\(\int \frac {(a+b x^2)^2}{x (c+d x^2)^3} \, dx\) [194]

Optimal result

Integrand size = 22, antiderivative size = 86 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^3} \, dx=\frac {(b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}+\frac {\frac {a^2}{c^2}-\frac {b^2}{d^2}}{2 \left (c+d x^2\right )}+\frac {a^2 \log (x)}{c^3}-\frac {a^2 \log \left (c+d x^2\right )}{2 c^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 86, 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} \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^3} \, dx=\frac {\frac {a^2}{c^2}-\frac {b^2}{d^2}}{2 \left (c+d x^2\right )}-\frac {a^2 \log \left (c+d x^2\right )}{2 c^3}+\frac {a^2 \log (x)}{c^3}+\frac {(b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^3} \, dx=-\frac {\frac {c (b c-a d) \left (b c \left (c+2 d x^2\right )+a d \left (3 c+2 d x^2\right )\right )}{d^2 \left (c+d x^2\right )^2}-4 a^2 \log (x)+2 a^2 \log \left (c+d x^2\right )}{4 c^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.62 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02

method	result	size
norman	\(\frac {-\frac {\left (a^{2} d -a b c \right ) x^{2}}{c^{2}}-\frac {\left (3 a^{2} d^{2}-2 a b c d -b^{2} c^{2}\right ) x^{4}}{4 c^{3}}}{\left (d \,x^{2}+c \right )^{2}}+\frac {a^{2} \ln \left (x \right )}{c^{3}}-\frac {a^{2} \ln \left (d \,x^{2}+c \right )}{2 c^{3}}\)	\(88\)
risch	\(\frac {\frac {\left (a^{2} d^{2}-b^{2} c^{2}\right ) x^{2}}{2 c^{2} d}+\frac {3 a^{2} d^{2}-2 a b c d -b^{2} c^{2}}{4 c \,d^{2}}}{\left (d \,x^{2}+c \right )^{2}}+\frac {a^{2} \ln \left (x \right )}{c^{3}}-\frac {a^{2} \ln \left (d \,x^{2}+c \right )}{2 c^{3}}\)	\(96\)
default	\(\frac {a^{2} \ln \left (x \right )}{c^{3}}-\frac {-\frac {c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{2 d^{2} \left (d \,x^{2}+c \right )^{2}}+a^{2} \ln \left (d \,x^{2}+c \right )-\frac {c \left (a^{2} d^{2}-b^{2} c^{2}\right )}{d^{2} \left (d \,x^{2}+c \right )}}{2 c^{3}}\)	\(98\)
parallelrisch	\(\frac {4 \ln \left (x \right ) x^{4} a^{2} d^{2}-2 \ln \left (d \,x^{2}+c \right ) x^{4} a^{2} d^{2}-3 a^{2} d^{2} x^{4}+2 x^{4} a b c d +b^{2} c^{2} x^{4}+8 \ln \left (x \right ) x^{2} a^{2} c d -4 \ln \left (d \,x^{2}+c \right ) x^{2} a^{2} c d -4 a^{2} c d \,x^{2}+4 a b \,c^{2} x^{2}+4 a^{2} c^{2} \ln \left (x \right )-2 \ln \left (d \,x^{2}+c \right ) a^{2} c^{2}}{4 c^{3} \left (d \,x^{2}+c \right )^{2}}\)	\(154\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (80) = 160\).

Time = 0.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.90 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^3} \, dx=-\frac {b^{2} c^{4} + 2 \, a b c^{3} d - 3 \, a^{2} c^{2} d^{2} + 2 \, {\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x^{2} + 2 \, {\left (a^{2} d^{4} x^{4} + 2 \, a^{2} c d^{3} x^{2} + a^{2} c^{2} d^{2}\right )} \log \left (d x^{2} + c\right ) - 4 \, {\left (a^{2} d^{4} x^{4} + 2 \, a^{2} c d^{3} x^{2} + a^{2} c^{2} d^{2}\right )} \log \left (x\right )}{4 \, {\left (c^{3} d^{4} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{5} d^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^3} \, dx=\frac {a^{2} \log {\left (x \right )}}{c^{3}} - \frac {a^{2} \log {\left (\frac {c}{d} + x^{2} \right )}}{2 c^{3}} + \frac {3 a^{2} c d^{2} - 2 a b c^{2} d - b^{2} c^{3} + x^{2} \cdot \left (2 a^{2} d^{3} - 2 b^{2} c^{2} d\right )}{4 c^{4} d^{2} + 8 c^{3} d^{3} x^{2} + 4 c^{2} d^{4} x^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^3} \, dx=-\frac {b^{2} c^{3} + 2 \, a b c^{2} d - 3 \, a^{2} c d^{2} + 2 \, {\left (b^{2} c^{2} d - a^{2} d^{3}\right )} x^{2}}{4 \, {\left (c^{2} d^{4} x^{4} + 2 \, c^{3} d^{3} x^{2} + c^{4} d^{2}\right )}} - \frac {a^{2} \log \left (d x^{2} + c\right )}{2 \, c^{3}} + \frac {a^{2} \log \left (x^{2}\right )}{2 \, c^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^3} \, dx=\frac {a^{2} \log \left (x^{2}\right )}{2 \, c^{3}} - \frac {a^{2} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, c^{3}} + \frac {3 \, a^{2} d^{4} x^{4} - 2 \, b^{2} c^{3} d x^{2} + 8 \, a^{2} c d^{3} x^{2} - b^{2} c^{4} - 2 \, a b c^{3} d + 6 \, a^{2} c^{2} d^{2}}{4 \, {\left (d x^{2} + c\right )}^{2} c^{3} d^{2}} \]

[In]

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

[Out]

1/2*a^2*log(x^2)/c^3 - 1/2*a^2*log(abs(d*x^2 + c))/c^3 + 1/4*(3*a^2*d^4*x^4 - 2*b^2*c^3*d*x^2 + 8*a^2*c*d^3*x^
2 - b^2*c^4 - 2*a*b*c^3*d + 6*a^2*c^2*d^2)/((d*x^2 + c)^2*c^3*d^2)

Mupad [B] (verification not implemented)

Time = 5.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^2\right )^2}{x \left (c+d x^2\right )^3} \, dx=\frac {a^2\,\ln \left (x\right )}{c^3}-\frac {a^2\,\ln \left (d\,x^2+c\right )}{2\,c^3}-\frac {\frac {-3\,a^2\,d^2+2\,a\,b\,c\,d+b^2\,c^2}{4\,c\,d^2}-\frac {x^2\,\left (a^2\,d^2-b^2\,c^2\right )}{2\,c^2\,d}}{c^2+2\,c\,d\,x^2+d^2\,x^4} \]

[In]

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

[Out]

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