\(\int \frac {A+B x+C x^2}{(d+e x) (a+c x^2)^2} \, dx\) [54]

Optimal result

Integrand size = 27, antiderivative size = 226 \[ \int \frac {A+B x+C x^2}{(d+e x) \left (a+c x^2\right )^2} \, dx=-\frac {a (B c d-A c e+a C e)-c (A c d-a C d+a B e) x}{2 a c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (a (C d-B e) \left (c d^2-a e^2\right )+A c d \left (c d^2+3 a e^2\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c} \left (c d^2+a e^2\right )^2}+\frac {e \left (C d^2-B d e+A e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^2}-\frac {e \left (C d^2-B d e+A e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^2} \]

[Out]

1/2*(-a*(-A*c*e+B*c*d+C*a*e)+c*(A*c*d+B*a*e-C*a*d)*x)/a/c/(a*e^2+c*d^2)/(c*x^2+a)+e*(A*e^2-B*d*e+C*d^2)*ln(e*x
+d)/(a*e^2+c*d^2)^2-1/2*e*(A*e^2-B*d*e+C*d^2)*ln(c*x^2+a)/(a*e^2+c*d^2)^2+1/2*(a*(-B*e+C*d)*(-a*e^2+c*d^2)+A*c
*d*(3*a*e^2+c*d^2))*arctan(x*c^(1/2)/a^(1/2))/a^(3/2)/(a*e^2+c*d^2)^2/c^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1661, 815, 649, 211, 266} \[ \int \frac {A+B x+C x^2}{(d+e x) \left (a+c x^2\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (3 a e^2+c d^2\right )+a \left (c d^2-a e^2\right ) (C d-B e)\right )}{2 a^{3/2} \sqrt {c} \left (a e^2+c d^2\right )^2}-\frac {a (a C e-A c e+B c d)-c x (a B e-a C d+A c d)}{2 a c \left (a+c x^2\right ) \left (a e^2+c d^2\right )}-\frac {e \log \left (a+c x^2\right ) \left (A e^2-B d e+C d^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac {e \log (d+e x) \left (A e^2-B d e+C d^2\right )}{\left (a e^2+c d^2\right )^2} \]

[In]

Int[(A + B*x + C*x^2)/((d + e*x)*(a + c*x^2)^2),x]

[Out]

-1/2*(a*(B*c*d - A*c*e + a*C*e) - c*(A*c*d - a*C*d + a*B*e)*x)/(a*c*(c*d^2 + a*e^2)*(a + c*x^2)) + ((a*(C*d -
B*e)*(c*d^2 - a*e^2) + A*c*d*(c*d^2 + 3*a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*Sqrt[c]*(c*d^2 + a*e^2
)^2) + (e*(C*d^2 - B*d*e + A*e^2)*Log[d + e*x])/(c*d^2 + a*e^2)^2 - (e*(C*d^2 - B*d*e + A*e^2)*Log[a + c*x^2])
/(2*(c*d^2 + a*e^2)^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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1661

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[(a*g - c*f*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {a (B c d-A c e+a C e)-c (A c d-a C d+a B e) x}{2 a c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \frac {-\frac {c \left (a d (C d-B e)+A \left (c d^2+2 a e^2\right )\right )}{c d^2+a e^2}-\frac {c e (A c d-a C d+a B e) x}{c d^2+a e^2}}{(d+e x) \left (a+c x^2\right )} \, dx}{2 a c} \\ & = -\frac {a (B c d-A c e+a C e)-c (A c d-a C d+a B e) x}{2 a c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \left (-\frac {2 a c e^2 \left (C d^2-B d e+A e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)}+\frac {c \left (-a (C d-B e) \left (c d^2-a e^2\right )-A c d \left (c d^2+3 a e^2\right )+2 a c e \left (C d^2-B d e+A e^2\right ) x\right )}{\left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}\right ) \, dx}{2 a c} \\ & = -\frac {a (B c d-A c e+a C e)-c (A c d-a C d+a B e) x}{2 a c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {e \left (C d^2-B d e+A e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^2}-\frac {\int \frac {-a (C d-B e) \left (c d^2-a e^2\right )-A c d \left (c d^2+3 a e^2\right )+2 a c e \left (C d^2-B d e+A e^2\right ) x}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^2} \\ & = -\frac {a (B c d-A c e+a C e)-c (A c d-a C d+a B e) x}{2 a c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {e \left (C d^2-B d e+A e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^2}-\frac {\left (c e \left (C d^2-B d e+A e^2\right )\right ) \int \frac {x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^2}+\frac {\left (a (C d-B e) \left (c d^2-a e^2\right )+A c d \left (c d^2+3 a e^2\right )\right ) \int \frac {1}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^2} \\ & = -\frac {a (B c d-A c e+a C e)-c (A c d-a C d+a B e) x}{2 a c \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (a (C d-B e) \left (c d^2-a e^2\right )+A c d \left (c d^2+3 a e^2\right )\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c} \left (c d^2+a e^2\right )^2}+\frac {e \left (C d^2-B d e+A e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^2}-\frac {e \left (C d^2-B d e+A e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x+C x^2}{(d+e x) \left (a+c x^2\right )^2} \, dx=\frac {\frac {\left (c d^2+a e^2\right ) \left (-a^2 C e+A c^2 d x+a c (-B d+A e-C d x+B e x)\right )}{a c \left (a+c x^2\right )}+\frac {\left (a (C d-B e) \left (c d^2-a e^2\right )+A c d \left (c d^2+3 a e^2\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {c}}+2 e \left (C d^2+e (-B d+A e)\right ) \log (d+e x)-e \left (C d^2+e (-B d+A e)\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^2} \]

