3.73 \(\int \frac{x^{14} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=140 \[ \frac{b^2 x (17 b B-13 A c)}{8 c^5 \left (b+c x^2\right )}-\frac{b^3 x (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}-\frac{7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{11/2}}-\frac{x^3 (3 b B-A c)}{3 c^4}+\frac{3 b x (2 b B-A c)}{c^5}+\frac{B x^5}{5 c^3} \]

[Out]

(3*b*(2*b*B - A*c)*x)/c^5 - ((3*b*B - A*c)*x^3)/(3*c^4) + (B*x^5)/(5*c^3) - (b^3*(b*B - A*c)*x)/(4*c^5*(b + c*
x^2)^2) + (b^2*(17*b*B - 13*A*c)*x)/(8*c^5*(b + c*x^2)) - (7*b^(3/2)*(9*b*B - 5*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b
]])/(8*c^(11/2))

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Rubi [A]  time = 0.234623, antiderivative size = 140, normalized size of antiderivative = 1., 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, 455, 1814, 1810, 205} \[ \frac{b^2 x (17 b B-13 A c)}{8 c^5 \left (b+c x^2\right )}-\frac{b^3 x (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}-\frac{7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{11/2}}-\frac{x^3 (3 b B-A c)}{3 c^4}+\frac{3 b x (2 b B-A c)}{c^5}+\frac{B x^5}{5 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(3*b*(2*b*B - A*c)*x)/c^5 - ((3*b*B - A*c)*x^3)/(3*c^4) + (B*x^5)/(5*c^3) - (b^3*(b*B - A*c)*x)/(4*c^5*(b + c*
x^2)^2) + (b^2*(17*b*B - 13*A*c)*x)/(8*c^5*(b + c*x^2)) - (7*b^(3/2)*(9*b*B - 5*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b
]])/(8*c^(11/2))

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]

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 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1810

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]

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]

Rubi steps

\begin{align*} \int \frac{x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{x^8 \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac{b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}-\frac{\int \frac{-b^3 (b B-A c)+4 b^2 c (b B-A c) x^2-4 b c^2 (b B-A c) x^4+4 c^3 (b B-A c) x^6-4 B c^4 x^8}{\left (b+c x^2\right )^2} \, dx}{4 c^5}\\ &=-\frac{b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}+\frac{b^2 (17 b B-13 A c) x}{8 c^5 \left (b+c x^2\right )}+\frac{\int \frac{-b^3 (15 b B-11 A c)+8 b^2 c (3 b B-2 A c) x^2-8 b c^2 (2 b B-A c) x^4+8 b B c^3 x^6}{b+c x^2} \, dx}{8 b c^5}\\ &=-\frac{b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}+\frac{b^2 (17 b B-13 A c) x}{8 c^5 \left (b+c x^2\right )}+\frac{\int \left (24 b^2 (2 b B-A c)-8 b c (3 b B-A c) x^2+8 b B c^2 x^4-\frac{7 \left (9 b^4 B-5 A b^3 c\right )}{b+c x^2}\right ) \, dx}{8 b c^5}\\ &=\frac{3 b (2 b B-A c) x}{c^5}-\frac{(3 b B-A c) x^3}{3 c^4}+\frac{B x^5}{5 c^3}-\frac{b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}+\frac{b^2 (17 b B-13 A c) x}{8 c^5 \left (b+c x^2\right )}-\frac{\left (7 b^2 (9 b B-5 A c)\right ) \int \frac{1}{b+c x^2} \, dx}{8 c^5}\\ &=\frac{3 b (2 b B-A c) x}{c^5}-\frac{(3 b B-A c) x^3}{3 c^4}+\frac{B x^5}{5 c^3}-\frac{b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}+\frac{b^2 (17 b B-13 A c) x}{8 c^5 \left (b+c x^2\right )}-\frac{7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.104349, size = 133, normalized size = 0.95 \[ \frac{x \left (7 b^2 c^2 x^2 \left (72 B x^2-125 A\right )-525 b^3 c \left (A-3 B x^2\right )-8 b c^3 x^4 \left (35 A+9 B x^2\right )+8 c^4 x^6 \left (5 A+3 B x^2\right )+945 b^4 B\right )}{120 c^5 \left (b+c x^2\right )^2}-\frac{7 b^{3/2} (9 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(x*(945*b^4*B - 525*b^3*c*(A - 3*B*x^2) + 8*c^4*x^6*(5*A + 3*B*x^2) - 8*b*c^3*x^4*(35*A + 9*B*x^2) + 7*b^2*c^2
*x^2*(-125*A + 72*B*x^2)))/(120*c^5*(b + c*x^2)^2) - (7*b^(3/2)*(9*b*B - 5*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(
8*c^(11/2))

