\(\int \frac {(a+b x^2)^2}{x^3 (c+d x^2)^3} \, dx\) [196]

Optimal result

Integrand size = 22, antiderivative size = 106 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^3} \, dx=-\frac {a^2}{2 c^3 x^2}-\frac {(b c-a d)^2}{4 c^2 d \left (c+d x^2\right )^2}+\frac {a (b c-a d)}{c^3 \left (c+d x^2\right )}+\frac {a (2 b c-3 a d) \log (x)}{c^4}-\frac {a (2 b c-3 a d) \log \left (c+d x^2\right )}{2 c^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 90} \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^3} \, dx=-\frac {a^2}{2 c^3 x^2}-\frac {a (2 b c-3 a d) \log \left (c+d x^2\right )}{2 c^4}+\frac {a \log (x) (2 b c-3 a d)}{c^4}+\frac {a (b c-a d)}{c^3 \left (c+d x^2\right )}-\frac {(b c-a d)^2}{4 c^2 d \left (c+d x^2\right )^2} \]

[In]

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

[Out]

-1/2*a^2/(c^3*x^2) - (b*c - a*d)^2/(4*c^2*d*(c + d*x^2)^2) + (a*(b*c - a*d))/(c^3*(c + d*x^2)) + (a*(2*b*c - 3
*a*d)*Log[x])/c^4 - (a*(2*b*c - 3*a*d)*Log[c + d*x^2])/(2*c^4)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2}{x^2 (c+d x)^3} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {a^2}{c^3 x^2}-\frac {a (-2 b c+3 a d)}{c^4 x}+\frac {(b c-a d)^2}{c^2 (c+d x)^3}+\frac {2 a d (-b c+a d)}{c^3 (c+d x)^2}+\frac {a d (-2 b c+3 a d)}{c^4 (c+d x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a^2}{2 c^3 x^2}-\frac {(b c-a d)^2}{4 c^2 d \left (c+d x^2\right )^2}+\frac {a (b c-a d)}{c^3 \left (c+d x^2\right )}+\frac {a (2 b c-3 a d) \log (x)}{c^4}-\frac {a (2 b c-3 a d) \log \left (c+d x^2\right )}{2 c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^3} \, dx=\frac {-\frac {2 a^2 c}{x^2}-\frac {c^2 (b c-a d)^2}{d \left (c+d x^2\right )^2}+\frac {4 a c (b c-a d)}{c+d x^2}+4 a (2 b c-3 a d) \log (x)+2 a (-2 b c+3 a d) \log \left (c+d x^2\right )}{4 c^4} \]

[In]

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

[Out]

((-2*a^2*c)/x^2 - (c^2*(b*c - a*d)^2)/(d*(c + d*x^2)^2) + (4*a*c*(b*c - a*d))/(c + d*x^2) + 4*a*(2*b*c - 3*a*d
)*Log[x] + 2*a*(-2*b*c + 3*a*d)*Log[c + d*x^2])/(4*c^4)