[In]

Integrate[(A + B*x + C*x^2)/((d + e*x)*(a + c*x^2)^2),x]

[Out]

(((c*d^2 + a*e^2)*(-(a^2*C*e) + A*c^2*d*x + a*c*(-(B*d) + A*e - C*d*x + B*e*x)))/(a*c*(a + c*x^2)) + ((a*(C*d
- B*e)*(c*d^2 - a*e^2) + A*c*d*(c*d^2 + 3*a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(a^(3/2)*Sqrt[c]) + 2*e*(C*d^2
+ e*(-(B*d) + A*e))*Log[d + e*x] - e*(C*d^2 + e*(-(B*d) + A*e))*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^2)

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {\frac {\frac {\left (A a c d \,e^{2}+A \,d^{3} c^{2}+a^{2} B \,e^{3}+B a c \,d^{2} e -C \,a^{2} d \,e^{2}-C a c \,d^{3}\right ) x}{2 a}+\frac {A a c \,e^{3}+A \,c^{2} d^{2} e -B a c d \,e^{2}-B \,c^{2} d^{3}-C \,a^{2} e^{3}-C a c \,d^{2} e}{2 c}}{c \,x^{2}+a}+\frac {\frac {\left (-2 A a c \,e^{3}+2 B a c d \,e^{2}-2 C a c \,d^{2} e \right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (3 A a c d \,e^{2}+A \,d^{3} c^{2}+a^{2} B \,e^{3}-B a c \,d^{2} e -C \,a^{2} d \,e^{2}+C a c \,d^{3}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{2 a}}{\left (e^{2} a +c \,d^{2}\right )^{2}}+\frac {e \left (A \,e^{2}-B d e +C \,d^{2}\right ) \ln \left (e x +d \right )}{\left (e^{2} a +c \,d^{2}\right )^{2}}\)	\(293\)
risch	\(\text {Expression too large to display}\)	\(15144\)

[In]

int((C*x^2+B*x+A)/(e*x+d)/(c*x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/(a*e^2+c*d^2)^2*((1/2*(A*a*c*d*e^2+A*c^2*d^3+B*a^2*e^3+B*a*c*d^2*e-C*a^2*d*e^2-C*a*c*d^3)/a*x+1/2*(A*a*c*e^3
+A*c^2*d^2*e-B*a*c*d*e^2-B*c^2*d^3-C*a^2*e^3-C*a*c*d^2*e)/c)/(c*x^2+a)+1/2/a*(1/2*(-2*A*a*c*e^3+2*B*a*c*d*e^2-
2*C*a*c*d^2*e)/c*ln(c*x^2+a)+(3*A*a*c*d*e^2+A*c^2*d^3+B*a^2*e^3-B*a*c*d^2*e-C*a^2*d*e^2+C*a*c*d^3)/(a*c)^(1/2)
*arctan(c*x/(a*c)^(1/2))))+e*(A*e^2-B*d*e+C*d^2)*ln(e*x+d)/(a*e^2+c*d^2)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 499 vs. \(2 (212) = 424\).

