\(\int \frac {x^2 (a+b x^2)^2}{c+d x^2} \, dx\) [170]

Optimal result

Integrand size = 22, antiderivative size = 83 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {(b c-a d)^2 x}{d^3}-\frac {b (b c-2 a d) x^3}{3 d^2}+\frac {b^2 x^5}{5 d}-\frac {\sqrt {c} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{d^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {472, 211} \[ \int \frac {x^2 \left (a+b x^2\right )^2}{c+d x^2} \, dx=-\frac {\sqrt {c} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{d^{7/2}}+\frac {x (b c-a d)^2}{d^3}-\frac {b x^3 (b c-2 a d)}{3 d^2}+\frac {b^2 x^5}{5 d} \]

[In]

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

[Out]

((b*c - a*d)^2*x)/d^3 - (b*(b*c - 2*a*d)*x^3)/(3*d^2) + (b^2*x^5)/(5*d) - (Sqrt[c]*(b*c - a*d)^2*ArcTan[(Sqrt[
d]*x)/Sqrt[c]])/d^(7/2)

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 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(b c-a d)^2}{d^3}-\frac {b (b c-2 a d) x^2}{d^2}+\frac {b^2 x^4}{d}+\frac {-b^2 c^3+2 a b c^2 d-a^2 c d^2}{d^3 \left (c+d x^2\right )}\right ) \, dx \\ & = \frac {(b c-a d)^2 x}{d^3}-\frac {b (b c-2 a d) x^3}{3 d^2}+\frac {b^2 x^5}{5 d}-\frac {\left (c (b c-a d)^2\right ) \int \frac {1}{c+d x^2} \, dx}{d^3} \\ & = \frac {(b c-a d)^2 x}{d^3}-\frac {b (b c-2 a d) x^3}{3 d^2}+\frac {b^2 x^5}{5 d}-\frac {\sqrt {c} (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{d^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {(-b c+a d)^2 x}{d^3}-\frac {b (b c-2 a d) x^3}{3 d^2}+\frac {b^2 x^5}{5 d}-\frac {\sqrt {c} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{d^{7/2}} \]

[In]

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

[Out]

((-(b*c) + a*d)^2*x)/d^3 - (b*(b*c - 2*a*d)*x^3)/(3*d^2) + (b^2*x^5)/(5*d) - (Sqrt[c]*(b*c - a*d)^2*ArcTan[(Sq
rt[d]*x)/Sqrt[c]])/d^(7/2)

Maple [A] (verified)

Time = 2.73 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {\frac {1}{5} b^{2} d^{2} x^{5}+\frac {2}{3} x^{3} a b \,d^{2}-\frac {1}{3} x^{3} b^{2} c d +a^{2} d^{2} x -2 a b c d x +b^{2} c^{2} x}{d^{3}}-\frac {c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{d^{3} \sqrt {c d}}\)	\(102\)
risch	\(\frac {b^{2} x^{5}}{5 d}+\frac {2 x^{3} a b}{3 d}-\frac {x^{3} b^{2} c}{3 d^{2}}+\frac {a^{2} x}{d}-\frac {2 a b c x}{d^{2}}+\frac {b^{2} c^{2} x}{d^{3}}+\frac {\sqrt {-c d}\, \ln \left (-\sqrt {-c d}\, x -c \right ) a^{2}}{2 d^{2}}-\frac {\sqrt {-c d}\, \ln \left (-\sqrt {-c d}\, x -c \right ) a b c}{d^{3}}+\frac {\sqrt {-c d}\, \ln \left (-\sqrt {-c d}\, x -c \right ) b^{2} c^{2}}{2 d^{4}}-\frac {\sqrt {-c d}\, \ln \left (\sqrt {-c d}\, x -c \right ) a^{2}}{2 d^{2}}+\frac {\sqrt {-c d}\, \ln \left (\sqrt {-c d}\, x -c \right ) a b c}{d^{3}}-\frac {\sqrt {-c d}\, \ln \left (\sqrt {-c d}\, x -c \right ) b^{2} c^{2}}{2 d^{4}}\)	\(233\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.75 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\left [\frac {6 \, b^{2} d^{2} x^{5} - 10 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} x^{3} + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{2} - 2 \, d x \sqrt {-\frac {c}{d}} - c}{d x^{2} + c}\right ) + 30 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x}{30 \, d^{3}}, \frac {3 \, b^{2} d^{2} x^{5} - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} x^{3} - 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {c}{d}} \arctan \left (\frac {d x \sqrt {\frac {c}{d}}}{c}\right ) + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x}{15 \, d^{3}}\right ] \]

