3.271 \(\int \frac {x^4 (c+d x^2)^2}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=145 \[ -\frac {\sqrt {a} (3 b c-7 a d) (b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{9/2}}+\frac {x (3 b c-7 a d) (b c-a d)}{2 b^4}-\frac {x^3 (3 b c-7 a d) (b c-a d)}{6 a b^3}+\frac {x^5 (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}+\frac {d^2 x^5}{5 b^2} \]

[Out]

1/2*(-7*a*d+3*b*c)*(-a*d+b*c)*x/b^4-1/6*(-7*a*d+3*b*c)*(-a*d+b*c)*x^3/a/b^3+1/5*d^2*x^5/b^2+1/2*(-a*d+b*c)^2*x
^5/a/b^2/(b*x^2+a)-1/2*(-7*a*d+3*b*c)*(-a*d+b*c)*arctan(x*b^(1/2)/a^(1/2))*a^(1/2)/b^(9/2)

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Rubi [A]  time = 0.13, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {463, 459, 302, 205} \[ \frac {x^5 (b c-a d)^2}{2 a b^2 \left (a+b x^2\right )}-\frac {x^3 (3 b c-7 a d) (b c-a d)}{6 a b^3}+\frac {x (3 b c-7 a d) (b c-a d)}{2 b^4}-\frac {\sqrt {a} (3 b c-7 a d) (b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{9/2}}+\frac {d^2 x^5}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(c + d*x^2)^2)/(a + b*x^2)^2,x]

[Out]

((3*b*c - 7*a*d)*(b*c - a*d)*x)/(2*b^4) - ((3*b*c - 7*a*d)*(b*c - a*d)*x^3)/(6*a*b^3) + (d^2*x^5)/(5*b^2) + ((
b*c - a*d)^2*x^5)/(2*a*b^2*(a + b*x^2)) - (Sqrt[a]*(3*b*c - 7*a*d)*(b*c - a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2
*b^(9/2))

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 302

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]

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^4 \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx &=\frac {(b c-a d)^2 x^5}{2 a b^2 \left (a+b x^2\right )}-\frac {\int \frac {x^4 \left (-2 b^2 c^2+5 (b c-a d)^2-2 a b d^2 x^2\right )}{a+b x^2} \, dx}{2 a b^2}\\ &=\frac {d^2 x^5}{5 b^2}+\frac {(b c-a d)^2 x^5}{2 a b^2 \left (a+b x^2\right )}-\frac {((3 b c-7 a d) (b c-a d)) \int \frac {x^4}{a+b x^2} \, dx}{2 a b^2}\\ &=\frac {d^2 x^5}{5 b^2}+\frac {(b c-a d)^2 x^5}{2 a b^2 \left (a+b x^2\right )}-\frac {((3 b c-7 a d) (b c-a d)) \int \left (-\frac {a}{b^2}+\frac {x^2}{b}+\frac {a^2}{b^2 \left (a+b x^2\right )}\right ) \, dx}{2 a b^2}\\ &=\frac {(3 b c-7 a d) (b c-a d) x}{2 b^4}-\frac {(3 b c-7 a d) (b c-a d) x^3}{6 a b^3}+\frac {d^2 x^5}{5 b^2}+\frac {(b c-a d)^2 x^5}{2 a b^2 \left (a+b x^2\right )}-\frac {(a (3 b c-7 a d) (b c-a d)) \int \frac {1}{a+b x^2} \, dx}{2 b^4}\\ &=\frac {(3 b c-7 a d) (b c-a d) x}{2 b^4}-\frac {(3 b c-7 a d) (b c-a d) x^3}{6 a b^3}+\frac {d^2 x^5}{5 b^2}+\frac {(b c-a d)^2 x^5}{2 a b^2 \left (a+b x^2\right )}-\frac {\sqrt {a} (3 b c-7 a d) (b c-a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 138, normalized size = 0.95 \[ -\frac {\sqrt {a} \left (7 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{9/2}}+\frac {x \left (3 a^2 d^2-4 a b c d+b^2 c^2\right )}{b^4}+\frac {a x (b c-a d)^2}{2 b^4 \left (a+b x^2\right )}+\frac {2 d x^3 (b c-a d)}{3 b^3}+\frac {d^2 x^5}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(c + d*x^2)^2)/(a + b*x^2)^2,x]

[Out]

((b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)/b^4 + (2*d*(b*c - a*d)*x^3)/(3*b^3) + (d^2*x^5)/(5*b^2) + (a*(b*c - a*d)
^2*x)/(2*b^4*(a + b*x^2)) - (Sqrt[a]*(3*b^2*c^2 - 10*a*b*c*d + 7*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(9
/2))