Time = 21.75 (sec) , antiderivative size = 1024, normalized size of antiderivative = 4.53 \[ \int \frac {A+B x+C x^2}{(d+e x) \left (a+c x^2\right )^2} \, dx=\left [-\frac {2 \, B a^{2} c^{2} d^{3} + 2 \, B a^{3} c d e^{2} + 2 \, {\left (C a^{3} c - A a^{2} c^{2}\right )} d^{2} e + 2 \, {\left (C a^{4} - A a^{3} c\right )} e^{3} - {\left (B a^{2} c d^{2} e - B a^{3} e^{3} - {\left (C a^{2} c + A a c^{2}\right )} d^{3} + {\left (C a^{3} - 3 \, A a^{2} c\right )} d e^{2} + {\left (B a c^{2} d^{2} e - B a^{2} c e^{3} - {\left (C a c^{2} + A c^{3}\right )} d^{3} + {\left (C a^{2} c - 3 \, A a c^{2}\right )} d e^{2}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (B a^{2} c^{2} d^{2} e + B a^{3} c e^{3} - {\left (C a^{2} c^{2} - A a c^{3}\right )} d^{3} - {\left (C a^{3} c - A a^{2} c^{2}\right )} d e^{2}\right )} x + 2 \, {\left (C a^{3} c d^{2} e - B a^{3} c d e^{2} + A a^{3} c e^{3} + {\left (C a^{2} c^{2} d^{2} e - B a^{2} c^{2} d e^{2} + A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (c x^{2} + a\right ) - 4 \, {\left (C a^{3} c d^{2} e - B a^{3} c d e^{2} + A a^{3} c e^{3} + {\left (C a^{2} c^{2} d^{2} e - B a^{2} c^{2} d e^{2} + A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (e x + d\right )}{4 \, {\left (a^{3} c^{3} d^{4} + 2 \, a^{4} c^{2} d^{2} e^{2} + a^{5} c e^{4} + {\left (a^{2} c^{4} d^{4} + 2 \, a^{3} c^{3} d^{2} e^{2} + a^{4} c^{2} e^{4}\right )} x^{2}\right )}}, -\frac {B a^{2} c^{2} d^{3} + B a^{3} c d e^{2} + {\left (C a^{3} c - A a^{2} c^{2}\right )} d^{2} e + {\left (C a^{4} - A a^{3} c\right )} e^{3} + {\left (B a^{2} c d^{2} e - B a^{3} e^{3} - {\left (C a^{2} c + A a c^{2}\right )} d^{3} + {\left (C a^{3} - 3 \, A a^{2} c\right )} d e^{2} + {\left (B a c^{2} d^{2} e - B a^{2} c e^{3} - {\left (C a c^{2} + A c^{3}\right )} d^{3} + {\left (C a^{2} c - 3 \, A a c^{2}\right )} d e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (B a^{2} c^{2} d^{2} e + B a^{3} c e^{3} - {\left (C a^{2} c^{2} - A a c^{3}\right )} d^{3} - {\left (C a^{3} c - A a^{2} c^{2}\right )} d e^{2}\right )} x + {\left (C a^{3} c d^{2} e - B a^{3} c d e^{2} + A a^{3} c e^{3} + {\left (C a^{2} c^{2} d^{2} e - B a^{2} c^{2} d e^{2} + A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (C a^{3} c d^{2} e - B a^{3} c d e^{2} + A a^{3} c e^{3} + {\left (C a^{2} c^{2} d^{2} e - B a^{2} c^{2} d e^{2} + A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (a^{3} c^{3} d^{4} + 2 \, a^{4} c^{2} d^{2} e^{2} + a^{5} c e^{4} + {\left (a^{2} c^{4} d^{4} + 2 \, a^{3} c^{3} d^{2} e^{2} + a^{4} c^{2} e^{4}\right )} x^{2}\right )}}\right ] \]

[In]

integrate((C*x^2+B*x+A)/(e*x+d)/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