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Maple [A]  time = 0.012, size = 174, normalized size = 1.2 \begin{align*}{\frac{B{x}^{5}}{5\,{c}^{3}}}+{\frac{A{x}^{3}}{3\,{c}^{3}}}-{\frac{B{x}^{3}b}{{c}^{4}}}-3\,{\frac{Abx}{{c}^{4}}}+6\,{\frac{B{b}^{2}x}{{c}^{5}}}-{\frac{13\,A{b}^{2}{x}^{3}}{8\,{c}^{3} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{17\,B{b}^{3}{x}^{3}}{8\,{c}^{4} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{11\,A{b}^{3}x}{8\,{c}^{4} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{15\,B{b}^{4}x}{8\,{c}^{5} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{35\,A{b}^{2}}{8\,{c}^{4}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}-{\frac{63\,B{b}^{3}}{8\,{c}^{5}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

1/5*B*x^5/c^3+1/3/c^3*A*x^3-1/c^4*B*x^3*b-3/c^4*A*b*x+6/c^5*B*b^2*x-13/8*b^2/c^3/(c*x^2+b)^2*A*x^3+17/8*b^3/c^
4/(c*x^2+b)^2*B*x^3-11/8*b^3/c^4/(c*x^2+b)^2*A*x+15/8*b^4/c^5/(c*x^2+b)^2*B*x+35/8*b^2/c^4/(b*c)^(1/2)*arctan(
x*c/(b*c)^(1/2))*A-63/8*b^3/c^5/(b*c)^(1/2)*arctan(x*c/(b*c)^(1/2))*B

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.875004, size = 892, normalized size = 6.37 \begin{align*} \left [\frac{48 \, B c^{4} x^{9} - 16 \,{\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{7} + 112 \,{\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{5} + 350 \,{\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{3} - 105 \,{\left (9 \, B b^{4} - 5 \, A b^{3} c +{\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} + 2 \,{\left (9 \, B b^{3} c - 5 \, A b^{2} 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 ) + 210 \,{\left (9 \, B b^{4} - 5 \, A b^{3} c\right )} x}{240 \,{\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}}, \frac{24 \, B c^{4} x^{9} - 8 \,{\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{7} + 56 \,{\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{5} + 175 \,{\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{3} - 105 \,{\left (9 \, B b^{4} - 5 \, A b^{3} c +{\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} + 2 \,{\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c x \sqrt{\frac{b}{c}}}{b}\right ) + 105 \,{\left (9 \, B b^{4} - 5 \, A b^{3} c\right )} x}{120 \,{\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

