3.1.60 \(\int \frac {x^{10} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [60]

Optimal. Leaf size=110 \[ \frac {b (3 b B-2 A c) x}{c^4}-\frac {(2 b B-A c) x^3}{3 c^3}+\frac {B x^5}{5 c^2}+\frac {b^2 (b B-A c) x}{2 c^4 \left (b+c x^2\right )}-\frac {b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{9/2}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1598, 466, 1824, 211} \begin {gather*} -\frac {b^{3/2} (7 b B-5 A c) \text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{9/2}}+\frac {b^2 x (b B-A c)}{2 c^4 \left (b+c x^2\right )}+\frac {b x (3 b B-2 A c)}{c^4}-\frac {x^3 (2 b B-A c)}{3 c^3}+\frac {B x^5}{5 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 466

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 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]

Rule 1824

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^6 \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=\frac {b^2 (b B-A c) x}{2 c^4 \left (b+c x^2\right )}-\frac {\int \frac {b^2 (b B-A c)-2 b c (b B-A c) x^2+2 c^2 (b B-A c) x^4-2 B c^3 x^6}{b+c x^2} \, dx}{2 c^4}\\ &=\frac {b^2 (b B-A c) x}{2 c^4 \left (b+c x^2\right )}-\frac {\int \left (-2 b (3 b B-2 A c)+2 c (2 b B-A c) x^2-2 B c^2 x^4+\frac {7 b^3 B-5 A b^2 c}{b+c x^2}\right ) \, dx}{2 c^4}\\ &=\frac {b (3 b B-2 A c) x}{c^4}-\frac {(2 b B-A c) x^3}{3 c^3}+\frac {B x^5}{5 c^2}+\frac {b^2 (b B-A c) x}{2 c^4 \left (b+c x^2\right )}-\frac {\left (b^2 (7 b B-5 A c)\right ) \int \frac {1}{b+c x^2} \, dx}{2 c^4}\\ &=\frac {b (3 b B-2 A c) x}{c^4}-\frac {(2 b B-A c) x^3}{3 c^3}+\frac {B x^5}{5 c^2}+\frac {b^2 (b B-A c) x}{2 c^4 \left (b+c x^2\right )}-\frac {b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 111, normalized size = 1.01 \begin {gather*} \frac {b (3 b B-2 A c) x}{c^4}+\frac {(-2 b B+A c) x^3}{3 c^3}+\frac {B x^5}{5 c^2}-\frac {\left (-b^3 B+A b^2 c\right ) x}{2 c^4 \left (b+c x^2\right )}-\frac {b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.39, size = 100, normalized size = 0.91

