\(\int \frac {x^2 (a+b x^2)^2}{(c+d x^2)^3} \, dx\) [191]

Optimal result

Integrand size = 22, antiderivative size = 127 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b^2 x}{d^3}+\frac {(b c-a d)^2 x^3}{4 c d^2 \left (c+d x^2\right )^2}+\frac {(b c-a d) (7 b c+a d) x}{8 c d^3 \left (c+d x^2\right )}-\frac {\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{3/2} d^{7/2}} \]

[Out]

b^2*x/d^3+1/4*(-a*d+b*c)^2*x^3/c/d^2/(d*x^2+c)^2+1/8*(-a*d+b*c)*(a*d+7*b*c)*x/c/d^3/(d*x^2+c)-1/8*(-a^2*d^2-6*
a*b*c*d+15*b^2*c^2)*arctan(x*d^(1/2)/c^(1/2))/c^(3/2)/d^(7/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {474, 466, 396, 211} \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=-\frac {\left (-a^2 d^2-6 a b c d+15 b^2 c^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{3/2} d^{7/2}}+\frac {x (b c-a d) (a d+7 b c)}{8 c d^3 \left (c+d x^2\right )}+\frac {x^3 (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}+\frac {b^2 x}{d^3} \]

[In]

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

[Out]

(b^2*x)/d^3 + ((b*c - a*d)^2*x^3)/(4*c*d^2*(c + d*x^2)^2) + ((b*c - a*d)*(7*b*c + a*d)*x)/(8*c*d^3*(c + d*x^2)
) - ((15*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(3/2)*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 396

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

Rule 466

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 474

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

Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d)^2 x^3}{4 c d^2 \left (c+d x^2\right )^2}-\frac {\int \frac {x^2 \left (-4 a^2 d^2+3 (b c-a d)^2-4 b^2 c d x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c d^2} \\ & = \frac {(b c-a d)^2 x^3}{4 c d^2 \left (c+d x^2\right )^2}+\frac {(b c-a d) (7 b c+a d) x}{8 c d^3 \left (c+d x^2\right )}+\frac {\int \frac {-d (b c-a d) (7 b c+a d)+8 b^2 c d^2 x^2}{c+d x^2} \, dx}{8 c d^4} \\ & = \frac {b^2 x}{d^3}+\frac {(b c-a d)^2 x^3}{4 c d^2 \left (c+d x^2\right )^2}+\frac {(b c-a d) (7 b c+a d) x}{8 c d^3 \left (c+d x^2\right )}-\frac {\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \int \frac {1}{c+d x^2} \, dx}{8 c d^3} \\ & = \frac {b^2 x}{d^3}+\frac {(b c-a d)^2 x^3}{4 c d^2 \left (c+d x^2\right )^2}+\frac {(b c-a d) (7 b c+a d) x}{8 c d^3 \left (c+d x^2\right )}-\frac {\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{3/2} d^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {x \left (a^2 d^2 \left (-c+d x^2\right )-2 a b c d \left (3 c+5 d x^2\right )+b^2 c \left (15 c^2+25 c d x^2+8 d^2 x^4\right )\right )}{8 c d^3 \left (c+d x^2\right )^2}-\frac {\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{3/2} d^{7/2}} \]

[In]

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

[Out]

(x*(a^2*d^2*(-c + d*x^2) - 2*a*b*c*d*(3*c + 5*d*x^2) + b^2*c*(15*c^2 + 25*c*d*x^2 + 8*d^2*x^4)))/(8*c*d^3*(c +
d*x^2)^2) - ((15*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(3/2)*d^(7/2))

Maple [A] (verified)

Time = 2.70 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {b^{2} x}{d^{3}}+\frac {\frac {\frac {d \left (a^{2} d^{2}-10 a b c d +9 b^{2} c^{2}\right ) x^{3}}{8 c}+\left (-\frac {1}{8} a^{2} d^{2}-\frac {3}{4} a b c d +\frac {7}{8} b^{2} c^{2}\right ) x}{\left (d \,x^{2}+c \right )^{2}}+\frac {\left (a^{2} d^{2}+6 a b c d -15 b^{2} c^{2}\right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 c \sqrt {c d}}}{d^{3}}\)	\(123\)
risch	\(\frac {b^{2} x}{d^{3}}+\frac {\frac {d \left (a^{2} d^{2}-10 a b c d +9 b^{2} c^{2}\right ) x^{3}}{8 c}+\left (-\frac {1}{8} a^{2} d^{2}-\frac {3}{4} a b c d +\frac {7}{8} b^{2} c^{2}\right ) x}{d^{3} \left (d \,x^{2}+c \right )^{2}}-\frac {\ln \left (d x +\sqrt {-c d}\right ) a^{2}}{16 d \sqrt {-c d}\, c}-\frac {3 \ln \left (d x +\sqrt {-c d}\right ) a b}{8 d^{2} \sqrt {-c d}}+\frac {15 c \ln \left (d x +\sqrt {-c d}\right ) b^{2}}{16 d^{3} \sqrt {-c d}}+\frac {\ln \left (-d x +\sqrt {-c d}\right ) a^{2}}{16 d \sqrt {-c d}\, c}+\frac {3 \ln \left (-d x +\sqrt {-c d}\right ) a b}{8 d^{2} \sqrt {-c d}}-\frac {15 c \ln \left (-d x +\sqrt {-c d}\right ) b^{2}}{16 d^{3} \sqrt {-c d}}\)	\(239\)