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fricas [A]  time = 0.50, size = 400, normalized size = 2.76 \[ \left [\frac {12 \, b^{3} d^{2} x^{7} + 4 \, {\left (10 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{5} + 20 \, {\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{3} + 15 \, {\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 30 \, {\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} x}{60 \, {\left (b^{5} x^{2} + a b^{4}\right )}}, \frac {6 \, b^{3} d^{2} x^{7} + 2 \, {\left (10 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{5} + 10 \, {\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{3} - 15 \, {\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2} + {\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 15 \, {\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} x}{30 \, {\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(12*b^3*d^2*x^7 + 4*(10*b^3*c*d - 7*a*b^2*d^2)*x^5 + 20*(3*b^3*c^2 - 10*a*b^2*c*d + 7*a^2*b*d^2)*x^3 + 1
5*(3*a*b^2*c^2 - 10*a^2*b*c*d + 7*a^3*d^2 + (3*b^3*c^2 - 10*a*b^2*c*d + 7*a^2*b*d^2)*x^2)*sqrt(-a/b)*log((b*x^
2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + 30*(3*a*b^2*c^2 - 10*a^2*b*c*d + 7*a^3*d^2)*x)/(b^5*x^2 + a*b^4), 1/3
0*(6*b^3*d^2*x^7 + 2*(10*b^3*c*d - 7*a*b^2*d^2)*x^5 + 10*(3*b^3*c^2 - 10*a*b^2*c*d + 7*a^2*b*d^2)*x^3 - 15*(3*
a*b^2*c^2 - 10*a^2*b*c*d + 7*a^3*d^2 + (3*b^3*c^2 - 10*a*b^2*c*d + 7*a^2*b*d^2)*x^2)*sqrt(a/b)*arctan(b*x*sqrt
(a/b)/a) + 15*(3*a*b^2*c^2 - 10*a^2*b*c*d + 7*a^3*d^2)*x)/(b^5*x^2 + a*b^4)]

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giac [A]  time = 0.45, size = 156, normalized size = 1.08 \[ -\frac {{\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{4}} + \frac {a b^{2} c^{2} x - 2 \, a^{2} b c d x + a^{3} d^{2} x}{2 \, {\left (b x^{2} + a\right )} b^{4}} + \frac {3 \, b^{8} d^{2} x^{5} + 10 \, b^{8} c d x^{3} - 10 \, a b^{7} d^{2} x^{3} + 15 \, b^{8} c^{2} x - 60 \, a b^{7} c d x + 45 \, a^{2} b^{6} d^{2} x}{15 \, b^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*a*b^2*c^2 - 10*a^2*b*c*d + 7*a^3*d^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + 1/2*(a*b^2*c^2*x - 2*a^2
*b*c*d*x + a^3*d^2*x)/((b*x^2 + a)*b^4) + 1/15*(3*b^8*d^2*x^5 + 10*b^8*c*d*x^3 - 10*a*b^7*d^2*x^3 + 15*b^8*c^2
*x - 60*a*b^7*c*d*x + 45*a^2*b^6*d^2*x)/b^10

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maple [A]  time = 0.01, size = 196, normalized size = 1.35 \[ \frac {d^{2} x^{5}}{5 b^{2}}-\frac {2 a \,d^{2} x^{3}}{3 b^{3}}+\frac {2 c d \,x^{3}}{3 b^{2}}+\frac {a^{3} d^{2} x}{2 \left (b \,x^{2}+a \right ) b^{4}}-\frac {7 a^{3} d^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{4}}-\frac {a^{2} c d x}{\left (b \,x^{2}+a \right ) b^{3}}+\frac {5 a^{2} c d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{3}}+\frac {a \,c^{2} x}{2 \left (b \,x^{2}+a \right ) b^{2}}-\frac {3 a \,c^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{2}}+\frac {3 a^{2} d^{2} x}{b^{4}}-\frac {4 a c d x}{b^{3}}+\frac {c^{2} x}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(d*x^2+c)^2/(b*x^2+a)^2,x)

[Out]

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

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maxima [A]  time = 2.42, size = 149, normalized size = 1.03 \[ \frac {{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} - \frac {{\left (3 \, a b^{2} c^{2} - 10 \, a^{2} b c d + 7 \, a^{3} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{4}} + \frac {3 \, b^{2} d^{2} x^{5} + 10 \, {\left (b^{2} c d - a b d^{2}\right )} x^{3} + 15 \, {\left (b^{2} c^{2} - 4 \, a b c d + 3 \, a^{2} d^{2}\right )} x}{15 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x/(b^5*x^2 + a*b^4) - 1/2*(3*a*b^2*c^2 - 10*a^2*b*c*d + 7*a^3*d^2)*arc
tan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + 1/15*(3*b^2*d^2*x^5 + 10*(b^2*c*d - a*b*d^2)*x^3 + 15*(b^2*c^2 - 4*a*b*c*
d + 3*a^2*d^2)*x)/b^4

