\(\int \frac {(a+b x^2)^2 (c+d x^2)}{x^4} \, dx\) [149]

Optimal result

Integrand size = 20, antiderivative size = 48 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^4} \, dx=-\frac {a^2 c}{3 x^3}-\frac {a (2 b c+a d)}{x}+b (b c+2 a d) x+\frac {1}{3} b^2 d x^3 \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^4} \, dx=-\frac {a^2 c}{3 x^3}+b x (2 a d+b c)-\frac {a (a d+2 b c)}{x}+\frac {1}{3} b^2 d x^3 \]

[In]

Int[((a + b*x^2)^2*(c + d*x^2))/x^4,x]

[Out]

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

Rule 459

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^4} \, dx=-\frac {a^2 c}{3 x^3}+\frac {-2 a b c-a^2 d}{x}+b (b c+2 a d) x+\frac {1}{3} b^2 d x^3 \]

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2))/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {d \,x^{3} b^{2}}{3}+2 a x b d +b^{2} c x -\frac {a^{2} c}{3 x^{3}}-\frac {a \left (a d +2 b c \right )}{x}\)	\(46\)
risch	\(\frac {d \,x^{3} b^{2}}{3}+2 a x b d +b^{2} c x +\frac {\left (-a^{2} d -2 a b c \right ) x^{2}-\frac {a^{2} c}{3}}{x^{3}}\)	\(50\)
norman	\(\frac {\frac {b^{2} d \,x^{6}}{3}+\left (2 a b d +b^{2} c \right ) x^{4}+\left (-a^{2} d -2 a b c \right ) x^{2}-\frac {a^{2} c}{3}}{x^{3}}\)	\(52\)
gosper	\(-\frac {-b^{2} d \,x^{6}-6 a b d \,x^{4}-3 b^{2} c \,x^{4}+3 a^{2} d \,x^{2}+6 a b c \,x^{2}+a^{2} c}{3 x^{3}}\)	\(55\)
parallelrisch	\(\frac {b^{2} d \,x^{6}+6 a b d \,x^{4}+3 b^{2} c \,x^{4}-3 a^{2} d \,x^{2}-6 a b c \,x^{2}-a^{2} c}{3 x^{3}}\)	\(55\)

[In]

int((b*x^2+a)^2*(d*x^2+c)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^4} \, dx=\frac {b^{2} d x^{6} + 3 \, {\left (b^{2} c + 2 \, a b d\right )} x^{4} - a^{2} c - 3 \, {\left (2 \, a b c + a^{2} d\right )} x^{2}}{3 \, x^{3}} \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^4} \, dx=\frac {b^{2} d x^{3}}{3} + x \left (2 a b d + b^{2} c\right ) + \frac {- a^{2} c + x^{2} \left (- 3 a^{2} d - 6 a b c\right )}{3 x^{3}} \]

[In]

integrate((b*x**2+a)**2*(d*x**2+c)/x**4,x)

[Out]

b**2*d*x**3/3 + x*(2*a*b*d + b**2*c) + (-a**2*c + x**2*(-3*a**2*d - 6*a*b*c))/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^4} \, dx=\frac {1}{3} \, b^{2} d x^{3} + {\left (b^{2} c + 2 \, a b d\right )} x - \frac {a^{2} c + 3 \, {\left (2 \, a b c + a^{2} d\right )} x^{2}}{3 \, x^{3}} \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)/x^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^4} \, dx=\frac {1}{3} \, b^{2} d x^{3} + b^{2} c x + 2 \, a b d x - \frac {6 \, a b c x^{2} + 3 \, a^{2} d x^{2} + a^{2} c}{3 \, x^{3}} \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c)/x^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^4} \, dx=x\,\left (c\,b^2+2\,a\,d\,b\right )-\frac {\frac {a^2\,c}{3}+x^2\,\left (d\,a^2+2\,b\,c\,a\right )}{x^3}+\frac {b^2\,d\,x^3}{3} \]

[In]

int(((a + b*x^2)^2*(c + d*x^2))/x^4,x)

[Out]

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