2}+\frac {b^{2} d \,x^{6}}{4}}{x^{2}}+\left (a^{2} d +2 a b c \right ) \ln \left (x \right )\)	\(51\)
risch	\(\frac {b^{2} d \,x^{4}}{4}+a b d \,x^{2}+\frac {b^{2} c \,x^{2}}{2}+a^{2} d +a b c +\frac {b^{2} c^{2}}{4 d}-\frac {a^{2} c}{2 x^{2}}+\ln \left (x \right ) a^{2} d +2 \ln \left (x \right ) a b c\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 52, normalized size = 1.02 \begin {gather*} \frac {1}{4} \, b^{2} d x^{4} + \frac {1}{2} \, {\left (b^{2} c + 2 \, a b d\right )} x^{2} + \frac {1}{2} \, {\left (2 \, a b c + a^{2} d\right )} \log \left (x^{2}\right ) - \frac {a^{2} c}{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)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.98, size = 54, normalized size = 1.06 \begin {gather*} \frac {b^{2} d x^{6} + 2 \, {\left (b^{2} c + 2 \, a b d\right )} x^{4} + 4 \, {\left (2 \, a b c + a^{2} d\right )} x^{2} \log \left (x\right ) - 2 \, a^{2} c}{4 \, 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^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.10, size = 48, normalized size = 0.94 \begin {gather*} - \frac {a^{2} c}{2 x^{2}} + a \left (a d + 2 b c\right ) \log {\left (x \right )} + \frac {b^{2} d x^{4}}{4} + x^{2} \left (a b d + \frac {b^{2} c}{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**3,x)

[Out]

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

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Giac [A]
time = 0.80, size = 70, normalized size = 1.37 \begin {gather*} \frac {1}{4} \, b^{2} d x^{4} + \frac {1}{2} \, b^{2} c x^{2} + a b d x^{2} + \frac {1}{2} \, {\left (2 \, a b c + a^{2} d\right )} \log \left (x^{2}\right ) - \frac {2 \, a b c x^{2} + a^{2} d x^{2} + a^{2} c}{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)/x^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.05, size = 48, normalized size = 0.94 \begin {gather*} x^2\,\left (\frac {c\,b^2}{2}+a\,d\,b\right )+\ln \left (x\right )\,\left (d\,a^2+2\,b\,c\,a\right )-\frac {a^2\,c}{2\,x^2}+\frac {b^2\,d\,x^4}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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