[-1/4*(2*B*a^2*c^2*d^3 + 2*B*a^3*c*d*e^2 + 2*(C*a^3*c - A*a^2*c^2)*d^2*e + 2*(C*a^4 - A*a^3*c)*e^3 - (B*a^2*c*
d^2*e - B*a^3*e^3 - (C*a^2*c + A*a*c^2)*d^3 + (C*a^3 - 3*A*a^2*c)*d*e^2 + (B*a*c^2*d^2*e - B*a^2*c*e^3 - (C*a*
c^2 + A*c^3)*d^3 + (C*a^2*c - 3*A*a*c^2)*d*e^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a))
- 2*(B*a^2*c^2*d^2*e + B*a^3*c*e^3 - (C*a^2*c^2 - A*a*c^3)*d^3 - (C*a^3*c - A*a^2*c^2)*d*e^2)*x + 2*(C*a^3*c*d
^2*e - B*a^3*c*d*e^2 + A*a^3*c*e^3 + (C*a^2*c^2*d^2*e - B*a^2*c^2*d*e^2 + A*a^2*c^2*e^3)*x^2)*log(c*x^2 + a) -
4*(C*a^3*c*d^2*e - B*a^3*c*d*e^2 + A*a^3*c*e^3 + (C*a^2*c^2*d^2*e - B*a^2*c^2*d*e^2 + A*a^2*c^2*e^3)*x^2)*log
(e*x + d))/(a^3*c^3*d^4 + 2*a^4*c^2*d^2*e^2 + a^5*c*e^4 + (a^2*c^4*d^4 + 2*a^3*c^3*d^2*e^2 + a^4*c^2*e^4)*x^2)
, -1/2*(B*a^2*c^2*d^3 + B*a^3*c*d*e^2 + (C*a^3*c - A*a^2*c^2)*d^2*e + (C*a^4 - A*a^3*c)*e^3 + (B*a^2*c*d^2*e -
B*a^3*e^3 - (C*a^2*c + A*a*c^2)*d^3 + (C*a^3 - 3*A*a^2*c)*d*e^2 + (B*a*c^2*d^2*e - B*a^2*c*e^3 - (C*a*c^2 + A
*c^3)*d^3 + (C*a^2*c - 3*A*a*c^2)*d*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - (B*a^2*c^2*d^2*e + B*a^3*c*e^3
- (C*a^2*c^2 - A*a*c^3)*d^3 - (C*a^3*c - A*a^2*c^2)*d*e^2)*x + (C*a^3*c*d^2*e - B*a^3*c*d*e^2 + A*a^3*c*e^3 +
(C*a^2*c^2*d^2*e - B*a^2*c^2*d*e^2 + A*a^2*c^2*e^3)*x^2)*log(c*x^2 + a) - 2*(C*a^3*c*d^2*e - B*a^3*c*d*e^2 +
A*a^3*c*e^3 + (C*a^2*c^2*d^2*e - B*a^2*c^2*d*e^2 + A*a^2*c^2*e^3)*x^2)*log(e*x + d))/(a^3*c^3*d^4 + 2*a^4*c^2*
d^2*e^2 + a^5*c*e^4 + (a^2*c^4*d^4 + 2*a^3*c^3*d^2*e^2 + a^4*c^2*e^4)*x^2)]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2}{(d+e x) \left (a+c x^2\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((C*x**2+B*x+A)/(e*x+d)/(c*x**2+a)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.30 \[ \int \frac {A+B x+C x^2}{(d+e x) \left (a+c x^2\right )^2} \, dx=-\frac {{\left (C d^{2} e - B d e^{2} + A e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {{\left (C d^{2} e - B d e^{2} + A e^{3}\right )} \log \left (e x + d\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} - \frac {{\left (B a c d^{2} e - B a^{2} e^{3} - {\left (C a c + A c^{2}\right )} d^{3} + {\left (C a^{2} - 3 \, A a c\right )} d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {a c}} - \frac {B a c d + {\left (C a^{2} - A a c\right )} e - {\left (B a c e - {\left (C a c - A c^{2}\right )} d\right )} x}{2 \, {\left (a^{2} c^{2} d^{2} + a^{3} c e^{2} + {\left (a c^{3} d^{2} + a^{2} c^{2} e^{2}\right )} x^{2}\right )}} \]

[In]

integrate((C*x^2+B*x+A)/(e*x+d)/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