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {a^{2}}{2 c^{3} x^{2}}-\frac {a \left (3 a d -2 b c \right ) \ln \left (x \right )}{c^{4}}+\frac {-\frac {c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{2 d \left (d \,x^{2}+c \right )^{2}}+a \left (3 a d -2 b c \right ) \ln \left (d \,x^{2}+c \right )-\frac {2 a c \left (a d -b c \right )}{d \,x^{2}+c}}{2 c^{4}}\)	\(114\)
norman	\(\frac {-\frac {a^{2}}{2 c}+\frac {\left (6 a^{2} d^{2}-4 a b c d +b^{2} c^{2}\right ) x^{4}}{2 c^{3}}+\frac {d \left (9 a^{2} d^{2}-6 a b c d +b^{2} c^{2}\right ) x^{6}}{4 c^{4}}}{x^{2} \left (d \,x^{2}+c \right )^{2}}-\frac {a \left (3 a d -2 b c \right ) \ln \left (x \right )}{c^{4}}+\frac {a \left (3 a d -2 b c \right ) \ln \left (d \,x^{2}+c \right )}{2 c^{4}}\)	\(125\)
risch	\(\frac {-\frac {d a \left (3 a d -2 b c \right ) x^{4}}{2 c^{3}}-\frac {\left (9 a^{2} d^{2}-6 a b c d +b^{2} c^{2}\right ) x^{2}}{4 c^{2} d}-\frac {a^{2}}{2 c}}{x^{2} \left (d \,x^{2}+c \right )^{2}}-\frac {3 a^{2} \ln \left (x \right ) d}{c^{4}}+\frac {2 a \ln \left (x \right ) b}{c^{3}}+\frac {3 a^{2} \ln \left (-d \,x^{2}-c \right ) d}{2 c^{4}}-\frac {a \ln \left (-d \,x^{2}-c \right ) b}{c^{3}}\)	\(134\)
parallelrisch	\(-\frac {12 \ln \left (x \right ) x^{6} a^{2} d^{3}-8 \ln \left (x \right ) x^{6} a b c \,d^{2}-6 \ln \left (d \,x^{2}+c \right ) x^{6} a^{2} d^{3}+4 \ln \left (d \,x^{2}+c \right ) x^{6} a b c \,d^{2}-9 a^{2} d^{3} x^{6}+6 x^{6} d^{2} a b c -b^{2} c^{2} d \,x^{6}+24 \ln \left (x \right ) x^{4} a^{2} c \,d^{2}-16 \ln \left (x \right ) x^{4} a b \,c^{2} d -12 \ln \left (d \,x^{2}+c \right ) x^{4} a^{2} c \,d^{2}+8 \ln \left (d \,x^{2}+c \right ) x^{4} a b \,c^{2} d -12 a^{2} c \,d^{2} x^{4}+8 a b \,c^{2} d \,x^{4}-2 b^{2} c^{3} x^{4}+12 \ln \left (x \right ) x^{2} a^{2} c^{2} d -8 \ln \left (x \right ) x^{2} a b \,c^{3}-6 \ln \left (d \,x^{2}+c \right ) x^{2} a^{2} c^{2} d +4 \ln \left (d \,x^{2}+c \right ) x^{2} a b \,c^{3}+2 a^{2} c^{3}}{4 c^{4} x^{2} \left (d \,x^{2}+c \right )^{2}}\)	\(289\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (100) = 200\).

Time = 0.27 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.42 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^3} \, dx=-\frac {2 \, a^{2} c^{3} d - 2 \, {\left (2 \, a b c^{2} d^{2} - 3 \, a^{2} c d^{3}\right )} x^{4} + {\left (b^{2} c^{4} - 6 \, a b c^{3} d + 9 \, a^{2} c^{2} d^{2}\right )} x^{2} + 2 \, {\left ({\left (2 \, a b c d^{3} - 3 \, a^{2} d^{4}\right )} x^{6} + 2 \, {\left (2 \, a b c^{2} d^{2} - 3 \, a^{2} c d^{3}\right )} x^{4} + {\left (2 \, a b c^{3} d - 3 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \log \left (d x^{2} + c\right ) - 4 \, {\left ({\left (2 \, a b c d^{3} - 3 \, a^{2} d^{4}\right )} x^{6} + 2 \, {\left (2 \, a b c^{2} d^{2} - 3 \, a^{2} c d^{3}\right )} x^{4} + {\left (2 \, a b c^{3} d - 3 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \log \left (x\right )}{4 \, {\left (c^{4} d^{3} x^{6} + 2 \, c^{5} d^{2} x^{4} + c^{6} d x^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.22 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^3} \, dx=- \frac {a \left (3 a d - 2 b c\right ) \log {\left (x \right )}}{c^{4}} + \frac {a \left (3 a d - 2 b c\right ) \log {\left (\frac {c}{d} + x^{2} \right )}}{2 c^{4}} + \frac {- 2 a^{2} c^{2} d + x^{4} \left (- 6 a^{2} d^{3} + 4 a b c d^{2}\right ) + x^{2} \left (- 9 a^{2} c d^{2} + 6 a b c^{2} d - b^{2} c^{3}\right )}{4 c^{5} d x^{2} + 8 c^{4} d^{2} x^{4} + 4 c^{3} d^{3} x^{6}} \]

