3.1.15 \(\int \frac {(a+b x^2) (c+d x^2)^2}{(e+f x^2)^4} \, dx\) [15]

Optimal. Leaf size=240 \[ -\frac {(b e-a f) x \left (c+d x^2\right )^2}{6 e f \left (e+f x^2\right )^3}-\frac {(d e (5 b e+a f)-c f (b e+5 a f)) x \left (c+d x^2\right )}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {\left (a f \left (3 d^2 e^2+4 c d e f-15 c^2 f^2\right )+b e \left (15 d^2 e^2-4 c d e f-3 c^2 f^2\right )\right ) x}{48 e^3 f^3 \left (e+f x^2\right )}+\frac {\left (b e \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{7/2}} \]

[Out]

-1/6*(-a*f+b*e)*x*(d*x^2+c)^2/e/f/(f*x^2+e)^3-1/24*(d*e*(a*f+5*b*e)-c*f*(5*a*f+b*e))*x*(d*x^2+c)/e^2/f^2/(f*x^
2+e)^2-1/48*(a*f*(-15*c^2*f^2+4*c*d*e*f+3*d^2*e^2)+b*e*(-3*c^2*f^2-4*c*d*e*f+15*d^2*e^2))*x/e^3/f^3/(f*x^2+e)+
1/16*(b*e*(c^2*f^2+2*c*d*e*f+5*d^2*e^2)+a*f*(5*c^2*f^2+2*c*d*e*f+d^2*e^2))*arctan(x*f^(1/2)/e^(1/2))/e^(7/2)/f
^(7/2)

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Rubi [A]
time = 0.18, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {540, 393, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a f \left (5 c^2 f^2+2 c d e f+d^2 e^2\right )+b e \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{16 e^{7/2} f^{7/2}}-\frac {x \left (a f \left (-15 c^2 f^2+4 c d e f+3 d^2 e^2\right )+b e \left (-3 c^2 f^2-4 c d e f+15 d^2 e^2\right )\right )}{48 e^3 f^3 \left (e+f x^2\right )}-\frac {x \left (c+d x^2\right ) (d e (a f+5 b e)-c f (5 a f+b e))}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {x \left (c+d x^2\right )^2 (b e-a 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)^2)/(e + f*x^2)^4,x]

[Out]

-1/6*((b*e - a*f)*x*(c + d*x^2)^2)/(e*f*(e + f*x^2)^3) - ((d*e*(5*b*e + a*f) - c*f*(b*e + 5*a*f))*x*(c + d*x^2
))/(24*e^2*f^2*(e + f*x^2)^2) - ((a*f*(3*d^2*e^2 + 4*c*d*e*f - 15*c^2*f^2) + b*e*(15*d^2*e^2 - 4*c*d*e*f - 3*c
^2*f^2))*x)/(48*e^3*f^3*(e + f*x^2)) + ((b*e*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5*
c^2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*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 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 )^2}{\left (e+f x^2\right )^4} \, dx &=-\frac {(b e-a f) x \left (c+d x^2\right )^2}{6 e f \left (e+f x^2\right )^3}-\frac {\int \frac {\left (c+d x^2\right ) \left (-c (b e+5 a f)-d (5 b e+a f) x^2\right )}{\left (e+f x^2\right )^3} \, dx}{6 e f}\\ &=-\frac {(b e-a f) x \left (c+d x^2\right )^2}{6 e f \left (e+f x^2\right )^3}-\frac {(d e (5 b e+a f)-c f (b e+5 a f)) x \left (c+d x^2\right )}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {\int \frac {c (d e (5 b e+a f)+3 c f (b e+5 a f))+d (b e (15 d e+c f)+a f (3 d e+5 c f)) x^2}{\left (e+f x^2\right )^2} \, dx}{24 e^2 f^2}\\ &=-\frac {(b e-a f) x \left (c+d x^2\right )^2}{6 e f \left (e+f x^2\right )^3}-\frac {(d e (5 b e+a f)-c f (b e+5 a f)) x \left (c+d x^2\right )}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {\left (a f \left (3 d^2 e^2+4 c d e f-15 c^2 f^2\right )+b e \left (15 d^2 e^2-4 c d e f-3 c^2 f^2\right )\right ) x}{48 e^3 f^3 \left (e+f x^2\right )}+\frac {\left (b e \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) \int \frac {1}{e+f x^2} \, dx}{16 e^3 f^3}\\ &=-\frac {(b e-a f) x \left (c+d x^2\right )^2}{6 e f \left (e+f x^2\right )^3}-\frac {(d e (5 b e+a f)-c f (b e+5 a f)) x \left (c+d x^2\right )}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {\left (a f \left (3 d^2 e^2+4 c d e f-15 c^2 f^2\right )+b e \left (15 d^2 e^2-4 c d e f-3 c^2 f^2\right )\right ) x}{48 e^3 f^3 \left (e+f x^2\right )}+\frac {\left (b e \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 242, normalized size = 1.01 \begin {gather*} -\frac {(b e-a f) (d e-c f)^2 x}{6 e f^3 \left (e+f x^2\right )^3}+\frac {(d e-c f) (b e (13 d e-c f)-a f (7 d e+5 c f)) x}{24 e^2 f^3 \left (e+f x^2\right )^2}+\frac {\left (b e \left (-11 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) x}{16 e^3 f^3 \left (e+f x^2\right )}+\frac {\left (b e \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((b*e - a*f)*(d*e - c*f)^2*x)/(e*f^3*(e + f*x^2)^3) + ((d*e - c*f)*(b*e*(13*d*e - c*f) - a*f*(7*d*e + 5*c
*f))*x)/(24*e^2*f^3*(e + f*x^2)^2) + ((b*e*(-11*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5*
c^2*f^2))*x)/(16*e^3*f^3*(e + f*x^2)) + ((b*e*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5
*c^2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(7/2))

