3.1.8 \(\int \frac {(a+b x^2) (c+d x^2)}{(e+f x^2)^4} \, dx\) [8]

Optimal. Leaf size=171 \[ -\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{5/2}} \]

[Out]

-1/6*(-c*f+d*e)*x*(b*x^2+a)/e/f/(f*x^2+e)^3-1/24*(3*b*e*(c*f+d*e)-a*f*(5*c*f+d*e))*x/e^2/f^2/(f*x^2+e)^2+1/16*
(b*e*(c*f+d*e)+a*f*(5*c*f+d*e))*x/e^3/f^2/(f*x^2+e)+1/16*(b*e*(c*f+d*e)+a*f*(5*c*f+d*e))*arctan(x*f^(1/2)/e^(1
/2))/e^(7/2)/f^(5/2)

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Rubi [A]
time = 0.11, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {540, 393, 205, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (a f (5 c f+d e)+b e (c f+d e))}{16 e^{7/2} f^{5/2}}+\frac {x (a f (5 c f+d e)+b e (c f+d e))}{16 e^3 f^2 \left (e+f x^2\right )}-\frac {x (3 b e (c f+d e)-a f (5 c f+d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x]

[Out]

-1/6*((d*e - c*f)*x*(a + b*x^2))/(e*f*(e + f*x^2)^3) - ((3*b*e*(d*e + c*f) - a*f*(d*e + 5*c*f))*x)/(24*e^2*f^2
*(e + f*x^2)^2) + ((b*e*(d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(16*e^3*f^2*(e + f*x^2)) + ((b*e*(d*e + c*f) + a*f
*(d*e + 5*c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(5/2))

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx &=-\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {\int \frac {-a (d e+5 c f)-3 b (d e+c f) x^2}{\left (e+f x^2\right )^3} \, dx}{6 e f}\\ &=-\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \int \frac {1}{\left (e+f x^2\right )^2} \, dx}{8 e^2 f^2}\\ &=-\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \int \frac {1}{e+f x^2} \, dx}{16 e^3 f^2}\\ &=-\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 171, normalized size = 1.00 \begin {gather*} \frac {(b e-a f) (d e-c f) x}{6 e f^2 \left (e+f x^2\right )^3}+\frac {(b e (-7 d e+c f)+a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x]

[Out]

((b*e - a*f)*(d*e - c*f)*x)/(6*e*f^2*(e + f*x^2)^3) + ((b*e*(-7*d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(24*e^2*f^2
*(e + f*x^2)^2) + ((b*e*(d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(16*e^3*f^2*(e + f*x^2)) + ((b*e*(d*e + c*f) + a*f
*(d*e + 5*c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(5/2))

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Maple [A]
time = 0.14, size = 163, normalized size = 0.95

method	result	size
default	\(\frac {\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) x^{5}}{16 e^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f -b d \,e^{2}\right ) x^{3}}{6 e^{2} f}+\frac {\left (11 a c \,f^{2}-a d e f -b c e f -b d \,e^{2}\right ) x}{16 e \,f^{2}}}{\left (f \,x^{2}+e \right )^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) \arctan \left (\frac {f x}{\sqrt {f e}}\right )}{16 e^{3} f^{2} \sqrt {f e}}\)	\(163\)
risch	\(\frac {\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) x^{5}}{16 e^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f -b d \,e^{2}\right ) x^{3}}{6 e^{2} f}+\frac {\left (11 a c \,f^{2}-a d e f -b c e f -b d \,e^{2}\right ) x}{16 e \,f^{2}}}{\left (f \,x^{2}+e \right )^{3}}-\frac {5 \ln \left (f x +\sqrt {-f e}\right ) a c}{32 \sqrt {-f e}\, e^{3}}-\frac {\ln \left (f x +\sqrt {-f e}\right ) a d}{32 \sqrt {-f e}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-f e}\right ) b c}{32 \sqrt {-f e}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-f e}\right ) b d}{32 \sqrt {-f e}\, f^{2} e}+\frac {5 \ln \left (-f x +\sqrt {-f e}\right ) a c}{32 \sqrt {-f e}\, e^{3}}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) a d}{32 \sqrt {-f e}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) b c}{32 \sqrt {-f e}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) b d}{32 \sqrt {-f e}\, f^{2} e}\)	\(331\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x,method=_RETURNVERBOSE)

[Out]

