\(\int \frac {(a+b x^2)^2}{x^4 (c+d x^2)^2} \, dx\) [188]

Optimal result

Integrand size = 22, antiderivative size = 126 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^2} \, dx=-\frac {a (6 b c-5 a d)}{3 c^3 x}-\frac {a^2}{3 c x^3 \left (c+d x^2\right )}+\frac {\left (3 b^2 c^2-6 a b c d+5 a^2 d^2\right ) x}{6 c^3 \left (c+d x^2\right )}+\frac {(b c-5 a d) (b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} \sqrt {d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {473, 467, 464, 211} \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^2} \, dx=\frac {x \left (5 a^2 d^2-6 a b c d+3 b^2 c^2\right )}{6 c^3 \left (c+d x^2\right )}-\frac {a^2}{3 c x^3 \left (c+d x^2\right )}+\frac {(b c-5 a d) (b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} \sqrt {d}}-\frac {a (6 b c-5 a d)}{3 c^3 x} \]

[In]

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

[Out]

-1/3*(a*(6*b*c - 5*a*d))/(c^3*x) - a^2/(3*c*x^3*(c + d*x^2)) + ((3*b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*x)/(6*c^3*
(c + d*x^2)) + ((b*c - 5*a*d)*(b*c - a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(7/2)*Sqrt[d])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 464

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

Rule 467

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 473

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^2} \, dx=-\frac {a^2}{3 c^2 x^3}+\frac {2 a (-b c+a d)}{c^3 x}+\frac {(b c-a d)^2 x}{2 c^3 \left (c+d x^2\right )}+\frac {\left (b^2 c^2-6 a b c d+5 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} \sqrt {d}} \]

[In]

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

[Out]

-1/3*a^2/(c^2*x^3) + (2*a*(-(b*c) + a*d))/(c^3*x) + ((b*c - a*d)^2*x)/(2*c^3*(c + d*x^2)) + ((b^2*c^2 - 6*a*b*
c*d + 5*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(7/2)*Sqrt[d])

Maple [A] (verified)

Time = 2.69 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {a^{2}}{3 c^{2} x^{3}}+\frac {2 \left (a d -b c \right ) a}{c^{3} x}+\frac {\frac {\left (\frac {1}{2} a^{2} d^{2}-a b c d +\frac {1}{2} b^{2} c^{2}\right ) x}{d \,x^{2}+c}+\frac {\left (5 a^{2} d^{2}-6 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \sqrt {c d}}}{c^{3}}\)	\(107\)
risch	\(\frac {\frac {\left (5 a^{2} d^{2}-6 a b c d +b^{2} c^{2}\right ) x^{4}}{2 c^{3}}+\frac {a \left (5 a d -6 b c \right ) x^{2}}{3 c^{2}}-\frac {a^{2}}{3 c}}{x^{3} \left (d \,x^{2}+c \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{7} d \,\textit {\_Z}^{2}+25 a^{4} d^{4}-60 a^{3} b c \,d^{3}+46 a^{2} b^{2} c^{2} d^{2}-12 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} c^{7} d +50 a^{4} d^{4}-120 a^{3} b c \,d^{3}+92 a^{2} b^{2} c^{2} d^{2}-24 a \,b^{3} c^{3} d +2 b^{4} c^{4}\right ) x +\left (-5 a^{2} c^{4} d^{2}+6 a b \,c^{5} d -b^{2} c^{6}\right ) \textit {\_R} \right )\right )}{4}\)	\(232\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.83 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^2} \, dx=\left [-\frac {4 \, a^{2} c^{3} d - 6 \, {\left (b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + 5 \, a^{2} c d^{3}\right )} x^{4} + 4 \, {\left (6 \, a b c^{3} d - 5 \, a^{2} c^{2} d^{2}\right )} x^{2} + 3 \, {\left ({\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{5} + {\left (b^{2} c^{3} - 6 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {-c d} \log \left (\frac {d x^{2} - 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right )}{12 \, {\left (c^{4} d^{2} x^{5} + c^{5} d x^{3}\right )}}, -\frac {2 \, a^{2} c^{3} d - 3 \, {\left (b^{2} c^{3} d - 6 \, a b c^{2} d^{2} + 5 \, a^{2} c d^{3}\right )} x^{4} + 2 \, {\left (6 \, a b c^{3} d - 5 \, a^{2} c^{2} d^{2}\right )} x^{2} - 3 \, {\left ({\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} x^{5} + {\left (b^{2} c^{3} - 6 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right )}{6 \, {\left (c^{4} d^{2} x^{5} + c^{5} d x^{3}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (114) = 228\).

Time = 0.55 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.97 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^2} \, dx=- \frac {\sqrt {- \frac {1}{c^{7} d}} \left (a d - b c\right ) \left (5 a d - b c\right ) \log {\left (- \frac {c^{4} \sqrt {- \frac {1}{c^{7} d}} \left (a d - b c\right ) \left (5 a d - b c\right )}{5 a^{2} d^{2} - 6 a b c d + b^{2} c^{2}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{c^{7} d}} \left (a d - b c\right ) \left (5 a d - b c\right ) \log {\left (\frac {c^{4} \sqrt {- \frac {1}{c^{7} d}} \left (a d - b c\right ) \left (5 a d - b c\right )}{5 a^{2} d^{2} - 6 a b c d + b^{2} c^{2}} + x \right )}}{4} + \frac {- 2 a^{2} c^{2} + x^{4} \cdot \left (15 a^{2} d^{2} - 18 a b c d + 3 b^{2} c^{2}\right ) + x^{2} \cdot \left (10 a^{2} c d - 12 a b c^{2}\right )}{6 c^{4} x^{3} + 6 c^{3} d x^{5}} \]

[In]

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

[Out]

-sqrt(-1/(c**7*d))*(a*d - b*c)*(5*a*d - b*c)*log(-c**4*sqrt(-1/(c**7*d))*(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/(c**7*d))*(a*d - b*c)*(5*a*d - b*c)*log(c**4*sqrt(-1/(c**7*d))*(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 + (-2*a**2*c**2 + x**4*(15*a**2*d**2 - 18*
a*b*c*d + 3*b**2*c**2) + x**2*(10*a**2*c*d - 12*a*b*c**2))/(6*c**4*x**3 + 6*c**3*d*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^2} \, dx=\frac {3 \, {\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} x^{4} - 2 \, a^{2} c^{2} - 2 \, {\left (6 \, a b c^{2} - 5 \, a^{2} c d\right )} x^{2}}{6 \, {\left (c^{3} d x^{5} + c^{4} x^{3}\right )}} + \frac {{\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {c d} c^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^2} \, dx=\frac {{\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {c d} c^{3}} + \frac {b^{2} c^{2} x - 2 \, a b c d x + a^{2} d^{2} x}{2 \, {\left (d x^{2} + c\right )} c^{3}} - \frac {6 \, a b c x^{2} - 6 \, a^{2} d x^{2} + a^{2} c}{3 \, c^{3} x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^2} \, dx=\frac {\frac {x^4\,\left (5\,a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2\right )}{2\,c^3}-\frac {a^2}{3\,c}+\frac {a\,x^2\,\left (5\,a\,d-6\,b\,c\right )}{3\,c^2}}{d\,x^5+c\,x^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,x\,\left (a\,d-b\,c\right )\,\left (5\,a\,d-b\,c\right )}{\sqrt {c}\,\left (5\,a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2\right )}\right )\,\left (a\,d-b\,c\right )\,\left (5\,a\,d-b\,c\right )}{2\,c^{7/2}\,\sqrt {d}} \]

[In]

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

[Out]

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