3.2.31 \(\int \frac {x^4}{a+b x^2} \, dx\) [131]

Optimal. Leaf size=42 \[ -\frac {a x}{b^2}+\frac {x^3}{3 b}+\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {308, 211} \begin {gather*} \frac {a^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}-\frac {a x}{b^2}+\frac {x^3}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b*x^2),x]

[Out]

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

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 308

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

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 42, normalized size = 1.00 \begin {gather*} -\frac {a x}{b^2}+\frac {x^3}{3 b}+\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b*x^2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.04, size = 38, normalized size = 0.90

method	result	size
default	\(-\frac {-\frac {1}{3} b \,x^{3}+a x}{b^{2}}+\frac {a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(38\)
risch	\(\frac {x^{3}}{3 b}-\frac {a x}{b^{2}}+\frac {\sqrt {-a b}\, a \ln \left (-\sqrt {-a b}\, x +a \right )}{2 b^{3}}-\frac {\sqrt {-a b}\, a \ln \left (\sqrt {-a b}\, x +a \right )}{2 b^{3}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.57, size = 37, normalized size = 0.88 \begin {gather*} \frac {a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {b x^{3} - 3 \, a x}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 1.10, size = 99, normalized size = 2.36 \begin {gather*} \left [\frac {2 \, b x^{3} + 3 \, a \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 6 \, a x}{6 \, b^{2}}, \frac {b x^{3} + 3 \, a \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 3 \, a x}{3 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (36) = 72\).
time = 0.06, size = 80, normalized size = 1.90 \begin {gather*} - \frac {a x}{b^{2}} - \frac {\sqrt {- \frac {a^{3}}{b^{5}}} \log {\left (x - \frac {b^{2} \sqrt {- \frac {a^{3}}{b^{5}}}}{a} \right )}}{2} + \frac {\sqrt {- \frac {a^{3}}{b^{5}}} \log {\left (x + \frac {b^{2} \sqrt {- \frac {a^{3}}{b^{5}}}}{a} \right )}}{2} + \frac {x^{3}}{3 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.60, size = 40, normalized size = 0.95 \begin {gather*} \frac {a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {b^{2} x^{3} - 3 \, a b x}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.07, size = 32, normalized size = 0.76 \begin {gather*} \frac {x^3}{3\,b}+\frac {a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{b^{5/2}}-\frac {a\,x}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b*x^2),x)

[Out]

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

________________________________________________________________________________________