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

Optimal. Leaf size=374 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+a \left (a^2 C e^4-6 a c d e^2 (C d-B e)+c^2 d^3 (C d-2 B e)\right )\right )}{2 a^{3/2} \sqrt {c} \left (a e^2+c d^2\right )^3}-\frac {a \left (-a B e^2+2 a C d e-2 A c d e+B c d^2\right )-x \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right )}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}+\frac {e \log \left (a+c x^2\right ) \left (a e^2 (2 C d-B e)-c d \left (2 C d^2-e (3 B d-4 A e)\right )\right )}{2 \left (a e^2+c d^2\right )^3}-\frac {e \left (A e^2-B d e+C d^2\right )}{(d+e x) \left (a e^2+c d^2\right )^2}-\frac {e \log (d+e x) \left (a e^2 (2 C d-B e)-c d \left (2 C d^2-e (3 B d-4 A e)\right )\right )}{\left (a e^2+c d^2\right )^3} \]

[Out]

-e*(A*e^2-B*d*e+C*d^2)/(a*e^2+c*d^2)^2/(e*x+d)+1/2*(-a*(-2*A*c*d*e-B*a*e^2+B*c*d^2+2*C*a*d*e)+(A*c*(-a*e^2+c*d
^2)+a*(a*C*e^2-c*d*(-2*B*e+C*d)))*x)/a/(a*e^2+c*d^2)^2/(c*x^2+a)-e*(a*e^2*(-B*e+2*C*d)-c*d*(2*C*d^2-e*(-4*A*e+
3*B*d)))*ln(e*x+d)/(a*e^2+c*d^2)^3+1/2*e*(a*e^2*(-B*e+2*C*d)-c*d*(2*C*d^2-e*(-4*A*e+3*B*d)))*ln(c*x^2+a)/(a*e^
2+c*d^2)^3+1/2*(A*c*(-3*a^2*e^4+6*a*c*d^2*e^2+c^2*d^4)+a*(a^2*C*e^4+c^2*d^3*(-2*B*e+C*d)-6*a*c*d*e^2*(-B*e+C*d
)))*arctan(x*c^(1/2)/a^(1/2))/a^(3/2)/(a*e^2+c*d^2)^3/c^(1/2)

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Rubi [A]  time = 0.95, antiderivative size = 371, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1647, 1629, 635, 205, 260} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+a \left (a^2 C e^4-6 a c d e^2 (C d-B e)+c^2 d^3 (C d-2 B e)\right )\right )}{2 a^{3/2} \sqrt {c} \left (a e^2+c d^2\right )^3}-\frac {a \left (-a B e^2+2 a C d e-2 A c d e+B c d^2\right )-x \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right )}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}-\frac {e \log \left (a+c x^2\right ) \left (-a e^2 (2 C d-B e)-c d e (3 B d-4 A e)+2 c C d^3\right )}{2 \left (a e^2+c d^2\right )^3}-\frac {e \left (A e^2-B d e+C d^2\right )}{(d+e x) \left (a e^2+c d^2\right )^2}+\frac {e \log (d+e x) \left (-a e^2 (2 C d-B e)-c d e (3 B d-4 A e)+2 c C d^3\right )}{\left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*(C*d^2 - B*d*e + A*e^2))/((c*d^2 + a*e^2)^2*(d + e*x))) - (a*(B*c*d^2 - 2*A*c*d*e + 2*a*C*d*e - a*B*e^2)
- (A*c*(c*d^2 - a*e^2) + a*(a*C*e^2 - c*d*(C*d - 2*B*e)))*x)/(2*a*(c*d^2 + a*e^2)^2*(a + c*x^2)) + ((A*c*(c^2*