(1/16*(5*a*c*f^2+a*d*e*f+b*c*e*f+b*d*e^2)/e^3*x^5+1/6*(5*a*c*f^2+a*d*e*f+b*c*e*f-b*d*e^2)/e^2/f*x^3+1/16*(11*a
*c*f^2-a*d*e*f-b*c*e*f-b*d*e^2)/e/f^2*x)/(f*x^2+e)^3+1/16*(5*a*c*f^2+a*d*e*f+b*c*e*f+b*d*e^2)/e^3/f^2/(f*e)^(1
/2)*arctan(f*x/(f*e)^(1/2))

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Maxima [A]
time = 0.52, size = 189, normalized size = 1.11 \begin {gather*} \frac {{\left (5 \, a c f^{2} + b d e^{2} + {\left (b c e + a d e\right )} f\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {7}{2}\right )}}{16 \, f^{\frac {5}{2}}} + \frac {3 \, {\left (5 \, a c f^{4} + b d f^{2} e^{2} + {\left (b c e + a d e\right )} f^{3}\right )} x^{5} + 8 \, {\left (5 \, a c f^{3} e - b d f e^{3} + {\left (b c e^{2} + a d e^{2}\right )} f^{2}\right )} x^{3} + 3 \, {\left (11 \, a c f^{2} e^{2} - b d e^{4} - {\left (b c e^{3} + a d e^{3}\right )} f\right )} x}{48 \, {\left (f^{5} x^{6} e^{3} + 3 \, f^{4} x^{4} e^{4} + 3 \, f^{3} x^{2} e^{5} + f^{2} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x, algorithm="maxima")

[Out]

1/16*(5*a*c*f^2 + b*d*e^2 + (b*c*e + a*d*e)*f)*arctan(sqrt(f)*x*e^(-1/2))*e^(-7/2)/f^(5/2) + 1/48*(3*(5*a*c*f^
4 + b*d*f^2*e^2 + (b*c*e + a*d*e)*f^3)*x^5 + 8*(5*a*c*f^3*e - b*d*f*e^3 + (b*c*e^2 + a*d*e^2)*f^2)*x^3 + 3*(11
*a*c*f^2*e^2 - b*d*e^4 - (b*c*e^3 + a*d*e^3)*f)*x)/(f^5*x^6*e^3 + 3*f^4*x^4*e^4 + 3*f^3*x^2*e^5 + f^2*e^6)

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Fricas [A]
time = 0.94, size = 647, normalized size = 3.78 \begin {gather*} \left [\frac {30 \, a c f^{5} x^{5} e - 6 \, b d f x e^{5} - 3 \, {\left (5 \, a c f^{5} x^{6} + b d e^{5} + {\left (3 \, b d f x^{2} + {\left (b c + a d\right )} f\right )} e^{4} + {\left (3 \, b d f^{2} x^{4} + 3 \, {\left (b c + a d\right )} f^{2} x^{2} + 5 \, a c f^{2}\right )} e^{3} + {\left (b d f^{3} x^{6} + 3 \, {\left (b c + a d\right )} f^{3} x^{4} + 15 \, a c f^{3} x^{2}\right )} e^{2} + {\left ({\left (b c + a d\right )} f^{4} x^{6} + 15 \, a c f^{4} x^{4}\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 (8 \, b d f^{2} x^{3} + 3 \, {\left (b c + a d\right )} f^{2} x\right )} e^{4} + 2 \, {\left (3 \, b d f^{3} x^{5} + 8 \, {\left (b c + a d\right )} f^{3} x^{3} + 33 \, a c f^{3} x\right )} e^{3} + 2 \, {\left (3 \, {\left (b c + a d\right )} f^{4} x^{5} + 40 \, a c f^{4} x^{3}\right )} e^{2}}{96 \, {\left (f^{6} x^{6} e^{4} + 3 \, f^{5} x^{4} e^{5} + 3 \, f^{4} x^{2} e^{6} + f^{3} e^{7}\right )}}, \frac {15 \, a c f^{5} x^{5} e - 3 \, b d f x e^{5} + 3 \, {\left (5 \, a c f^{5} x^{6} + b d e^{5} + {\left (3 \, b d f x^{2} + {\left (b c + a d\right )} f\right )} e^{4} + {\left (3 \, b d f^{2} x^{4} + 3 \, {\left (b c + a d\right )} f^{2} x^{2} + 5 \, a c f^{2}\right )} e^{3} + {\left (b d f^{3} x^{6} + 3 \, {\left (b c + a d\right )} f^{3} x^{4} + 15 \, a c f^{3} x^{2}\right )} e^{2} + {\left ({\left (b c + a d\right )} f^{4} x^{6} + 15 \, a c f^{4} x^{4}\right )} e\right )} \sqrt {f} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {1}{2}} - {\left (8 \, b d f^{2} x^{3} + 3 \, {\left (b c + a d\right )} f^{2} x\right )} e^{4} + {\left (3 \, b d f^{3} x^{5} + 8 \, {\left (b c + a d\right )} f^{3} x^{3} + 33 \, a c f^{3} x\right )} e^{3} + {\left (3 \, {\left (b c + a d\right )} f^{4} x^{5} + 40 \, a c f^{4} x^{3}\right )} e^{2}}{48 \, {\left (f^{6} x^{6} e^{4} + 3 \, f^{5} x^{4} e^{5} + 3 \, f^{4} x^{2} e^{6} + f^{3} e^{7}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x, algorithm="fricas")

