\(\int \frac {(c+d x^2)^3}{x^2 (a+b x^2)^2} \, dx\) [286]

Optimal result

Integrand size = 22, antiderivative size = 131 \[ \int \frac {\left (c+d x^2\right )^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {c^2 (3 b c-a d)}{2 a^2 b x}-\frac {d^2 (b c-3 a d) x}{2 a b^2}+\frac {(b c-a d) \left (c+d x^2\right )^2}{2 a b x \left (a+b x^2\right )}-\frac {3 (b c-a d)^2 (b c+a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {479, 584, 211} \[ \int \frac {\left (c+d x^2\right )^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (b c-a d)^2 (a d+b c)}{2 a^{5/2} b^{5/2}}-\frac {c^2 (3 b c-a d)}{2 a^2 b x}-\frac {d^2 x (b c-3 a d)}{2 a b^2}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{2 a b x \left (a+b x^2\right )} \]

[In]

Int[(c + d*x^2)^3/(x^2*(a + b*x^2)^2),x]

[Out]

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

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 479

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

Rule 584

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72 \[ \int \frac {\left (c+d x^2\right )^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {c^3}{a^2 x}+\frac {d^3 x}{b^2}+\frac {(-b c+a d)^3 x}{2 a^2 b^2 \left (a+b x^2\right )}-\frac {3 (-b c+a d)^2 (b c+a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} b^{5/2}} \]

[In]

Integrate[(c + d*x^2)^3/(x^2*(a + b*x^2)^2),x]

[Out]

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

Maple [A] (verified)

Time = 2.75 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {d^{3} x}{b^{2}}-\frac {c^{3}}{x \,a^{2}}-\frac {\frac {\left (-\frac {1}{2} a^{3} d^{3}+\frac {3}{2} a^{2} b c \,d^{2}-\frac {3}{2} a \,b^{2} c^{2} d +\frac {1}{2} b^{3} c^{3}\right ) x}{b \,x^{2}+a}+\frac {3 \left (a^{3} d^{3}-a^{2} b c \,d^{2}-a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}}{a^{2} b^{2}}\)	\(129\)
risch	\(\frac {d^{3} x}{b^{2}}+\frac {\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -3 b^{3} c^{3}\right ) x^{2}}{2 a^{2}}-\frac {b^{2} c^{3}}{a}}{b^{2} x \left (b \,x^{2}+a \right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{6} d^{6}-2 a^{5} b c \,d^{5}-a^{4} b^{2} c^{2} d^{4}+4 a^{3} b^{3} c^{3} d^{3}-a^{2} b^{4} c^{4} d^{2}-2 a \,b^{5} c^{5} d +b^{6} c^{6}+a^{5} \textit {\_Z}^{2} b \right )}{\sum }\textit {\_R} \ln \left (\left (2 a^{6} d^{6}-4 a^{5} b c \,d^{5}-2 a^{4} b^{2} c^{2} d^{4}+8 a^{3} b^{3} c^{3} d^{3}-2 a^{2} b^{4} c^{4} d^{2}-4 a \,b^{5} c^{5} d +2 b^{6} c^{6}+3 \textit {\_R}^{2} a^{5} b \right ) x +\left (a^{6} d^{3}-a^{5} b c \,d^{2}-a^{4} b^{2} c^{2} d +a^{3} b^{3} c^{3}\right ) \textit {\_R} \right )\right )}{4 b^{2}}\)	\(312\)

[In]

int((d*x^2+c)^3/x^2/(b*x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 412, normalized size of antiderivative = 3.15 \[ \int \frac {\left (c+d x^2\right )^3}{x^2 \left (a+b x^2\right )^2} \, dx=\left [\frac {4 \, a^{3} b^{2} d^{3} x^{4} - 4 \, a^{2} b^{3} c^{3} - 6 \, {\left (a b^{4} c^{3} - a^{2} b^{3} c^{2} d + a^{3} b^{2} c d^{2} - a^{4} b d^{3}\right )} x^{2} - 3 \, {\left ({\left (b^{4} c^{3} - a b^{3} c^{2} d - a^{2} b^{2} c d^{2} + a^{3} b d^{3}\right )} x^{3} + {\left (a b^{3} c^{3} - a^{2} b^{2} c^{2} d - a^{3} b c d^{2} + a^{4} d^{3}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{4 \, {\left (a^{3} b^{4} x^{3} + a^{4} b^{3} x\right )}}, \frac {2 \, a^{3} b^{2} d^{3} x^{4} - 2 \, a^{2} b^{3} c^{3} - 3 \, {\left (a b^{4} c^{3} - a^{2} b^{3} c^{2} d + a^{3} b^{2} c d^{2} - a^{4} b d^{3}\right )} x^{2} - 3 \, {\left ({\left (b^{4} c^{3} - a b^{3} c^{2} d - a^{2} b^{2} c d^{2} + a^{3} b d^{3}\right )} x^{3} + {\left (a b^{3} c^{3} - a^{2} b^{2} c^{2} d - a^{3} b c d^{2} + a^{4} d^{3}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{2 \, {\left (a^{3} b^{4} x^{3} + a^{4} b^{3} x\right )}}\right ] \]