d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4) + a*(a^2*C*e^4 + c^2*d^3*(C*d - 2*B*e) - 6*a*c*d*e^2*(C*d - B*e)))*ArcTan[(Sq
rt[c]*x)/Sqrt[a]])/(2*a^(3/2)*Sqrt[c]*(c*d^2 + a*e^2)^3) + (e*(2*c*C*d^3 - c*d*e*(3*B*d - 4*A*e) - a*e^2*(2*C*
d - B*e))*Log[d + e*x])/(c*d^2 + a*e^2)^3 - (e*(2*c*C*d^3 - c*d*e*(3*B*d - 4*A*e) - a*e^2*(2*C*d - B*e))*Log[a
+ c*x^2])/(2*(c*d^2 + a*e^2)^3)

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 260

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 635

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 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1647

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*} \int \frac {A+B x+C x^2}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx &=-\frac {a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}-\frac {\int \frac {-\frac {c \left (A \left (c^2 d^4+5 a c d^2 e^2+2 a^2 e^4\right )-a d^2 \left (a C e^2-c d (C d-2 B e)\right )\right )}{\left (c d^2+a e^2\right )^2}-\frac {2 c e (A c d-a C d+a B e) x}{c d^2+a e^2}-\frac {c e^2 \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x^2}{\left (c d^2+a e^2\right )^2}}{(d+e x)^2 \left (a+c x^2\right )} \, dx}{2 a c}\\ &=-\frac {a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \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)^2}+\frac {2 a c e^2 \left (-2 c C d^3+c d e (3 B d-4 A e)+a e^2 (2 C d-B e)\right )}{\left (c d^2+a e^2\right )^3 (d+e x)}+\frac {c \left (-A c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )-a \left (a^2 C e^4+c^2 d^3 (C d-2 B e)-6 a c d e^2 (C d-B e)\right )+2 a c e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) x\right )}{\left (c d^2+a e^2\right )^3 \left (a+c x^2\right )}\right ) \, dx}{2 a c}\\ &=-\frac {e \left (C d^2-B d e+A e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac {a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac {e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac {\int \frac {-A c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )-a \left (a^2 C e^4+c^2 d^3 (C d-2 B e)-6 a c d e^2 (C d-B e)\right )+2 a c e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) x}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=-\frac {e \left (C d^2-B d e+A e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac {a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac {e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac {\left (c e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right )\right ) \int \frac {x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}+\frac {\left (A c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+a \left (a^2 C e^4+c^2 d^3 (C d-2 B e)-6 a c d e^2 (C d-B e)\right )\right ) \int \frac {1}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=-\frac {e \left (C d^2-B d e+A e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac {a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac {\left (A c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+a \left (a^2 C e^4+c^2 d^3 (C d-2 B e)-6 a c d e^2 (C d-B e)\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 )^3}+\frac {e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac {e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 320, normalized size = 0.86 \[ \frac {\frac {\left (a e^2+c d^2\right ) \left (a^2 e (B e-2 C d+C e x)-a c \left (A e (e x-2 d)+B d (d-2 e x)+C d^2 x\right )+A c^2 d^2 x\right )}{a \left (a+c x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+a \left (a^2 C e^4+6 a c d e^2 (B e-C d)+c^2 d^3 (C d-2 B e)\right )\right )}{a^{3/2} \sqrt {c}}-e \log \left (a+c x^2\right ) \left (a e^2 (B e-2 C d)+c d e (4 A e-3 B d)+2 c C d^3\right )+2 e \log (d+e x) \left (a e^2 (B e-2 C d)+c d e (4 A e-3 B d)+2 c C d^3\right )-\frac {2 e \left (a e^2+c d^2\right ) \left (e (A e-B d)+C d^2\right )}{d+e x}}{2 \left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*e*(c*d^2 + a*e^2)*(C*d^2 + e*(-(B*d) + A*e)))/(d + e*x) + ((c*d^2 + a*e^2)*(A*c^2*d^2*x + a^2*e*(-2*C*d +
B*e + C*e*x) - a*c*(C*d^2*x + B*d*(d - 2*e*x) + A*e*(-2*d + e*x))))/(a*(a + c*x^2)) + ((A*c*(c^2*d^4 + 6*a*c*
d^2*e^2 - 3*a^2*e^4) + a*(a^2*C*e^4 + c^2*d^3*(C*d - 2*B*e) + 6*a*c*d*e^2*(-(C*d) + B*e)))*ArcTan[(Sqrt[c]*x)/
Sqrt[a]])/(a^(3/2)*Sqrt[c]) + 2*e*(2*c*C*d^3 + c*d*e*(-3*B*d + 4*A*e) + a*e^2*(-2*C*d + B*e))*Log[d + e*x] - e
*(2*c*C*d^3 + c*d*e*(-3*B*d + 4*A*e) + a*e^2*(-2*C*d + B*e))*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^3)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.19, size = 608, normalized size = 1.63 \[ \frac {{\left (C a c^{2} d^{4} e^{2} + A c^{3} d^{4} e^{2} - 2 \, B a c^{2} d^{3} e^{3} - 6 \, C a^{2} c d^{2} e^{4} + 6 \, A a c^{2} d^{2} e^{4} + 6 \, B a^{2} c d e^{5} + C a^{3} e^{6} - 3 \, A a^{2} c e^{6}\right )} \arctan \left (\frac {{\left (c d - \frac {c d^{2}}{x e + d} - \frac {a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt {a c}}\right ) e^{\left (-2\right )}}{2 \, {\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt {a c}} - \frac {{\left (2 \, C c d^{3} e - 3 \, B c d^{2} e^{2} - 2 \, C a d e^{3} + 4 \, A c d e^{3} + B a e^{4}\right )} \log \left (c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} - \frac {\frac {C d^{2} e^{5}}{x e + d} - \frac {B d e^{6}}{x e + d} + \frac {A e^{7}}{x e + d}}{c^{2} d^{4} e^{4} + 2 \, a c d^{2} e^{6} + a^{2} e^{8}} - \frac {\frac {C a c^{2} d^{3} e - A c^{3} d^{3} e - 3 \, B a c^{2} d^{2} e^{2} - 3 \, C a^{2} c d e^{3} + 3 \, A a c^{2} d e^{3} + B a^{2} c e^{4}}{c d^{2} + a e^{2}} - \frac {{\left (C a c^{2} d^{4} e^{2} - A c^{3} d^{4} e^{2} - 4 \, B a c^{2} d^{3} e^{3} - 6 \, C a^{2} c d^{2} e^{4} + 6 \, A a c^{2} d^{2} e^{4} + 4 \, B a^{2} c d e^{5} + C a^{3} e^{6} - A a^{2} c e^{6}\right )} e^{\left (-1\right )}}{{\left (c d^{2} + a e^{2}\right )} {\left (x e + d\right )}}}{2 \, {\left (c d^{2} + a e^{2}\right )}^{2} a {\left (c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(C*a*c^2*d^4*e^2 + A*c^3*d^4*e^2 - 2*B*a*c^2*d^3*e^3 - 6*C*a^2*c*d^2*e^4 + 6*A*a*c^2*d^2*e^4 + 6*B*a^2*c*d
*e^5 + C*a^3*e^6 - 3*A*a^2*c*e^6)*arctan((c*d - c*d^2/(x*e + d) - a*e^2/(x*e + d))*e^(-1)/sqrt(a*c))*e^(-2)/((
a*c^3*d^6 + 3*a^2*c^2*d^4*e^2 + 3*a^3*c*d^2*e^4 + a^4*e^6)*sqrt(a*c)) - 1/2*(2*C*c*d^3*e - 3*B*c*d^2*e^2 - 2*C
*a*d*e^3 + 4*A*c*d*e^3 + B*a*e^4)*log(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2)/(c^3*d^6 +
3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6) - (C*d^2*e^5/(x*e + d) - B*d*e^6/(x*e + d) + A*e^7/(x*e + d))/(c^
2*d^4*e^4 + 2*a*c*d^2*e^6 + a^2*e^8) - 1/2*((C*a*c^2*d^3*e - A*c^3*d^3*e - 3*B*a*c^2*d^2*e^2 - 3*C*a^2*c*d*e^3
+ 3*A*a*c^2*d*e^3 + B*a^2*c*e^4)/(c*d^2 + a*e^2) - (C*a*c^2*d^4*e^2 - A*c^3*d^4*e^2 - 4*B*a*c^2*d^3*e^3 - 6*C
*a^2*c*d^2*e^4 + 6*A*a*c^2*d^2*e^4 + 4*B*a^2*c*d*e^5 + C*a^3*e^6 - A*a^2*c*e^6)*e^(-1)/((c*d^2 + a*e^2)*(x*e +
d)))/((c*d^2 + a*e^2)^2*a*(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2))

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maple [B]  time = 0.02, size = 1036, normalized size = 2.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(a*e^2+c*d^2)^3*c*ln(c*x^2+a)*d*A*e^3+3/2/(a*e^2+c*d^2)^3*c*ln(c*x^2+a)*e^2*d^2*B-1/(a*e^2+c*d^2)^3*c*ln(c*
x^2+a)*C*d^3*e+1/2/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*C*c^2*d^4+1/(a*e^2+c*d^2)^3*a*ln(c*x^
2+a)*C*d*e^3+1/2/(a*e^2+c*d^2)^3*a^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*C*e^4+2*e/(a*e^2+c*d^2)^3*ln(e*x+d)
