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

Optimal. Leaf size=118 \[ \frac{x^3 (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}-\frac{x (b c-a d) (5 b c-a d)}{2 c d^3}+\frac{(b c-a d) (5 b c-a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 \sqrt{c} d^{7/2}}+\frac{b^2 x^3}{3 d^2} \]

[Out]

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

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Rubi [A]  time = 0.110162, antiderivative size = 118, 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} \[ \frac{x^3 (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}-\frac{x (b c-a d) (5 b c-a d)}{2 c d^3}+\frac{(b c-a d) (5 b c-a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 \sqrt{c} d^{7/2}}+\frac{b^2 x^3}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0713394, size = 105, normalized size = 0.89 \[ \frac{\left (a^2 d^2-6 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 \sqrt{c} d^{7/2}}-\frac{x (b c-a d)^2}{2 d^3 \left (c+d x^2\right )}-\frac{2 b x (b c-a d)}{d^3}+\frac{b^2 x^3}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 156, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}{x}^{3}}{3\,{d}^{2}}}+2\,{\frac{abx}{{d}^{2}}}-2\,{\frac{{b}^{2}cx}{{d}^{3}}}-{\frac{x{a}^{2}}{2\,d \left ( d{x}^{2}+c \right ) }}+{\frac{abcx}{{d}^{2} \left ( d{x}^{2}+c \right ) }}-{\frac{{b}^{2}{c}^{2}x}{2\,{d}^{3} \left ( d{x}^{2}+c \right ) }}+{\frac{{a}^{2}}{2\,d}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-3\,{\frac{abc}{{d}^{2}\sqrt{cd}}\arctan \left ({\frac{dx}{\sqrt{cd}}} \right ) }+{\frac{5\,{b}^{2}{c}^{2}}{2\,{d}^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50136, size = 713, normalized size = 6.04 \begin{align*} \left [\frac{4 \, b^{2} c d^{3} x^{5} - 4 \,{\left (5 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3}\right )} x^{3} - 3 \,{\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} +{\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2}\right )} \sqrt{-c d} \log \left (\frac{d x^{2} - 2 \, \sqrt{-c d} x - c}{d x^{2} + c}\right ) - 6 \,{\left (5 \, b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x}{12 \,{\left (c d^{5} x^{2} + c^{2} d^{4}\right )}}, \frac{2 \, b^{2} c d^{3} x^{5} - 2 \,{\left (5 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3}\right )} x^{3} + 3 \,{\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} +{\left (5 \, b^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2}\right )} \sqrt{c d} \arctan \left (\frac{\sqrt{c d} x}{c}\right ) - 3 \,{\left (5 \, b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x}{6 \,{\left (c d^{5} x^{2} + c^{2} d^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.00835, size = 245, normalized size = 2.08 \begin{align*} \frac{b^{2} x^{3}}{3 d^{2}} - \frac{x \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{2 c d^{3} + 2 d^{4} x^{2}} - \frac{\sqrt{- \frac{1}{c d^{7}}} \left (a d - 5 b c\right ) \left (a d - b c\right ) \log{\left (- \frac{c d^{3} \sqrt{- \frac{1}{c d^{7}}} \left (a d - 5 b c\right ) \left (a d - b c\right )}{a^{2} d^{2} - 6 a b c d + 5 b^{2} c^{2}} + x \right )}}{4} + \frac{\sqrt{- \frac{1}{c d^{7}}} \left (a d - 5 b c\right ) \left (a d - b c\right ) \log{\left (\frac{c d^{3} \sqrt{- \frac{1}{c d^{7}}} \left (a d - 5 b c\right ) \left (a d - b c\right )}{a^{2} d^{2} - 6 a b c d + 5 b^{2} c^{2}} + x \right )}}{4} + \frac{x \left (2 a b d - 2 b^{2} c\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17881, size = 154, normalized size = 1.31 \begin{align*} \frac{{\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{2 \, \sqrt{c d} d^{3}} - \frac{b^{2} c^{2} x - 2 \, a b c d x + a^{2} d^{2} x}{2 \,{\left (d x^{2} + c\right )} d^{3}} + \frac{b^{2} d^{4} x^{3} - 6 \, b^{2} c d^{3} x + 6 \, a b d^{4} x}{3 \, d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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