method	result	size
default	\(-\frac {-\frac {1}{5} B \,c^{2} x^{5}-\frac {1}{3} A \,c^{2} x^{3}+\frac {2}{3} B b c \,x^{3}+2 A b c x -3 b^{2} B x}{c^{4}}+\frac {b^{2} \left (\frac {\left (-\frac {A c}{2}+\frac {B b}{2}\right ) x}{c \,x^{2}+b}+\frac {\left (5 A c -7 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{c^{4}}\)	\(100\)
risch	\(\frac {B \,x^{5}}{5 c^{2}}+\frac {A \,x^{3}}{3 c^{2}}-\frac {2 B b \,x^{3}}{3 c^{3}}-\frac {2 A b x}{c^{3}}+\frac {3 b^{2} B x}{c^{4}}+\frac {\left (-\frac {1}{2} A \,b^{2} c +\frac {1}{2} B \,b^{3}\right ) x}{c^{4} \left (c \,x^{2}+b \right )}+\frac {5 \sqrt {-b c}\, b \ln \left (-\sqrt {-b c}\, x +b \right ) A}{4 c^{4}}-\frac {7 \sqrt {-b c}\, b^{2} \ln \left (-\sqrt {-b c}\, x +b \right ) B}{4 c^{5}}-\frac {5 \sqrt {-b c}\, b \ln \left (\sqrt {-b c}\, x +b \right ) A}{4 c^{4}}+\frac {7 \sqrt {-b c}\, b^{2} \ln \left (\sqrt {-b c}\, x +b \right ) B}{4 c^{5}}\)	\(178\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 112, normalized size = 1.02 \begin {gather*} \frac {{\left (B b^{3} - A b^{2} c\right )} x}{2 \, {\left (c^{5} x^{2} + b c^{4}\right )}} - \frac {{\left (7 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{4}} + \frac {3 \, B c^{2} x^{5} - 5 \, {\left (2 \, B b c - A c^{2}\right )} x^{3} + 15 \, {\left (3 \, B b^{2} - 2 \, A b c\right )} x}{15 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.84, size = 298, normalized size = 2.71 \begin {gather*} \left [\frac {12 \, B c^{3} x^{7} - 4 \, {\left (7 \, B b c^{2} - 5 \, A c^{3}\right )} x^{5} + 20 \, {\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3} - 15 \, {\left (7 \, B b^{3} - 5 \, A b^{2} c + {\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2}\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} + 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) + 30 \, {\left (7 \, B b^{3} - 5 \, A b^{2} c\right )} x}{60 \, {\left (c^{5} x^{2} + b c^{4}\right )}}, \frac {6 \, B c^{3} x^{7} - 2 \, {\left (7 \, B b c^{2} - 5 \, A c^{3}\right )} x^{5} + 10 \, {\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3} - 15 \, {\left (7 \, B b^{3} - 5 \, A b^{2} c + {\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2}\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) + 15 \, {\left (7 \, B b^{3} - 5 \, A b^{2} c\right )} x}{30 \, {\left (c^{5} x^{2} + b c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(12*B*c^3*x^7 - 4*(7*B*b*c^2 - 5*A*c^3)*x^5 + 20*(7*B*b^2*c - 5*A*b*c^2)*x^3 - 15*(7*B*b^3 - 5*A*b^2*c +
(7*B*b^2*c - 5*A*b*c^2)*x^2)*sqrt(-b/c)*log((c*x^2 + 2*c*x*sqrt(-b/c) - b)/(c*x^2 + b)) + 30*(7*B*b^3 - 5*A*b
^2*c)*x)/(c^5*x^2 + b*c^4), 1/30*(6*B*c^3*x^7 - 2*(7*B*b*c^2 - 5*A*c^3)*x^5 + 10*(7*B*b^2*c - 5*A*b*c^2)*x^3 -
15*(7*B*b^3 - 5*A*b^2*c + (7*B*b^2*c - 5*A*b*c^2)*x^2)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b) + 15*(7*B*b^3 - 5*A*
b^2*c)*x)/(c^5*x^2 + b*c^4)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (104) = 208\).
time = 0.38, size = 211, normalized size = 1.92 \begin {gather*} \frac {B x^{5}}{5 c^{2}} + x^{3} \left (\frac {A}{3 c^{2}} - \frac {2 B b}{3 c^{3}}\right ) + x \left (- \frac {2 A b}{c^{3}} + \frac {3 B b^{2}}{c^{4}}\right ) + \frac {x \left (- A b^{2} c + B b^{3}\right )}{2 b c^{4} + 2 c^{5} x^{2}} + \frac {\sqrt {- \frac {b^{3}}{c^{9}}} \left (- 5 A c + 7 B b\right ) \log {\left (- \frac {c^{4} \sqrt {- \frac {b^{3}}{c^{9}}} \left (- 5 A c + 7 B b\right )}{- 5 A b c + 7 B b^{2}} + x \right )}}{4} - \frac {\sqrt {- \frac {b^{3}}{c^{9}}} \left (- 5 A c + 7 B b\right ) \log {\left (\frac {c^{4} \sqrt {- \frac {b^{3}}{c^{9}}} \left (- 5 A c + 7 B b\right )}{- 5 A b c + 7 B b^{2}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.89, size = 115, normalized size = 1.05 \begin {gather*} -\frac {{\left (7 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{4}} + \frac {B b^{3} x - A b^{2} c x}{2 \, {\left (c x^{2} + b\right )} c^{4}} + \frac {3 \, B c^{8} x^{5} - 10 \, B b c^{7} x^{3} + 5 \, A c^{8} x^{3} + 45 \, B b^{2} c^{6} x - 30 \, A b c^{7} x}{15 \, c^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(7*B*b^3 - 5*A*b^2*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) + 1/2*(B*b^3*x - A*b^2*c*x)/((c*x^2 + b)*c^4)
+ 1/15*(3*B*c^8*x^5 - 10*B*b*c^7*x^3 + 5*A*c^8*x^3 + 45*B*b^2*c^6*x - 30*A*b*c^7*x)/c^10

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Mupad [B]
time = 0.09, size = 141, normalized size = 1.28 \begin {gather*} x^3\,\left (\frac {A}{3\,c^2}-\frac {2\,B\,b}{3\,c^3}\right )-x\,\left (\frac {2\,b\,\left (\frac {A}{c^2}-\frac {2\,B\,b}{c^3}\right )}{c}+\frac {B\,b^2}{c^4}\right )+\frac {B\,x^5}{5\,c^2}+\frac {x\,\left (\frac {B\,b^3}{2}-\frac {A\,b^2\,c}{2}\right )}{c^5\,x^2+b\,c^4}-\frac {b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,x\,\left (5\,A\,c-7\,B\,b\right )}{7\,B\,b^3-5\,A\,b^2\,c}\right )\,\left (5\,A\,c-7\,B\,b\right )}{2\,c^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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