*C*c*d^3+4*e^3/(a*e^2+c*d^2)^3*ln(e*x+d)*A*c*d-3*e^2/(a*e^2+c*d^2)^3*ln(e*x+d)*B*c*d^2-2*e^3/(a*e^2+c*d^2)^3*l
n(e*x+d)*C*a*d+1/2/(a*e^2+c*d^2)^3/(c*x^2+a)*a^2*C*e^4*x-1/2/(a*e^2+c*d^2)^3/(c*x^2+a)*C*c^2*d^4*x+1/(a*e^2+c*
d^2)^3/(c*x^2+a)*A*c^2*d^3*e-1/(a*e^2+c*d^2)^3/(c*x^2+a)*C*a^2*d*e^3-e^3/(a*e^2+c*d^2)^2/(e*x+d)*A-1/(a*e^2+c*
d^2)^3/(c*x^2+a)*C*a*c*d^3*e-3/2/(a*e^2+c*d^2)^3*a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*c*e^4+1/2/(a*e^2+c*
d^2)^3/a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*c^3*d^4+3/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*
x)*A*c^2*d^2*e^2-1/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*c^2*d^3*e-1/2/(a*e^2+c*d^2)^3/(c*x^
2+a)*A*a*c*e^4*x+1/2/(a*e^2+c*d^2)^3/(c*x^2+a)/a*x*A*c^3*d^4+1/(a*e^2+c*d^2)^3/(c*x^2+a)*B*c^2*d^3*e*x+1/(a*e^
2+c*d^2)^3/(c*x^2+a)*A*a*c*d*e^3+1/(a*e^2+c*d^2)^3/(c*x^2+a)*d*a*c*B*e^3*x-3/(a*e^2+c*d^2)^3*a/(a*c)^(1/2)*arc
tan(1/(a*c)^(1/2)*c*x)*C*c*d^2*e^2+3/(a*e^2+c*d^2)^3*a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*c*d*e^3-1/2/(a*
e^2+c*d^2)^3*a*ln(c*x^2+a)*e^4*B+e^4/(a*e^2+c*d^2)^3*ln(e*x+d)*B*a+1/2/(a*e^2+c*d^2)^3/(c*x^2+a)*B*a^2*e^4-1/2
/(a*e^2+c*d^2)^3/(c*x^2+a)*B*c^2*d^4+e^2/(a*e^2+c*d^2)^2/(e*x+d)*B*d-e/(a*e^2+c*d^2)^2/(e*x+d)*C*d^2

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maxima [A]  time = 1.04, size = 604, normalized size = 1.61 \[ -\frac {{\left (2 \, C c d^{3} e - 3 \, B c d^{2} e^{2} + B a e^{4} - 2 \, {\left (C a - 2 \, A c\right )} d e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} + \frac {{\left (2 \, C c d^{3} e - 3 \, B c d^{2} e^{2} + B a e^{4} - 2 \, {\left (C a - 2 \, A c\right )} d e^{3}\right )} \log \left (e x + d\right )}{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}} - \frac {{\left (2 \, B a c^{2} d^{3} e - 6 \, B a^{2} c d e^{3} - {\left (C a c^{2} + A c^{3}\right )} d^{4} + 6 \, {\left (C a^{2} c - A a c^{2}\right )} d^{2} e^{2} - {\left (C a^{3} - 3 \, A a^{2} c\right )} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt {a c}} - \frac {B a c d^{3} - 3 \, B a^{2} d e^{2} + 2 \, A a^{2} e^{3} + 2 \, {\left (2 \, C a^{2} - A a c\right )} d^{2} e - {\left (4 \, B a c d e^{2} - {\left (3 \, C a c - A c^{2}\right )} d^{2} e + {\left (C a^{2} - 3 \, A a c\right )} e^{3}\right )} x^{2} - {\left (B a c d^{2} e + B a^{2} e^{3} - {\left (C a c - A c^{2}\right )} d^{3} - {\left (C a^{2} - A a c\right )} d e^{2}\right )} x}{2 \, {\left (a^{2} c^{2} d^{5} + 2 \, a^{3} c d^{3} e^{2} + a^{4} d e^{4} + {\left (a c^{3} d^{4} e + 2 \, a^{2} c^{2} d^{2} e^{3} + a^{3} c e^{5}\right )} x^{3} + {\left (a c^{3} d^{5} + 2 \, a^{2} c^{2} d^{3} e^{2} + a^{3} c d e^{4}\right )} x^{2} + {\left (a^{2} c^{2} d^{4} e + 2 \, a^{3} c d^{2} e^{3} + a^{4} e^{5}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*C*c*d^3*e - 3*B*c*d^2*e^2 + B*a*e^4 - 2*(C*a - 2*A*c)*d*e^3)*log(c*x^2 + a)/(c^3*d^6 + 3*a*c^2*d^4*e^2
+ 3*a^2*c*d^2*e^4 + a^3*e^6) + (2*C*c*d^3*e - 3*B*c*d^2*e^2 + B*a*e^4 - 2*(C*a - 2*A*c)*d*e^3)*log(e*x + d)/(
c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6) - 1/2*(2*B*a*c^2*d^3*e - 6*B*a^2*c*d*e^3 - (C*a*c^2 + A
*c^3)*d^4 + 6*(C*a^2*c - A*a*c^2)*d^2*e^2 - (C*a^3 - 3*A*a^2*c)*e^4)*arctan(c*x/sqrt(a*c))/((a*c^3*d^6 + 3*a^2
*c^2*d^4*e^2 + 3*a^3*c*d^2*e^4 + a^4*e^6)*sqrt(a*c)) - 1/2*(B*a*c*d^3 - 3*B*a^2*d*e^2 + 2*A*a^2*e^3 + 2*(2*C*a
^2 - A*a*c)*d^2*e - (4*B*a*c*d*e^2 - (3*C*a*c - A*c^2)*d^2*e + (C*a^2 - 3*A*a*c)*e^3)*x^2 - (B*a*c*d^2*e + B*a
^2*e^3 - (C*a*c - A*c^2)*d^3 - (C*a^2 - A*a*c)*d*e^2)*x)/(a^2*c^2*d^5 + 2*a^3*c*d^3*e^2 + a^4*d*e^4 + (a*c^3*d
^4*e + 2*a^2*c^2*d^2*e^3 + a^3*c*e^5)*x^3 + (a*c^3*d^5 + 2*a^2*c^2*d^3*e^2 + a^3*c*d*e^4)*x^2 + (a^2*c^2*d^4*e
+ 2*a^3*c*d^2*e^3 + a^4*e^5)*x)

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mupad [B]  time = 9.91, size = 2094, normalized size = 5.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x^2*(C*a^2*e^3 - 3*A*a*c*e^3 + A*c^2*d^2*e + 4*B*a*c*d*e^2 - 3*C*a*c*d^2*e))/(2*a*(a^2*e^4 + c^2*d^4 + 2*a*c
*d^2*e^2)) - (2*A*a*e^3 + B*c*d^3 - 3*B*a*d*e^2 - 2*A*c*d^2*e + 4*C*a*d^2*e)/(2*(a*e^2 + c*d^2)^2) + (x*(A*c*d
+ B*a*e - C*a*d))/(2*a*(a*e^2 + c*d^2)))/(a*d + a*e*x + c*d*x^2 + c*e*x^3) - (log(3*A*e^6*(-a^3*c)^(3/2) - A*
c^4*d^6*(-a^3*c)^(1/2) + C*a^4*e^6*(-a^3*c)^(1/2) + 31*C*d^2*e^4*(-a^3*c)^(3/2) + 6*B*a^5*c*e^6 - 18*B*d*e^5*(
-a^3*c)^(3/2) - 6*B*e^6*x*(-a^3*c)^(3/2) - C*a^5*c*e^6*x + 14*C*d*e^5*x*(-a^3*c)^(3/2) - 2*A*a^2*c^4*d^5*e + 3
0*A*a^4*c^2*d*e^5 - 14*C*a^3*c^3*d^5*e + 3*A*a^4*c^2*e^6*x + C*a^2*c^4*d^6*x - C*a*c^3*d^6*(-a^3*c)^(1/2) - 36
*A*a^3*c^3*d^3*e^3 + 22*B*a^3*c^3*d^4*e^2 - 36*B*a^4*c^2*d^2*e^4 + 36*C*a^4*c^2*d^3*e^3 - 14*C*a^5*c*d*e^5 + A
*a*c^5*d^6*x + 5*A*a^2*c^4*d^4*e^2*x - 57*A*a^3*c^3*d^2*e^4*x + 44*B*a^3*c^3*d^3*e^3*x - 31*C*a^3*c^3*d^4*e^2*
x + 31*C*a^4*c^2*d^2*e^4*x - 5*A*a*c^3*d^4*e^2*(-a^3*c)^(1/2) + 57*A*a^2*c^2*d^2*e^4*(-a^3*c)^(1/2) - 44*B*a^2
*c^2*d^3*e^3*(-a^3*c)^(1/2) + 31*C*a^2*c^2*d^4*e^2*(-a^3*c)^(1/2) - 2*B*a^2*c^4*d^5*e*x - 18*B*a^4*c^2*d*e^5*x
