3.77 \(\int \frac {x^{10} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=95 \[ -\frac {3 (5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 \sqrt {b} c^{7/2}}+\frac {x (9 b B-5 A c)}{8 c^3 \left (b+c x^2\right )}-\frac {b x (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}+\frac {B x}{c^3} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1584, 455, 1157, 388, 205} \[ \frac {x (9 b B-5 A c)}{8 c^3 \left (b+c x^2\right )}-\frac {b x (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}-\frac {3 (5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 \sqrt {b} c^{7/2}}+\frac {B x}{c^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^10*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

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

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 455

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[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {x^4 \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac {b (b B-A c) x}{4 c^3 \left (b+c x^2\right )^2}-\frac {\int \frac {-b (b B-A c)+4 c (b B-A c) x^2-4 B c^2 x^4}{\left (b+c x^2\right )^2} \, dx}{4 c^3}\\ &=-\frac {b (b B-A c) x}{4 c^3 \left (b+c x^2\right )^2}+\frac {(9 b B-5 A c) x}{8 c^3 \left (b+c x^2\right )}+\frac {\int \frac {-b (7 b B-3 A c)+8 b B c x^2}{b+c x^2} \, dx}{8 b c^3}\\ &=\frac {B x}{c^3}-\frac {b (b B-A c) x}{4 c^3 \left (b+c x^2\right )^2}+\frac {(9 b B-5 A c) x}{8 c^3 \left (b+c x^2\right )}-\frac {(3 (5 b B-A c)) \int \frac {1}{b+c x^2} \, dx}{8 c^3}\\ &=\frac {B x}{c^3}-\frac {b (b B-A c) x}{4 c^3 \left (b+c x^2\right )^2}+\frac {(9 b B-5 A c) x}{8 c^3 \left (b+c x^2\right )}-\frac {3 (5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 \sqrt {b} c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 92, normalized size = 0.97 \[ \frac {x \left (b \left (25 B c x^2-3 A c\right )+c^2 x^2 \left (8 B x^2-5 A\right )+15 b^2 B\right )}{8 c^3 \left (b+c x^2\right )^2}-\frac {3 (5 b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 \sqrt {b} c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^10*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(x*(15*b^2*B + c^2*x^2*(-5*A + 8*B*x^2) + b*(-3*A*c + 25*B*c*x^2)))/(8*c^3*(b + c*x^2)^2) - (3*(5*b*B - A*c)*A
rcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*Sqrt[b]*c^(7/2))

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fricas [A]  time = 0.82, size = 328, normalized size = 3.45 \[ \left [\frac {16 \, B b c^{3} x^{5} + 10 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x^{3} + 3 \, {\left ({\left (5 \, B b c^{2} - A c^{3}\right )} x^{4} + 5 \, B b^{3} - A b^{2} c + 2 \, {\left (5 \, B b^{2} c - A b c^{2}\right )} x^{2}\right )} \sqrt {-b c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right ) + 6 \, {\left (5 \, B b^{3} c - A b^{2} c^{2}\right )} x}{16 \, {\left (b c^{6} x^{4} + 2 \, b^{2} c^{5} x^{2} + b^{3} c^{4}\right )}}, \frac {8 \, B b c^{3} x^{5} + 5 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x^{3} - 3 \, {\left ({\left (5 \, B b c^{2} - A c^{3}\right )} x^{4} + 5 \, B b^{3} - A b^{2} c + 2 \, {\left (5 \, B b^{2} c - A b c^{2}\right )} x^{2}\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right ) + 3 \, {\left (5 \, B b^{3} c - A b^{2} c^{2}\right )} x}{8 \, {\left (b c^{6} x^{4} + 2 \, b^{2} c^{5} x^{2} + b^{3} c^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

[1/16*(16*B*b*c^3*x^5 + 10*(5*B*b^2*c^2 - A*b*c^3)*x^3 + 3*((5*B*b*c^2 - A*c^3)*x^4 + 5*B*b^3 - A*b^2*c + 2*(5
*B*b^2*c - A*b*c^2)*x^2)*sqrt(-b*c)*log((c*x^2 - 2*sqrt(-b*c)*x - b)/(c*x^2 + b)) + 6*(5*B*b^3*c - A*b^2*c^2)*
x)/(b*c^6*x^4 + 2*b^2*c^5*x^2 + b^3*c^4), 1/8*(8*B*b*c^3*x^5 + 5*(5*B*b^2*c^2 - A*b*c^3)*x^3 - 3*((5*B*b*c^2 -
A*c^3)*x^4 + 5*B*b^3 - A*b^2*c + 2*(5*B*b^2*c - A*b*c^2)*x^2)*sqrt(b*c)*arctan(sqrt(b*c)*x/b) + 3*(5*B*b^3*c
- A*b^2*c^2)*x)/(b*c^6*x^4 + 2*b^2*c^5*x^2 + b^3*c^4)]