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Maple [A]
time = 0.16, size = 289, normalized size = 1.20

method	result	size
default	\(\frac {\frac {\left (5 c^{2} a \,f^{3}+2 a c d e \,f^{2}+a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}+2 b c d \,e^{2} f -11 b \,d^{2} e^{3}\right ) x^{5}}{16 e^{3} f}+\frac {\left (5 c^{2} a \,f^{3}+2 a c d e \,f^{2}-a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}-2 b c d \,e^{2} f -5 b \,d^{2} e^{3}\right ) x^{3}}{6 e^{2} f^{2}}+\frac {\left (11 c^{2} a \,f^{3}-2 a c d e \,f^{2}-a \,d^{2} e^{2} f -b \,c^{2} e \,f^{2}-2 b c d \,e^{2} f -5 b \,d^{2} e^{3}\right ) x}{16 f^{3} e}}{\left (f \,x^{2}+e \right )^{3}}+\frac {\left (5 c^{2} a \,f^{3}+2 a c d e \,f^{2}+a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}+2 b c d \,e^{2} f +5 b \,d^{2} e^{3}\right ) \arctan \left (\frac {f x}{\sqrt {f e}}\right )}{16 e^{3} f^{3} \sqrt {f e}}\)	\(289\)
risch	\(\frac {\frac {\left (5 c^{2} a \,f^{3}+2 a c d e \,f^{2}+a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}+2 b c d \,e^{2} f -11 b \,d^{2} e^{3}\right ) x^{5}}{16 e^{3} f}+\frac {\left (5 c^{2} a \,f^{3}+2 a c d e \,f^{2}-a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}-2 b c d \,e^{2} f -5 b \,d^{2} e^{3}\right ) x^{3}}{6 e^{2} f^{2}}+\frac {\left (11 c^{2} a \,f^{3}-2 a c d e \,f^{2}-a \,d^{2} e^{2} f -b \,c^{2} e \,f^{2}-2 b c d \,e^{2} f -5 b \,d^{2} e^{3}\right ) x}{16 f^{3} e}}{\left (f \,x^{2}+e \right )^{3}}-\frac {5 \ln \left (f x +\sqrt {-f e}\right ) c^{2} a}{32 \sqrt {-f e}\, e^{3}}-\frac {\ln \left (f x +\sqrt {-f e}\right ) a c d}{16 \sqrt {-f e}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-f e}\right ) a \,d^{2}}{32 \sqrt {-f e}\, f^{2} e}-\frac {\ln \left (f x +\sqrt {-f e}\right ) b \,c^{2}}{32 \sqrt {-f e}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-f e}\right ) b c d}{16 \sqrt {-f e}\, f^{2} e}-\frac {5 \ln \left (f x +\sqrt {-f e}\right ) b \,d^{2}}{32 \sqrt {-f e}\, f^{3}}+\frac {5 \ln \left (-f x +\sqrt {-f e}\right ) c^{2} a}{32 \sqrt {-f e}\, e^{3}}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) a c d}{16 \sqrt {-f e}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) a \,d^{2}}{32 \sqrt {-f e}\, f^{2} e}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) b \,c^{2}}{32 \sqrt {-f e}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) b c d}{16 \sqrt {-f e}\, f^{2} e}+\frac {5 \ln \left (-f x +\sqrt {-f e}\right ) b \,d^{2}}{32 \sqrt {-f e}\, f^{3}}\)	\(550\)