-1/2*(C*d^2*e - B*d*e^2 + A*e^3)*log(c*x^2 + a)/(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4) + (C*d^2*e - B*d*e^2 + A*e
^3)*log(e*x + d)/(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4) - 1/2*(B*a*c*d^2*e - B*a^2*e^3 - (C*a*c + A*c^2)*d^3 + (C
*a^2 - 3*A*a*c)*d*e^2)*arctan(c*x/sqrt(a*c))/((a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4)*sqrt(a*c)) - 1/2*(B*a*c*
d + (C*a^2 - A*a*c)*e - (B*a*c*e - (C*a*c - A*c^2)*d)*x)/(a^2*c^2*d^2 + a^3*c*e^2 + (a*c^3*d^2 + a^2*c^2*e^2)*
x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.61 \[ \int \frac {A+B x+C x^2}{(d+e x) \left (a+c x^2\right )^2} \, dx=-\frac {{\left (C d^{2} e - B d e^{2} + A e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac {{\left (C d^{2} e^{2} - B d e^{3} + A e^{4}\right )} \log \left ({\left | e x + d \right |}\right )}{c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}} + \frac {{\left (C a c d^{3} + A c^{2} d^{3} - B a c d^{2} e - C a^{2} d e^{2} + 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} \sqrt {a c}} - \frac {B a c^{2} d^{3} + C a^{2} c d^{2} e - A a c^{2} d^{2} e + B a^{2} c d e^{2} + C a^{3} e^{3} - A a^{2} c e^{3} + {\left (C a c^{2} d^{3} - A c^{3} d^{3} - B a c^{2} d^{2} e + C a^{2} c d e^{2} - A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} x}{2 \, {\left (c d^{2} + a e^{2}\right )}^{2} {\left (c x^{2} + a\right )} a c} \]

[In]

integrate((C*x^2+B*x+A)/(e*x+d)/(c*x^2+a)^2,x, algorithm="giac")

[Out]

-1/2*(C*d^2*e - B*d*e^2 + A*e^3)*log(c*x^2 + a)/(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4) + (C*d^2*e^2 - B*d*e^3 + A
*e^4)*log(abs(e*x + d))/(c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5) + 1/2*(C*a*c*d^3 + A*c^2*d^3 - B*a*c*d^2*e - C*a
^2*d*e^2 + 3*A*a*c*d*e^2 + B*a^2*e^3)*arctan(c*x/sqrt(a*c))/((a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4)*sqrt(a*c)
) - 1/2*(B*a*c^2*d^3 + C*a^2*c*d^2*e - A*a*c^2*d^2*e + B*a^2*c*d*e^2 + C*a^3*e^3 - A*a^2*c*e^3 + (C*a*c^2*d^3
- A*c^3*d^3 - B*a*c^2*d^2*e + C*a^2*c*d*e^2 - A*a*c^2*d*e^2 - B*a^2*c*e^3)*x)/((c*d^2 + a*e^2)^2*(c*x^2 + a)*a
*c)

Mupad [B] (verification not implemented)

Time = 17.72 (sec) , antiderivative size = 1493, normalized size of antiderivative = 6.61 \[ \int \frac {A+B x+C x^2}{(d+e x) \left (a+c x^2\right )^2} \, dx=\text {Too large to display} \]

[In]

int((A + B*x + C*x^2)/((a + c*x^2)^2*(d + e*x)),x)

[Out]