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giac [A]  time = 0.18, size = 80, normalized size = 0.84 \[ \frac {B x}{c^{3}} - \frac {3 \, {\left (5 \, B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{3}} + \frac {9 \, B b c x^{3} - 5 \, A c^{2} x^{3} + 7 \, B b^{2} x - 3 \, A b c x}{8 \, {\left (c x^{2} + b\right )}^{2} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 122, normalized size = 1.28 \[ -\frac {5 A \,x^{3}}{8 \left (c \,x^{2}+b \right )^{2} c}+\frac {9 B b \,x^{3}}{8 \left (c \,x^{2}+b \right )^{2} c^{2}}-\frac {3 A b x}{8 \left (c \,x^{2}+b \right )^{2} c^{2}}+\frac {7 B \,b^{2} x}{8 \left (c \,x^{2}+b \right )^{2} c^{3}}+\frac {3 A \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, c^{2}}-\frac {15 B b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, c^{3}}+\frac {B x}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

B*x/c^3-5/8/c/(c*x^2+b)^2*A*x^3+9/8/c^2/(c*x^2+b)^2*B*x^3*b-3/8/c^2/(c*x^2+b)^2*A*b*x+7/8/c^3/(c*x^2+b)^2*B*b^
2*x+3/8/c^2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*A-15/8/c^3/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*b*B

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maxima [A]  time = 2.94, size = 94, normalized size = 0.99 \[ \frac {{\left (9 \, B b c - 5 \, A c^{2}\right )} x^{3} + {\left (7 \, B b^{2} - 3 \, A b c\right )} x}{8 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} + \frac {B x}{c^{3}} - \frac {3 \, {\left (5 \, B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.15, size = 92, normalized size = 0.97 \[ \frac {B\,x}{c^3}-\frac {x^3\,\left (\frac {5\,A\,c^2}{8}-\frac {9\,B\,b\,c}{8}\right )-x\,\left (\frac {7\,B\,b^2}{8}-\frac {3\,A\,b\,c}{8}\right )}{b^2\,c^3+2\,b\,c^4\,x^2+c^5\,x^4}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c-5\,B\,b\right )}{8\,\sqrt {b}\,c^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^10*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

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

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sympy [B]  time = 1.11, size = 194, normalized size = 2.04 \[ \frac {B x}{c^{3}} + \frac {3 \sqrt {- \frac {1}{b c^{7}}} \left (- A c + 5 B b\right ) \log {\left (- \frac {3 b c^{3} \sqrt {- \frac {1}{b c^{7}}} \left (- A c + 5 B b\right )}{- 3 A c + 15 B b} + x \right )}}{16} - \frac {3 \sqrt {- \frac {1}{b c^{7}}} \left (- A c + 5 B b\right ) \log {\left (\frac {3 b c^{3} \sqrt {- \frac {1}{b c^{7}}} \left (- A c + 5 B b\right )}{- 3 A c + 15 B b} + x \right )}}{16} + \frac {x^{3} \left (- 5 A c^{2} + 9 B b c\right ) + x \left (- 3 A b c + 7 B b^{2}\right )}{8 b^{2} c^{3} + 16 b c^{4} x^{2} + 8 c^{5} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

B*x/c**3 + 3*sqrt(-1/(b*c**7))*(-A*c + 5*B*b)*log(-3*b*c**3*sqrt(-1/(b*c**7))*(-A*c + 5*B*b)/(-3*A*c + 15*B*b)
+ x)/16 - 3*sqrt(-1/(b*c**7))*(-A*c + 5*B*b)*log(3*b*c**3*sqrt(-1/(b*c**7))*(-A*c + 5*B*b)/(-3*A*c + 15*B*b)
+ x)/16 + (x**3*(-5*A*c**2 + 9*B*b*c) + x*(-3*A*b*c + 7*B*b**2))/(8*b**2*c**3 + 16*b*c**4*x**2 + 8*c**5*x**4)

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