\(\int \frac {(a+b x^2)^2}{(c+d x^2)^3} \, dx\) [587]

Optimal result

Integrand size = 19, antiderivative size = 118 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {(b c-a d)^2 x}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (5 b c+3 a d) x}{8 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{5/2}} \] Output:

1/4*(-a*d+b*c)^2*x/c/d^2/(d*x^2+c)^2-1/8*(-a*d+b*c)*(3*a*d+5*b*c)*x/c^2/d^ 
2/(d*x^2+c)+1/8*(3*a^2*d^2+2*a*b*c*d+3*b^2*c^2)*arctan(d^(1/2)*x/c^(1/2))/ 
c^(5/2)/d^(5/2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {(b c-a d) x \left (a d \left (5 c+3 d x^2\right )+b c \left (3 c+5 d x^2\right )\right )}{8 c^2 d^2 \left (c+d x^2\right )^2}+\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{5/2}} \] Input:

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

Output:

-1/8*((b*c - a*d)*x*(a*d*(5*c + 3*d*x^2) + b*c*(3*c + 5*d*x^2)))/(c^2*d^2* 
(c + d*x^2)^2) + ((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[d]*x)/S 
qrt[c]])/(8*c^(5/2)*d^(5/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {315, 298, 218}

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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx\)

\(\Big \downarrow \) 315

\(\displaystyle \frac {\int \frac {b (3 b c+a d) x^2+a (b c+3 a d)}{\left (d x^2+c\right )^2}dx}{4 c d}-\frac {x \left (a+b x^2\right ) (b c-a d)}{4 c d \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 a^2 d}{c}+2 a b+\frac {3 b^2 c}{d}\right ) \int \frac {1}{d x^2+c}dx-\frac {3 x \left (\frac {b^2 c}{d}-\frac {a^2 d}{c}\right )}{2 \left (c+d x^2\right )}}{4 c d}-\frac {x \left (a+b x^2\right ) (b c-a d)}{4 c d \left (c+d x^2\right )^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\left (\frac {3 a^2 d}{c}+2 a b+\frac {3 b^2 c}{d}\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 \sqrt {c} \sqrt {d}}-\frac {3 x \left (\frac {b^2 c}{d}-\frac {a^2 d}{c}\right )}{2 \left (c+d x^2\right )}}{4 c d}-\frac {x \left (a+b x^2\right ) (b c-a d)}{4 c d \left (c+d x^2\right )^2}\)

Input:

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

Output:

-1/4*((b*c - a*d)*x*(a + b*x^2))/(c*d*(c + d*x^2)^2) + ((-3*((b^2*c)/d - ( 
a^2*d)/c)*x)/(2*(c + d*x^2)) + ((2*a*b + (3*b^2*c)/d + (3*a^2*d)/c)*ArcTan 
[(Sqrt[d]*x)/Sqrt[c]])/(2*Sqrt[c]*Sqrt[d]))/(4*c*d)
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 315

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {\frac {\left (3 a^{2} d^{2}+2 a b c d -5 b^{2} c^{2}\right ) x^{3}}{8 c^{2} d}+\frac {\left (5 a^{2} d^{2}-2 a b c d -3 b^{2} c^{2}\right ) x}{8 c \,d^{2}}}{\left (x^{2} d +c \right )^{2}}+\frac {\left (3 a^{2} d^{2}+2 a b c d +3 b^{2} c^{2}\right ) \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{8 c^{2} d^{2} \sqrt {c d}}\)	\(124\)
risch	\(\frac {\frac {\left (3 a^{2} d^{2}+2 a b c d -5 b^{2} c^{2}\right ) x^{3}}{8 c^{2} d}+\frac {\left (5 a^{2} d^{2}-2 a b c d -3 b^{2} c^{2}\right ) x}{8 c \,d^{2}}}{\left (x^{2} d +c \right )^{2}}-\frac {3 \ln \left (x d +\sqrt {-c d}\right ) a^{2}}{16 \sqrt {-c d}\, c^{2}}-\frac {\ln \left (x d +\sqrt {-c d}\right ) a b}{8 \sqrt {-c d}\, d c}-\frac {3 \ln \left (x d +\sqrt {-c d}\right ) b^{2}}{16 \sqrt {-c d}\, d^{2}}+\frac {3 \ln \left (-x d +\sqrt {-c d}\right ) a^{2}}{16 \sqrt {-c d}\, c^{2}}+\frac {\ln \left (-x d +\sqrt {-c d}\right ) a b}{8 \sqrt {-c d}\, d c}+\frac {3 \ln \left (-x d +\sqrt {-c d}\right ) b^{2}}{16 \sqrt {-c d}\, d^{2}}\)	\(236\)

Input:

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

Output:

(1/8*(3*a^2*d^2+2*a*b*c*d-5*b^2*c^2)/c^2/d*x^3+1/8*(5*a^2*d^2-2*a*b*c*d-3* 
b^2*c^2)/c/d^2*x)/(d*x^2+c)^2+1/8*(3*a^2*d^2+2*a*b*c*d+3*b^2*c^2)/c^2/d^2/ 
(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (104) = 208\).

