\(\int \frac {x (a+b x^2)^2}{(c+d x^2)^2} \, dx\) [183]

Optimal result

Integrand size = 20, antiderivative size = 62 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {b^2 x^2}{2 d^2}-\frac {(b c-a d)^2}{2 d^3 \left (c+d x^2\right )}-\frac {b (b c-a d) \log \left (c+d x^2\right )}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 62, 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} \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=-\frac {(b c-a d)^2}{2 d^3 \left (c+d x^2\right )}-\frac {b (b c-a d) \log \left (c+d x^2\right )}{d^3}+\frac {b^2 x^2}{2 d^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.90 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {b^2 d x^2-\frac {(b c-a d)^2}{c+d x^2}+2 b (-b c+a d) \log \left (c+d x^2\right )}{2 d^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {b^{2} x^{2}}{2 d^{2}}+\frac {\left (a d -b c \right ) \left (\frac {2 b \ln \left (d \,x^{2}+c \right )}{d}-\frac {a d -b c}{d \left (d \,x^{2}+c \right )}\right )}{2 d^{2}}\)	\(63\)
norman	\(\frac {-\frac {a^{2} d^{2}-2 a b c d +2 b^{2} c^{2}}{2 d^{3}}+\frac {b^{2} x^{4}}{2 d}}{d \,x^{2}+c}+\frac {\left (a d -b c \right ) b \ln \left (d \,x^{2}+c \right )}{d^{3}}\)	\(72\)
risch	\(\frac {b^{2} x^{2}}{2 d^{2}}-\frac {a^{2}}{2 d \left (d \,x^{2}+c \right )}+\frac {a b c}{d^{2} \left (d \,x^{2}+c \right )}-\frac {b^{2} c^{2}}{2 d^{3} \left (d \,x^{2}+c \right )}+\frac {b \ln \left (d \,x^{2}+c \right ) a}{d^{2}}-\frac {b^{2} \ln \left (d \,x^{2}+c \right ) c}{d^{3}}\)	\(97\)
parallelrisch	\(\frac {b^{2} d^{2} x^{4}+2 \ln \left (d \,x^{2}+c \right ) x^{2} a b \,d^{2}-2 \ln \left (d \,x^{2}+c \right ) x^{2} b^{2} c d +2 \ln \left (d \,x^{2}+c \right ) a b c d -2 \ln \left (d \,x^{2}+c \right ) b^{2} c^{2}-a^{2} d^{2}+2 a b c d -2 b^{2} c^{2}}{2 d^{3} \left (d \,x^{2}+c \right )}\)	\(114\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.63 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {b^{2} d^{2} x^{4} + b^{2} c d x^{2} - b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} - 2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, {\left (d^{4} x^{2} + c d^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.10 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {b^{2} x^{2}}{2 d^{2}} + \frac {b \left (a d - b c\right ) \log {\left (c + d x^{2} \right )}}{d^{3}} + \frac {- a^{2} d^{2} + 2 a b c d - b^{2} c^{2}}{2 c d^{3} + 2 d^{4} x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.19 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {b^{2} x^{2}}{2 \, d^{2}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{2 \, {\left (d^{4} x^{2} + c d^{3}\right )}} - \frac {{\left (b^{2} c - a b d\right )} \log \left (d x^{2} + c\right )}{d^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.77 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {{\left (d x^{2} + c\right )} b^{2}}{2 \, d^{3}} + \frac {{\left (b^{2} c - a b d\right )} \log \left (\frac {{\left | d x^{2} + c \right |}}{{\left (d x^{2} + c\right )}^{2} {\left | d \right |}}\right )}{d^{3}} - \frac {\frac {b^{2} c^{2} d}{d x^{2} + c} - \frac {2 \, a b c d^{2}}{d x^{2} + c} + \frac {a^{2} d^{3}}{d x^{2} + c}}{2 \, d^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.24 \[ \int \frac {x \left (a+b x^2\right )^2}{\left (c+d x^2\right )^2} \, dx=\frac {b^2\,x^2}{2\,d^2}-\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{2\,d\,\left (d^3\,x^2+c\,d^2\right )}-\frac {\ln \left (d\,x^2+c\right )\,\left (b^2\,c-a\,b\,d\right )}{d^3} \]

[In]

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

[Out]

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