Integrand size = 24, antiderivative size = 99 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {(b c-a d)^2 x}{3 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{c^2 x \sqrt {c+d x^2}}+\frac {\left (b^2 c^2+4 a b c d-8 a^2 d^2\right ) x}{3 c^3 d \sqrt {c+d x^2}} \] Output:
-1/3*(-a*d+b*c)^2*x/c^2/d/(d*x^2+c)^(3/2)-a^2/c^2/x/(d*x^2+c)^(1/2)+1/3*(- 8*a^2*d^2+4*a*b*c*d+b^2*c^2)*x/c^3/d/(d*x^2+c)^(1/2)
Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {b^2 c^2 x^4+2 a b c x^2 \left (3 c+2 d x^2\right )-a^2 \left (3 c^2+12 c d x^2+8 d^2 x^4\right )}{3 c^3 x \left (c+d x^2\right )^{3/2}} \] Input:
Integrate[(a + b*x^2)^2/(x^2*(c + d*x^2)^(5/2)),x]
Output:
(b^2*c^2*x^4 + 2*a*b*c*x^2*(3*c + 2*d*x^2) - a^2*(3*c^2 + 12*c*d*x^2 + 8*d ^2*x^4))/(3*c^3*x*(c + d*x^2)^(3/2))
Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {365, 292, 208}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 365 |
\(\displaystyle \frac {\int \frac {b^2 c x^2+2 a (b c-2 a d)}{\left (d x^2+c\right )^{5/2}}dx}{c}-\frac {a^2}{c x \left (c+d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 292 |
\(\displaystyle \frac {\frac {4 a (b c-2 a d) \int \frac {1}{\left (d x^2+c\right )^{3/2}}dx}{3 c}+\frac {x \left (2 a (b c-2 a d)+b^2 c x^2\right )}{3 c \left (c+d x^2\right )^{3/2}}}{c}-\frac {a^2}{c x \left (c+d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {x \left (2 a (b c-2 a d)+b^2 c x^2\right )}{3 c \left (c+d x^2\right )^{3/2}}+\frac {4 a x (b c-2 a d)}{3 c^2 \sqrt {c+d x^2}}}{c}-\frac {a^2}{c x \left (c+d x^2\right )^{3/2}}\) |
Input:
Int[(a + b*x^2)^2/(x^2*(c + d*x^2)^(5/2)),x]
Output:
-(a^2/(c*x*(c + d*x^2)^(3/2))) + ((x*(2*a*(b*c - 2*a*d) + b^2*c*x^2))/(3*c *(c + d*x^2)^(3/2)) + (4*a*(b*c - 2*a*d)*x)/(3*c^2*Sqrt[c + d*x^2]))/c
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( a*(p + 1))) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ {a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt Q[q, 0] && NeQ[p, -1]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Time = 0.53 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71
method | result | size |
pseudoelliptic | \(\frac {\left (-8 d^{2} x^{4}-12 c d \,x^{2}-3 c^{2}\right ) a^{2}+6 c \left (\frac {2 x^{2} d}{3}+c \right ) b \,x^{2} a +b^{2} c^{2} x^{4}}{3 \left (x^{2} d +c \right )^{\frac {3}{2}} x \,c^{3}}\) | \(70\) |
gosper | \(-\frac {8 a^{2} d^{2} x^{4}-4 a b c d \,x^{4}-b^{2} c^{2} x^{4}+12 a^{2} c d \,x^{2}-6 a b \,c^{2} x^{2}+3 a^{2} c^{2}}{3 x \left (x^{2} d +c \right )^{\frac {3}{2}} c^{3}}\) | \(78\) |
