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

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

[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 = 0.23, 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 = {526, 528, 388, 205} \[ \frac {(d e-c f) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (b e (5 d e-c f)-a f (c f+3 d e))}{2 e^{3/2} f^{7/2}}+\frac {d x \left (c+d x^2\right ) (5 b e-3 a f)}{6 e f^2}-\frac {d x (b e (15 d e-13 c f)-3 a f (3 d e-c f))}{6 e f^3}-\frac {x \left (c+d x^2\right )^2 (b e-a f)}{2 e f \left (e+f x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(b*e*(15*d*e - 13*c*f) - 3*a*f*(3*d*e - c*f))*x)/(6*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))*ArcT
an[(Sqrt[f]*x)/Sqrt[e]])/(2*e^(3/2)*f^(7/2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 388

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 526

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 528

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 = 0.10, size = 134, normalized size = 0.82 \[ \frac {(d e-c f) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (b e (5 d e-c f)-a f (c f+3 d e))}{2 e^{3/2} f^{7/2}}-\frac {x (b e-a f) (d e-c f)^2}{2 e f^3 \left (e+f x^2\right )}+\frac {d x (a d f+2 b c f-2 b d e)}{f^3}+\frac {b d^2 x^3}{3 f^2} \]

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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fricas [A]  time = 0.84, size = 552, normalized size = 3.37 \[ \left [\frac {4 \, b d^{2} e^{2} f^{3} x^{5} - 4 \, {\left (5 \, b d^{2} e^{3} f^{2} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{2} f^{3}\right )} x^{3} - 3 \, {\left (5 \, b d^{2} e^{4} + a c^{2} e f^{3} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{3} f + {\left (b c^{2} + 2 \, a c d\right )} e^{2} f^{2} + {\left (5 \, b d^{2} e^{3} f + a c^{2} f^{4} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{2} f^{2} + {\left (b c^{2} + 2 \, a c d\right )} e f^{3}\right )} x^{2}\right )} \sqrt {-e f} \log \left (\frac {f x^{2} - 2 \, \sqrt {-e f} x - e}{f x^{2} + e}\right ) - 6 \, {\left (5 \, b d^{2} e^{4} f - a c^{2} e f^{4} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{3} f^{2} + {\left (b c^{2} + 2 \, a c d\right )} e^{2} f^{3}\right )} x}{12 \, {\left (e^{2} f^{5} x^{2} + e^{3} f^{4}\right )}}, \frac {2 \, b d^{2} e^{2} f^{3} x^{5} - 2 \, {\left (5 \, b d^{2} e^{3} f^{2} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{2} f^{3}\right )} x^{3} + 3 \, {\left (5 \, b d^{2} e^{4} + a c^{2} e f^{3} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{3} f + {\left (b c^{2} + 2 \, a c d\right )} e^{2} f^{2} + {\left (5 \, b d^{2} e^{3} f + a c^{2} f^{4} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{2} f^{2} + {\left (b c^{2} + 2 \, a c d\right )} e f^{3}\right )} x^{2}\right )} \sqrt {e f} \arctan \left (\frac {\sqrt {e f} x}{e}\right ) - 3 \, {\left (5 \, b d^{2} e^{4} f - a c^{2} e f^{4} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{3} f^{2} + {\left (b c^{2} + 2 \, a c d\right )} e^{2} f^{3}\right )} x}{6 \, {\left (e^{2} f^{5} x^{2} + e^{3} f^{4}\right )}}\right ] \]

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

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giac [A]  time = 0.33, size = 195, normalized size = 1.19 \[ \frac {{\left (a c^{2} f^{3} + b c^{2} f^{2} e + 2 \, a c d f^{2} e - 6 \, b c d f e^{2} - 3 \, a d^{2} f e^{2} + 5 \, b d^{2} e^{3}\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {3}{2}\right )}}{2 \, f^{\frac {7}{2}}} + \frac {{\left (a c^{2} f^{3} x - b c^{2} f^{2} x e - 2 \, a c d f^{2} x e + 2 \, b c d f x e^{2} + a d^{2} f x e^{2} - b d^{2} x e^{3}\right )} e^{\left (-1\right )}}{2 \, {\left (f x^{2} + e\right )} f^{3}} + \frac {b d^{2} f^{4} x^{3} + 6 \, b c d f^{4} x + 3 \, a d^{2} f^{4} x - 6 \, b d^{2} f^{3} x e}{3 \, f^{6}} \]

