3.1.68 \(\int \frac {x^2 (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [68]

Optimal. Leaf size=70 \[ -\frac {A}{b^2 x}+\frac {(b B-A c) x}{2 b^2 \left (b+c x^2\right )}+\frac {(b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{5/2} \sqrt {c}} \]

[Out]

-A/b^2/x+1/2*(-A*c+B*b)*x/b^2/(c*x^2+b)+1/2*(-3*A*c+B*b)*arctan(x*c^(1/2)/b^(1/2))/b^(5/2)/c^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1598, 467, 464, 211} \begin {gather*} \frac {(b B-3 A c) \text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{5/2} \sqrt {c}}+\frac {x (b B-A c)}{2 b^2 \left (b+c x^2\right )}-\frac {A}{b^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(b^2*x)) + ((b*B - A*c)*x)/(2*b^2*(b + c*x^2)) + ((b*B - 3*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(5/2)*Sq
rt[c])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 467

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1598

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]

Rubi steps

\begin {align*} \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {A+B x^2}{x^2 \left (b+c x^2\right )^2} \, dx\\ &=\frac {(b B-A c) x}{2 b^2 \left (b+c x^2\right )}-\frac {1}{2} \int \frac {-\frac {2 A}{b}-\frac {(b B-A c) x^2}{b^2}}{x^2 \left (b+c x^2\right )} \, dx\\ &=-\frac {A}{b^2 x}+\frac {(b B-A c) x}{2 b^2 \left (b+c x^2\right )}+\frac {(b B-3 A c) \int \frac {1}{b+c x^2} \, dx}{2 b^2}\\ &=-\frac {A}{b^2 x}+\frac {(b B-A c) x}{2 b^2 \left (b+c x^2\right )}+\frac {(b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{5/2} \sqrt {c}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 70, normalized size = 1.00 \begin {gather*} -\frac {A}{b^2 x}+\frac {(b B-A c) x}{2 b^2 \left (b+c x^2\right )}+\frac {(b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{5/2} \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(b^2*x)) + ((b*B - A*c)*x)/(2*b^2*(b + c*x^2)) + ((b*B - 3*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(5/2)*Sq
rt[c])

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Maple [A]
time = 0.40, size = 62, normalized size = 0.89

method	result	size
default	\(-\frac {\frac {\left (\frac {A c}{2}-\frac {B b}{2}\right ) x}{c \,x^{2}+b}+\frac {\left (3 A c -B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}}{b^{2}}-\frac {A}{b^{2} x}\)	\(62\)
risch	\(\frac {-\frac {\left (3 A c -B b \right ) x^{2}}{2 b^{2}}-\frac {A}{b}}{x \left (c \,x^{2}+b \right )}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (b^{5} c \,\textit {\_Z}^{2}+9 A^{2} c^{2}-6 A B b c +B^{2} b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{5} c +18 A^{2} c^{2}-12 A B b c +2 B^{2} b^{2}\right ) x +\left (3 A \,b^{3} c -B \,b^{4}\right ) \textit {\_R} \right )\right )}{4}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x^2+A)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-1/b^2*((1/2*A*c-1/2*B*b)*x/(c*x^2+b)+1/2*(3*A*c-B*b)/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))-A/b^2/x

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Maxima [A]
time = 0.49, size = 63, normalized size = 0.90 \begin {gather*} \frac {{\left (B b - 3 \, A c\right )} x^{2} - 2 \, A b}{2 \, {\left (b^{2} c x^{3} + b^{3} x\right )}} + \frac {{\left (B b - 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

1/2*((B*b - 3*A*c)*x^2 - 2*A*b)/(b^2*c*x^3 + b^3*x) + 1/2*(B*b - 3*A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^2)

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Fricas [A]
time = 1.87, size = 210, normalized size = 3.00 \begin {gather*} \left [-\frac {4 \, A b^{2} c - 2 \, {\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{2} - {\left ({\left (B b c - 3 \, A c^{2}\right )} x^{3} + {\left (B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right )}{4 \, {\left (b^{3} c^{2} x^{3} + b^{4} c x\right )}}, -\frac {2 \, A b^{2} c - {\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{2} - {\left ({\left (B b c - 3 \, A c^{2}\right )} x^{3} + {\left (B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right )}{2 \, {\left (b^{3} c^{2} x^{3} + b^{4} c x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

[-1/4*(4*A*b^2*c - 2*(B*b^2*c - 3*A*b*c^2)*x^2 - ((B*b*c - 3*A*c^2)*x^3 + (B*b^2 - 3*A*b*c)*x)*sqrt(-b*c)*log(
(c*x^2 + 2*sqrt(-b*c)*x - b)/(c*x^2 + b)))/(b^3*c^2*x^3 + b^4*c*x), -1/2*(2*A*b^2*c - (B*b^2*c - 3*A*b*c^2)*x^
2 - ((B*b*c - 3*A*c^2)*x^3 + (B*b^2 - 3*A*b*c)*x)*sqrt(b*c)*arctan(sqrt(b*c)*x/b))/(b^3*c^2*x^3 + b^4*c*x)]

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Sympy [A]
time = 0.25, size = 114, normalized size = 1.63 \begin {gather*} - \frac {\sqrt {- \frac {1}{b^{5} c}} \left (- 3 A c + B b\right ) \log {\left (- b^{3} \sqrt {- \frac {1}{b^{5} c}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{b^{5} c}} \left (- 3 A c + B b\right ) \log {\left (b^{3} \sqrt {- \frac {1}{b^{5} c}} + x \right )}}{4} + \frac {- 2 A b + x^{2} \left (- 3 A c + B b\right )}{2 b^{3} x + 2 b^{2} c x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

-sqrt(-1/(b**5*c))*(-3*A*c + B*b)*log(-b**3*sqrt(-1/(b**5*c)) + x)/4 + sqrt(-1/(b**5*c))*(-3*A*c + B*b)*log(b*
*3*sqrt(-1/(b**5*c)) + x)/4 + (-2*A*b + x**2*(-3*A*c + B*b))/(2*b**3*x + 2*b**2*c*x**3)

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Giac [A]
time = 1.63, size = 62, normalized size = 0.89 \begin {gather*} \frac {{\left (B b - 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{2}} + \frac {B b x^{2} - 3 \, A c x^{2} - 2 \, A b}{2 \, {\left (c x^{3} + b x\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

1/2*(B*b - 3*A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^2) + 1/2*(B*b*x^2 - 3*A*c*x^2 - 2*A*b)/((c*x^3 + b*x)*b^2
)

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Mupad [B]
time = 0.13, size = 63, normalized size = 0.90 \begin {gather*} -\frac {\frac {A}{b}+\frac {x^2\,\left (3\,A\,c-B\,b\right )}{2\,b^2}}{c\,x^3+b\,x}-\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (3\,A\,c-B\,b\right )}{2\,b^{5/2}\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)

[Out]

- (A/b + (x^2*(3*A*c - B*b))/(2*b^2))/(b*x + c*x^3) - (atan((c^(1/2)*x)/b^(1/2))*(3*A*c - B*b))/(2*b^(5/2)*c^(
1/2))

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