(log(A*c^3*d^5*(-a^3*c)^(1/2) - B*a^3*e^5*(-a^3*c)^(1/2) + 6*A*a^4*c*e^5 - B*a^4*c*e^5*x - 2*A*a^2*c^3*d^4*e -
8*C*a^3*c^2*d^4*e + 8*C*a^4*c*d^2*e^3 + C*a^2*c^3*d^5*x + C*a*c^2*d^5*(-a^3*c)^(1/2) + C*a^3*d*e^4*(-a^3*c)^(
1/2) - 12*A*a^3*c^2*d^2*e^3 + 8*B*a^3*c^2*d^3*e^2 - 8*B*a^4*c*d*e^4 + A*a*c^4*d^5*x + 2*A*a^2*c^3*d^3*e^2*x +
14*B*a^3*c^2*d^2*e^3*x - 14*C*a^3*c^2*d^3*e^2*x + 2*A*a*c^2*d^3*e^2*(-a^3*c)^(1/2) + 14*B*a^2*c*d^2*e^3*(-a^3*
c)^(1/2) - 14*C*a^2*c*d^3*e^2*(-a^3*c)^(1/2) + C*a^4*c*d*e^4*x - 15*A*a^3*c^2*d*e^4*x - B*a^2*c^3*d^4*e*x - 15
*A*a^2*c*d*e^4*(-a^3*c)^(1/2) - B*a*c^2*d^4*e*(-a^3*c)^(1/2) - 6*A*a^2*c*e^5*x*(-a^3*c)^(1/2) + 2*A*c^3*d^4*e*
x*(-a^3*c)^(1/2) + 8*B*a^2*c*d*e^4*x*(-a^3*c)^(1/2) + 8*C*a*c^2*d^4*e*x*(-a^3*c)^(1/2) + 12*A*a*c^2*d^2*e^3*x*
(-a^3*c)^(1/2) - 8*B*a*c^2*d^3*e^2*x*(-a^3*c)^(1/2) - 8*C*a^2*c*d^2*e^3*x*(-a^3*c)^(1/2))*(a^2*((B*e^3*(-a^3*c
)^(1/2))/4 - (C*d*e^2*(-a^3*c)^(1/2))/4) - c*(a^3*((A*e^3)/2 - (B*d*e^2)/2 + (C*d^2*e)/2) - a*((C*d^3*(-a^3*c)
^(1/2))/4 + (3*A*d*e^2*(-a^3*c)^(1/2))/4 - (B*d^2*e*(-a^3*c)^(1/2))/4)) + (A*c^2*d^3*(-a^3*c)^(1/2))/4))/(a^5*
c*e^4 + a^3*c^3*d^4 + 2*a^4*c^2*d^2*e^2) - (log(A*c^3*d^5*(-a^3*c)^(1/2) - B*a^3*e^5*(-a^3*c)^(1/2) - 6*A*a^4*
c*e^5 + B*a^4*c*e^5*x + 2*A*a^2*c^3*d^4*e + 8*C*a^3*c^2*d^4*e - 8*C*a^4*c*d^2*e^3 - C*a^2*c^3*d^5*x + C*a*c^2*
d^5*(-a^3*c)^(1/2) + C*a^3*d*e^4*(-a^3*c)^(1/2) + 12*A*a^3*c^2*d^2*e^3 - 8*B*a^3*c^2*d^3*e^2 + 8*B*a^4*c*d*e^4
- A*a*c^4*d^5*x - 2*A*a^2*c^3*d^3*e^2*x - 14*B*a^3*c^2*d^2*e^3*x + 14*C*a^3*c^2*d^3*e^2*x + 2*A*a*c^2*d^3*e^2
*(-a^3*c)^(1/2) + 14*B*a^2*c*d^2*e^3*(-a^3*c)^(1/2) - 14*C*a^2*c*d^3*e^2*(-a^3*c)^(1/2) - C*a^4*c*d*e^4*x + 15
*A*a^3*c^2*d*e^4*x + B*a^2*c^3*d^4*e*x - 15*A*a^2*c*d*e^4*(-a^3*c)^(1/2) - B*a*c^2*d^4*e*(-a^3*c)^(1/2) - 6*A*
a^2*c*e^5*x*(-a^3*c)^(1/2) + 2*A*c^3*d^4*e*x*(-a^3*c)^(1/2) + 8*B*a^2*c*d*e^4*x*(-a^3*c)^(1/2) + 8*C*a*c^2*d^4
*e*x*(-a^3*c)^(1/2) + 12*A*a*c^2*d^2*e^3*x*(-a^3*c)^(1/2) - 8*B*a*c^2*d^3*e^2*x*(-a^3*c)^(1/2) - 8*C*a^2*c*d^2
*e^3*x*(-a^3*c)^(1/2))*(c*(a^3*((A*e^3)/2 - (B*d*e^2)/2 + (C*d^2*e)/2) + a*((C*d^3*(-a^3*c)^(1/2))/4 + (3*A*d*
e^2*(-a^3*c)^(1/2))/4 - (B*d^2*e*(-a^3*c)^(1/2))/4)) + a^2*((B*e^3*(-a^3*c)^(1/2))/4 - (C*d*e^2*(-a^3*c)^(1/2)
)/4) + (A*c^2*d^3*(-a^3*c)^(1/2))/4))/(a^5*c*e^4 + a^3*c^3*d^4 + 2*a^4*c^2*d^2*e^2) - ((B*c*d - A*c*e + C*a*e)
/(2*c*(a*e^2 + c*d^2)) - (x*(A*c*d + B*a*e - C*a*d))/(2*a*(a*e^2 + c*d^2)))/(a + c*x^2) + (e*log(d + e*x)*(A*e
^2 + C*d^2 - B*d*e))/(a*e^2 + c*d^2)^2