[In]

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

[Out]

-a*(3*a*d - 2*b*c)*log(x)/c**4 + a*(3*a*d - 2*b*c)*log(c/d + x**2)/(2*c**4) + (-2*a**2*c**2*d + x**4*(-6*a**2*
d**3 + 4*a*b*c*d**2) + x**2*(-9*a**2*c*d**2 + 6*a*b*c**2*d - b**2*c**3))/(4*c**5*d*x**2 + 8*c**4*d**2*x**4 + 4
*c**3*d**3*x**6)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.34 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^3} \, dx=-\frac {2 \, a^{2} c^{2} d - 2 \, {\left (2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{4} + {\left (b^{2} c^{3} - 6 \, a b c^{2} d + 9 \, a^{2} c d^{2}\right )} x^{2}}{4 \, {\left (c^{3} d^{3} x^{6} + 2 \, c^{4} d^{2} x^{4} + c^{5} d x^{2}\right )}} - \frac {{\left (2 \, a b c - 3 \, a^{2} d\right )} \log \left (d x^{2} + c\right )}{2 \, c^{4}} + \frac {{\left (2 \, a b c - 3 \, a^{2} d\right )} \log \left (x^{2}\right )}{2 \, c^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.67 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^3} \, dx=\frac {{\left (2 \, a b c - 3 \, a^{2} d\right )} \log \left (x^{2}\right )}{2 \, c^{4}} - \frac {{\left (2 \, a b c d - 3 \, a^{2} d^{2}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, c^{4} d} - \frac {2 \, a b c x^{2} - 3 \, a^{2} d x^{2} + a^{2} c}{2 \, c^{4} x^{2}} + \frac {6 \, a b c d^{3} x^{4} - 9 \, a^{2} d^{4} x^{4} + 16 \, a b c^{2} d^{2} x^{2} - 22 \, a^{2} c d^{3} x^{2} - b^{2} c^{4} + 12 \, a b c^{3} d - 14 \, a^{2} c^{2} d^{2}}{4 \, {\left (d x^{2} + c\right )}^{2} c^{4} d} \]

[In]

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

[Out]

1/2*(2*a*b*c - 3*a^2*d)*log(x^2)/c^4 - 1/2*(2*a*b*c*d - 3*a^2*d^2)*log(abs(d*x^2 + c))/(c^4*d) - 1/2*(2*a*b*c*
x^2 - 3*a^2*d*x^2 + a^2*c)/(c^4*x^2) + 1/4*(6*a*b*c*d^3*x^4 - 9*a^2*d^4*x^4 + 16*a*b*c^2*d^2*x^2 - 22*a^2*c*d^
3*x^2 - b^2*c^4 + 12*a*b*c^3*d - 14*a^2*c^2*d^2)/((d*x^2 + c)^2*c^4*d)

Mupad [B] (verification not implemented)

Time = 5.41 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^3} \, dx=\frac {\ln \left (d\,x^2+c\right )\,\left (3\,a^2\,d-2\,a\,b\,c\right )}{2\,c^4}-\frac {\frac {a^2}{2\,c}+\frac {x^2\,\left (9\,a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2\right )}{4\,c^2\,d}+\frac {a\,d\,x^4\,\left (3\,a\,d-2\,b\,c\right )}{2\,c^3}}{c^2\,x^2+2\,c\,d\,x^4+d^2\,x^6}-\frac {\ln \left (x\right )\,\left (3\,a^2\,d-2\,a\,b\,c\right )}{c^4} \]

[In]

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

[Out]

(log(c + d*x^2)*(3*a^2*d - 2*a*b*c))/(2*c^4) - (a^2/(2*c) + (x^2*(9*a^2*d^2 + b^2*c^2 - 6*a*b*c*d))/(4*c^2*d)
+ (a*d*x^4*(3*a*d - 2*b*c))/(2*c^3))/(c^2*x^2 + d^2*x^6 + 2*c*d*x^4) - (log(x)*(3*a^2*d - 2*a*b*c))/c^4