3.1.85 \(\int \frac {x^2 (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=117 \[ -\frac {5 \sqrt {c} (3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{9/2}}-\frac {c x (7 b B-11 A c)}{8 b^4 \left (b+c x^2\right )}-\frac {b B-3 A c}{b^4 x}-\frac {c x (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}-\frac {A}{3 b^3 x^3} \]

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Rubi [A]  time = 0.18, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1584, 456, 1259, 1261, 205} \begin {gather*} -\frac {c x (7 b B-11 A c)}{8 b^4 \left (b+c x^2\right )}-\frac {c x (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}-\frac {b B-3 A c}{b^4 x}-\frac {5 \sqrt {c} (3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{9/2}}-\frac {A}{3 b^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

-A/(3*b^3*x^3) - (b*B - 3*A*c)/(b^4*x) - (c*(b*B - A*c)*x)/(4*b^3*(b + c*x^2)^2) - (c*(7*b*B - 11*A*c)*x)/(8*b
^4*(b + c*x^2)) - (5*Sqrt[c]*(3*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(9/2))

Rule 205

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

Rule 456

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 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 1584

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 )^3} \, dx &=\int \frac {A+B x^2}{x^4 \left (b+c x^2\right )^3} \, dx\\ &=-\frac {c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac {1}{4} c \int \frac {-\frac {4 A}{b c}-\frac {4 (b B-A c) x^2}{b^2 c}+\frac {3 (b B-A c) x^4}{b^3}}{x^4 \left (b+c x^2\right )^2} \, dx\\ &=-\frac {c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac {c (7 b B-11 A c) x}{8 b^4 \left (b+c x^2\right )}-\frac {\int \frac {-8 A b c-8 c (b B-2 A c) x^2+\frac {c^2 (7 b B-11 A c) x^4}{b}}{x^4 \left (b+c x^2\right )} \, dx}{8 b^3 c}\\ &=-\frac {c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac {c (7 b B-11 A c) x}{8 b^4 \left (b+c x^2\right )}-\frac {\int \left (-\frac {8 A c}{x^4}-\frac {8 c (b B-3 A c)}{b x^2}+\frac {5 c^2 (3 b B-7 A c)}{b \left (b+c x^2\right )}\right ) \, dx}{8 b^3 c}\\ &=-\frac {A}{3 b^3 x^3}-\frac {b B-3 A c}{b^4 x}-\frac {c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac {c (7 b B-11 A c) x}{8 b^4 \left (b+c x^2\right )}-\frac {(5 c (3 b B-7 A c)) \int \frac {1}{b+c x^2} \, dx}{8 b^4}\\ &=-\frac {A}{3 b^3 x^3}-\frac {b B-3 A c}{b^4 x}-\frac {c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac {c (7 b B-11 A c) x}{8 b^4 \left (b+c x^2\right )}-\frac {5 \sqrt {c} (3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 119, normalized size = 1.02 \begin {gather*} -\frac {5 \sqrt {c} (3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{9/2}}-\frac {x \left (7 b B c-11 A c^2\right )}{8 b^4 \left (b+c x^2\right )}+\frac {3 A c-b B}{b^4 x}-\frac {c x (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}-\frac {A}{3 b^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

-1/3*A/(b^3*x^3) + (-(b*B) + 3*A*c)/(b^4*x) - (c*(b*B - A*c)*x)/(4*b^3*(b + c*x^2)^2) - ((7*b*B*c - 11*A*c^2)*
x)/(8*b^4*(b + c*x^2)) - (5*Sqrt[c]*(3*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(9/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

IntegrateAlgebraic[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^3, x]

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fricas [A]  time = 0.41, size = 368, normalized size = 3.15 \begin {gather*} \left [-\frac {30 \, {\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 50 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} + 16 \, A b^{3} + 16 \, {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} + 15 \, {\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} + 2 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5} + {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} + 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{48 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}}, -\frac {15 \, {\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 25 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} + 8 \, A b^{3} + 8 \, {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} + 15 \, {\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} + 2 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5} + {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{24 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\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)^3,x, algorithm="fricas")

[Out]

[-1/48*(30*(3*B*b*c^2 - 7*A*c^3)*x^6 + 50*(3*B*b^2*c - 7*A*b*c^2)*x^4 + 16*A*b^3 + 16*(3*B*b^3 - 7*A*b^2*c)*x^
2 + 15*((3*B*b*c^2 - 7*A*c^3)*x^7 + 2*(3*B*b^2*c - 7*A*b*c^2)*x^5 + (3*B*b^3 - 7*A*b^2*c)*x^3)*sqrt(-c/b)*log(
(c*x^2 + 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3), -1/24*(15*(3*B*b*c^2 - 7*A
*c^3)*x^6 + 25*(3*B*b^2*c - 7*A*b*c^2)*x^4 + 8*A*b^3 + 8*(3*B*b^3 - 7*A*b^2*c)*x^2 + 15*((3*B*b*c^2 - 7*A*c^3)
*x^7 + 2*(3*B*b^2*c - 7*A*b*c^2)*x^5 + (3*B*b^3 - 7*A*b^2*c)*x^3)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^4*c^2*x^7
+ 2*b^5*c*x^5 + b^6*x^3)]