+ 2*B*a*c^3*d^5*e*(-a^3*c)^(1/2) - 2*A*c^4*d^5*e*x*(-a^3*c)^(1/2) - 36*B*a^2*c^2*d^2*e^4*x*(-a^3*c)^(1/2) + 3
6*C*a^2*c^2*d^3*e^3*x*(-a^3*c)^(1/2) - 14*C*a*c^3*d^5*e*x*(-a^3*c)^(1/2) - 36*A*a*c^3*d^3*e^3*x*(-a^3*c)^(1/2)
+ 30*A*a^2*c^2*d*e^5*x*(-a^3*c)^(1/2) + 22*B*a*c^3*d^4*e^2*x*(-a^3*c)^(1/2))*(c^2*(a*((C*d^4*(-a^3*c)^(1/2))/
4 + (3*A*d^2*e^2*(-a^3*c)^(1/2))/2 - (B*d^3*e*(-a^3*c)^(1/2))/2) + a^3*(2*A*d*e^3 - (3*B*d^2*e^2)/2 + C*d^3*e)
) - c*(a^2*((3*A*e^4*(-a^3*c)^(1/2))/4 + (3*C*d^2*e^2*(-a^3*c)^(1/2))/2 - (3*B*d*e^3*(-a^3*c)^(1/2))/2) - a^4*
((B*e^4)/2 - C*d*e^3)) + (A*c^3*d^4*(-a^3*c)^(1/2))/4 + (C*a^3*e^4*(-a^3*c)^(1/2))/4))/(a^6*c*e^6 + a^3*c^4*d^
6 + 3*a^4*c^3*d^4*e^2 + 3*a^5*c^2*d^2*e^4) + (log(3*A*e^6*(-a^3*c)^(3/2) - A*c^4*d^6*(-a^3*c)^(1/2) + C*a^4*e^
6*(-a^3*c)^(1/2) + 31*C*d^2*e^4*(-a^3*c)^(3/2) - 6*B*a^5*c*e^6 - 18*B*d*e^5*(-a^3*c)^(3/2) - 6*B*e^6*x*(-a^3*c
)^(3/2) + C*a^5*c*e^6*x + 14*C*d*e^5*x*(-a^3*c)^(3/2) + 2*A*a^2*c^4*d^5*e - 30*A*a^4*c^2*d*e^5 + 14*C*a^3*c^3*
d^5*e - 3*A*a^4*c^2*e^6*x - C*a^2*c^4*d^6*x - C*a*c^3*d^6*(-a^3*c)^(1/2) + 36*A*a^3*c^3*d^3*e^3 - 22*B*a^3*c^3
*d^4*e^2 + 36*B*a^4*c^2*d^2*e^4 - 36*C*a^4*c^2*d^3*e^3 + 14*C*a^5*c*d*e^5 - A*a*c^5*d^6*x - 5*A*a^2*c^4*d^4*e^
2*x + 57*A*a^3*c^3*d^2*e^4*x - 44*B*a^3*c^3*d^3*e^3*x + 31*C*a^3*c^3*d^4*e^2*x - 31*C*a^4*c^2*d^2*e^4*x - 5*A*
a*c^3*d^4*e^2*(-a^3*c)^(1/2) + 57*A*a^2*c^2*d^2*e^4*(-a^3*c)^(1/2) - 44*B*a^2*c^2*d^3*e^3*(-a^3*c)^(1/2) + 31*
C*a^2*c^2*d^4*e^2*(-a^3*c)^(1/2) + 2*B*a^2*c^4*d^5*e*x + 18*B*a^4*c^2*d*e^5*x + 2*B*a*c^3*d^5*e*(-a^3*c)^(1/2)
- 2*A*c^4*d^5*e*x*(-a^3*c)^(1/2) - 36*B*a^2*c^2*d^2*e^4*x*(-a^3*c)^(1/2) + 36*C*a^2*c^2*d^3*e^3*x*(-a^3*c)^(1
/2) - 14*C*a*c^3*d^5*e*x*(-a^3*c)^(1/2) - 36*A*a*c^3*d^3*e^3*x*(-a^3*c)^(1/2) + 30*A*a^2*c^2*d*e^5*x*(-a^3*c)^
(1/2) + 22*B*a*c^3*d^4*e^2*x*(-a^3*c)^(1/2))*(c^2*(a*((C*d^4*(-a^3*c)^(1/2))/4 + (3*A*d^2*e^2*(-a^3*c)^(1/2))/
2 - (B*d^3*e*(-a^3*c)^(1/2))/2) - a^3*(2*A*d*e^3 - (3*B*d^2*e^2)/2 + C*d^3*e)) - c*(a^2*((3*A*e^4*(-a^3*c)^(1/
2))/4 + (3*C*d^2*e^2*(-a^3*c)^(1/2))/2 - (3*B*d*e^3*(-a^3*c)^(1/2))/2) + a^4*((B*e^4)/2 - C*d*e^3)) + (A*c^3*d
^4*(-a^3*c)^(1/2))/4 + (C*a^3*e^4*(-a^3*c)^(1/2))/4))/(a^6*c*e^6 + a^3*c^4*d^6 + 3*a^4*c^3*d^4*e^2 + 3*a^5*c^2
*d^2*e^4) + (log(d + e*x)*(a*(B*e^4 - 2*C*d*e^3) + c*(4*A*d*e^3 - 3*B*d^2*e^2 + 2*C*d^3*e)))/(a^3*e^6 + c^3*d^
6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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