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

Optimal. Leaf size=290 \[ -\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )\right )}{2 a^{3/2} \left (a e^2+c d^2\right )^3}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}+\frac {e^2 \log \left (a+c x^2\right ) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{2 \left (a e^2+c d^2\right )^3}+\frac {e \left (-3 a A e^2+4 a B d e+A c d^2\right )}{2 a (d+e x) \left (a e^2+c d^2\right )^2}-\frac {e^2 \log (d+e x) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{\left (a e^2+c d^2\right )^3} \]

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Rubi [A]  time = 0.41, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {823, 801, 635, 205, 260} \begin {gather*} -\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )\right )}{2 a^{3/2} \left (a e^2+c d^2\right )^3}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}+\frac {e^2 \log \left (a+c x^2\right ) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{2 \left (a e^2+c d^2\right )^3}+\frac {e \left (-3 a A e^2+4 a B d e+A c d^2\right )}{2 a (d+e x) \left (a e^2+c d^2\right )^2}-\frac {e^2 \log (d+e x) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{\left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(A*c*d^2 + 4*a*B*d*e - 3*a*A*e^2))/(2*a*(c*d^2 + a*e^2)^2*(d + e*x)) - (a*(B*d - A*e) - (A*c*d + a*B*e)*x)/
(2*a*(c*d^2 + a*e^2)*(d + e*x)*(a + c*x^2)) - (Sqrt[c]*(2*a*B*d*e*(c*d^2 - 3*a*e^2) - A*(c^2*d^4 + 6*a*c*d^2*e
^2 - 3*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*(c*d^2 + a*e^2)^3) - (e^2*(3*B*c*d^2 - 4*A*c*d*e - a*
B*e^2)*Log[d + e*x])/(c*d^2 + a*e^2)^3 + (e^2*(3*B*c*d^2 - 4*A*c*d*e - a*B*e^2)*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 801

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx &=-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac {\int \frac {-c \left (A c d^2-2 a B d e+3 a A e^2\right )-2 c e (A c d+a B e) x}{(d+e x)^2 \left (a+c x^2\right )} \, dx}{2 a c \left (c d^2+a e^2\right )}\\ &=-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac {\int \left (\frac {c e^2 \left (A c d^2+4 a B d e-3 a A e^2\right )}{\left (c d^2+a e^2\right ) (d+e x)^2}-\frac {2 a c e^3 \left (-3 B c d^2+4 A c d e+a B e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)}+\frac {c^2 \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )-2 a e^2 \left (3 B c d^2-4 A c d e-a B 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 \left (c d^2+a e^2\right )}\\ &=\frac {e \left (A c d^2+4 a B d e-3 a A e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac {e^2 \left (3 B c d^2-4 A c d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac {c \int \frac {2 a B d e \left (c d^2-3 a e^2\right )-A \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )-2 a e^2 \left (3 B c d^2-4 A c d e-a B e^2\right ) x}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=\frac {e \left (A c d^2+4 a B d e-3 a A e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac {e^2 \left (3 B c d^2-4 A c d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac {\left (c e^2 \left (3 B c d^2-4 A c d e-a B e^2\right )\right ) \int \frac {x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}-\frac {\left (c \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )\right )\right ) \int \frac {1}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=\frac {e \left (A c d^2+4 a B d e-3 a A e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac {\sqrt {c} \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \left (c d^2+a e^2\right )^3}-\frac {e^2 \left (3 B c d^2-4 A c d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac {e^2 \left (3 B c d^2-4 A c d e-a B e^2\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 = 0.36, size = 251, normalized size = 0.87 \begin {gather*} \frac {\frac {\left (a e^2+c d^2\right ) \left (a^2 B e^2-a c (A e (e x-2 d)+B d (d-2 e x))+A c^2 d^2 x\right )}{a \left (a+c x^2\right )}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+2 a B d e \left (3 a e^2-c d^2\right )\right )}{a^{3/2}}-e^2 \log \left (a+c x^2\right ) \left (a B e^2+4 A c d e-3 B c d^2\right )-\frac {2 e^2 \left (a e^2+c d^2\right ) (A e-B d)}{d+e x}+2 e^2 \log (d+e x) \left (a B e^2+4 A c d e-3 B c d^2\right )}{2 \left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 57.64, size = 1940, normalized size = 6.69

