\(\int \frac {(a+b x^2)^2 (c+d x^2)^3}{x^2} \, dx\) [165]

Optimal result

Integrand size = 22, antiderivative size = 120 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^2} \, dx=-\frac {a^2 c^3}{x}+a c^2 (2 b c+3 a d) x+\frac {1}{3} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^3+\frac {1}{5} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^5+\frac {1}{7} b d^2 (3 b c+2 a d) x^7+\frac {1}{9} b^2 d^3 x^9 \]

[Out]

-a^2*c^3/x+a*c^2*(3*a*d+2*b*c)*x+1/3*c*(3*a^2*d^2+6*a*b*c*d+b^2*c^2)*x^3+1/5*d*(a^2*d^2+6*a*b*c*d+3*b^2*c^2)*x
^5+1/7*b*d^2*(2*a*d+3*b*c)*x^7+1/9*b^2*d^3*x^9

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 120, 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} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^2} \, dx=\frac {1}{5} d x^5 \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+\frac {1}{3} c x^3 \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )-\frac {a^2 c^3}{x}+a c^2 x (3 a d+2 b c)+\frac {1}{7} b d^2 x^7 (2 a d+3 b c)+\frac {1}{9} b^2 d^3 x^9 \]

[In]

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

[Out]

-((a^2*c^3)/x) + a*c^2*(2*b*c + 3*a*d)*x + (c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^3)/3 + (d*(3*b^2*c^2 + 6*a*b
*c*d + a^2*d^2)*x^5)/5 + (b*d^2*(3*b*c + 2*a*d)*x^7)/7 + (b^2*d^3*x^9)/9

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 c^2 (2 b c+3 a d)+\frac {a^2 c^3}{x^2}+c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^2+d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^4+b d^2 (3 b c+2 a d) x^6+b^2 d^3 x^8\right ) \, dx \\ & = -\frac {a^2 c^3}{x}+a c^2 (2 b c+3 a d) x+\frac {1}{3} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^3+\frac {1}{5} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^5+\frac {1}{7} b d^2 (3 b c+2 a d) x^7+\frac {1}{9} b^2 d^3 x^9 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^2} \, dx=-\frac {a^2 c^3}{x}+a c^2 (2 b c+3 a d) x+\frac {1}{3} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^3+\frac {1}{5} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^5+\frac {1}{7} b d^2 (3 b c+2 a d) x^7+\frac {1}{9} b^2 d^3 x^9 \]

[In]

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

[Out]

-((a^2*c^3)/x) + a*c^2*(2*b*c + 3*a*d)*x + (c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^3)/3 + (d*(3*b^2*c^2 + 6*a*b
*c*d + a^2*d^2)*x^5)/5 + (b*d^2*(3*b*c + 2*a*d)*x^7)/7 + (b^2*d^3*x^9)/9

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.05

method	result	size
norman	\(\frac {\frac {b^{2} d^{3} x^{10}}{9}+\left (\frac {2}{7} a b \,d^{3}+\frac {3}{7} b^{2} c \,d^{2}\right ) x^{8}+\left (\frac {1}{5} a^{2} d^{3}+\frac {6}{5} a b c \,d^{2}+\frac {3}{5} b^{2} c^{2} d \right ) x^{6}+\left (c \,a^{2} d^{2}+2 a b \,c^{2} d +\frac {1}{3} b^{2} c^{3}\right ) x^{4}+\left (3 a^{2} c^{2} d +2 a b \,c^{3}\right ) x^{2}-a^{2} c^{3}}{x}\)	\(126\)
default	\(\frac {b^{2} d^{3} x^{9}}{9}+\frac {2 a b \,d^{3} x^{7}}{7}+\frac {3 b^{2} c \,d^{2} x^{7}}{7}+\frac {a^{2} d^{3} x^{5}}{5}+\frac {6 x^{5} d^{2} a b c}{5}+\frac {3 b^{2} c^{2} d \,x^{5}}{5}+a^{2} c \,d^{2} x^{3}+2 a b \,c^{2} d \,x^{3}+\frac {b^{2} c^{3} x^{3}}{3}+3 a^{2} c^{2} d x +2 a b \,c^{3} x -\frac {a^{2} c^{3}}{x}\)	\(131\)
risch	\(\frac {b^{2} d^{3} x^{9}}{9}+\frac {2 a b \,d^{3} x^{7}}{7}+\frac {3 b^{2} c \,d^{2} x^{7}}{7}+\frac {a^{2} d^{3} x^{5}}{5}+\frac {6 x^{5} d^{2} a b c}{5}+\frac {3 b^{2} c^{2} d \,x^{5}}{5}+a^{2} c \,d^{2} x^{3}+2 a b \,c^{2} d \,x^{3}+\frac {b^{2} c^{3} x^{3}}{3}+3 a^{2} c^{2} d x +2 a b \,c^{3} x -\frac {a^{2} c^{3}}{x}\)	\(131\)
gosper	\(-\frac {-35 b^{2} d^{3} x^{10}-90 a b \,d^{3} x^{8}-135 b^{2} c \,d^{2} x^{8}-63 a^{2} d^{3} x^{6}-378 x^{6} d^{2} a b c -189 b^{2} c^{2} d \,x^{6}-315 a^{2} c \,d^{2} x^{4}-630 a b \,c^{2} d \,x^{4}-105 b^{2} c^{3} x^{4}-945 a^{2} c^{2} d \,x^{2}-630 a b \,c^{3} x^{2}+315 a^{2} c^{3}}{315 x}\)	\(138\)
parallelrisch	\(\frac {35 b^{2} d^{3} x^{10}+90 a b \,d^{3} x^{8}+135 b^{2} c \,d^{2} x^{8}+63 a^{2} d^{3} x^{6}+378 x^{6} d^{2} a b c +189 b^{2} c^{2} d \,x^{6}+315 a^{2} c \,d^{2} x^{4}+630 a b \,c^{2} d \,x^{4}+105 b^{2} c^{3} x^{4}+945 a^{2} c^{2} d \,x^{2}+630 a b \,c^{3} x^{2}-315 a^{2} c^{3}}{315 x}\)	\(138\)

