\(\int \frac {(a+b x^2) (c+d x^2)^2}{e+f x^2} \, dx\) [12]

Optimal result

Integrand size = 26, antiderivative size = 142 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{e+f x^2} \, dx=-\frac {\left (5 a d f (3 d e-5 c f)-b \left (15 d^2 e^2-25 c d e f+8 c^2 f^2\right )\right ) x}{15 f^3}-\frac {(5 b d e-4 b c f-5 a d f) x \left (c+d x^2\right )}{15 f^2}+\frac {b x \left (c+d x^2\right )^2}{5 f}-\frac {(b e-a f) (d e-c f)^2 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{7/2}} \]

[Out]

-1/15*(5*a*d*f*(-5*c*f+3*d*e)-b*(8*c^2*f^2-25*c*d*e*f+15*d^2*e^2))*x/f^3-1/15*(-5*a*d*f-4*b*c*f+5*b*d*e)*x*(d*
x^2+c)/f^2+1/5*b*x*(d*x^2+c)^2/f-(-a*f+b*e)*(-c*f+d*e)^2*arctan(x*f^(1/2)/e^(1/2))/f^(7/2)/e^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {542, 396, 211} \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{e+f x^2} \, dx=-\frac {(b e-a f) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (d e-c f)^2}{\sqrt {e} f^{7/2}}-\frac {x \left (5 a d f (3 d e-5 c f)-b \left (8 c^2 f^2-25 c d e f+15 d^2 e^2\right )\right )}{15 f^3}-\frac {x \left (c+d x^2\right ) (-5 a d f-4 b c f+5 b d e)}{15 f^2}+\frac {b x \left (c+d x^2\right )^2}{5 f} \]

[In]

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

[Out]

-1/15*((5*a*d*f*(3*d*e - 5*c*f) - b*(15*d^2*e^2 - 25*c*d*e*f + 8*c^2*f^2))*x)/f^3 - ((5*b*d*e - 4*b*c*f - 5*a*
d*f)*x*(c + d*x^2))/(15*f^2) + (b*x*(c + d*x^2)^2)/(5*f) - ((b*e - a*f)*(d*e - c*f)^2*ArcTan[(Sqrt[f]*x)/Sqrt[
e]])/(Sqrt[e]*f^(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 542

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

Rubi steps \begin{align*} \text {integral}& = \frac {b x \left (c+d x^2\right )^2}{5 f}+\frac {\int \frac {\left (c+d x^2\right ) \left (-c (b e-5 a f)+(-5 b d e+4 b c f+5 a d f) x^2\right )}{e+f x^2} \, dx}{5 f} \\ & = -\frac {(5 b d e-4 b c f-5 a d f) x \left (c+d x^2\right )}{15 f^2}+\frac {b x \left (c+d x^2\right )^2}{5 f}+\frac {\int \frac {c (b e (5 d e-7 c f)-5 a f (d e-3 c f))-\left (5 a d f (3 d e-5 c f)-b \left (15 d^2 e^2-25 c d e f+8 c^2 f^2\right )\right ) x^2}{e+f x^2} \, dx}{15 f^2} \\ & = -\frac {\left (5 a d f (3 d e-5 c f)-b \left (15 d^2 e^2-25 c d e f+8 c^2 f^2\right )\right ) x}{15 f^3}-\frac {(5 b d e-4 b c f-5 a d f) x \left (c+d x^2\right )}{15 f^2}+\frac {b x \left (c+d x^2\right )^2}{5 f}-\frac {\left ((b e-a f) (d e-c f)^2\right ) \int \frac {1}{e+f x^2} \, dx}{f^3} \\ & = -\frac {\left (5 a d f (3 d e-5 c f)-b \left (15 d^2 e^2-25 c d e f+8 c^2 f^2\right )\right ) x}{15 f^3}-\frac {(5 b d e-4 b c f-5 a d f) x \left (c+d x^2\right )}{15 f^2}+\frac {b x \left (c+d x^2\right )^2}{5 f}-\frac {(b e-a f) (d e-c f)^2 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{e+f x^2} \, dx=\frac {\left (b (d e-c f)^2+a d f (-d e+2 c f)\right ) x}{f^3}+\frac {d (-b d e+2 b c f+a d f) x^3}{3 f^2}+\frac {b d^2 x^5}{5 f}-\frac {(b e-a f) (d e-c f)^2 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{7/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.20

