3.1.13 \(\int \frac {(a+b x^2) (c+d x^2)^2}{(e+f x^2)^2} \, dx\) [13]

Optimal. Leaf size=164 \[ -\frac {d (b e (15 d e-13 c f)-3 a f (3 d e-c f)) x}{6 e f^3}+\frac {d (5 b e-3 a f) x \left (c+d x^2\right )}{6 e f^2}-\frac {(b e-a f) x \left (c+d x^2\right )^2}{2 e f \left (e+f x^2\right )}+\frac {(d e-c f) (b e (5 d e-c f)-a f (3 d e+c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{2 e^{3/2} f^{7/2}} \]

[Out]

-1/6*d*(b*e*(-13*c*f+15*d*e)-3*a*f*(-c*f+3*d*e))*x/e/f^3+1/6*d*(-3*a*f+5*b*e)*x*(d*x^2+c)/e/f^2-1/2*(-a*f+b*e)
*x*(d*x^2+c)^2/e/f/(f*x^2+e)+1/2*(-c*f+d*e)*(b*e*(-c*f+5*d*e)-a*f*(c*f+3*d*e))*arctan(x*f^(1/2)/e^(1/2))/e^(3/
2)/f^(7/2)

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Rubi [A]
time = 0.15, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {540, 542, 396, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (d e-c f) (b e (5 d e-c f)-a f (c f+3 d e))}{2 e^{3/2} f^{7/2}}-\frac {d x (b e (15 d e-13 c f)-3 a f (3 d e-c f))}{6 e f^3}+\frac {d x \left (c+d x^2\right ) (5 b e-3 a f)}{6 e f^2}-\frac {x \left (c+d x^2\right )^2 (b e-a f)}{2 e f \left (e+f x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(d*(b*e*(15*d*e - 13*c*f) - 3*a*f*(3*d*e - c*f))*x)/(e*f^3) + (d*(5*b*e - 3*a*f)*x*(c + d*x^2))/(6*e*f^2)
- ((b*e - a*f)*x*(c + d*x^2)^2)/(2*e*f*(e + f*x^2)) + ((d*e - c*f)*(b*e*(5*d*e - c*f) - a*f*(3*d*e + c*f))*Ar
cTan[(Sqrt[f]*x)/Sqrt[e]])/(2*e^(3/2)*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 540

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(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 - 1)*Simp[c*(b*e*n*(p + 1) + b*e - a*f) + d*(b*e*n*(p + 1) + (b*e - a*f)*(n*q + 1))
*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && GtQ[q, 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*} \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^2} \, dx &=-\frac {(b e-a f) x \left (c+d x^2\right )^2}{2 e f \left (e+f x^2\right )}-\frac {\int \frac {\left (c+d x^2\right ) \left (-c (b e+a f)-d (5 b e-3 a f) x^2\right )}{e+f x^2} \, dx}{2 e f}\\ &=\frac {d (5 b e-3 a f) x \left (c+d x^2\right )}{6 e f^2}-\frac {(b e-a f) x \left (c+d x^2\right )^2}{2 e f \left (e+f x^2\right )}-\frac {\int \frac {c (b e (5 d e-3 c f)-3 a f (d e+c f))+d (b e (15 d e-13 c f)-3 a f (3 d e-c f)) x^2}{e+f x^2} \, dx}{6 e f^2}\\ &=-\frac {d (b e (15 d e-13 c f)-3 a f (3 d e-c f)) x}{6 e f^3}+\frac {d (5 b e-3 a f) x \left (c+d x^2\right )}{6 e f^2}-\frac {(b e-a f) x \left (c+d x^2\right )^2}{2 e f \left (e+f x^2\right )}+\frac {((d e-c f) (b e (5 d e-c f)-a f (3 d e+c f))) \int \frac {1}{e+f x^2} \, dx}{2 e f^3}\\ &=-\frac {d (b e (15 d e-13 c f)-3 a f (3 d e-c f)) x}{6 e f^3}+\frac {d (5 b e-3 a f) x \left (c+d x^2\right )}{6 e f^2}-\frac {(b e-a f) x \left (c+d x^2\right )^2}{2 e f \left (e+f x^2\right )}+\frac {(d e-c f) (b e (5 d e-c f)-a f (3 d e+c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{2 e^{3/2} f^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 134, normalized size = 0.82 \begin {gather*} \frac {d (-2 b d e+2 b c f+a d f) x}{f^3}+\frac {b d^2 x^3}{3 f^2}-\frac {(b e-a f) (d e-c f)^2 x}{2 e f^3 \left (e+f x^2\right )}+\frac {(d e-c f) (b e (5 d e-c f)-a f (3 d e+c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{2 e^{3/2} f^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.16, size = 182, normalized size = 1.11