[In]

integrate((d*x^2+c)^3/x^2/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (112) = 224\).

Time = 0.89 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.36 \[ \int \frac {\left (c+d x^2\right )^3}{x^2 \left (a+b x^2\right )^2} \, dx=\frac {3 \sqrt {- \frac {1}{a^{5} b^{5}}} \left (a d - b c\right )^{2} \left (a d + b c\right ) \log {\left (- \frac {3 a^{3} b^{2} \sqrt {- \frac {1}{a^{5} b^{5}}} \left (a d - b c\right )^{2} \left (a d + b c\right )}{3 a^{3} d^{3} - 3 a^{2} b c d^{2} - 3 a b^{2} c^{2} d + 3 b^{3} c^{3}} + x \right )}}{4} - \frac {3 \sqrt {- \frac {1}{a^{5} b^{5}}} \left (a d - b c\right )^{2} \left (a d + b c\right ) \log {\left (\frac {3 a^{3} b^{2} \sqrt {- \frac {1}{a^{5} b^{5}}} \left (a d - b c\right )^{2} \left (a d + b c\right )}{3 a^{3} d^{3} - 3 a^{2} b c d^{2} - 3 a b^{2} c^{2} d + 3 b^{3} c^{3}} + x \right )}}{4} + \frac {- 2 a b^{2} c^{3} + x^{2} \left (a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - 3 b^{3} c^{3}\right )}{2 a^{3} b^{2} x + 2 a^{2} b^{3} x^{3}} + \frac {d^{3} x}{b^{2}} \]

[In]

integrate((d*x**2+c)**3/x**2/(b*x**2+a)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.07 \[ \int \frac {\left (c+d x^2\right )^3}{x^2 \left (a+b x^2\right )^2} \, dx=\frac {d^{3} x}{b^{2}} - \frac {2 \, a b^{2} c^{3} + {\left (3 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{2}}{2 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2} x\right )}} - \frac {3 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2} b^{2}} \]

[In]

integrate((d*x^2+c)^3/x^2/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.09 \[ \int \frac {\left (c+d x^2\right )^3}{x^2 \left (a+b x^2\right )^2} \, dx=\frac {d^{3} x}{b^{2}} - \frac {3 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2} b^{2}} - \frac {3 \, b^{3} c^{3} x^{2} - 3 \, a b^{2} c^{2} d x^{2} + 3 \, a^{2} b c d^{2} x^{2} - a^{3} d^{3} x^{2} + 2 \, a b^{2} c^{3}}{2 \, {\left (b x^{3} + a x\right )} a^{2} b^{2}} \]

[In]

integrate((d*x^2+c)^3/x^2/(b*x^2+a)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.84 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.32 \[ \int \frac {\left (c+d x^2\right )^3}{x^2 \left (a+b x^2\right )^2} \, dx=\frac {\frac {x^2\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-3\,b^3\,c^3\right )}{2\,a^2}-\frac {b^2\,c^3}{a}}{b^3\,x^3+a\,b^2\,x}+\frac {d^3\,x}{b^2}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {b}\,x\,\left (a\,d+b\,c\right )\,{\left (a\,d-b\,c\right )}^2}{\sqrt {a}\,\left (3\,a^3\,d^3-3\,a^2\,b\,c\,d^2-3\,a\,b^2\,c^2\,d+3\,b^3\,c^3\right )}\right )\,\left (a\,d+b\,c\right )\,{\left (a\,d-b\,c\right )}^2}{2\,a^{5/2}\,b^{5/2}} \]

[In]

int((c + d*x^2)^3/(x^2*(a + b*x^2)^2),x)

[Out]

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