3.1845 \(\int \frac {x^4}{a+\frac {b}{x^2}} \, dx\)

Optimal. Leaf size=55 \[ -\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{a^{7/2}}+\frac {b^2 x}{a^3}-\frac {b x^3}{3 a^2}+\frac {x^5}{5 a} \]

[Out]

b^2*x/a^3-1/3*b*x^3/a^2+1/5*x^5/a-b^(5/2)*arctan(x*a^(1/2)/b^(1/2))/a^(7/2)

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {263, 302, 205} \[ \frac {b^2 x}{a^3}-\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{a^{7/2}}-\frac {b x^3}{3 a^2}+\frac {x^5}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b/x^2),x]

[Out]

(b^2*x)/a^3 - (b*x^3)/(3*a^2) + x^5/(5*a) - (b^(5/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/a^(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 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 55, normalized size = 1.00 \[ -\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{a^{7/2}}+\frac {b^2 x}{a^3}-\frac {b x^3}{3 a^2}+\frac {x^5}{5 a} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b/x^2),x]

[Out]

(b^2*x)/a^3 - (b*x^3)/(3*a^2) + x^5/(5*a) - (b^(5/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/a^(7/2)

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fricas [A]  time = 0.98, size = 126, normalized size = 2.29 \[ \left [\frac {6 \, a^{2} x^{5} - 10 \, a b x^{3} + 15 \, b^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - b}{a x^{2} + b}\right ) + 30 \, b^{2} x}{30 \, a^{3}}, \frac {3 \, a^{2} x^{5} - 5 \, a b x^{3} - 15 \, b^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a x \sqrt {\frac {b}{a}}}{b}\right ) + 15 \, b^{2} x}{15 \, a^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/x^2),x, algorithm="fricas")

[Out]

[1/30*(6*a^2*x^5 - 10*a*b*x^3 + 15*b^2*sqrt(-b/a)*log((a*x^2 - 2*a*x*sqrt(-b/a) - b)/(a*x^2 + b)) + 30*b^2*x)/
a^3, 1/15*(3*a^2*x^5 - 5*a*b*x^3 - 15*b^2*sqrt(b/a)*arctan(a*x*sqrt(b/a)/b) + 15*b^2*x)/a^3]

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giac [A]  time = 0.18, size = 55, normalized size = 1.00 \[ -\frac {b^{3} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {3 \, a^{4} x^{5} - 5 \, a^{3} b x^{3} + 15 \, a^{2} b^{2} x}{15 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/x^2),x, algorithm="giac")

[Out]

-b^3*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/15*(3*a^4*x^5 - 5*a^3*b*x^3 + 15*a^2*b^2*x)/a^5

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maple [A]  time = 0.00, size = 49, normalized size = 0.89 \[ \frac {x^{5}}{5 a}-\frac {b \,x^{3}}{3 a^{2}}-\frac {b^{3} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{3}}+\frac {b^{2} x}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a+b/x^2),x)

[Out]

1/5/a*x^5-1/3/a^2*b*x^3+1/a^3*b^2*x-b^3/a^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*a*x)

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maxima [A]  time = 2.00, size = 50, normalized size = 0.91 \[ -\frac {b^{3} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {3 \, a^{2} x^{5} - 5 \, a b x^{3} + 15 \, b^{2} x}{15 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/x^2),x, algorithm="maxima")

[Out]

-b^3*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/15*(3*a^2*x^5 - 5*a*b*x^3 + 15*b^2*x)/a^3

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mupad [B]  time = 0.06, size = 43, normalized size = 0.78 \[ \frac {x^5}{5\,a}-\frac {b\,x^3}{3\,a^2}+\frac {b^2\,x}{a^3}-\frac {b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b/x^2),x)

[Out]

x^5/(5*a) - (b*x^3)/(3*a^2) + (b^2*x)/a^3 - (b^(5/2)*atan((a^(1/2)*x)/b^(1/2)))/a^(7/2)

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sympy [A]  time = 0.19, size = 95, normalized size = 1.73 \[ \frac {\sqrt {- \frac {b^{5}}{a^{7}}} \log {\left (- \frac {a^{3} \sqrt {- \frac {b^{5}}{a^{7}}}}{b^{2}} + x \right )}}{2} - \frac {\sqrt {- \frac {b^{5}}{a^{7}}} \log {\left (\frac {a^{3} \sqrt {- \frac {b^{5}}{a^{7}}}}{b^{2}} + x \right )}}{2} + \frac {x^{5}}{5 a} - \frac {b x^{3}}{3 a^{2}} + \frac {b^{2} x}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(a+b/x**2),x)

[Out]

sqrt(-b**5/a**7)*log(-a**3*sqrt(-b**5/a**7)/b**2 + x)/2 - sqrt(-b**5/a**7)*log(a**3*sqrt(-b**5/a**7)/b**2 + x)
/2 + x**5/(5*a) - b*x**3/(3*a**2) + b**2*x/a**3

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