\(\int \frac {x (a+b x^2)^2}{(c+d x^2)^3} \, dx\) [192]

Optimal result

Integrand size = 20, antiderivative size = 67 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {(b c-a d)^2}{4 d^3 \left (c+d x^2\right )^2}+\frac {b (b c-a d)}{d^3 \left (c+d x^2\right )}+\frac {b^2 \log \left (c+d x^2\right )}{2 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {455, 45} \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b (b c-a d)}{d^3 \left (c+d x^2\right )}-\frac {(b c-a d)^2}{4 d^3 \left (c+d x^2\right )^2}+\frac {b^2 \log \left (c+d x^2\right )}{2 d^3} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.81 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {\frac {(b c-a d) \left (3 b c+a d+4 b d x^2\right )}{\left (c+d x^2\right )^2}+2 b^2 \log \left (c+d x^2\right )}{4 d^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.65 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09

method	result	size
risch	\(\frac {-\frac {b \left (a d -b c \right ) x^{2}}{d^{2}}-\frac {a^{2} d^{2}+2 a b c d -3 b^{2} c^{2}}{4 d^{3}}}{\left (d \,x^{2}+c \right )^{2}}+\frac {b^{2} \ln \left (d \,x^{2}+c \right )}{2 d^{3}}\)	\(73\)
norman	\(\frac {-\frac {a^{2} d^{2}+2 a b c d -3 b^{2} c^{2}}{4 d^{3}}-\frac {\left (a b d -b^{2} c \right ) x^{2}}{d^{2}}}{\left (d \,x^{2}+c \right )^{2}}+\frac {b^{2} \ln \left (d \,x^{2}+c \right )}{2 d^{3}}\)	\(75\)
default	\(-\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{4 d^{3} \left (d \,x^{2}+c \right )^{2}}+\frac {b^{2} \ln \left (d \,x^{2}+c \right )}{2 d^{3}}-\frac {\left (a d -b c \right ) b}{d^{3} \left (d \,x^{2}+c \right )}\)	\(76\)
parallelrisch	\(\frac {2 \ln \left (d \,x^{2}+c \right ) x^{4} b^{2} d^{2}+4 \ln \left (d \,x^{2}+c \right ) x^{2} b^{2} c d -4 x^{2} a b \,d^{2}+4 x^{2} b^{2} c d +2 \ln \left (d \,x^{2}+c \right ) b^{2} c^{2}-a^{2} d^{2}-2 a b c d +3 b^{2} c^{2}}{4 d^{3} \left (d \,x^{2}+c \right )^{2}}\)	\(111\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.61 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2} + 4 \, {\left (b^{2} c d - a b d^{2}\right )} x^{2} + 2 \, {\left (b^{2} d^{2} x^{4} + 2 \, b^{2} c d x^{2} + b^{2} c^{2}\right )} \log \left (d x^{2} + c\right )}{4 \, {\left (d^{5} x^{4} + 2 \, c d^{4} x^{2} + c^{2} d^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.30 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b^{2} \log {\left (c + d x^{2} \right )}}{2 d^{3}} + \frac {- a^{2} d^{2} - 2 a b c d + 3 b^{2} c^{2} + x^{2} \left (- 4 a b d^{2} + 4 b^{2} c d\right )}{4 c^{2} d^{3} + 8 c d^{4} x^{2} + 4 d^{5} x^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.30 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2} + 4 \, {\left (b^{2} c d - a b d^{2}\right )} x^{2}}{4 \, {\left (d^{5} x^{4} + 2 \, c d^{4} x^{2} + c^{2} d^{3}\right )}} + \frac {b^{2} \log \left (d x^{2} + c\right )}{2 \, d^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.13 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b^{2} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, d^{3}} + \frac {4 \, {\left (b^{2} c - a b d\right )} x^{2} + \frac {3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}}{d}}{4 \, {\left (d x^{2} + c\right )}^{2} d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.24 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b^2\,\ln \left (d\,x^2+c\right )}{2\,d^3}-\frac {\frac {a^2\,d^2+2\,a\,b\,c\,d-3\,b^2\,c^2}{4\,d^3}+\frac {b\,x^2\,\left (a\,d-b\,c\right )}{d^2}}{c^2+2\,c\,d\,x^2+d^2\,x^4} \]

[In]

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

[Out]

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