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 494, normalized size = 1.70 \begin {gather*} \frac {{\left (A c^{3} d^{4} e^{2} - 2 \, B a c^{2} d^{3} e^{3} + 6 \, A a c^{2} d^{2} e^{4} + 6 \, B a^{2} c d e^{5} - 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 (3 \, B c d^{2} e^{2} - 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 {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 {A c^{3} d^{3} e + 3 \, B a c^{2} d^{2} e^{2} - 3 \, A a c^{2} d e^{3} - B a^{2} c e^{4}}{c d^{2} + a e^{2}} - \frac {{\left (A c^{3} d^{4} e^{2} + 4 \, B a c^{2} d^{3} e^{3} - 6 \, A a c^{2} d^{2} e^{4} - 4 \, B a^{2} c d e^{5} + 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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(A*c^3*d^4*e^2 - 2*B*a*c^2*d^3*e^3 + 6*A*a*c^2*d^2*e^4 + 6*B*a^2*c*d*e^5 - 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*(3*B*c*d^2*e^2 - 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) + (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*((A*c^3*d^3*e + 3*B*a*c^2*d^2*e^2 - 3*A*a*c^2*d*e^3 - B*a
^2*c*e^4)/(c*d^2 + a*e^2) - (A*c^3*d^4*e^2 + 4*B*a*c^2*d^3*e^3 - 6*A*a*c^2*d^2*e^4 - 4*B*a^2*c*d*e^5 + 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 = 0.07, size = 661, normalized size = 2.28 \begin {gather*} -\frac {A a c \,e^{4} x}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}-\frac {3 A a c \,e^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {A \,c^{3} d^{4} x}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right ) a}+\frac {A \,c^{3} d^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}\, a}+\frac {3 A \,c^{2} d^{2} e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {B a c d \,e^{3} x}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}+\frac {3 B a c d \,e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {B \,c^{2} d^{3} e x}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}-\frac {B \,c^{2} d^{3} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {A a c d \,e^{3}}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}+\frac {A \,c^{2} d^{3} e}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}-\frac {2 A c d \,e^{3} \ln \left (c \,x^{2}+a \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}+\frac {4 A c d \,e^{3} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}+\frac {B \,a^{2} e^{4}}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}-\frac {B a \,e^{4} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3}}+\frac {B a \,e^{4} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {B \,c^{2} d^{4}}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3} \left (c \,x^{2}+a \right )}+\frac {3 B c \,d^{2} e^{2} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {3 B c \,d^{2} e^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {A \,e^{3}}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \left (e x +d \right )}+\frac {B d \,e^{2}}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \left (e x +d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 511, normalized size = 1.76 \begin {gather*} \frac {{\left (3 \, B c d^{2} e^{2} - 4 \, A c d e^{3} - B a e^{4}\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 (3 \, B c d^{2} e^{2} - 4 \, A c d e^{3} - B a e^{4}\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 (A c^{3} d^{4} - 2 \, B a c^{2} d^{3} e + 6 \, A a c^{2} d^{2} e^{2} + 6 \, B a^{2} c d e^{3} - 3 \, A a^{2} c 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} - 2 \, A a c d^{2} e - 3 \, B a^{2} d e^{2} + 2 \, A a^{2} e^{3} - {\left (A c^{2} d^{2} e + 4 \, B a c d e^{2} - 3 \, A a c e^{3}\right )} x^{2} - {\left (A c^{2} d^{3} + B a c d^{2} e + A a c d e^{2} + B a^{2} e^{3}\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*B*c*d^2*e^2 - 4*A*c*d*e^3 - B*a*e^4)*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) - (3*B*c*d^2*e^2 - 4*A*c*d*e^3 - B*a*e^4)*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*(A*c^3*d^4 - 2*B*a*c^2*d^3*e + 6*A*a*c^2*d^2*e^2 + 6*B*a^2*c*d*e^3 - 3*A*a^2*c*e^4)*arctan(c*x/sqr
t(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 - 2*A*a*c*d^2
*e - 3*B*a^2*d*e^2 + 2*A*a^2*e^3 - (A*c^2*d^2*e + 4*B*a*c*d*e^2 - 3*A*a*c*e^3)*x^2 - (A*c^2*d^3 + B*a*c*d^2*e
+ A*a*c*d*e^2 + B*a^2*e^3)*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 = 4.05, size = 2029, normalized size = 7.00