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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maple [B]  time = 0.01, size = 299, normalized size = 1.82 \[ \frac {b \,d^{2} x^{3}}{3 f^{2}}+\frac {a \,c^{2} x}{2 \left (f \,x^{2}+e \right ) e}+\frac {a \,c^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{2 \sqrt {e f}\, e}-\frac {a c d x}{\left (f \,x^{2}+e \right ) f}+\frac {a c d \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f}+\frac {a \,d^{2} e x}{2 \left (f \,x^{2}+e \right ) f^{2}}-\frac {3 a \,d^{2} e \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{2 \sqrt {e f}\, f^{2}}-\frac {b \,c^{2} x}{2 \left (f \,x^{2}+e \right ) f}+\frac {b \,c^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{2 \sqrt {e f}\, f}+\frac {b c d e x}{\left (f \,x^{2}+e \right ) f^{2}}-\frac {3 b c d e \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f^{2}}-\frac {b \,d^{2} e^{2} x}{2 \left (f \,x^{2}+e \right ) f^{3}}+\frac {5 b \,d^{2} e^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{2 \sqrt {e f}\, f^{3}}+\frac {a \,d^{2} x}{f^{2}}+\frac {2 b c d x}{f^{2}}-\frac {2 b \,d^{2} e x}{f^{3}} \]

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)

[Out]

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

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maxima [A]  time = 1.94, size = 186, normalized size = 1.13 \[ -\frac {{\left (b d^{2} e^{3} - a c^{2} f^{3} - {\left (2 \, b c d + a d^{2}\right )} e^{2} f + {\left (b c^{2} + 2 \, a c d\right )} e f^{2}\right )} x}{2 \, {\left (e f^{4} x^{2} + e^{2} f^{3}\right )}} + \frac {b d^{2} f x^{3} - 3 \, {\left (2 \, b d^{2} e - {\left (2 \, b c d + a d^{2}\right )} f\right )} x}{3 \, f^{3}} + \frac {{\left (5 \, b d^{2} e^{3} + a c^{2} f^{3} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{2} f + {\left (b c^{2} + 2 \, a c d\right )} e f^{2}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{2 \, \sqrt {e f} e f^{3}} \]

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

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

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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sympy [B]  time = 2.83, size = 483, normalized size = 2.95 \[ \frac {b d^{2} x^{3}}{3 f^{2}} + x \left (\frac {a d^{2}}{f^{2}} + \frac {2 b c d}{f^{2}} - \frac {2 b d^{2} e}{f^{3}}\right ) + \frac {x \left (a c^{2} f^{3} - 2 a c d e f^{2} + a d^{2} e^{2} f - b c^{2} e f^{2} + 2 b c d e^{2} f - b d^{2} e^{3}\right )}{2 e^{2} f^{3} + 2 e f^{4} x^{2}} - \frac {\sqrt {- \frac {1}{e^{3} f^{7}}} \left (c f - d e\right ) \left (a c f^{2} + 3 a d e f + b c e f - 5 b d e^{2}\right ) \log {\left (- \frac {e^{2} f^{3} \sqrt {- \frac {1}{e^{3} f^{7}}} \left (c f - d e\right ) \left (a c f^{2} + 3 a d e f + b c e f - 5 b d e^{2}\right )}{a c^{2} f^{3} + 2 a c d e f^{2} - 3 a d^{2} e^{2} f + b c^{2} e f^{2} - 6 b c d e^{2} f + 5 b d^{2} e^{3}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{e^{3} f^{7}}} \left (c f - d e\right ) \left (a c f^{2} + 3 a d e f + b c e f - 5 b d e^{2}\right ) \log {\left (\frac {e^{2} f^{3} \sqrt {- \frac {1}{e^{3} f^{7}}} \left (c f - d e\right ) \left (a c f^{2} + 3 a d e f + b c e f - 5 b d e^{2}\right )}{a c^{2} f^{3} + 2 a c d e f^{2} - 3 a d^{2} e^{2} f + b c^{2} e f^{2} - 6 b c d e^{2} f + 5 b d^{2} e^{3}} + x \right )}}{4} \]

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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