[Out]

[1/96*(30*a*c*f^5*x^5*e - 6*b*d*f*x*e^5 - 3*(5*a*c*f^5*x^6 + b*d*e^5 + (3*b*d*f*x^2 + (b*c + a*d)*f)*e^4 + (3*
b*d*f^2*x^4 + 3*(b*c + a*d)*f^2*x^2 + 5*a*c*f^2)*e^3 + (b*d*f^3*x^6 + 3*(b*c + a*d)*f^3*x^4 + 15*a*c*f^3*x^2)*
e^2 + ((b*c + a*d)*f^4*x^6 + 15*a*c*f^4*x^4)*e)*sqrt(-f*e)*log((f*x^2 - 2*sqrt(-f*e)*x - e)/(f*x^2 + e)) - 2*(
8*b*d*f^2*x^3 + 3*(b*c + a*d)*f^2*x)*e^4 + 2*(3*b*d*f^3*x^5 + 8*(b*c + a*d)*f^3*x^3 + 33*a*c*f^3*x)*e^3 + 2*(3
*(b*c + a*d)*f^4*x^5 + 40*a*c*f^4*x^3)*e^2)/(f^6*x^6*e^4 + 3*f^5*x^4*e^5 + 3*f^4*x^2*e^6 + f^3*e^7), 1/48*(15*
a*c*f^5*x^5*e - 3*b*d*f*x*e^5 + 3*(5*a*c*f^5*x^6 + b*d*e^5 + (3*b*d*f*x^2 + (b*c + a*d)*f)*e^4 + (3*b*d*f^2*x^
4 + 3*(b*c + a*d)*f^2*x^2 + 5*a*c*f^2)*e^3 + (b*d*f^3*x^6 + 3*(b*c + a*d)*f^3*x^4 + 15*a*c*f^3*x^2)*e^2 + ((b*
c + a*d)*f^4*x^6 + 15*a*c*f^4*x^4)*e)*sqrt(f)*arctan(sqrt(f)*x*e^(-1/2))*e^(1/2) - (8*b*d*f^2*x^3 + 3*(b*c + a
*d)*f^2*x)*e^4 + (3*b*d*f^3*x^5 + 8*(b*c + a*d)*f^3*x^3 + 33*a*c*f^3*x)*e^3 + (3*(b*c + a*d)*f^4*x^5 + 40*a*c*
f^4*x^3)*e^2)/(f^6*x^6*e^4 + 3*f^5*x^4*e^5 + 3*f^4*x^2*e^6 + f^3*e^7)]

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Sympy [A]
time = 3.55, size = 313, normalized size = 1.83 \begin {gather*} - \frac {\sqrt {- \frac {1}{e^{7} f^{5}}} \cdot \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log {\left (- e^{4} f^{2} \sqrt {- \frac {1}{e^{7} f^{5}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{e^{7} f^{5}}} \cdot \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log {\left (e^{4} f^{2} \sqrt {- \frac {1}{e^{7} f^{5}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a c f^{4} + 3 a d e f^{3} + 3 b c e f^{3} + 3 b d e^{2} f^{2}\right ) + x^{3} \cdot \left (40 a c e f^{3} + 8 a d e^{2} f^{2} + 8 b c e^{2} f^{2} - 8 b d e^{3} f\right ) + x \left (33 a c e^{2} f^{2} - 3 a d e^{3} f - 3 b c e^{3} f - 3 b d e^{4}\right )}{48 e^{6} f^{2} + 144 e^{5} f^{3} x^{2} + 144 e^{4} f^{4} x^{4} + 48 e^{3} f^{5} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(d*x**2+c)/(f*x**2+e)**4,x)