[In]

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

[Out]

1/x*(1/9*b^2*d^3*x^10+(2/7*a*b*d^3+3/7*b^2*c*d^2)*x^8+(1/5*a^2*d^3+6/5*a*b*c*d^2+3/5*b^2*c^2*d)*x^6+(c*a^2*d^2
+2*a*b*c^2*d+1/3*b^2*c^3)*x^4+(3*a^2*c^2*d+2*a*b*c^3)*x^2-a^2*c^3)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^2} \, dx=\frac {35 \, b^{2} d^{3} x^{10} + 45 \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{8} + 63 \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{6} - 315 \, a^{2} c^{3} + 105 \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{4} + 315 \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}}{315 \, x} \]

[In]

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

[Out]

1/315*(35*b^2*d^3*x^10 + 45*(3*b^2*c*d^2 + 2*a*b*d^3)*x^8 + 63*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^6 - 315
*a^2*c^3 + 105*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^4 + 315*(2*a*b*c^3 + 3*a^2*c^2*d)*x^2)/x

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^2} \, dx=- \frac {a^{2} c^{3}}{x} + \frac {b^{2} d^{3} x^{9}}{9} + x^{7} \cdot \left (\frac {2 a b d^{3}}{7} + \frac {3 b^{2} c d^{2}}{7}\right ) + x^{5} \left (\frac {a^{2} d^{3}}{5} + \frac {6 a b c d^{2}}{5} + \frac {3 b^{2} c^{2} d}{5}\right ) + x^{3} \left (a^{2} c d^{2} + 2 a b c^{2} d + \frac {b^{2} c^{3}}{3}\right ) + x \left (3 a^{2} c^{2} d + 2 a b c^{3}\right ) \]

[In]

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

[Out]

-a**2*c**3/x + b**2*d**3*x**9/9 + x**7*(2*a*b*d**3/7 + 3*b**2*c*d**2/7) + x**5*(a**2*d**3/5 + 6*a*b*c*d**2/5 +
3*b**2*c**2*d/5) + x**3*(a**2*c*d**2 + 2*a*b*c**2*d + b**2*c**3/3) + x*(3*a**2*c**2*d + 2*a*b*c**3)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^2} \, dx=\frac {1}{9} \, b^{2} d^{3} x^{9} + \frac {1}{7} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{7} + \frac {1}{5} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{5} - \frac {a^{2} c^{3}}{x} + \frac {1}{3} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{3} + {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x \]

[In]

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

[Out]

1/9*b^2*d^3*x^9 + 1/7*(3*b^2*c*d^2 + 2*a*b*d^3)*x^7 + 1/5*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^5 - a^2*c^3/
x + 1/3*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^3 + (2*a*b*c^3 + 3*a^2*c^2*d)*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^2} \, dx=\frac {1}{9} \, b^{2} d^{3} x^{9} + \frac {3}{7} \, b^{2} c d^{2} x^{7} + \frac {2}{7} \, a b d^{3} x^{7} + \frac {3}{5} \, b^{2} c^{2} d x^{5} + \frac {6}{5} \, a b c d^{2} x^{5} + \frac {1}{5} \, a^{2} d^{3} x^{5} + \frac {1}{3} \, b^{2} c^{3} x^{3} + 2 \, a b c^{2} d x^{3} + a^{2} c d^{2} x^{3} + 2 \, a b c^{3} x + 3 \, a^{2} c^{2} d x - \frac {a^{2} c^{3}}{x} \]

[In]

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

[Out]

1/9*b^2*d^3*x^9 + 3/7*b^2*c*d^2*x^7 + 2/7*a*b*d^3*x^7 + 3/5*b^2*c^2*d*x^5 + 6/5*a*b*c*d^2*x^5 + 1/5*a^2*d^3*x^
5 + 1/3*b^2*c^3*x^3 + 2*a*b*c^2*d*x^3 + a^2*c*d^2*x^3 + 2*a*b*c^3*x + 3*a^2*c^2*d*x - a^2*c^3/x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^2} \, dx=x^3\,\left (a^2\,c\,d^2+2\,a\,b\,c^2\,d+\frac {b^2\,c^3}{3}\right )+x^5\,\left (\frac {a^2\,d^3}{5}+\frac {6\,a\,b\,c\,d^2}{5}+\frac {3\,b^2\,c^2\,d}{5}\right )-\frac {a^2\,c^3}{x}+\frac {b^2\,d^3\,x^9}{9}+\frac {b\,d^2\,x^7\,\left (2\,a\,d+3\,b\,c\right )}{7}+a\,c^2\,x\,\left (3\,a\,d+2\,b\,c\right ) \]

[In]

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

[Out]

x^3*((b^2*c^3)/3 + a^2*c*d^2 + 2*a*b*c^2*d) + x^5*((a^2*d^3)/5 + (3*b^2*c^2*d)/5 + (6*a*b*c*d^2)/5) - (a^2*c^3
)/x + (b^2*d^3*x^9)/9 + (b*d^2*x^7*(2*a*d + 3*b*c))/7 + a*c^2*x*(3*a*d + 2*b*c)