method	result	size
default	\(\frac {d \left (\frac {1}{3} b d \,x^{3} f +a d f x +2 b c f x -2 b d e x \right )}{f^{3}}+\frac {\frac {\left (c^{2} a \,f^{3}-2 a c d e \,f^{2}+a \,d^{2} e^{2} f -b \,c^{2} e \,f^{2}+2 b c d \,e^{2} f -b \,d^{2} e^{3}\right ) x}{2 e \left (f \,x^{2}+e \right )}+\frac {\left (c^{2} a \,f^{3}+2 a c d e \,f^{2}-3 a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}-6 b c d \,e^{2} f +5 b \,d^{2} e^{3}\right ) \arctan \left (\frac {f x}{\sqrt {f e}}\right )}{2 e \sqrt {f e}}}{f^{3}}\)	\(182\)
risch	\(\frac {d^{2} b \,x^{3}}{3 f^{2}}+\frac {d^{2} a x}{f^{2}}+\frac {2 d b c x}{f^{2}}-\frac {2 d^{2} b e x}{f^{3}}+\frac {\left (c^{2} a \,f^{3}-2 a c d e \,f^{2}+a \,d^{2} e^{2} f -b \,c^{2} e \,f^{2}+2 b c d \,e^{2} f -b \,d^{2} e^{3}\right ) x}{2 e \,f^{3} \left (f \,x^{2}+e \right )}-\frac {\ln \left (f x +\sqrt {-f e}\right ) c^{2} a}{4 \sqrt {-f e}\, e}-\frac {\ln \left (f x +\sqrt {-f e}\right ) a c d}{2 f \sqrt {-f e}}+\frac {3 e \ln \left (f x +\sqrt {-f e}\right ) a \,d^{2}}{4 f^{2} \sqrt {-f e}}-\frac {\ln \left (f x +\sqrt {-f e}\right ) b \,c^{2}}{4 f \sqrt {-f e}}+\frac {3 e \ln \left (f x +\sqrt {-f e}\right ) b c d}{2 f^{2} \sqrt {-f e}}-\frac {5 e^{2} \ln \left (f x +\sqrt {-f e}\right ) b \,d^{2}}{4 f^{3} \sqrt {-f e}}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) c^{2} a}{4 \sqrt {-f e}\, e}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) a c d}{2 f \sqrt {-f e}}-\frac {3 e \ln \left (-f x +\sqrt {-f e}\right ) a \,d^{2}}{4 f^{2} \sqrt {-f e}}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) b \,c^{2}}{4 f \sqrt {-f e}}-\frac {3 e \ln \left (-f x +\sqrt {-f e}\right ) b c d}{2 f^{2} \sqrt {-f e}}+\frac {5 e^{2} \ln \left (-f x +\sqrt {-f e}\right ) b \,d^{2}}{4 f^{3} \sqrt {-f e}}\)	\(440\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d/f^3*(1/3*b*d*x^3*f+a*d*f*x+2*b*c*f*x-2*b*d*e*x)+1/f^3*(1/2*(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)/e*x/(f*x^2+e)+1/2*(a*c^2*f^3+2*a*c*d*e*f^2-3*a*d^2*e^2*f+b*c^2*e*f^2-6*b*c*d*e^2*f+5*
b*d^2*e^3)/e/(f*e)^(1/2)*arctan(f*x/(f*e)^(1/2)))