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)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 301, normalized size = 1.25 \begin {gather*} \frac {3 \, {\left (5 \, a c^{2} f^{5} - 11 \, b d^{2} f^{2} e^{3} + {\left (b c^{2} e + 2 \, a c d e\right )} f^{4} + {\left (2 \, b c d e^{2} + a d^{2} e^{2}\right )} f^{3}\right )} x^{5} + 8 \, {\left (5 \, a c^{2} f^{4} e - 5 \, b d^{2} f e^{4} + {\left (b c^{2} e^{2} + 2 \, a c d e^{2}\right )} f^{3} - {\left (2 \, b c d e^{3} + a d^{2} e^{3}\right )} f^{2}\right )} x^{3} + 3 \, {\left (11 \, a c^{2} f^{3} e^{2} - 5 \, b d^{2} e^{5} - {\left (b c^{2} e^{3} + 2 \, a c d e^{3}\right )} f^{2} - {\left (2 \, b c d e^{4} + a d^{2} e^{4}\right )} f\right )} x}{48 \, {\left (f^{6} x^{6} e^{3} + 3 \, f^{5} x^{4} e^{4} + 3 \, f^{4} x^{2} e^{5} + f^{3} e^{6}\right )}} + \frac {{\left (5 \, a c^{2} f^{3} + 5 \, 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 )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {7}{2}\right )}}{16 \, f^{\frac {7}{2}}} \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)^4,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (228) = 456\).
time = 1.22, size = 1033, normalized size = 4.30 \begin {gather*} \left [\frac {30 \, a c^{2} f^{6} x^{5} e - 30 \, b d^{2} f x e^{6} - 3 \, {\left (5 \, a c^{2} f^{6} x^{6} + 5 \, b d^{2} e^{6} + {\left (15 \, b d^{2} f x^{2} + {\left (2 \, b c d + a d^{2}\right )} f\right )} e^{5} + {\left (15 \, b d^{2} f^{2} x^{4} + 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^{4} + {\left (5 \, b d^{2} f^{3} x^{6} + 3 \, {\left (2 \, b c d + a d^{2}\right )} f^{3} x^{4} + 5 \, a c^{2} f^{3} + 3 \, {\left (b c^{2} + 2 \, a c d\right )} f^{3} x^{2}\right )} e^{3} + {\left ({\left (2 \, b c d + a d^{2}\right )} f^{4} x^{6} + 15 \, a c^{2} f^{4} x^{2} + 3 \, {\left (b c^{2} + 2 \, a c d\right )} f^{4} x^{4}\right )} e^{2} + {\left (15 \, a c^{2} f^{5} x^{4} + {\left (b c^{2} + 2 \, a c d\right )} f^{5} x^{6}\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 (40 \, b d^{2} f^{2} x^{3} + 3 \, {\left (2 \, b c d + a d^{2}\right )} f^{2} x\right )} e^{5} - 2 \, {\left (33 \, b d^{2} f^{3} x^{5} + 8 \, {\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^{4} + 2 \, {\left (3 \, {\left (2 \, b c d + a d^{2}\right )} f^{4} x^{5} + 33 \, a c^{2} f^{4} x + 8 \, {\left (b c^{2} + 2 \, a c d\right )} f^{4} x^{3}\right )} e^{3} + 2 \, {\left (40 \, a c^{2} f^{5} x^{3} + 3 \, {\left (b c^{2} + 2 \, a c d\right )} f^{5} x^{5}\right )} e^{2}}{96 \, {\left (f^{7} x^{6} e^{4} + 3 \, f^{6} x^{4} e^{5} + 3 \, f^{5} x^{2} e^{6} + f^{4} e^{7}\right )}}, \frac {15 \, a c^{2} f^{6} x^{5} e - 15 \, b d^{2} f x e^{6} + 3 \, {\left (5 \, a c^{2} f^{6} x^{6} + 5 \, b d^{2} e^{6} + {\left (15 \, b d^{2} f x^{2} + {\left (2 \, b c d + a d^{2}\right )} f\right )} e^{5} + {\left (15 \, b d^{2} f^{2} x^{4} + 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^{4} + {\left (5 \, b d^{2} f^{3} x^{6} + 3 \, {\left (2 \, b c d + a d^{2}\right )} f^{3} x^{4} + 5 \, a c^{2} f^{3} + 3 \, {\left (b c^{2} + 2 \, a c d\right )} f^{3} x^{2}\right )} e^{3} + {\left ({\left (2 \, b c d + a d^{2}\right )} f^{4} x^{6} + 15 \, a c^{2} f^{4} x^{2} + 3 \, {\left (b c^{2} + 2 \, a c d\right )} f^{4} x^{4}\right )} e^{2} + {\left (15 \, a c^{2} f^{5} x^{4} + {\left (b c^{2} + 2 \, a c d\right )} f^{5} x^{6}\right )} e\right )} \sqrt {f} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {1}{2}} - {\left (40 \, b d^{2} f^{2} x^{3} + 3 \, {\left (2 \, b c d + a d^{2}\right )} f^{2} x\right )} e^{5} - {\left (33 \, b d^{2} f^{3} x^{5} + 8 \, {\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^{4} + {\left (3 \, {\left (2 \, b c d + a d^{2}\right )} f^{4} x^{5} + 33 \, a c^{2} f^{4} x + 8 \, {\left (b c^{2} + 2 \, a c d\right )} f^{4} x^{3}\right )} e^{3} + {\left (40 \, a c^{2} f^{5} x^{3} + 3 \, {\left (b c^{2} + 2 \, a c d\right )} f^{5} x^{5}\right )} e^{2}}{48 \, {\left (f^{7} x^{6} e^{4} + 3 \, f^{6} x^{4} e^{5} + 3 \, f^{5} x^{2} e^{6} + f^{4} 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)^2/(f*x^2+e)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**4,x)