trager | \(-\frac {8 a^{2} d^{2} x^{4}-4 a b c d \,x^{4}-b^{2} c^{2} x^{4}+12 a^{2} c d \,x^{2}-6 a b \,c^{2} x^{2}+3 a^{2} c^{2}}{3 x \left (x^{2} d +c \right )^{\frac {3}{2}} c^{3}}\) | \(78\) |
orering | \(-\frac {8 a^{2} d^{2} x^{4}-4 a b c d \,x^{4}-b^{2} c^{2} x^{4}+12 a^{2} c d \,x^{2}-6 a b \,c^{2} x^{2}+3 a^{2} c^{2}}{3 x \left (x^{2} d +c \right )^{\frac {3}{2}} c^{3}}\) | \(78\) |
risch | \(-\frac {a^{2} \sqrt {x^{2} d +c}}{c^{3} x}-\frac {\left (a d -b c \right ) \left (5 a d \,x^{2}+x^{2} b c +6 a c \right ) x \sqrt {x^{2} d +c}}{3 \left (d^{2} x^{4}+2 c d \,x^{2}+c^{2}\right ) c^{3}}\) | \(83\) |
default | \(a^{2} \left (-\frac {1}{c x \left (x^{2} d +c \right )^{\frac {3}{2}}}-\frac {4 d \left (\frac {x}{3 c \left (x^{2} d +c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {x^{2} d +c}}\right )}{c}\right )+b^{2} \left (-\frac {x}{2 d \left (x^{2} d +c \right )^{\frac {3}{2}}}+\frac {c \left (\frac {x}{3 c \left (x^{2} d +c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {x^{2} d +c}}\right )}{2 d}\right )+2 a b \left (\frac {x}{3 c \left (x^{2} d +c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {x^{2} d +c}}\right )\) | \(153\) |
Input:
int((b*x^2+a)^2/x^2/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3*((-8*d^2*x^4-12*c*d*x^2-3*c^2)*a^2+6*c*(2/3*x^2*d+c)*b*x^2*a+b^2*c^2*x ^4)/(d*x^2+c)^(3/2)/x/c^3
Time = 0.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {{\left ({\left (b^{2} c^{2} + 4 \, a b c d - 8 \, a^{2} d^{2}\right )} x^{4} - 3 \, a^{2} c^{2} + 6 \, {\left (a b c^{2} - 2 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{3} d^{2} x^{5} + 2 \, c^{4} d x^{3} + c^{5} x\right )}} \] Input:
integrate((b*x^2+a)^2/x^2/(d*x^2+c)^(5/2),x, algorithm="fricas")
Output:
1/3*((b^2*c^2 + 4*a*b*c*d - 8*a^2*d^2)*x^4 - 3*a^2*c^2 + 6*(a*b*c^2 - 2*a^ 2*c*d)*x^2)*sqrt(d*x^2 + c)/(c^3*d^2*x^5 + 2*c^4*d*x^3 + c^5*x)
\[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{x^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((b*x**2+a)**2/x**2/(d*x**2+c)**(5/2),x)
Output:
Integral((a + b*x**2)**2/(x**2*(c + d*x**2)**(5/2)), x)
Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {4 \, a b x}{3 \, \sqrt {d x^{2} + c} c^{2}} + \frac {2 \, a b x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {b^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {b^{2} x}{3 \, \sqrt {d x^{2} + c} c d} - \frac {8 \, a^{2} d x}{3 \, \sqrt {d x^{2} + c} c^{3}} - \frac {4 \, a^{2} d x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} - \frac {a^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c x} \] Input:
integrate((b*x^2+a)^2/x^2/(d*x^2+c)^(5/2),x, algorithm="maxima")
Output:
4/3*a*b*x/(sqrt(d*x^2 + c)*c^2) + 2/3*a*b*x/((d*x^2 + c)^(3/2)*c) - 1/3*b^ 2*x/((d*x^2 + c)^(3/2)*d) + 1/3*b^2*x/(sqrt(d*x^2 + c)*c*d) - 8/3*a^2*d*x/ (sqrt(d*x^2 + c)*c^3) - 4/3*a^2*d*x/((d*x^2 + c)^(3/2)*c^2) - a^2/((d*x^2 + c)^(3/2)*c*x)
Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {x {\left (\frac {{\left (b^{2} c^{4} d + 4 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3}\right )} x^{2}}{c^{5} d} + \frac {6 \, {\left (a b c^{4} d - a^{2} c^{3} d^{2}\right )}}{c^{5} d}\right )}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {2 \, a^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )} c^{2}} \] Input:
integrate((b*x^2+a)^2/x^2/(d*x^2+c)^(5/2),x, algorithm="giac")
Output:
1/3*x*((b^2*c^4*d + 4*a*b*c^3*d^2 - 5*a^2*c^2*d^3)*x^2/(c^5*d) + 6*(a*b*c^ 4*d - a^2*c^3*d^2)/(c^5*d))/(d*x^2 + c)^(3/2) + 2*a^2*sqrt(d)/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)*c^2)
Time = 0.99 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {3\,a^2\,c^2+12\,a^2\,c\,d\,x^2+8\,a^2\,d^2\,x^4-6\,a\,b\,c^2\,x^2-4\,a\,b\,c\,d\,x^4-b^2\,c^2\,x^4}{3\,c^3\,x\,{\left (d\,x^2+c\right )}^{3/2}} \] Input:
int((a + b*x^2)^2/(x^2*(c + d*x^2)^(5/2)),x)
Output:
-(3*a^2*c^2 + 8*a^2*d^2*x^4 - b^2*c^2*x^4 - 6*a*b*c^2*x^2 + 12*a^2*c*d*x^2 - 4*a*b*c*d*x^4)/(3*c^3*x*(c + d*x^2)^(3/2))
Time = 0.22 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.75 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {-3 \sqrt {d \,x^{2}+c}\, a^{2} c^{2} d^{2}-12 \sqrt {d \,x^{2}+c}\, a^{2} c \,d^{3} x^{2}-8 \sqrt {d \,x^{2}+c}\, a^{2} d^{4} x^{4}+6 \sqrt {d \,x^{2}+c}\, a b \,c^{2} d^{2} x^{2}+4 \sqrt {d \,x^{2}+c}\, a b c \,d^{3} x^{4}+\sqrt {d \,x^{2}+c}\, b^{2} c^{2} d^{2} x^{4}+8 \sqrt {d}\, a^{2} c^{2} d^{2} x +16 \sqrt {d}\, a^{2} c \,d^{3} x^{3}+8 \sqrt {d}\, a^{2} d^{4} x^{5}-4 \sqrt {d}\, a b \,c^{3} d x -8 \sqrt {d}\, a b \,c^{2} d^{2} x^{3}-4 \sqrt {d}\, a b c \,d^{3} x^{5}+2 \sqrt {d}\, b^{2} c^{4} x +4 \sqrt {d}\, b^{2} c^{3} d \,x^{3}+2 \sqrt {d}\, b^{2} c^{2} d^{2} x^{5}}{3 c^{3} d^{2} x \left (d^{2} x^{4}+2 c d \,x^{2}+c^{2}\right )} \] Input:
int((b*x^2+a)^2/x^2/(d*x^2+c)^(5/2),x)
Output:
( - 3*sqrt(c + d*x**2)*a**2*c**2*d**2 - 12*sqrt(c + d*x**2)*a**2*c*d**3*x* *2 - 8*sqrt(c + d*x**2)*a**2*d**4*x**4 + 6*sqrt(c + d*x**2)*a*b*c**2*d**2* x**2 + 4*sqrt(c + d*x**2)*a*b*c*d**3*x**4 + sqrt(c + d*x**2)*b**2*c**2*d** 2*x**4 + 8*sqrt(d)*a**2*c**2*d**2*x + 16*sqrt(d)*a**2*c*d**3*x**3 + 8*sqrt (d)*a**2*d**4*x**5 - 4*sqrt(d)*a*b*c**3*d*x - 8*sqrt(d)*a*b*c**2*d**2*x**3 - 4*sqrt(d)*a*b*c*d**3*x**5 + 2*sqrt(d)*b**2*c**4*x + 4*sqrt(d)*b**2*c**3 *d*x**3 + 2*sqrt(d)*b**2*c**2*d**2*x**5)/(3*c**3*d**2*x*(c**2 + 2*c*d*x**2 + d**2*x**4))