method	result	size
default	\(\frac {\frac {1}{5} b \,d^{2} x^{5} f^{2}+\frac {1}{3} a \,d^{2} f^{2} x^{3}+\frac {2}{3} b c d \,f^{2} x^{3}-\frac {1}{3} b \,d^{2} e f \,x^{3}+2 a c d \,f^{2} x -a \,d^{2} e f x +b \,c^{2} f^{2} x -2 b c d e f x +b \,d^{2} e^{2} x}{f^{3}}+\frac {\left (c^{2} a \,f^{3}-2 a c d e \,f^{2}+a \,d^{2} e^{2} f -b \,c^{2} e \,f^{2}+2 b c d \,e^{2} f -b \,d^{2} e^{3}\right ) \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{f^{3} \sqrt {e f}}\)	\(170\)
risch	\(\frac {b \,d^{2} x^{5}}{5 f}+\frac {a \,d^{2} x^{3}}{3 f}+\frac {2 b c d \,x^{3}}{3 f}-\frac {b \,d^{2} e \,x^{3}}{3 f^{2}}+\frac {2 a c d x}{f}-\frac {a \,d^{2} e x}{f^{2}}+\frac {b \,c^{2} x}{f}-\frac {2 b c d e x}{f^{2}}+\frac {b \,d^{2} e^{2} x}{f^{3}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) c^{2} a}{2 \sqrt {-e f}}+\frac {\ln \left (f x +\sqrt {-e f}\right ) a c d e}{f \sqrt {-e f}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) a \,d^{2} e^{2}}{2 f^{2} \sqrt {-e f}}+\frac {\ln \left (f x +\sqrt {-e f}\right ) b \,c^{2} e}{2 f \sqrt {-e f}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) b c d \,e^{2}}{f^{2} \sqrt {-e f}}+\frac {\ln \left (f x +\sqrt {-e f}\right ) b \,d^{2} e^{3}}{2 f^{3} \sqrt {-e f}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) c^{2} a}{2 \sqrt {-e f}}-\frac {\ln \left (-f x +\sqrt {-e f}\right ) a c d e}{f \sqrt {-e f}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) a \,d^{2} e^{2}}{2 f^{2} \sqrt {-e f}}-\frac {\ln \left (-f x +\sqrt {-e f}\right ) b \,c^{2} e}{2 f \sqrt {-e f}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) b c d \,e^{2}}{f^{2} \sqrt {-e f}}-\frac {\ln \left (-f x +\sqrt {-e f}\right ) b \,d^{2} e^{3}}{2 f^{3} \sqrt {-e f}}\)	\(429\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.58 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{e+f x^2} \, dx=\left [\frac {6 \, b d^{2} e f^{3} x^{5} - 10 \, {\left (b d^{2} e^{2} f^{2} - {\left (2 \, b c d + a d^{2}\right )} e f^{3}\right )} x^{3} + 15 \, {\left (b d^{2} e^{3} - a c^{2} f^{3} - {\left (2 \, b c d + a d^{2}\right )} e^{2} f + {\left (b c^{2} + 2 \, a c d\right )} e f^{2}\right )} \sqrt {-e f} \log \left (\frac {f x^{2} - 2 \, \sqrt {-e f} x - e}{f x^{2} + e}\right ) + 30 \, {\left (b d^{2} e^{3} f - {\left (2 \, b c d + a d^{2}\right )} e^{2} f^{2} + {\left (b c^{2} + 2 \, a c d\right )} e f^{3}\right )} x}{30 \, e f^{4}}, \frac {3 \, b d^{2} e f^{3} x^{5} - 5 \, {\left (b d^{2} e^{2} f^{2} - {\left (2 \, b c d + a d^{2}\right )} e f^{3}\right )} x^{3} - 15 \, {\left (b d^{2} e^{3} - a c^{2} f^{3} - {\left (2 \, b c d + a d^{2}\right )} e^{2} f + {\left (b c^{2} + 2 \, a c d\right )} e f^{2}\right )} \sqrt {e f} \arctan \left (\frac {\sqrt {e f} x}{e}\right ) + 15 \, {\left (b d^{2} e^{3} f - {\left (2 \, b c d + a d^{2}\right )} e^{2} f^{2} + {\left (b c^{2} + 2 \, a c d\right )} e f^{3}\right )} x}{15 \, e f^{4}}\right ] \]

[In]

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

[Out]

[1/30*(6*b*d^2*e*f^3*x^5 - 10*(b*d^2*e^2*f^2 - (2*b*c*d + a*d^2)*e*f^3)*x^3 + 15*(b*d^2*e^3 - a*c^2*f^3 - (2*b
*c*d + a*d^2)*e^2*f + (b*c^2 + 2*a*c*d)*e*f^2)*sqrt(-e*f)*log((f*x^2 - 2*sqrt(-e*f)*x - e)/(f*x^2 + e)) + 30*(
b*d^2*e^3*f - (2*b*c*d + a*d^2)*e^2*f^2 + (b*c^2 + 2*a*c*d)*e*f^3)*x)/(e*f^4), 1/15*(3*b*d^2*e*f^3*x^5 - 5*(b*
d^2*e^2*f^2 - (2*b*c*d + a*d^2)*e*f^3)*x^3 - 15*(b*d^2*e^3 - a*c^2*f^3 - (2*b*c*d + a*d^2)*e^2*f + (b*c^2 + 2*
a*c*d)*e*f^2)*sqrt(e*f)*arctan(sqrt(e*f)*x/e) + 15*(b*d^2*e^3*f - (2*b*c*d + a*d^2)*e^2*f^2 + (b*c^2 + 2*a*c*d
)*e*f^3)*x)/(e*f^4)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (138) = 276\).