[1/240*(48*B*c^4*x^9 - 16*(9*B*b*c^3 - 5*A*c^4)*x^7 + 112*(9*B*b^2*c^2 - 5*A*b*c^3)*x^5 + 350*(9*B*b^3*c - 5*A
*b^2*c^2)*x^3 - 105*(9*B*b^4 - 5*A*b^3*c + (9*B*b^2*c^2 - 5*A*b*c^3)*x^4 + 2*(9*B*b^3*c - 5*A*b^2*c^2)*x^2)*sq
rt(-b/c)*log((c*x^2 + 2*c*x*sqrt(-b/c) - b)/(c*x^2 + b)) + 210*(9*B*b^4 - 5*A*b^3*c)*x)/(c^7*x^4 + 2*b*c^6*x^2
+ b^2*c^5), 1/120*(24*B*c^4*x^9 - 8*(9*B*b*c^3 - 5*A*c^4)*x^7 + 56*(9*B*b^2*c^2 - 5*A*b*c^3)*x^5 + 175*(9*B*b
^3*c - 5*A*b^2*c^2)*x^3 - 105*(9*B*b^4 - 5*A*b^3*c + (9*B*b^2*c^2 - 5*A*b*c^3)*x^4 + 2*(9*B*b^3*c - 5*A*b^2*c^
2)*x^2)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b) + 105*(9*B*b^4 - 5*A*b^3*c)*x)/(c^7*x^4 + 2*b*c^6*x^2 + b^2*c^5)]

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Sympy [A]  time = 1.63422, size = 250, normalized size = 1.79 \begin{align*} \frac{B x^{5}}{5 c^{3}} + \frac{7 \sqrt{- \frac{b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right ) \log{\left (- \frac{7 c^{5} \sqrt{- \frac{b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right )}{- 35 A b c + 63 B b^{2}} + x \right )}}{16} - \frac{7 \sqrt{- \frac{b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right ) \log{\left (\frac{7 c^{5} \sqrt{- \frac{b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right )}{- 35 A b c + 63 B b^{2}} + x \right )}}{16} + \frac{x^{3} \left (- 13 A b^{2} c^{2} + 17 B b^{3} c\right ) + x \left (- 11 A b^{3} c + 15 B b^{4}\right )}{8 b^{2} c^{5} + 16 b c^{6} x^{2} + 8 c^{7} x^{4}} - \frac{x^{3} \left (- A c + 3 B b\right )}{3 c^{4}} + \frac{x \left (- 3 A b c + 6 B b^{2}\right )}{c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**14*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

B*x**5/(5*c**3) + 7*sqrt(-b**3/c**11)*(-5*A*c + 9*B*b)*log(-7*c**5*sqrt(-b**3/c**11)*(-5*A*c + 9*B*b)/(-35*A*b
*c + 63*B*b**2) + x)/16 - 7*sqrt(-b**3/c**11)*(-5*A*c + 9*B*b)*log(7*c**5*sqrt(-b**3/c**11)*(-5*A*c + 9*B*b)/(
-35*A*b*c + 63*B*b**2) + x)/16 + (x**3*(-13*A*b**2*c**2 + 17*B*b**3*c) + x*(-11*A*b**3*c + 15*B*b**4))/(8*b**2
*c**5 + 16*b*c**6*x**2 + 8*c**7*x**4) - x**3*(-A*c + 3*B*b)/(3*c**4) + x*(-3*A*b*c + 6*B*b**2)/c**5

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Giac [A]  time = 1.15251, size = 186, normalized size = 1.33 \begin{align*} -\frac{7 \,{\left (9 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} c^{5}} + \frac{17 \, B b^{3} c x^{3} - 13 \, A b^{2} c^{2} x^{3} + 15 \, B b^{4} x - 11 \, A b^{3} c x}{8 \,{\left (c x^{2} + b\right )}^{2} c^{5}} + \frac{3 \, B c^{12} x^{5} - 15 \, B b c^{11} x^{3} + 5 \, A c^{12} x^{3} + 90 \, B b^{2} c^{10} x - 45 \, A b c^{11} x}{15 \, c^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

-7/8*(9*B*b^3 - 5*A*b^2*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^5) + 1/8*(17*B*b^3*c*x^3 - 13*A*b^2*c^2*x^3 + 15
*B*b^4*x - 11*A*b^3*c*x)/((c*x^2 + b)^2*c^5) + 1/15*(3*B*c^12*x^5 - 15*B*b*c^11*x^3 + 5*A*c^12*x^3 + 90*B*b^2*
c^10*x - 45*A*b*c^11*x)/c^15