[In]

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

[Out]

[1/30*(6*b^2*d^2*x^5 - 10*(b^2*c*d - 2*a*b*d^2)*x^3 + 15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-c/d)*log((d*x^2
- 2*d*x*sqrt(-c/d) - c)/(d*x^2 + c)) + 30*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x)/d^3, 1/15*(3*b^2*d^2*x^5 - 5*(b^
2*c*d - 2*a*b*d^2)*x^3 - 15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(c/d)*arctan(d*x*sqrt(c/d)/c) + 15*(b^2*c^2 -
2*a*b*c*d + a^2*d^2)*x)/d^3]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (73) = 146\).

Time = 0.28 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.34 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {b^{2} x^{5}}{5 d} + x^{3} \cdot \left (\frac {2 a b}{3 d} - \frac {b^{2} c}{3 d^{2}}\right ) + x \left (\frac {a^{2}}{d} - \frac {2 a b c}{d^{2}} + \frac {b^{2} c^{2}}{d^{3}}\right ) + \frac {\sqrt {- \frac {c}{d^{7}}} \left (a d - b c\right )^{2} \log {\left (- \frac {d^{3} \sqrt {- \frac {c}{d^{7}}} \left (a d - b c\right )^{2}}{a^{2} d^{2} - 2 a b c d + b^{2} c^{2}} + x \right )}}{2} - \frac {\sqrt {- \frac {c}{d^{7}}} \left (a d - b c\right )^{2} \log {\left (\frac {d^{3} \sqrt {- \frac {c}{d^{7}}} \left (a d - b c\right )^{2}}{a^{2} d^{2} - 2 a b c d + b^{2} c^{2}} + x \right )}}{2} \]

[In]

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

[Out]

b**2*x**5/(5*d) + x**3*(2*a*b/(3*d) - b**2*c/(3*d**2)) + x*(a**2/d - 2*a*b*c/d**2 + b**2*c**2/d**3) + sqrt(-c/
d**7)*(a*d - b*c)**2*log(-d**3*sqrt(-c/d**7)*(a*d - b*c)**2/(a**2*d**2 - 2*a*b*c*d + b**2*c**2) + x)/2 - sqrt(
-c/d**7)*(a*d - b*c)**2*log(d**3*sqrt(-c/d**7)*(a*d - b*c)**2/(a**2*d**2 - 2*a*b*c*d + b**2*c**2) + x)/2

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.25 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{c+d x^2} \, dx=-\frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {c d} d^{3}} + \frac {3 \, b^{2} d^{2} x^{5} - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} x^{3} + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x}{15 \, d^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.36 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{c+d x^2} \, dx=-\frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {c d} d^{3}} + \frac {3 \, b^{2} d^{4} x^{5} - 5 \, b^{2} c d^{3} x^{3} + 10 \, a b d^{4} x^{3} + 15 \, b^{2} c^{2} d^{2} x - 30 \, a b c d^{3} x + 15 \, a^{2} d^{4} x}{15 \, d^{5}} \]

[In]

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

[Out]

-(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*d^3) + 1/15*(3*b^2*d^4*x^5 - 5*b^2*c*d^3
*x^3 + 10*a*b*d^4*x^3 + 15*b^2*c^2*d^2*x - 30*a*b*c*d^3*x + 15*a^2*d^4*x)/d^5

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.54 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{c+d x^2} \, dx=x\,\left (\frac {a^2}{d}+\frac {c\,\left (\frac {b^2\,c}{d^2}-\frac {2\,a\,b}{d}\right )}{d}\right )-x^3\,\left (\frac {b^2\,c}{3\,d^2}-\frac {2\,a\,b}{3\,d}\right )+\frac {b^2\,x^5}{5\,d}-\frac {\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,x\,{\left (a\,d-b\,c\right )}^2}{a^2\,c\,d^2-2\,a\,b\,c^2\,d+b^2\,c^3}\right )\,{\left (a\,d-b\,c\right )}^2}{d^{7/2}} \]

[In]

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

[Out]

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