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

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

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 116, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.49, size = 342, normalized size = 2.95 \[ \left [\frac {4 \, a b^{3} d^{2} x^{5} + 4 \, {\left (6 \, a b^{3} c d - 5 \, a^{2} b^{2} d^{2}\right )} x^{3} - 3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 6 \, {\left (a b^{3} c^{2} - 6 \, a^{2} b^{2} c d + 5 \, a^{3} b d^{2}\right )} x}{12 \, {\left (a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac {2 \, a b^{3} d^{2} x^{5} + 2 \, {\left (6 \, a b^{3} c d - 5 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 3 \, {\left (a b^{3} c^{2} - 6 \, a^{2} b^{2} c d + 5 \, a^{3} b d^{2}\right )} x}{6 \, {\left (a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \]

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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giac [A]  time = 0.36, size = 114, normalized size = 0.98 \[ \frac {{\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} - \frac {b^{2} c^{2} x - 2 \, a b c d x + a^{2} d^{2} x}{2 \, {\left (b x^{2} + a\right )} b^{3}} + \frac {b^{4} d^{2} x^{3} + 6 \, b^{4} c d x - 6 \, a b^{3} d^{2} x}{3 \, b^{6}} \]

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

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maple [A]  time = 0.01, size = 156, normalized size = 1.34 \[ \frac {d^{2} x^{3}}{3 b^{2}}-\frac {a^{2} d^{2} x}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {5 a^{2} d^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{3}}+\frac {a c d x}{\left (b \,x^{2}+a \right ) b^{2}}-\frac {3 a c d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}-\frac {c^{2} x}{2 \left (b \,x^{2}+a \right ) b}+\frac {c^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b}-\frac {2 a \,d^{2} x}{b^{3}}+\frac {2 c d x}{b^{2}} \]

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

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maxima [A]  time = 2.41, size = 109, normalized size = 0.94 \[ -\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} + \frac {{\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {b d^{2} x^{3} + 6 \, {\left (b c d - a d^{2}\right )} x}{3 \, b^{3}} \]

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]

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

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mupad [B]  time = 0.18, size = 148, normalized size = 1.28 \[ \frac {d^2\,x^3}{3\,b^2}-\frac {x\,\left (\frac {a^2\,d^2}{2}-a\,b\,c\,d+\frac {b^2\,c^2}{2}\right )}{b^4\,x^2+a\,b^3}-x\,\left (\frac {2\,a\,d^2}{b^3}-\frac {2\,c\,d}{b^2}\right )+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x\,\left (a\,d-b\,c\right )\,\left (5\,a\,d-b\,c\right )}{\sqrt {a}\,\left (5\,a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2\right )}\right )\,\left (a\,d-b\,c\right )\,\left (5\,a\,d-b\,c\right )}{2\,\sqrt {a}\,b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.86, size = 246, normalized size = 2.12 \[ x \left (- \frac {2 a d^{2}}{b^{3}} + \frac {2 c d}{b^{2}}\right ) + \frac {x \left (- a^{2} d^{2} + 2 a b c d - b^{2} c^{2}\right )}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {\sqrt {- \frac {1}{a b^{7}}} \left (a d - b c\right ) \left (5 a d - b c\right ) \log {\left (- \frac {a b^{3} \sqrt {- \frac {1}{a b^{7}}} \left (a d - b c\right ) \left (5 a d - b c\right )}{5 a^{2} d^{2} - 6 a b c d + b^{2} c^{2}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a b^{7}}} \left (a d - b c\right ) \left (5 a d - b c\right ) \log {\left (\frac {a b^{3} \sqrt {- \frac {1}{a b^{7}}} \left (a d - b c\right ) \left (5 a d - b c\right )}{5 a^{2} d^{2} - 6 a b c d + b^{2} c^{2}} + x \right )}}{4} + \frac {d^{2} x^{3}}{3 b^{2}} \]

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*(-2*a*d**2/b**3 + 2*c*d/b**2) + 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)

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