3.201 \(\int \frac {x^2 (c+d x^2)}{a+b x^2} \, dx\)

Optimal. Leaf size=58 \[ -\frac {\sqrt {a} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}+\frac {x (b c-a d)}{b^2}+\frac {d x^3}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {459, 321, 205} \[ \frac {x (b c-a d)}{b^2}-\frac {\sqrt {a} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}+\frac {d x^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(c + d*x^2))/(a + b*x^2),x]

[Out]

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

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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.98 \[ \frac {\sqrt {a} (a d-b c) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}+\frac {x (b c-a d)}{b^2}+\frac {d x^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(c + d*x^2))/(a + b*x^2),x]

[Out]

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

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fricas [A]  time = 0.47, size = 129, normalized size = 2.22 \[ \left [\frac {2 \, b d x^{3} - 3 \, {\left (b c - a d\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 6 \, {\left (b c - a d\right )} x}{6 \, b^{2}}, \frac {b d x^{3} - 3 \, {\left (b c - a d\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 3 \, {\left (b c - a d\right )} x}{3 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.44, size = 58, normalized size = 1.00 \[ -\frac {{\left (a b c - a^{2} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {b^{2} d x^{3} + 3 \, b^{2} c x - 3 \, a b d x}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 68, normalized size = 1.17 \[ \frac {d \,x^{3}}{3 b}+\frac {a^{2} d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}-\frac {a c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b}-\frac {a d x}{b^{2}}+\frac {c x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(d*x^2+c)/(b*x^2+a),x)

[Out]

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

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maxima [A]  time = 2.16, size = 54, normalized size = 0.93 \[ -\frac {{\left (a b c - a^{2} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {b d x^{3} + 3 \, {\left (b c - a d\right )} x}{3 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 70, normalized size = 1.21 \[ x\,\left (\frac {c}{b}-\frac {a\,d}{b^2}\right )+\frac {d\,x^3}{3\,b}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,x\,\left (a\,d-b\,c\right )}{a^2\,d-a\,b\,c}\right )\,\left (a\,d-b\,c\right )}{b^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(c + d*x^2))/(a + b*x^2),x)

[Out]

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

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sympy [A]  time = 0.33, size = 90, normalized size = 1.55 \[ x \left (- \frac {a d}{b^{2}} + \frac {c}{b}\right ) - \frac {\sqrt {- \frac {a}{b^{5}}} \left (a d - b c\right ) \log {\left (- b^{2} \sqrt {- \frac {a}{b^{5}}} + x \right )}}{2} + \frac {\sqrt {- \frac {a}{b^{5}}} \left (a d - b c\right ) \log {\left (b^{2} \sqrt {- \frac {a}{b^{5}}} + x \right )}}{2} + \frac {d x^{3}}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(d*x**2+c)/(b*x**2+a),x)

[Out]

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

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