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

Optimal. Leaf size=116 \[ \frac{x^3 (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}-\frac{x (b c-5 a d) (b c-a d)}{2 a b^3}+\frac{(b c-5 a d) (b c-a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 \sqrt{a} b^{7/2}}+\frac{d^2 x^3}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.111412, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {463, 459, 321, 205} \[ \frac{x^3 (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}-\frac{x (b c-5 a d) (b c-a d)}{2 a b^3}+\frac{(b c-5 a d) (b c-a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 \sqrt{a} b^{7/2}}+\frac{d^2 x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

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]

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

Mathematica [A]  time = 0.0738896, size = 105, normalized size = 0.91 \[ \frac{\left (5 a^2 d^2-6 a b c d+b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 \sqrt{a} b^{7/2}}-\frac{x (b c-a d)^2}{2 b^3 \left (a+b x^2\right )}+\frac{2 d x (b c-a d)}{b^3}+\frac{d^2 x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(b*c - a*d)*x)/b^3 + (d^2*x^3)/(3*b^2) - ((b*c - a*d)^2*x)/(2*b^3*(a + b*x^2)) + ((b^2*c^2 - 6*a*b*c*d +
5*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[a]*b^(7/2))

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Maple [A]  time = 0.008, size = 156, normalized size = 1.3 \begin{align*}{\frac{{d}^{2}{x}^{3}}{3\,{b}^{2}}}-2\,{\frac{a{d}^{2}x}{{b}^{3}}}+2\,{\frac{dxc}{{b}^{2}}}-{\frac{{a}^{2}{d}^{2}x}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{cxad}{{b}^{2} \left ( b{x}^{2}+a \right ) }}-{\frac{x{c}^{2}}{2\,b \left ( b{x}^{2}+a \right ) }}+{\frac{5\,{a}^{2}{d}^{2}}{2\,{b}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-3\,{\frac{acd}{{b}^{2}\sqrt{ab}}\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ) }+{\frac{{c}^{2}}{2\,b}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*d^2*x^3/b^2-2*d^2/b^3*a*x+2*d/b^2*x*c-1/2/b^3*x/(b*x^2+a)*a^2*d^2+1/b^2*x/(b*x^2+a)*c*a*d-1/2/b*x/(b*x^2+a
)*c^2+5/2/b^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*a^2*d^2-3/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c*a*d+1/2/
b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c^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^2*(d*x^2+c)^2/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51846, size = 713, normalized size = 6.15 \begin{align*} \left [\frac{4 \, a b^{3} d^{2} x^{5} + 4 \,{\left (6 \, a b^{3} c d - 5 \, a^{2} b^{2} d^{2}\right )} x^{3} - 3 \,{\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} +{\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right ) - 6 \,{\left (a b^{3} c^{2} - 6 \, a^{2} b^{2} c d + 5 \, a^{3} b d^{2}\right )} x}{12 \,{\left (a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac{2 \, a b^{3} d^{2} x^{5} + 2 \,{\left (6 \, a b^{3} c d - 5 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \,{\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} +{\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right ) - 3 \,{\left (a b^{3} c^{2} - 6 \, a^{2} b^{2} c d + 5 \, a^{3} b d^{2}\right )} x}{6 \,{\left (a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(4*a*b^3*d^2*x^5 + 4*(6*a*b^3*c*d - 5*a^2*b^2*d^2)*x^3 - 3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c
^2 - 6*a*b^2*c*d + 5*a^2*b*d^2)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 6*(a*b^3*c^2 -
6*a^2*b^2*c*d + 5*a^3*b*d^2)*x)/(a*b^5*x^2 + a^2*b^4), 1/6*(2*a*b^3*d^2*x^5 + 2*(6*a*b^3*c*d - 5*a^2*b^2*d^2)
*x^3 + 3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d + 5*a^2*b*d^2)*x^2)*sqrt(a*b)*arctan(sq
rt(a*b)*x/a) - 3*(a*b^3*c^2 - 6*a^2*b^2*c*d + 5*a^3*b*d^2)*x)/(a*b^5*x^2 + a^2*b^4)]

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Sympy [B]  time = 1.02119, size = 245, normalized size = 2.11 \begin{align*} - \frac{x \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{2 a b^{3} + 2 b^{4} x^{2}} - \frac{\sqrt{- \frac{1}{a b^{7}}} \left (a d - b c\right ) \left (5 a d - b c\right ) \log{\left (- \frac{a b^{3} \sqrt{- \frac{1}{a b^{7}}} \left (a d - b c\right ) \left (5 a d - b c\right )}{5 a^{2} d^{2} - 6 a b c d + b^{2} c^{2}} + x \right )}}{4} + \frac{\sqrt{- \frac{1}{a b^{7}}} \left (a d - b c\right ) \left (5 a d - b c\right ) \log{\left (\frac{a b^{3} \sqrt{- \frac{1}{a b^{7}}} \left (a d - b c\right ) \left (5 a d - b c\right )}{5 a^{2} d^{2} - 6 a b c d + b^{2} c^{2}} + x \right )}}{4} + \frac{d^{2} x^{3}}{3 b^{2}} - \frac{x \left (2 a d^{2} - 2 b c d\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*(a**2*d**2 - 2*a*b*c*d + b**2*c**2)/(2*a*b**3 + 2*b**4*x**2) - sqrt(-1/(a*b**7))*(a*d - b*c)*(5*a*d - b*c)*
log(-a*b**3*sqrt(-1/(a*b**7))*(a*d - b*c)*(5*a*d - b*c)/(5*a**2*d**2 - 6*a*b*c*d + b**2*c**2) + x)/4 + sqrt(-1
/(a*b**7))*(a*d - b*c)*(5*a*d - b*c)*log(a*b**3*sqrt(-1/(a*b**7))*(a*d - b*c)*(5*a*d - b*c)/(5*a**2*d**2 - 6*a
*b*c*d + b**2*c**2) + x)/4 + d**2*x**3/(3*b**2) - x*(2*a*d**2 - 2*b*c*d)/b**3

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Giac [A]  time = 1.1219, size = 154, normalized size = 1.33 \begin{align*} \frac{{\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} b^{3}} - \frac{b^{2} c^{2} x - 2 \, a b c d x + a^{2} d^{2} x}{2 \,{\left (b x^{2} + a\right )} b^{3}} + \frac{b^{4} d^{2} x^{3} + 6 \, b^{4} c d x - 6 \, a b^{3} d^{2} x}{3 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/2*(b^2*c^2*x - 2*a*b*c*d*x + a
^2*d^2*x)/((b*x^2 + a)*b^3) + 1/3*(b^4*d^2*x^3 + 6*b^4*c*d*x - 6*a*b^3*d^2*x)/b^6