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

Optimal. Leaf size=85 \[ \frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{9/2}}-\frac{7 x^5}{8 c^2 \left (b+c x^2\right )}-\frac{35 b x}{8 c^4}-\frac{x^7}{4 c \left (b+c x^2\right )^2}+\frac{35 x^3}{24 c^3} \]

[Out]

(-35*b*x)/(8*c^4) + (35*x^3)/(24*c^3) - x^7/(4*c*(b + c*x^2)^2) - (7*x^5)/(8*c^2*(b + c*x^2)) + (35*b^(3/2)*Ar
cTan[(Sqrt[c]*x)/Sqrt[b]])/(8*c^(9/2))

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Rubi [A]  time = 0.0424448, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {1584, 288, 302, 205} \[ \frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{9/2}}-\frac{7 x^5}{8 c^2 \left (b+c x^2\right )}-\frac{35 b x}{8 c^4}-\frac{x^7}{4 c \left (b+c x^2\right )^2}+\frac{35 x^3}{24 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^14/(b*x^2 + c*x^4)^3,x]

[Out]

(-35*b*x)/(8*c^4) + (35*x^3)/(24*c^3) - x^7/(4*c*(b + c*x^2)^2) - (7*x^5)/(8*c^2*(b + c*x^2)) + (35*b^(3/2)*Ar
cTan[(Sqrt[c]*x)/Sqrt[b]])/(8*c^(9/2))

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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 (b x^2+c x^4\right )^3} \, dx &=\int \frac{x^8}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac{x^7}{4 c \left (b+c x^2\right )^2}+\frac{7 \int \frac{x^6}{\left (b+c x^2\right )^2} \, dx}{4 c}\\ &=-\frac{x^7}{4 c \left (b+c x^2\right )^2}-\frac{7 x^5}{8 c^2 \left (b+c x^2\right )}+\frac{35 \int \frac{x^4}{b+c x^2} \, dx}{8 c^2}\\ &=-\frac{x^7}{4 c \left (b+c x^2\right )^2}-\frac{7 x^5}{8 c^2 \left (b+c x^2\right )}+\frac{35 \int \left (-\frac{b}{c^2}+\frac{x^2}{c}+\frac{b^2}{c^2 \left (b+c x^2\right )}\right ) \, dx}{8 c^2}\\ &=-\frac{35 b x}{8 c^4}+\frac{35 x^3}{24 c^3}-\frac{x^7}{4 c \left (b+c x^2\right )^2}-\frac{7 x^5}{8 c^2 \left (b+c x^2\right )}+\frac{\left (35 b^2\right ) \int \frac{1}{b+c x^2} \, dx}{8 c^4}\\ &=-\frac{35 b x}{8 c^4}+\frac{35 x^3}{24 c^3}-\frac{x^7}{4 c \left (b+c x^2\right )^2}-\frac{7 x^5}{8 c^2 \left (b+c x^2\right )}+\frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0470508, size = 77, normalized size = 0.91 \[ \frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{9/2}}-\frac{175 b^2 c x^3+105 b^3 x+56 b c^2 x^5-8 c^3 x^7}{24 c^4 \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^14/(b*x^2 + c*x^4)^3,x]

[Out]

-(105*b^3*x + 175*b^2*c*x^3 + 56*b*c^2*x^5 - 8*c^3*x^7)/(24*c^4*(b + c*x^2)^2) + (35*b^(3/2)*ArcTan[(Sqrt[c]*x
)/Sqrt[b]])/(8*c^(9/2))

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Maple [A]  time = 0.051, size = 77, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}}{3\,{c}^{3}}}-3\,{\frac{bx}{{c}^{4}}}-{\frac{13\,{b}^{2}{x}^{3}}{8\,{c}^{3} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{11\,{b}^{3}x}{8\,{c}^{4} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{35\,{b}^{2}}{8\,{c}^{4}}\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/(c*x^4+b*x^2)^3,x)

[Out]

1/3*x^3/c^3-3*b*x/c^4-13/8/c^3*b^2/(c*x^2+b)^2*x^3-11/8/c^4*b^3/(c*x^2+b)^2*x+35/8/c^4*b^2/(b*c)^(1/2)*arctan(
x*c/(b*c)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53735, size = 493, normalized size = 5.8 \begin{align*} \left [\frac{16 \, c^{3} x^{7} - 112 \, b c^{2} x^{5} - 350 \, b^{2} c x^{3} - 210 \, b^{3} x + 105 \,{\left (b c^{2} x^{4} + 2 \, b^{2} c x^{2} + b^{3}\right )} \sqrt{-\frac{b}{c}} \log \left (\frac{c x^{2} + 2 \, c x \sqrt{-\frac{b}{c}} - b}{c x^{2} + b}\right )}{48 \,{\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}}, \frac{8 \, c^{3} x^{7} - 56 \, b c^{2} x^{5} - 175 \, b^{2} c x^{3} - 105 \, b^{3} x + 105 \,{\left (b c^{2} x^{4} + 2 \, b^{2} c x^{2} + b^{3}\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c x \sqrt{\frac{b}{c}}}{b}\right )}{24 \,{\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.636313, size = 131, normalized size = 1.54 \begin{align*} - \frac{3 b x}{c^{4}} - \frac{35 \sqrt{- \frac{b^{3}}{c^{9}}} \log{\left (x - \frac{c^{4} \sqrt{- \frac{b^{3}}{c^{9}}}}{b} \right )}}{16} + \frac{35 \sqrt{- \frac{b^{3}}{c^{9}}} \log{\left (x + \frac{c^{4} \sqrt{- \frac{b^{3}}{c^{9}}}}{b} \right )}}{16} - \frac{11 b^{3} x + 13 b^{2} c x^{3}}{8 b^{2} c^{4} + 16 b c^{5} x^{2} + 8 c^{6} x^{4}} + \frac{x^{3}}{3 c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*x/c**4 - 35*sqrt(-b**3/c**9)*log(x - c**4*sqrt(-b**3/c**9)/b)/16 + 35*sqrt(-b**3/c**9)*log(x + c**4*sqrt(
-b**3/c**9)/b)/16 - (11*b**3*x + 13*b**2*c*x**3)/(8*b**2*c**4 + 16*b*c**5*x**2 + 8*c**6*x**4) + x**3/(3*c**3)

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Giac [A]  time = 1.22774, size = 99, normalized size = 1.16 \begin{align*} \frac{35 \, b^{2} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} c^{4}} - \frac{13 \, b^{2} c x^{3} + 11 \, b^{3} x}{8 \,{\left (c x^{2} + b\right )}^{2} c^{4}} + \frac{c^{6} x^{3} - 9 \, b c^{5} x}{3 \, c^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/8*b^2*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) - 1/8*(13*b^2*c*x^3 + 11*b^3*x)/((c*x^2 + b)^2*c^4) + 1/3*(c^6*
x^3 - 9*b*c^5*x)/c^9