[In]

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

[Out]

b^2*x/d^3+1/d^3*((1/8*d*(a^2*d^2-10*a*b*c*d+9*b^2*c^2)/c*x^3+(-1/8*a^2*d^2-3/4*a*b*c*d+7/8*b^2*c^2)*x)/(d*x^2+
c)^2+1/8*(a^2*d^2+6*a*b*c*d-15*b^2*c^2)/c/(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. 227 vs. \(2 (113) = 226\).

Time = 0.25 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.74 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\left [\frac {16 \, b^{2} c^{2} d^{3} x^{5} + 2 \, {\left (25 \, b^{2} c^{3} d^{2} - 10 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{3} + {\left (15 \, b^{2} c^{4} - 6 \, a b c^{3} d - a^{2} c^{2} d^{2} + {\left (15 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3} - a^{2} d^{4}\right )} x^{4} + 2 \, {\left (15 \, b^{2} c^{3} d - 6 \, a b c^{2} d^{2} - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {-c d} \log \left (\frac {d x^{2} - 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right ) + 2 \, {\left (15 \, b^{2} c^{4} d - 6 \, a b c^{3} d^{2} - a^{2} c^{2} d^{3}\right )} x}{16 \, {\left (c^{2} d^{6} x^{4} + 2 \, c^{3} d^{5} x^{2} + c^{4} d^{4}\right )}}, \frac {8 \, b^{2} c^{2} d^{3} x^{5} + {\left (25 \, b^{2} c^{3} d^{2} - 10 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{3} - {\left (15 \, b^{2} c^{4} - 6 \, a b c^{3} d - a^{2} c^{2} d^{2} + {\left (15 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3} - a^{2} d^{4}\right )} x^{4} + 2 \, {\left (15 \, b^{2} c^{3} d - 6 \, a b c^{2} d^{2} - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right ) + {\left (15 \, b^{2} c^{4} d - 6 \, a b c^{3} d^{2} - a^{2} c^{2} d^{3}\right )} x}{8 \, {\left (c^{2} d^{6} x^{4} + 2 \, c^{3} d^{5} x^{2} + c^{4} d^{4}\right )}}\right ] \]

[In]

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

[Out]

[1/16*(16*b^2*c^2*d^3*x^5 + 2*(25*b^2*c^3*d^2 - 10*a*b*c^2*d^3 + a^2*c*d^4)*x^3 + (15*b^2*c^4 - 6*a*b*c^3*d -
a^2*c^2*d^2 + (15*b^2*c^2*d^2 - 6*a*b*c*d^3 - a^2*d^4)*x^4 + 2*(15*b^2*c^3*d - 6*a*b*c^2*d^2 - 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*(15*b^2*c^4*d - 6*a*b*c^3*d^2 - a^2*c^2*d^3)*x)/
(c^2*d^6*x^4 + 2*c^3*d^5*x^2 + c^4*d^4), 1/8*(8*b^2*c^2*d^3*x^5 + (25*b^2*c^3*d^2 - 10*a*b*c^2*d^3 + a^2*c*d^4
)*x^3 - (15*b^2*c^4 - 6*a*b*c^3*d - a^2*c^2*d^2 + (15*b^2*c^2*d^2 - 6*a*b*c*d^3 - a^2*d^4)*x^4 + 2*(15*b^2*c^3
*d - 6*a*b*c^2*d^2 - a^2*c*d^3)*x^2)*sqrt(c*d)*arctan(sqrt(c*d)*x/c) + (15*b^2*c^4*d - 6*a*b*c^3*d^2 - a^2*c^2
*d^3)*x)/(c^2*d^6*x^4 + 2*c^3*d^5*x^2 + c^4*d^4)]