[Out]

Timed out

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Giac [A]
time = 0.63, size = 311, normalized size = 1.30 \begin {gather*} \frac {{\left (5 \, a c^{2} f^{3} + b c^{2} f^{2} e + 2 \, a c d f^{2} e + 2 \, b c d f e^{2} + 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 {7}{2}\right )}}{16 \, f^{\frac {7}{2}}} + \frac {{\left (15 \, a c^{2} f^{5} x^{5} + 3 \, b c^{2} f^{4} x^{5} e + 6 \, a c d f^{4} x^{5} e + 6 \, b c d f^{3} x^{5} e^{2} + 3 \, a d^{2} f^{3} x^{5} e^{2} - 33 \, b d^{2} f^{2} x^{5} e^{3} + 40 \, a c^{2} f^{4} x^{3} e + 8 \, b c^{2} f^{3} x^{3} e^{2} + 16 \, a c d f^{3} x^{3} e^{2} - 16 \, b c d f^{2} x^{3} e^{3} - 8 \, a d^{2} f^{2} x^{3} e^{3} - 40 \, b d^{2} f x^{3} e^{4} + 33 \, a c^{2} f^{3} x e^{2} - 3 \, b c^{2} f^{2} x e^{3} - 6 \, a c d f^{2} x e^{3} - 6 \, b c d f x e^{4} - 3 \, a d^{2} f x e^{4} - 15 \, b d^{2} x e^{5}\right )} e^{\left (-3\right )}}{48 \, {\left (f x^{2} + e\right )}^{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)^4,x, algorithm="giac")

[Out]

1/16*(5*a*c^2*f^3 + b*c^2*f^2*e + 2*a*c*d*f^2*e + 2*b*c*d*f*e^2 + a*d^2*f*e^2 + 5*b*d^2*e^3)*arctan(sqrt(f)*x*
e^(-1/2))*e^(-7/2)/f^(7/2) + 1/48*(15*a*c^2*f^5*x^5 + 3*b*c^2*f^4*x^5*e + 6*a*c*d*f^4*x^5*e + 6*b*c*d*f^3*x^5*
e^2 + 3*a*d^2*f^3*x^5*e^2 - 33*b*d^2*f^2*x^5*e^3 + 40*a*c^2*f^4*x^3*e + 8*b*c^2*f^3*x^3*e^2 + 16*a*c*d*f^3*x^3
*e^2 - 16*b*c*d*f^2*x^3*e^3 - 8*a*d^2*f^2*x^3*e^3 - 40*b*d^2*f*x^3*e^4 + 33*a*c^2*f^3*x*e^2 - 3*b*c^2*f^2*x*e^
3 - 6*a*c*d*f^2*x*e^3 - 6*b*c*d*f*x*e^4 - 3*a*d^2*f*x*e^4 - 15*b*d^2*x*e^5)*e^(-3)/((f*x^2 + e)^3*f^3)

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Mupad [B]
time = 0.98, size = 303, normalized size = 1.26 \begin {gather*} \frac {\frac {x^3\,\left (b\,c^2\,e\,f^2+5\,a\,c^2\,f^3-2\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2-5\,b\,d^2\,e^3-a\,d^2\,e^2\,f\right )}{6\,e^2\,f^2}-\frac {x\,\left (b\,c^2\,e\,f^2-11\,a\,c^2\,f^3+2\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2+5\,b\,d^2\,e^3+a\,d^2\,e^2\,f\right )}{16\,e\,f^3}+\frac {x^5\,\left (b\,c^2\,e\,f^2+5\,a\,c^2\,f^3+2\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2-11\,b\,d^2\,e^3+a\,d^2\,e^2\,f\right )}{16\,e^3\,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 (b\,c^2\,e\,f^2+5\,a\,c^2\,f^3+2\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2+5\,b\,d^2\,e^3+a\,d^2\,e^2\,f\right )}{16\,e^{7/2}\,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)^4,x)

[Out]

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

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