Time = 0.21 (sec) , antiderivative size = 449, normalized size of antiderivative = 3.81 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\left [-\frac {2 \, {\left (5 \, b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} - 3 \, a^{2} c d^{4}\right )} x^{3} + {\left (3 \, b^{2} c^{4} + 2 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (3 \, b^{2} c^{2} d^{2} + 2 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{2} c^{3} d + 2 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {-c d} \log \left (\frac {d x^{2} - 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right ) + 2 \, {\left (3 \, b^{2} c^{4} d + 2 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3}\right )} x}{16 \, {\left (c^{3} d^{5} x^{4} + 2 \, c^{4} d^{4} x^{2} + c^{5} d^{3}\right )}}, -\frac {{\left (5 \, b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} - 3 \, a^{2} c d^{4}\right )} x^{3} - {\left (3 \, b^{2} c^{4} + 2 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (3 \, b^{2} c^{2} d^{2} + 2 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{2} c^{3} d + 2 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right ) + {\left (3 \, b^{2} c^{4} d + 2 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3}\right )} x}{8 \, {\left (c^{3} d^{5} x^{4} + 2 \, c^{4} d^{4} x^{2} + c^{5} d^{3}\right )}}\right ] \] Input:

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

Output:

[-1/16*(2*(5*b^2*c^3*d^2 - 2*a*b*c^2*d^3 - 3*a^2*c*d^4)*x^3 + (3*b^2*c^4 + 
 2*a*b*c^3*d + 3*a^2*c^2*d^2 + (3*b^2*c^2*d^2 + 2*a*b*c*d^3 + 3*a^2*d^4)*x 
^4 + 2*(3*b^2*c^3*d + 2*a*b*c^2*d^2 + 3*a^2*c*d^3)*x^2)*sqrt(-c*d)*log((d* 
x^2 - 2*sqrt(-c*d)*x - c)/(d*x^2 + c)) + 2*(3*b^2*c^4*d + 2*a*b*c^3*d^2 - 
5*a^2*c^2*d^3)*x)/(c^3*d^5*x^4 + 2*c^4*d^4*x^2 + c^5*d^3), -1/8*((5*b^2*c^ 
3*d^2 - 2*a*b*c^2*d^3 - 3*a^2*c*d^4)*x^3 - (3*b^2*c^4 + 2*a*b*c^3*d + 3*a^ 
2*c^2*d^2 + (3*b^2*c^2*d^2 + 2*a*b*c*d^3 + 3*a^2*d^4)*x^4 + 2*(3*b^2*c^3*d 
 + 2*a*b*c^2*d^2 + 3*a^2*c*d^3)*x^2)*sqrt(c*d)*arctan(sqrt(c*d)*x/c) + (3* 
b^2*c^4*d + 2*a*b*c^3*d^2 - 5*a^2*c^2*d^3)*x)/(c^3*d^5*x^4 + 2*c^4*d^4*x^2 
 + c^5*d^3)]
 

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.89 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{c^{5} d^{5}}} \cdot \left (3 a^{2} d^{2} + 2 a b c d + 3 b^{2} c^{2}\right ) \log {\left (- c^{3} d^{2} \sqrt {- \frac {1}{c^{5} d^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{c^{5} d^{5}}} \cdot \left (3 a^{2} d^{2} + 2 a b c d + 3 b^{2} c^{2}\right ) \log {\left (c^{3} d^{2} \sqrt {- \frac {1}{c^{5} d^{5}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a^{2} d^{3} + 2 a b c d^{2} - 5 b^{2} c^{2} d\right ) + x \left (5 a^{2} c d^{2} - 2 a b c^{2} d - 3 b^{2} c^{3}\right )}{8 c^{4} d^{2} + 16 c^{3} d^{3} x^{2} + 8 c^{2} d^{4} x^{4}} \] Input:

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

Output:

-sqrt(-1/(c**5*d**5))*(3*a**2*d**2 + 2*a*b*c*d + 3*b**2*c**2)*log(-c**3*d* 
*2*sqrt(-1/(c**5*d**5)) + x)/16 + sqrt(-1/(c**5*d**5))*(3*a**2*d**2 + 2*a* 
b*c*d + 3*b**2*c**2)*log(c**3*d**2*sqrt(-1/(c**5*d**5)) + x)/16 + (x**3*(3 
*a**2*d**3 + 2*a*b*c*d**2 - 5*b**2*c**2*d) + x*(5*a**2*c*d**2 - 2*a*b*c**2 
*d - 3*b**2*c**3))/(8*c**4*d**2 + 16*c**3*d**3*x**2 + 8*c**2*d**4*x**4)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {{\left (5 \, b^{2} c^{2} d - 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{3} + {\left (3 \, b^{2} c^{3} + 2 \, a b c^{2} d - 5 \, a^{2} c d^{2}\right )} x}{8 \, {\left (c^{2} d^{4} x^{4} + 2 \, c^{3} d^{3} x^{2} + c^{4} d^{2}\right )}} + \frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{2} d^{2}} \] Input:

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

Output:

-1/8*((5*b^2*c^2*d - 2*a*b*c*d^2 - 3*a^2*d^3)*x^3 + (3*b^2*c^3 + 2*a*b*c^2 
*d - 5*a^2*c*d^2)*x)/(c^2*d^4*x^4 + 2*c^3*d^3*x^2 + c^4*d^2) + 1/8*(3*b^2* 
c^2 + 2*a*b*c*d + 3*a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*c^2*d^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{2} d^{2}} - \frac {5 \, b^{2} c^{2} d x^{3} - 2 \, a b c d^{2} x^{3} - 3 \, a^{2} d^{3} x^{3} + 3 \, b^{2} c^{3} x + 2 \, a b c^{2} d x - 5 \, a^{2} c d^{2} x}{8 \, {\left (d x^{2} + c\right )}^{2} c^{2} d^{2}} \] Input:

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

Output:

1/8*(3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*c 
^2*d^2) - 1/8*(5*b^2*c^2*d*x^3 - 2*a*b*c*d^2*x^3 - 3*a^2*d^3*x^3 + 3*b^2*c 
^3*x + 2*a*b*c^2*d*x - 5*a^2*c*d^2*x)/((d*x^2 + c)^2*c^2*d^2)
 

Mupad [B] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )\,\left (3\,a^2\,d^2+2\,a\,b\,c\,d+3\,b^2\,c^2\right )}{8\,c^{5/2}\,d^{5/2}}-\frac {\frac {x\,\left (-5\,a^2\,d^2+2\,a\,b\,c\,d+3\,b^2\,c^2\right )}{8\,c\,d^2}-\frac {x^3\,\left (3\,a^2\,d^2+2\,a\,b\,c\,d-5\,b^2\,c^2\right )}{8\,c^2\,d}}{c^2+2\,c\,d\,x^2+d^2\,x^4} \] Input:

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

Output:

(atan((d^(1/2)*x)/c^(1/2))*(3*a^2*d^2 + 3*b^2*c^2 + 2*a*b*c*d))/(8*c^(5/2) 
*d^(5/2)) - ((x*(3*b^2*c^2 - 5*a^2*d^2 + 2*a*b*c*d))/(8*c*d^2) - (x^3*(3*a 
^2*d^2 - 5*b^2*c^2 + 2*a*b*c*d))/(8*c^2*d))/(c^2 + d^2*x^4 + 2*c*d*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.92 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {3 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} c^{2} d^{2}+6 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} c \,d^{3} x^{2}+3 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} d^{4} x^{4}+2 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a b \,c^{3} d +4 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a b \,c^{2} d^{2} x^{2}+2 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a b c \,d^{3} x^{4}+3 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{4}+6 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{3} d \,x^{2}+3 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{2} d^{2} x^{4}+5 a^{2} c^{2} d^{3} x +3 a^{2} c \,d^{4} x^{3}-2 a b \,c^{3} d^{2} x +2 a b \,c^{2} d^{3} x^{3}-3 b^{2} c^{4} d x -5 b^{2} c^{3} d^{2} x^{3}}{8 c^{3} d^{3} \left (d^{2} x^{4}+2 c d \,x^{2}+c^{2}\right )} \] Input:

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

Output:

(3*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*c**2*d**2 + 6*sqrt(d 
)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*c*d**3*x**2 + 3*sqrt(d)*sqrt( 
c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*d**4*x**4 + 2*sqrt(d)*sqrt(c)*atan(( 
d*x)/(sqrt(d)*sqrt(c)))*a*b*c**3*d + 4*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d) 
*sqrt(c)))*a*b*c**2*d**2*x**2 + 2*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt 
(c)))*a*b*c*d**3*x**4 + 3*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b* 
*2*c**4 + 6*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b**2*c**3*d*x**2 
 + 3*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b**2*c**2*d**2*x**4 + 5 
*a**2*c**2*d**3*x + 3*a**2*c*d**4*x**3 - 2*a*b*c**3*d**2*x + 2*a*b*c**2*d* 
*3*x**3 - 3*b**2*c**4*d*x - 5*b**2*c**3*d**2*x**3)/(8*c**3*d**3*(c**2 + 2* 
c*d*x**2 + d**2*x**4))