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Maxima [A]
time = 0.49, size = 186, normalized size = 1.13 \begin {gather*} \frac {{\left (a c^{2} f^{3} - b d^{2} e^{3} - {\left (b c^{2} e + 2 \, a c d e\right )} f^{2} + {\left (2 \, b c d e^{2} + a d^{2} e^{2}\right )} f\right )} x}{2 \, {\left (f^{4} x^{2} e + f^{3} e^{2}\right )}} + \frac {{\left (a c^{2} f^{3} + 5 \, b d^{2} e^{3} + {\left (b c^{2} e + 2 \, a c d e\right )} f^{2} - 3 \, {\left (2 \, b c d e^{2} + a d^{2} e^{2}\right )} f\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {3}{2}\right )}}{2 \, f^{\frac {7}{2}}} + \frac {b d^{2} f x^{3} - 3 \, {\left (2 \, b d^{2} e - {\left (2 \, b c d + a d^{2}\right )} f\right )} x}{3 \, f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.00, size = 552, normalized size = 3.37 \begin {gather*} \left [\frac {6 \, a c^{2} f^{4} x e - 30 \, b d^{2} f x e^{4} - 3 \, {\left (a c^{2} f^{4} x^{2} + 5 \, b d^{2} e^{4} + {\left (5 \, b d^{2} f x^{2} - 3 \, {\left (2 \, b c d + a d^{2}\right )} f\right )} e^{3} - {\left (3 \, {\left (2 \, b c d + a d^{2}\right )} f^{2} x^{2} - {\left (b c^{2} + 2 \, a c d\right )} f^{2}\right )} e^{2} + {\left (a c^{2} f^{3} + {\left (b c^{2} + 2 \, a c d\right )} f^{3} x^{2}\right )} e\right )} \sqrt {-f e} \log \left (\frac {f x^{2} - 2 \, \sqrt {-f e} x - e}{f x^{2} + e}\right ) - 2 \, {\left (10 \, b d^{2} f^{2} x^{3} - 9 \, {\left (2 \, b c d + a d^{2}\right )} f^{2} x\right )} e^{3} + 2 \, {\left (2 \, b d^{2} f^{3} x^{5} + 6 \, {\left (2 \, b c d + a d^{2}\right )} f^{3} x^{3} - 3 \, {\left (b c^{2} + 2 \, a c d\right )} f^{3} x\right )} e^{2}}{12 \, {\left (f^{5} x^{2} e^{2} + f^{4} e^{3}\right )}}, \frac {3 \, a c^{2} f^{4} x e - 15 \, b d^{2} f x e^{4} + 3 \, {\left (a c^{2} f^{4} x^{2} + 5 \, b d^{2} e^{4} + {\left (5 \, b d^{2} f x^{2} - 3 \, {\left (2 \, b c d + a d^{2}\right )} f\right )} e^{3} - {\left (3 \, {\left (2 \, b c d + a d^{2}\right )} f^{2} x^{2} - {\left (b c^{2} + 2 \, a c d\right )} f^{2}\right )} e^{2} + {\left (a c^{2} f^{3} + {\left (b c^{2} + 2 \, a c d\right )} f^{3} x^{2}\right )} e\right )} \sqrt {f} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {1}{2}} - {\left (10 \, b d^{2} f^{2} x^{3} - 9 \, {\left (2 \, b c d + a d^{2}\right )} f^{2} x\right )} e^{3} + {\left (2 \, b d^{2} f^{3} x^{5} + 6 \, {\left (2 \, b c d + a d^{2}\right )} f^{3} x^{3} - 3 \, {\left (b c^{2} + 2 \, a c d\right )} f^{3} x\right )} e^{2}}{6 \, {\left (f^{5} x^{2} e^{2} + f^{4} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(6*a*c^2*f^4*x*e - 30*b*d^2*f*x*e^4 - 3*(a*c^2*f^4*x^2 + 5*b*d^2*e^4 + (5*b*d^2*f*x^2 - 3*(2*b*c*d + a*d
^2)*f)*e^3 - (3*(2*b*c*d + a*d^2)*f^2*x^2 - (b*c^2 + 2*a*c*d)*f^2)*e^2 + (a*c^2*f^3 + (b*c^2 + 2*a*c*d)*f^3*x^
2)*e)*sqrt(-f*e)*log((f*x^2 - 2*sqrt(-f*e)*x - e)/(f*x^2 + e)) - 2*(10*b*d^2*f^2*x^3 - 9*(2*b*c*d + a*d^2)*f^2
*x)*e^3 + 2*(2*b*d^2*f^3*x^5 + 6*(2*b*c*d + a*d^2)*f^3*x^3 - 3*(b*c^2 + 2*a*c*d)*f^3*x)*e^2)/(f^5*x^2*e^2 + f^
4*e^3), 1/6*(3*a*c^2*f^4*x*e - 15*b*d^2*f*x*e^4 + 3*(a*c^2*f^4*x^2 + 5*b*d^2*e^4 + (5*b*d^2*f*x^2 - 3*(2*b*c*d
+ a*d^2)*f)*e^3 - (3*(2*b*c*d + a*d^2)*f^2*x^2 - (b*c^2 + 2*a*c*d)*f^2)*e^2 + (a*c^2*f^3 + (b*c^2 + 2*a*c*d)*
f^3*x^2)*e)*sqrt(f)*arctan(sqrt(f)*x*e^(-1/2))*e^(1/2) - (10*b*d^2*f^2*x^3 - 9*(2*b*c*d + a*d^2)*f^2*x)*e^3 +
(2*b*d^2*f^3*x^5 + 6*(2*b*c*d + a*d^2)*f^3*x^3 - 3*(b*c^2 + 2*a*c*d)*f^3*x)*e^2)/(f^5*x^2*e^2 + f^4*e^3)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 483 vs. \(2 (151) = 302\).
time = 1.50, size = 483, normalized size = 2.95 \begin {gather*} \frac {b d^{2} x^{3}}{3 f^{2}} + x \left (\frac {a d^{2}}{f^{2}} + \frac {2 b c d}{f^{2}} - \frac {2 b d^{2} e}{f^{3}}\right ) + \frac {x \left (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}\right )}{2 e^{2} f^{3} + 2 e f^{4} x^{2}} - \frac {\sqrt {- \frac {1}{e^{3} f^{7}}} \left (c f - d e\right ) \left (a c f^{2} + 3 a d e f + b c e f - 5 b d e^{2}\right ) \log {\left (- \frac {e^{2} f^{3} \sqrt {- \frac {1}{e^{3} f^{7}}} \left (c f - d e\right ) \left (a c f^{2} + 3 a d e f + b c e f - 5 b d e^{2}\right )}{a c^{2} f^{3} + 2 a c d e f^{2} - 3 a d^{2} e^{2} f + b c^{2} e f^{2} - 6 b c d e^{2} f + 5 b d^{2} e^{3}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{e^{3} f^{7}}} \left (c f - d e\right ) \left (a c f^{2} + 3 a d e f + b c e f - 5 b d e^{2}\right ) \log {\left (\frac {e^{2} f^{3} \sqrt {- \frac {1}{e^{3} f^{7}}} \left (c f - d e\right ) \left (a c f^{2} + 3 a d e f + b c e f - 5 b d e^{2}\right )}{a c^{2} f^{3} + 2 a c d e f^{2} - 3 a d^{2} e^{2} f + b c^{2} e f^{2} - 6 b c d e^{2} f + 5 b d^{2} e^{3}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d**2*x**3/(3*f**2) + x*(a*d**2/f**2 + 2*b*c*d/f**2 - 2*b*d**2*e/f**3) + 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)/(2*e**2*f**3 + 2*e*f**4*x**2) - sqrt(-1/(e**3*f**7
))*(c*f - d*e)*(a*c*f**2 + 3*a*d*e*f + b*c*e*f - 5*b*d*e**2)*log(-e**2*f**3*sqrt(-1/(e**3*f**7))*(c*f - d*e)*(
a*c*f**2 + 3*a*d*e*f + b*c*e*f - 5*b*d*e**2)/(a*c**2*f**3 + 2*a*c*d*e*f**2 - 3*a*d**2*e**2*f + b*c**2*e*f**2 -
6*b*c*d*e**2*f + 5*b*d**2*e**3) + x)/4 + sqrt(-1/(e**3*f**7))*(c*f - d*e)*(a*c*f**2 + 3*a*d*e*f + b*c*e*f - 5
*b*d*e**2)*log(e**2*f**3*sqrt(-1/(e**3*f**7))*(c*f - d*e)*(a*c*f**2 + 3*a*d*e*f + b*c*e*f - 5*b*d*e**2)/(a*c**
2*f**3 + 2*a*c*d*e*f**2 - 3*a*d**2*e**2*f + b*c**2*e*f**2 - 6*b*c*d*e**2*f + 5*b*d**2*e**3) + x)/4

