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

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

[Out]

1/2*(-3*a*d+7*b*c)*(-a*d+b*c)*x/d^4-1/6*(-3*a*d+7*b*c)*(-a*d+b*c)*x^3/c/d^3+1/5*b^2*x^5/d^2+1/2*(-a*d+b*c)^2*x
^5/c/d^2/(d*x^2+c)-1/2*(-3*a*d+7*b*c)*(-a*d+b*c)*arctan(x*d^(1/2)/c^(1/2))*c^(1/2)/d^(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} \[ \frac {x^5 (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}-\frac {x^3 (7 b c-3 a d) (b c-a d)}{6 c d^3}+\frac {x (7 b c-3 a d) (b c-a d)}{2 d^4}-\frac {\sqrt {c} (7 b c-3 a d) (b c-a d) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 d^{9/2}}+\frac {b^2 x^5}{5 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.10, size = 138, normalized size = 0.95 \[ -\frac {\sqrt {c} \left (3 a^2 d^2-10 a b c d+7 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 d^{9/2}}+\frac {x \left (a^2 d^2-4 a b c d+3 b^2 c^2\right )}{d^4}+\frac {c x (b c-a d)^2}{2 d^4 \left (c+d x^2\right )}-\frac {2 b x^3 (b c-a d)}{3 d^3}+\frac {b^2 x^5}{5 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 400, normalized size = 2.76 \[ \left [\frac {12 \, b^{2} d^{3} x^{7} - 4 \, {\left (7 \, b^{2} c d^{2} - 10 \, a b d^{3}\right )} x^{5} + 20 \, {\left (7 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{3} + 15 \, {\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2} + {\left (7 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{2} - 2 \, d x \sqrt {-\frac {c}{d}} - c}{d x^{2} + c}\right ) + 30 \, {\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x}{60 \, {\left (d^{5} x^{2} + c d^{4}\right )}}, \frac {6 \, b^{2} d^{3} x^{7} - 2 \, {\left (7 \, b^{2} c d^{2} - 10 \, a b d^{3}\right )} x^{5} + 10 \, {\left (7 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{3} - 15 \, {\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2} + {\left (7 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {\frac {c}{d}} \arctan \left (\frac {d x \sqrt {\frac {c}{d}}}{c}\right ) + 15 \, {\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x}{30 \, {\left (d^{5} x^{2} + c d^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.37, size = 156, normalized size = 1.08 \[ -\frac {{\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {c d} d^{4}} + \frac {b^{2} c^{3} x - 2 \, a b c^{2} d x + a^{2} c d^{2} x}{2 \, {\left (d x^{2} + c\right )} d^{4}} + \frac {3 \, b^{2} d^{8} x^{5} - 10 \, b^{2} c d^{7} x^{3} + 10 \, a b d^{8} x^{3} + 45 \, b^{2} c^{2} d^{6} x - 60 \, a b c d^{7} x + 15 \, a^{2} d^{8} x}{15 \, d^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.38, size = 149, normalized size = 1.03 \[ \frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} x}{2 \, {\left (d^{5} x^{2} + c d^{4}\right )}} - \frac {{\left (7 \, b^{2} c^{3} - 10 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {c d} d^{4}} + \frac {3 \, b^{2} d^{2} x^{5} - 10 \, {\left (b^{2} c d - a b d^{2}\right )} x^{3} + 15 \, {\left (3 \, b^{2} c^{2} - 4 \, a b c d + a^{2} d^{2}\right )} x}{15 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.14, size = 200, normalized size = 1.38 \[ x\,\left (\frac {a^2}{d^2}+\frac {2\,c\,\left (\frac {2\,b^2\,c}{d^3}-\frac {2\,a\,b}{d^2}\right )}{d}-\frac {b^2\,c^2}{d^4}\right )-x^3\,\left (\frac {2\,b^2\,c}{3\,d^3}-\frac {2\,a\,b}{3\,d^2}\right )+\frac {b^2\,x^5}{5\,d^2}+\frac {x\,\left (\frac {a^2\,c\,d^2}{2}-a\,b\,c^2\,d+\frac {b^2\,c^3}{2}\right )}{d^5\,x^2+c\,d^4}-\frac {\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,x\,\left (a\,d-b\,c\right )\,\left (3\,a\,d-7\,b\,c\right )}{3\,a^2\,c\,d^2-10\,a\,b\,c^2\,d+7\,b^2\,c^3}\right )\,\left (a\,d-b\,c\right )\,\left (3\,a\,d-7\,b\,c\right )}{2\,d^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.99, size = 286, normalized size = 1.97 \[ \frac {b^{2} x^{5}}{5 d^{2}} + x^{3} \left (\frac {2 a b}{3 d^{2}} - \frac {2 b^{2} c}{3 d^{3}}\right ) + x \left (\frac {a^{2}}{d^{2}} - \frac {4 a b c}{d^{3}} + \frac {3 b^{2} c^{2}}{d^{4}}\right ) + \frac {x \left (a^{2} c d^{2} - 2 a b c^{2} d + b^{2} c^{3}\right )}{2 c d^{4} + 2 d^{5} x^{2}} + \frac {\sqrt {- \frac {c}{d^{9}}} \left (a d - b c\right ) \left (3 a d - 7 b c\right ) \log {\left (- \frac {d^{4} \sqrt {- \frac {c}{d^{9}}} \left (a d - b c\right ) \left (3 a d - 7 b c\right )}{3 a^{2} d^{2} - 10 a b c d + 7 b^{2} c^{2}} + x \right )}}{4} - \frac {\sqrt {- \frac {c}{d^{9}}} \left (a d - b c\right ) \left (3 a d - 7 b c\right ) \log {\left (\frac {d^{4} \sqrt {- \frac {c}{d^{9}}} \left (a d - b c\right ) \left (3 a d - 7 b c\right )}{3 a^{2} d^{2} - 10 a b c d + 7 b^{2} c^{2}} + x \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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