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mupad [B]  time = 0.17, size = 200, normalized size = 1.38 \[ x\,\left (\frac {c^2}{b^2}+\frac {2\,a\,\left (\frac {2\,a\,d^2}{b^3}-\frac {2\,c\,d}{b^2}\right )}{b}-\frac {a^2\,d^2}{b^4}\right )-x^3\,\left (\frac {2\,a\,d^2}{3\,b^3}-\frac {2\,c\,d}{3\,b^2}\right )+\frac {d^2\,x^5}{5\,b^2}+\frac {x\,\left (\frac {a^3\,d^2}{2}-a^2\,b\,c\,d+\frac {a\,b^2\,c^2}{2}\right )}{b^5\,x^2+a\,b^4}-\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,x\,\left (a\,d-b\,c\right )\,\left (7\,a\,d-3\,b\,c\right )}{7\,a^3\,d^2-10\,a^2\,b\,c\,d+3\,a\,b^2\,c^2}\right )\,\left (a\,d-b\,c\right )\,\left (7\,a\,d-3\,b\,c\right )}{2\,b^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(c + d*x^2)^2)/(a + b*x^2)^2,x)

[Out]

x*(c^2/b^2 + (2*a*((2*a*d^2)/b^3 - (2*c*d)/b^2))/b - (a^2*d^2)/b^4) - x^3*((2*a*d^2)/(3*b^3) - (2*c*d)/(3*b^2)
) + (d^2*x^5)/(5*b^2) + (x*((a^3*d^2)/2 + (a*b^2*c^2)/2 - a^2*b*c*d))/(a*b^4 + b^5*x^2) - (a^(1/2)*atan((a^(1/
2)*b^(1/2)*x*(a*d - b*c)*(7*a*d - 3*b*c))/(7*a^3*d^2 + 3*a*b^2*c^2 - 10*a^2*b*c*d))*(a*d - b*c)*(7*a*d - 3*b*c
))/(2*b^(9/2))

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sympy [B]  time = 0.98, size = 286, normalized size = 1.97 \[ x^{3} \left (- \frac {2 a d^{2}}{3 b^{3}} + \frac {2 c d}{3 b^{2}}\right ) + x \left (\frac {3 a^{2} d^{2}}{b^{4}} - \frac {4 a c d}{b^{3}} + \frac {c^{2}}{b^{2}}\right ) + \frac {x \left (a^{3} d^{2} - 2 a^{2} b c d + a b^{2} c^{2}\right )}{2 a b^{4} + 2 b^{5} x^{2}} + \frac {\sqrt {- \frac {a}{b^{9}}} \left (a d - b c\right ) \left (7 a d - 3 b c\right ) \log {\left (- \frac {b^{4} \sqrt {- \frac {a}{b^{9}}} \left (a d - b c\right ) \left (7 a d - 3 b c\right )}{7 a^{2} d^{2} - 10 a b c d + 3 b^{2} c^{2}} + x \right )}}{4} - \frac {\sqrt {- \frac {a}{b^{9}}} \left (a d - b c\right ) \left (7 a d - 3 b c\right ) \log {\left (\frac {b^{4} \sqrt {- \frac {a}{b^{9}}} \left (a d - b c\right ) \left (7 a d - 3 b c\right )}{7 a^{2} d^{2} - 10 a b c d + 3 b^{2} c^{2}} + x \right )}}{4} + \frac {d^{2} x^{5}}{5 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*(-2*a*d**2/(3*b**3) + 2*c*d/(3*b**2)) + x*(3*a**2*d**2/b**4 - 4*a*c*d/b**3 + c**2/b**2) + x*(a**3*d**2 -
2*a**2*b*c*d + a*b**2*c**2)/(2*a*b**4 + 2*b**5*x**2) + sqrt(-a/b**9)*(a*d - b*c)*(7*a*d - 3*b*c)*log(-b**4*sqr
t(-a/b**9)*(a*d - b*c)*(7*a*d - 3*b*c)/(7*a**2*d**2 - 10*a*b*c*d + 3*b**2*c**2) + x)/4 - sqrt(-a/b**9)*(a*d -
b*c)*(7*a*d - 3*b*c)*log(b**4*sqrt(-a/b**9)*(a*d - b*c)*(7*a*d - 3*b*c)/(7*a**2*d**2 - 10*a*b*c*d + 3*b**2*c**
2) + x)/4 + d**2*x**5/(5*b**2)

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