\(\int \frac {(a+b x^2)^2 (c+d x^2)^2}{x^4} \, dx\) [158]

Optimal result

Integrand size = 22, antiderivative size = 80 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^2}{x^4} \, dx=-\frac {a^2 c^2}{3 x^3}-\frac {2 a c (b c+a d)}{x}+\left (b^2 c^2+4 a b c d+a^2 d^2\right ) x+\frac {2}{3} b d (b c+a d) x^3+\frac {1}{5} b^2 d^2 x^5 \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {459} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^2}{x^4} \, dx=x \left (a^2 d^2+4 a b c d+b^2 c^2\right )-\frac {a^2 c^2}{3 x^3}+\frac {2}{3} b d x^3 (a d+b c)-\frac {2 a c (a d+b c)}{x}+\frac {1}{5} b^2 d^2 x^5 \]

[In]

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

[Out]

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

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^2 c^2 \left (1+\frac {a d (4 b c+a d)}{b^2 c^2}\right )+\frac {a^2 c^2}{x^4}+\frac {2 a c (b c+a d)}{x^2}+2 b d (b c+a d) x^2+b^2 d^2 x^4\right ) \, dx \\ & = -\frac {a^2 c^2}{3 x^3}-\frac {2 a c (b c+a d)}{x}+\left (b^2 c^2+4 a b c d+a^2 d^2\right ) x+\frac {2}{3} b d (b c+a d) x^3+\frac {1}{5} b^2 d^2 x^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^2}{x^4} \, dx=-\frac {a^2 c^2}{3 x^3}-\frac {2 a c (b c+a d)}{x}+\left (b^2 c^2+4 a b c d+a^2 d^2\right ) x+\frac {2}{3} b d (b c+a d) x^3+\frac {1}{5} b^2 d^2 x^5 \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {b^{2} d^{2} x^{5}}{5}+\frac {2 x^{3} a b \,d^{2}}{3}+\frac {2 x^{3} b^{2} c d}{3}+a^{2} d^{2} x +4 a b c d x +b^{2} c^{2} x -\frac {a^{2} c^{2}}{3 x^{3}}-\frac {2 a c \left (a d +b c \right )}{x}\)	\(81\)
norman	\(\frac {\frac {b^{2} d^{2} x^{8}}{5}+\left (\frac {2}{3} a b \,d^{2}+\frac {2}{3} b^{2} c d \right ) x^{6}+\left (a^{2} d^{2}+4 a b c d +b^{2} c^{2}\right ) x^{4}+\left (-2 a^{2} c d -2 b \,c^{2} a \right ) x^{2}-\frac {a^{2} c^{2}}{3}}{x^{3}}\)	\(88\)
risch	\(\frac {b^{2} d^{2} x^{5}}{5}+\frac {2 x^{3} a b \,d^{2}}{3}+\frac {2 x^{3} b^{2} c d}{3}+a^{2} d^{2} x +4 a b c d x +b^{2} c^{2} x +\frac {\left (-2 a^{2} c d -2 b \,c^{2} a \right ) x^{2}-\frac {a^{2} c^{2}}{3}}{x^{3}}\)	\(88\)
gosper	\(-\frac {-3 b^{2} d^{2} x^{8}-10 a b \,d^{2} x^{6}-10 b^{2} c d \,x^{6}-15 a^{2} d^{2} x^{4}-60 x^{4} b d a c -15 b^{2} c^{2} x^{4}+30 a^{2} c d \,x^{2}+30 a b \,c^{2} x^{2}+5 a^{2} c^{2}}{15 x^{3}}\)	\(97\)
parallelrisch	\(\frac {3 b^{2} d^{2} x^{8}+10 a b \,d^{2} x^{6}+10 b^{2} c d \,x^{6}+15 a^{2} d^{2} x^{4}+60 x^{4} b d a c +15 b^{2} c^{2} x^{4}-30 a^{2} c d \,x^{2}-30 a b \,c^{2} x^{2}-5 a^{2} c^{2}}{15 x^{3}}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^2}{x^4} \, dx=\frac {3 \, b^{2} d^{2} x^{8} + 10 \, {\left (b^{2} c d + a b d^{2}\right )} x^{6} + 15 \, {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{4} - 5 \, a^{2} c^{2} - 30 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2}}{15 \, x^{3}} \]

[In]

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

[Out]

1/15*(3*b^2*d^2*x^8 + 10*(b^2*c*d + a*b*d^2)*x^6 + 15*(b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4 - 5*a^2*c^2 - 30*(a*
b*c^2 + a^2*c*d)*x^2)/x^3

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^2}{x^4} \, dx=\frac {b^{2} d^{2} x^{5}}{5} + x^{3} \cdot \left (\frac {2 a b d^{2}}{3} + \frac {2 b^{2} c d}{3}\right ) + x \left (a^{2} d^{2} + 4 a b c d + b^{2} c^{2}\right ) + \frac {- a^{2} c^{2} + x^{2} \left (- 6 a^{2} c d - 6 a b c^{2}\right )}{3 x^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^2}{x^4} \, dx=\frac {1}{5} \, b^{2} d^{2} x^{5} + \frac {2}{3} \, {\left (b^{2} c d + a b d^{2}\right )} x^{3} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x - \frac {a^{2} c^{2} + 6 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^2}{x^4} \, dx=\frac {1}{5} \, b^{2} d^{2} x^{5} + \frac {2}{3} \, b^{2} c d x^{3} + \frac {2}{3} \, a b d^{2} x^{3} + b^{2} c^{2} x + 4 \, a b c d x + a^{2} d^{2} x - \frac {6 \, a b c^{2} x^{2} + 6 \, a^{2} c d x^{2} + a^{2} c^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^2}{x^4} \, dx=x\,\left (a^2\,d^2+4\,a\,b\,c\,d+b^2\,c^2\right )-\frac {x^2\,\left (2\,d\,a^2\,c+2\,b\,a\,c^2\right )+\frac {a^2\,c^2}{3}}{x^3}+\frac {b^2\,d^2\,x^5}{5}+\frac {2\,b\,d\,x^3\,\left (a\,d+b\,c\right )}{3} \]

[In]

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

[Out]

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