Time = 0.52 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.44 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{e+f x^2} \, dx=\frac {b d^{2} x^{5}}{5 f} + x^{3} \left (\frac {a d^{2}}{3 f} + \frac {2 b c d}{3 f} - \frac {b d^{2} e}{3 f^{2}}\right ) + x \left (\frac {2 a c d}{f} - \frac {a d^{2} e}{f^{2}} + \frac {b c^{2}}{f} - \frac {2 b c d e}{f^{2}} + \frac {b d^{2} e^{2}}{f^{3}}\right ) - \frac {\sqrt {- \frac {1}{e f^{7}}} \left (a f - b e\right ) \left (c f - d e\right )^{2} \log {\left (- \frac {e f^{3} \sqrt {- \frac {1}{e f^{7}}} \left (a f - b e\right ) \left (c f - d e\right )^{2}}{a c^{2} f^{3} - 2 a c d e f^{2} + a d^{2} e^{2} f - b c^{2} e f^{2} + 2 b c d e^{2} f - b d^{2} e^{3}} + x \right )}}{2} + \frac {\sqrt {- \frac {1}{e f^{7}}} \left (a f - b e\right ) \left (c f - d e\right )^{2} \log {\left (\frac {e f^{3} \sqrt {- \frac {1}{e f^{7}}} \left (a f - b e\right ) \left (c f - d e\right )^{2}}{a c^{2} f^{3} - 2 a c d e f^{2} + a d^{2} e^{2} f - b c^{2} e f^{2} + 2 b c d e^{2} f - b d^{2} e^{3}} + x \right )}}{2} \]

[In]

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

[Out]

b*d**2*x**5/(5*f) + x**3*(a*d**2/(3*f) + 2*b*c*d/(3*f) - b*d**2*e/(3*f**2)) + x*(2*a*c*d/f - a*d**2*e/f**2 + b
*c**2/f - 2*b*c*d*e/f**2 + b*d**2*e**2/f**3) - sqrt(-1/(e*f**7))*(a*f - b*e)*(c*f - d*e)**2*log(-e*f**3*sqrt(-
1/(e*f**7))*(a*f - b*e)*(c*f - d*e)**2/(a*c**2*f**3 - 2*a*c*d*e*f**2 + a*d**2*e**2*f - b*c**2*e*f**2 + 2*b*c*d
*e**2*f - b*d**2*e**3) + x)/2 + sqrt(-1/(e*f**7))*(a*f - b*e)*(c*f - d*e)**2*log(e*f**3*sqrt(-1/(e*f**7))*(a*f
- b*e)*(c*f - d*e)**2/(a*c**2*f**3 - 2*a*c*d*e*f**2 + a*d**2*e**2*f - b*c**2*e*f**2 + 2*b*c*d*e**2*f - b*d**2
*e**3) + x)/2

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{e+f x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{e+f x^2} \, dx=-\frac {{\left (b d^{2} e^{3} - 2 \, b c d e^{2} f - a d^{2} e^{2} f + b c^{2} e f^{2} + 2 \, a c d e f^{2} - a c^{2} f^{3}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f} f^{3}} + \frac {3 \, b d^{2} f^{4} x^{5} - 5 \, b d^{2} e f^{3} x^{3} + 10 \, b c d f^{4} x^{3} + 5 \, a d^{2} f^{4} x^{3} + 15 \, b d^{2} e^{2} f^{2} x - 30 \, b c d e f^{3} x - 15 \, a d^{2} e f^{3} x + 15 \, b c^{2} f^{4} x + 30 \, a c d f^{4} x}{15 \, f^{5}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{e+f x^2} \, dx=x^3\,\left (\frac {a\,d^2+2\,b\,c\,d}{3\,f}-\frac {b\,d^2\,e}{3\,f^2}\right )+x\,\left (\frac {b\,c^2+2\,a\,d\,c}{f}-\frac {e\,\left (\frac {a\,d^2+2\,b\,c\,d}{f}-\frac {b\,d^2\,e}{f^2}\right )}{f}\right )+\frac {b\,d^2\,x^5}{5\,f}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x\,\left (a\,f-b\,e\right )\,{\left (c\,f-d\,e\right )}^2}{\sqrt {e}\,\left (-b\,c^2\,e\,f^2+a\,c^2\,f^3+2\,b\,c\,d\,e^2\,f-2\,a\,c\,d\,e\,f^2-b\,d^2\,e^3+a\,d^2\,e^2\,f\right )}\right )\,\left (a\,f-b\,e\right )\,{\left (c\,f-d\,e\right )}^2}{\sqrt {e}\,f^{7/2}} \]

[In]

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

[Out]

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