\(\int \frac {(a+b x^2)^2 (c+d x^2)}{x^2} \, dx\) [147]

Optimal result

Integrand size = 20, antiderivative size = 48 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=-\frac {a^2 c}{x}+a (2 b c+a d) x+\frac {1}{3} b (b c+2 a d) x^3+\frac {1}{5} b^2 d x^5 \]

[Out]

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

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

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=-\frac {a^2 c}{x}+a (2 b c+a d) x+\frac {1}{3} b (b c+2 a d) x^3+\frac {1}{5} b^2 d x^5 \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {b^{2} d \,x^{5}}{5}+\frac {2 x^{3} a b d}{3}+\frac {b^{2} c \,x^{3}}{3}+a^{2} d x +2 a b c x -\frac {a^{2} c}{x}\)	\(49\)
risch	\(\frac {b^{2} d \,x^{5}}{5}+\frac {2 x^{3} a b d}{3}+\frac {b^{2} c \,x^{3}}{3}+a^{2} d x +2 a b c x -\frac {a^{2} c}{x}\)	\(49\)
norman	\(\frac {\frac {b^{2} d \,x^{6}}{5}+\left (\frac {2}{3} a b d +\frac {1}{3} b^{2} c \right ) x^{4}+\left (a^{2} d +2 a b c \right ) x^{2}-a^{2} c}{x}\)	\(52\)
gosper	\(-\frac {-3 b^{2} d \,x^{6}-10 a b d \,x^{4}-5 b^{2} c \,x^{4}-15 a^{2} d \,x^{2}-30 a b c \,x^{2}+15 a^{2} c}{15 x}\)	\(56\)
parallelrisch	\(\frac {3 b^{2} d \,x^{6}+10 a b d \,x^{4}+5 b^{2} c \,x^{4}+15 a^{2} d \,x^{2}+30 a b c \,x^{2}-15 a^{2} c}{15 x}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=\frac {3 \, b^{2} d x^{6} + 5 \, {\left (b^{2} c + 2 \, a b d\right )} x^{4} - 15 \, a^{2} c + 15 \, {\left (2 \, a b c + a^{2} d\right )} x^{2}}{15 \, x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=- \frac {a^{2} c}{x} + \frac {b^{2} d x^{5}}{5} + x^{3} \cdot \left (\frac {2 a b d}{3} + \frac {b^{2} c}{3}\right ) + x \left (a^{2} d + 2 a b c\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=\frac {1}{5} \, b^{2} d x^{5} + \frac {1}{3} \, {\left (b^{2} c + 2 \, a b d\right )} x^{3} - \frac {a^{2} c}{x} + {\left (2 \, a b c + a^{2} d\right )} x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=\frac {1}{5} \, b^{2} d x^{5} + \frac {1}{3} \, b^{2} c x^{3} + \frac {2}{3} \, a b d x^{3} + 2 \, a b c x + a^{2} d x - \frac {a^{2} c}{x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^2} \, dx=x\,\left (d\,a^2+2\,b\,c\,a\right )+x^3\,\left (\frac {c\,b^2}{3}+\frac {2\,a\,d\,b}{3}\right )-\frac {a^2\,c}{x}+\frac {b^2\,d\,x^5}{5} \]

[In]

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

[Out]

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