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giac [A]  time = 0.16, size = 108, normalized size = 0.92 \begin {gather*} -\frac {5 \, {\left (3 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{4}} - \frac {7 \, B b c^{2} x^{3} - 11 \, A c^{3} x^{3} + 9 \, B b^{2} c x - 13 \, A b c^{2} x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{4}} - \frac {3 \, B b x^{2} - 9 \, A c x^{2} + A b}{3 \, b^{4} 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)^3,x, algorithm="giac")

[Out]

-5/8*(3*B*b*c - 7*A*c^2)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) - 1/8*(7*B*b*c^2*x^3 - 11*A*c^3*x^3 + 9*B*b^2*c
*x - 13*A*b*c^2*x)/((c*x^2 + b)^2*b^4) - 1/3*(3*B*b*x^2 - 9*A*c*x^2 + A*b)/(b^4*x^3)

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maple [A]  time = 0.06, size = 152, normalized size = 1.30 \begin {gather*} \frac {11 A \,c^{3} x^{3}}{8 \left (c \,x^{2}+b \right )^{2} b^{4}}-\frac {7 B \,c^{2} x^{3}}{8 \left (c \,x^{2}+b \right )^{2} b^{3}}+\frac {13 A \,c^{2} x}{8 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {9 B c x}{8 \left (c \,x^{2}+b \right )^{2} b^{2}}+\frac {35 A \,c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b^{4}}-\frac {15 B c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b^{3}}+\frac {3 A c}{b^{4} x}-\frac {B}{b^{3} x}-\frac {A}{3 b^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

11/8/b^4*c^3/(c*x^2+b)^2*A*x^3-7/8/b^3*c^2/(c*x^2+b)^2*B*x^3+13/8/b^3*c^2/(c*x^2+b)^2*A*x-9/8/b^2*c/(c*x^2+b)^
2*B*x+35/8/b^4*c^2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*A-15/8/b^3*c/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*B-
1/3*A/b^3/x^3+3/b^4/x*A*c-1/b^3/x*B

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maxima [A]  time = 3.11, size = 128, normalized size = 1.09 \begin {gather*} -\frac {15 \, {\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 25 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} + 8 \, A b^{3} + 8 \, {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2}}{24 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}} - \frac {5 \, {\left (3 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{4}} \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)^3,x, algorithm="maxima")

[Out]

-1/24*(15*(3*B*b*c^2 - 7*A*c^3)*x^6 + 25*(3*B*b^2*c - 7*A*b*c^2)*x^4 + 8*A*b^3 + 8*(3*B*b^3 - 7*A*b^2*c)*x^2)/
(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3) - 5/8*(3*B*b*c - 7*A*c^2)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4)

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mupad [B]  time = 0.18, size = 114, normalized size = 0.97 \begin {gather*} \frac {\frac {x^2\,\left (7\,A\,c-3\,B\,b\right )}{3\,b^2}-\frac {A}{3\,b}+\frac {5\,c^2\,x^6\,\left (7\,A\,c-3\,B\,b\right )}{8\,b^4}+\frac {25\,c\,x^4\,\left (7\,A\,c-3\,B\,b\right )}{24\,b^3}}{b^2\,x^3+2\,b\,c\,x^5+c^2\,x^7}+\frac {5\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (7\,A\,c-3\,B\,b\right )}{8\,b^{9/2}} \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)^3,x)

[Out]

((x^2*(7*A*c - 3*B*b))/(3*b^2) - A/(3*b) + (5*c^2*x^6*(7*A*c - 3*B*b))/(8*b^4) + (25*c*x^4*(7*A*c - 3*B*b))/(2
4*b^3))/(b^2*x^3 + c^2*x^7 + 2*b*c*x^5) + (5*c^(1/2)*atan((c^(1/2)*x)/b^(1/2))*(7*A*c - 3*B*b))/(8*b^(9/2))

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sympy [B]  time = 0.83, size = 226, normalized size = 1.93 \begin {gather*} \frac {5 \sqrt {- \frac {c}{b^{9}}} \left (- 7 A c + 3 B b\right ) \log {\left (- \frac {5 b^{5} \sqrt {- \frac {c}{b^{9}}} \left (- 7 A c + 3 B b\right )}{- 35 A c^{2} + 15 B b c} + x \right )}}{16} - \frac {5 \sqrt {- \frac {c}{b^{9}}} \left (- 7 A c + 3 B b\right ) \log {\left (\frac {5 b^{5} \sqrt {- \frac {c}{b^{9}}} \left (- 7 A c + 3 B b\right )}{- 35 A c^{2} + 15 B b c} + x \right )}}{16} + \frac {- 8 A b^{3} + x^{6} \left (105 A c^{3} - 45 B b c^{2}\right ) + x^{4} \left (175 A b c^{2} - 75 B b^{2} c\right ) + x^{2} \left (56 A b^{2} c - 24 B b^{3}\right )}{24 b^{6} x^{3} + 48 b^{5} c x^{5} + 24 b^{4} c^{2} x^{7}} \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)**3,x)

[Out]

5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)*log(-5*b**5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)/(-35*A*c**2 + 15*B*b*c) + x)/16 -
5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)*log(5*b**5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)/(-35*A*c**2 + 15*B*b*c) + x)/16 + (
-8*A*b**3 + x**6*(105*A*c**3 - 45*B*b*c**2) + x**4*(175*A*b*c**2 - 75*B*b**2*c) + x**2*(56*A*b**2*c - 24*B*b**
3))/(24*b**6*x**3 + 48*b**5*c*x**5 + 24*b**4*c**2*x**7)

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