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Giac [A]
time = 0.67, size = 195, normalized size = 1.19 \begin {gather*} \frac {{\left (a c^{2} f^{3} + b c^{2} f^{2} e + 2 \, a c d f^{2} e - 6 \, b c d f e^{2} - 3 \, a d^{2} f e^{2} + 5 \, b d^{2} e^{3}\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {3}{2}\right )}}{2 \, f^{\frac {7}{2}}} + \frac {{\left (a c^{2} f^{3} x - b c^{2} f^{2} x e - 2 \, a c d f^{2} x e + 2 \, b c d f x e^{2} + a d^{2} f x e^{2} - b d^{2} x e^{3}\right )} e^{\left (-1\right )}}{2 \, {\left (f x^{2} + e\right )} f^{3}} + \frac {b d^{2} f^{4} x^{3} + 6 \, b c d f^{4} x + 3 \, a d^{2} f^{4} x - 6 \, b d^{2} f^{3} x e}{3 \, f^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*c^2*f^3 + b*c^2*f^2*e + 2*a*c*d*f^2*e - 6*b*c*d*f*e^2 - 3*a*d^2*f*e^2 + 5*b*d^2*e^3)*arctan(sqrt(f)*x*e
^(-1/2))*e^(-3/2)/f^(7/2) + 1/2*(a*c^2*f^3*x - b*c^2*f^2*x*e - 2*a*c*d*f^2*x*e + 2*b*c*d*f*x*e^2 + a*d^2*f*x*e
^2 - b*d^2*x*e^3)*e^(-1)/((f*x^2 + e)*f^3) + 1/3*(b*d^2*f^4*x^3 + 6*b*c*d*f^4*x + 3*a*d^2*f^4*x - 6*b*d^2*f^3*
x*e)/f^6