Sympy [A] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.76 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b^{2} x}{d^{3}} - \frac {\sqrt {- \frac {1}{c^{3} d^{7}}} \left (a^{2} d^{2} + 6 a b c d - 15 b^{2} c^{2}\right ) \log {\left (- c^{2} d^{3} \sqrt {- \frac {1}{c^{3} d^{7}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{c^{3} d^{7}}} \left (a^{2} d^{2} + 6 a b c d - 15 b^{2} c^{2}\right ) \log {\left (c^{2} d^{3} \sqrt {- \frac {1}{c^{3} d^{7}}} + x \right )}}{16} + \frac {x^{3} \left (a^{2} d^{3} - 10 a b c d^{2} + 9 b^{2} c^{2} d\right ) + x \left (- a^{2} c d^{2} - 6 a b c^{2} d + 7 b^{2} c^{3}\right )}{8 c^{3} d^{3} + 16 c^{2} d^{4} x^{2} + 8 c d^{5} x^{4}} \]

[In]

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

[Out]

b**2*x/d**3 - sqrt(-1/(c**3*d**7))*(a**2*d**2 + 6*a*b*c*d - 15*b**2*c**2)*log(-c**2*d**3*sqrt(-1/(c**3*d**7))
+ x)/16 + sqrt(-1/(c**3*d**7))*(a**2*d**2 + 6*a*b*c*d - 15*b**2*c**2)*log(c**2*d**3*sqrt(-1/(c**3*d**7)) + x)/
16 + (x**3*(a**2*d**3 - 10*a*b*c*d**2 + 9*b**2*c**2*d) + x*(-a**2*c*d**2 - 6*a*b*c**2*d + 7*b**2*c**3))/(8*c**
3*d**3 + 16*c**2*d**4*x**2 + 8*c*d**5*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.13 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {{\left (9 \, b^{2} c^{2} d - 10 \, a b c d^{2} + a^{2} d^{3}\right )} x^{3} + {\left (7 \, b^{2} c^{3} - 6 \, a b c^{2} d - a^{2} c d^{2}\right )} x}{8 \, {\left (c d^{5} x^{4} + 2 \, c^{2} d^{4} x^{2} + c^{3} d^{3}\right )}} + \frac {b^{2} x}{d^{3}} - \frac {{\left (15 \, b^{2} c^{2} - 6 \, a b c d - a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c d^{3}} \]

[In]

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

[Out]

1/8*((9*b^2*c^2*d - 10*a*b*c*d^2 + a^2*d^3)*x^3 + (7*b^2*c^3 - 6*a*b*c^2*d - a^2*c*d^2)*x)/(c*d^5*x^4 + 2*c^2*
d^4*x^2 + c^3*d^3) + b^2*x/d^3 - 1/8*(15*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*c*d^3
)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.05 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b^{2} x}{d^{3}} - \frac {{\left (15 \, b^{2} c^{2} - 6 \, a b c d - a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c d^{3}} + \frac {9 \, b^{2} c^{2} d x^{3} - 10 \, a b c d^{2} x^{3} + a^{2} d^{3} x^{3} + 7 \, b^{2} c^{3} x - 6 \, a b c^{2} d x - a^{2} c d^{2} x}{8 \, {\left (d x^{2} + c\right )}^{2} c d^{3}} \]

[In]

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

[Out]

b^2*x/d^3 - 1/8*(15*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*c*d^3) + 1/8*(9*b^2*c^2*d*
x^3 - 10*a*b*c*d^2*x^3 + a^2*d^3*x^3 + 7*b^2*c^3*x - 6*a*b*c^2*d*x - a^2*c*d^2*x)/((d*x^2 + c)^2*c*d^3)

Mupad [B] (verification not implemented)

Time = 5.15 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06 \[ \int \frac {x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx=\frac {b^2\,x}{d^3}-\frac {x\,\left (\frac {a^2\,d^2}{8}+\frac {3\,a\,b\,c\,d}{4}-\frac {7\,b^2\,c^2}{8}\right )-\frac {x^3\,\left (a^2\,d^3-10\,a\,b\,c\,d^2+9\,b^2\,c^2\,d\right )}{8\,c}}{c^2\,d^3+2\,c\,d^4\,x^2+d^5\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )\,\left (a^2\,d^2+6\,a\,b\,c\,d-15\,b^2\,c^2\right )}{8\,c^{3/2}\,d^{7/2}} \]

[In]

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

[Out]

(b^2*x)/d^3 - (x*((a^2*d^2)/8 - (7*b^2*c^2)/8 + (3*a*b*c*d)/4) - (x^3*(a^2*d^3 + 9*b^2*c^2*d - 10*a*b*c*d^2))/
(8*c))/(c^2*d^3 + d^5*x^4 + 2*c*d^4*x^2) + (atan((d^(1/2)*x)/c^(1/2))*(a^2*d^2 - 15*b^2*c^2 + 6*a*b*c*d))/(8*c
^(3/2)*d^(7/2))