\(\int \frac {(a+b x^2)^2}{(c+d x^2)^3} \, dx\) [193]

Optimal result

Integrand size = 19, antiderivative size = 116 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {(b c-a d) x \left (a+b x^2\right )}{4 c d \left (c+d x^2\right )^2}+\frac {3 \left (\frac {a^2}{c^2}-\frac {b^2}{d^2}\right ) x}{8 \left (c+d x^2\right )}+\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {424, 393, 211} \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{5/2}}+\frac {3 x \left (\frac {a^2}{c^2}-\frac {b^2}{d^2}\right )}{8 \left (c+d x^2\right )}-\frac {x \left (a+b x^2\right ) (b c-a d)}{4 c d \left (c+d x^2\right )^2} \]

[In]

Int[(a + b*x^2)^2/(c + d*x^2)^3,x]

[Out]

-1/4*((b*c - a*d)*x*(a + b*x^2))/(c*d*(c + d*x^2)^2) + (3*(a^2/c^2 - b^2/d^2)*x)/(8*(c + d*x^2)) + ((3*b^2*c^2
+ 2*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(5/2)*d^(5/2))

Rule 211

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

Rule 393

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

Rule 424

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(b c-a d) x \left (a+b x^2\right )}{4 c d \left (c+d x^2\right )^2}+\frac {\int \frac {a (b c+3 a d)+b (3 b c+a d) x^2}{\left (c+d x^2\right )^2} \, dx}{4 c d} \\ & = -\frac {(b c-a d) x \left (a+b x^2\right )}{4 c d \left (c+d x^2\right )^2}+\frac {3 \left (\frac {a^2}{c^2}-\frac {b^2}{d^2}\right ) x}{8 \left (c+d x^2\right )}+\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \int \frac {1}{c+d x^2} \, dx}{8 c^2 d^2} \\ & = -\frac {(b c-a d) x \left (a+b x^2\right )}{4 c d \left (c+d x^2\right )^2}+\frac {3 \left (\frac {a^2}{c^2}-\frac {b^2}{d^2}\right ) x}{8 \left (c+d x^2\right )}+\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {(b c-a d) x \left (a d \left (5 c+3 d x^2\right )+b c \left (3 c+5 d x^2\right )\right )}{8 c^2 d^2 \left (c+d x^2\right )^2}+\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{5/2}} \]

[In]

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

[Out]

-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)/Sqrt[c]])/(8*c^(5/2)*d^(5/2))

Maple [A] (verified)

Time = 2.61 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.07

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

[In]

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

[Out]

(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(d*x/(c*d)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (102) = 204\).

Time = 0.37 (sec) , antiderivative size = 449, normalized size of antiderivative = 3.87 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\left [-\frac {2 \, {\left (5 \, b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} - 3 \, a^{2} c d^{4}\right )} x^{3} + {\left (3 \, b^{2} c^{4} + 2 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (3 \, b^{2} c^{2} d^{2} + 2 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{2} c^{3} d + 2 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {-c d} \log \left (\frac {d x^{2} - 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right ) + 2 \, {\left (3 \, b^{2} c^{4} d + 2 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3}\right )} x}{16 \, {\left (c^{3} d^{5} x^{4} + 2 \, c^{4} d^{4} x^{2} + c^{5} d^{3}\right )}}, -\frac {{\left (5 \, b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} - 3 \, a^{2} c d^{4}\right )} x^{3} - {\left (3 \, b^{2} c^{4} + 2 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (3 \, b^{2} c^{2} d^{2} + 2 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{2} c^{3} d + 2 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right ) + {\left (3 \, b^{2} c^{4} d + 2 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3}\right )} x}{8 \, {\left (c^{3} d^{5} x^{4} + 2 \, c^{4} d^{4} x^{2} + c^{5} d^{3}\right )}}\right ] \]

[In]

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

[Out]

[-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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (110) = 220\).

Time = 0.56 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.92 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{c^{5} d^{5}}} \cdot \left (3 a^{2} d^{2} + 2 a b c d + 3 b^{2} c^{2}\right ) \log {\left (- c^{3} d^{2} \sqrt {- \frac {1}{c^{5} d^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{c^{5} d^{5}}} \cdot \left (3 a^{2} d^{2} + 2 a b c d + 3 b^{2} c^{2}\right ) \log {\left (c^{3} d^{2} \sqrt {- \frac {1}{c^{5} d^{5}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a^{2} d^{3} + 2 a b c d^{2} - 5 b^{2} c^{2} d\right ) + x \left (5 a^{2} c d^{2} - 2 a b c^{2} d - 3 b^{2} c^{3}\right )}{8 c^{4} d^{2} + 16 c^{3} d^{3} x^{2} + 8 c^{2} d^{4} x^{4}} \]

[In]

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

[Out]

-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 + sq
rt(-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)

none

Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {{\left (5 \, b^{2} c^{2} d - 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{3} + {\left (3 \, b^{2} c^{3} + 2 \, a b c^{2} d - 5 \, a^{2} c d^{2}\right )} x}{8 \, {\left (c^{2} d^{4} x^{4} + 2 \, c^{3} d^{3} x^{2} + c^{4} d^{2}\right )}} + \frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{2} d^{2}} \]

[In]

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

[Out]

-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)

none

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{2} d^{2}} - \frac {5 \, b^{2} c^{2} d x^{3} - 2 \, a b c d^{2} x^{3} - 3 \, a^{2} d^{3} x^{3} + 3 \, b^{2} c^{3} x + 2 \, a b c^{2} d x - 5 \, a^{2} c d^{2} x}{8 \, {\left (d x^{2} + c\right )}^{2} c^{2} d^{2}} \]

[In]

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

[Out]

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 = 5.10 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )\,\left (3\,a^2\,d^2+2\,a\,b\,c\,d+3\,b^2\,c^2\right )}{8\,c^{5/2}\,d^{5/2}}-\frac {\frac {x\,\left (-5\,a^2\,d^2+2\,a\,b\,c\,d+3\,b^2\,c^2\right )}{8\,c\,d^2}-\frac {x^3\,\left (3\,a^2\,d^2+2\,a\,b\,c\,d-5\,b^2\,c^2\right )}{8\,c^2\,d}}{c^2+2\,c\,d\,x^2+d^2\,x^4} \]

[In]

int((a + b*x^2)^2/(c + d*x^2)^3,x)

[Out]

(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)