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Mupad [B]
time = 0.17, size = 257, normalized size = 1.57 \begin {gather*} x\,\left (\frac {a\,d^2+2\,b\,c\,d}{f^2}-\frac {2\,b\,d^2\,e}{f^3}\right )+\frac {b\,d^2\,x^3}{3\,f^2}+\frac {x\,\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 )}{2\,e\,\left (f^4\,x^2+e\,f^3\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x\,\left (c\,f-d\,e\right )\,\left (a\,c\,f^2-5\,b\,d\,e^2+3\,a\,d\,e\,f+b\,c\,e\,f\right )}{\sqrt {e}\,\left (b\,c^2\,e\,f^2+a\,c^2\,f^3-6\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2+5\,b\,d^2\,e^3-3\,a\,d^2\,e^2\,f\right )}\right )\,\left (c\,f-d\,e\right )\,\left (a\,c\,f^2-5\,b\,d\,e^2+3\,a\,d\,e\,f+b\,c\,e\,f\right )}{2\,e^{3/2}\,f^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((a*d^2 + 2*b*c*d)/f^2 - (2*b*d^2*e)/f^3) + (b*d^2*x^3)/(3*f^2) + (x*(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))/(2*e*(e*f^3 + f^4*x^2)) + (atan((f^(1/2)*x*(c*f - d*e)*(a*c*f^2
- 5*b*d*e^2 + 3*a*d*e*f + b*c*e*f))/(e^(1/2)*(a*c^2*f^3 + 5*b*d^2*e^3 - 3*a*d^2*e^2*f + b*c^2*e*f^2 + 2*a*c*d*
e*f^2 - 6*b*c*d*e^2*f)))*(c*f - d*e)*(a*c*f^2 - 5*b*d*e^2 + 3*a*d*e*f + b*c*e*f))/(2*e^(3/2)*f^(7/2))

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