[Out]

-sqrt(-1/(e**7*f**5))*(5*a*c*f**2 + a*d*e*f + b*c*e*f + b*d*e**2)*log(-e**4*f**2*sqrt(-1/(e**7*f**5)) + x)/32
+ sqrt(-1/(e**7*f**5))*(5*a*c*f**2 + a*d*e*f + b*c*e*f + b*d*e**2)*log(e**4*f**2*sqrt(-1/(e**7*f**5)) + x)/32
+ (x**5*(15*a*c*f**4 + 3*a*d*e*f**3 + 3*b*c*e*f**3 + 3*b*d*e**2*f**2) + x**3*(40*a*c*e*f**3 + 8*a*d*e**2*f**2
+ 8*b*c*e**2*f**2 - 8*b*d*e**3*f) + x*(33*a*c*e**2*f**2 - 3*a*d*e**3*f - 3*b*c*e**3*f - 3*b*d*e**4))/(48*e**6*
f**2 + 144*e**5*f**3*x**2 + 144*e**4*f**4*x**4 + 48*e**3*f**5*x**6)

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Giac [A]
time = 0.69, size = 184, normalized size = 1.08 \begin {gather*} \frac {{\left (5 \, a c f^{2} + b c f e + a d f e + b d e^{2}\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {7}{2}\right )}}{16 \, f^{\frac {5}{2}}} + \frac {{\left (15 \, a c f^{4} x^{5} + 3 \, b c f^{3} x^{5} e + 3 \, a d f^{3} x^{5} e + 3 \, b d f^{2} x^{5} e^{2} + 40 \, a c f^{3} x^{3} e + 8 \, b c f^{2} x^{3} e^{2} + 8 \, a d f^{2} x^{3} e^{2} - 8 \, b d f x^{3} e^{3} + 33 \, a c f^{2} x e^{2} - 3 \, b c f x e^{3} - 3 \, a d f x e^{3} - 3 \, b d x e^{4}\right )} e^{\left (-3\right )}}{48 \, {\left (f x^{2} + e\right )}^{3} f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x, algorithm="giac")

[Out]

1/16*(5*a*c*f^2 + b*c*f*e + a*d*f*e + b*d*e^2)*arctan(sqrt(f)*x*e^(-1/2))*e^(-7/2)/f^(5/2) + 1/48*(15*a*c*f^4*
x^5 + 3*b*c*f^3*x^5*e + 3*a*d*f^3*x^5*e + 3*b*d*f^2*x^5*e^2 + 40*a*c*f^3*x^3*e + 8*b*c*f^2*x^3*e^2 + 8*a*d*f^2
*x^3*e^2 - 8*b*d*f*x^3*e^3 + 33*a*c*f^2*x*e^2 - 3*b*c*f*x*e^3 - 3*a*d*f*x*e^3 - 3*b*d*x*e^4)*e^(-3)/((f*x^2 +
e)^3*f^2)

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Mupad [B]
time = 0.18, size = 176, normalized size = 1.03 \begin {gather*} \frac {\frac {x^5\,\left (5\,a\,c\,f^2+b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e^3}-\frac {x\,\left (b\,d\,e^2-11\,a\,c\,f^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e\,f^2}+\frac {x^3\,\left (5\,a\,c\,f^2-b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{6\,e^2\,f}}{e^3+3\,e^2\,f\,x^2+3\,e\,f^2\,x^4+f^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {e}}\right )\,\left (5\,a\,c\,f^2+b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e^{7/2}\,f^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x)

[Out]

((x^5*(5*a*c*f^2 + b*d*e^2 + a*d*e*f + b*c*e*f))/(16*e^3) - (x*(b*d*e^2 - 11*a*c*f^2 + a*d*e*f + b*c*e*f))/(16
*e*f^2) + (x^3*(5*a*c*f^2 - b*d*e^2 + a*d*e*f + b*c*e*f))/(6*e^2*f))/(e^3 + f^3*x^6 + 3*e^2*f*x^2 + 3*e*f^2*x^
4) + (atan((f^(1/2)*x)/e^(1/2))*(5*a*c*f^2 + b*d*e^2 + a*d*e*f + b*c*e*f))/(16*e^(7/2)*f^(5/2))

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