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*(A*c*d + B*a*e))/(2*a*(a*e^2 + c*d^2)) - (2*A*a*e^3 + B*c*d^3 - 3*B*a*d*e^2 - 2*A*c*d^2*e)/(2*(a*e^2 + c*d
^2)^2) + (x^2*(A*c^2*d^2*e - 3*A*a*c*e^3 + 4*B*a*c*d*e^2))/(2*a*(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2)))/(a*d + a
*e*x + c*d*x^2 + c*e*x^3) + (log(9*A^2*a^6*e^12*(-a^3*c)^(3/2) + A^2*c^6*d^12*(-a^3*c)^(3/2) - 36*B^2*a^10*e^1
2*(-a^3*c)^(1/2) - 558*A^2*a^2*d^2*e^10*(-a^3*c)^(5/2) + 24*B^2*a^2*d^4*e^8*(-a^3*c)^(5/2) - 108*B^2*a^6*d^2*e
^10*(-a^3*c)^(3/2) - 612*A^2*c^2*d^6*e^6*(-a^3*c)^(5/2) - 308*B^2*c^2*d^8*e^4*(-a^3*c)^(5/2) + 36*B^2*a^11*c*e
^12*x + A^2*a^4*c^8*d^12*x + 9*A^2*a^10*c^2*e^12*x + 276*A*B*a^2*d^3*e^9*(-a^3*c)^(5/2) + 808*A*B*c^2*d^7*e^5*
(-a^3*c)^(5/2) - 1119*A^2*a*c*d^4*e^8*(-a^3*c)^(5/2) - 424*B^2*a*c*d^6*e^6*(-a^3*c)^(5/2) + 14*A^2*a^5*c^7*d^1
0*e^2*x + 55*A^2*a^6*c^6*d^8*e^4*x + 612*A^2*a^7*c^5*d^6*e^6*x + 1119*A^2*a^8*c^4*d^4*e^8*x + 558*A^2*a^9*c^3*
d^2*e^10*x + 4*B^2*a^6*c^6*d^10*e^2*x + 308*B^2*a^7*c^5*d^8*e^4*x + 424*B^2*a^8*c^4*d^6*e^6*x - 24*B^2*a^9*c^3
*d^4*e^8*x - 108*B^2*a^10*c^2*d^2*e^10*x + 14*A^2*a*c^5*d^10*e^2*(-a^3*c)^(3/2) + 252*A*B*a^6*d*e^11*(-a^3*c)^
(3/2) + 55*A^2*a^2*c^4*d^8*e^4*(-a^3*c)^(3/2) + 4*B^2*a^2*c^4*d^10*e^2*(-a^3*c)^(3/2) - 4*A*B*a^5*c^7*d^11*e*x
+ 252*A*B*a^10*c^2*d*e^11*x + 1320*A*B*a*c*d^5*e^7*(-a^3*c)^(5/2) - 4*A*B*a*c^5*d^11*e*(-a^3*c)^(3/2) - 20*A*
B*a^6*c^6*d^9*e^3*x - 808*A*B*a^7*c^5*d^7*e^5*x - 1320*A*B*a^8*c^4*d^5*e^7*x - 276*A*B*a^9*c^3*d^3*e^9*x - 20*
A*B*a^2*c^4*d^9*e^3*(-a^3*c)^(3/2))*(c*(a^3*((3*B*d^2*e^2)/2 - 2*A*d*e^3) - a*((3*A*d^2*e^2*(-a^3*c)^(1/2))/2
- (B*d^3*e*(-a^3*c)^(1/2))/2)) + a^2*((3*A*e^4*(-a^3*c)^(1/2))/4 - (3*B*d*e^3*(-a^3*c)^(1/2))/2) - (B*a^4*e^4)
/2 - (A*c^2*d^4*(-a^3*c)^(1/2))/4))/(a^6*e^6 + a^3*c^3*d^6 + 3*a^5*c*d^2*e^4 + 3*a^4*c^2*d^4*e^2) + (log(36*B^
2*a^10*e^12*(-a^3*c)^(1/2) - A^2*c^6*d^12*(-a^3*c)^(3/2) - 9*A^2*a^6*e^12*(-a^3*c)^(3/2) + 558*A^2*a^2*d^2*e^1
0*(-a^3*c)^(5/2) - 24*B^2*a^2*d^4*e^8*(-a^3*c)^(5/2) + 108*B^2*a^6*d^2*e^10*(-a^3*c)^(3/2) + 612*A^2*c^2*d^6*e
^6*(-a^3*c)^(5/2) + 308*B^2*c^2*d^8*e^4*(-a^3*c)^(5/2) + 36*B^2*a^11*c*e^12*x + A^2*a^4*c^8*d^12*x + 9*A^2*a^1
0*c^2*e^12*x - 276*A*B*a^2*d^3*e^9*(-a^3*c)^(5/2) - 808*A*B*c^2*d^7*e^5*(-a^3*c)^(5/2) + 1119*A^2*a*c*d^4*e^8*
(-a^3*c)^(5/2) + 424*B^2*a*c*d^6*e^6*(-a^3*c)^(5/2) + 14*A^2*a^5*c^7*d^10*e^2*x + 55*A^2*a^6*c^6*d^8*e^4*x + 6
12*A^2*a^7*c^5*d^6*e^6*x + 1119*A^2*a^8*c^4*d^4*e^8*x + 558*A^2*a^9*c^3*d^2*e^10*x + 4*B^2*a^6*c^6*d^10*e^2*x
+ 308*B^2*a^7*c^5*d^8*e^4*x + 424*B^2*a^8*c^4*d^6*e^6*x - 24*B^2*a^9*c^3*d^4*e^8*x - 108*B^2*a^10*c^2*d^2*e^10
*x - 14*A^2*a*c^5*d^10*e^2*(-a^3*c)^(3/2) - 252*A*B*a^6*d*e^11*(-a^3*c)^(3/2) - 55*A^2*a^2*c^4*d^8*e^4*(-a^3*c
)^(3/2) - 4*B^2*a^2*c^4*d^10*e^2*(-a^3*c)^(3/2) - 4*A*B*a^5*c^7*d^11*e*x + 252*A*B*a^10*c^2*d*e^11*x - 1320*A*
B*a*c*d^5*e^7*(-a^3*c)^(5/2) + 4*A*B*a*c^5*d^11*e*(-a^3*c)^(3/2) - 20*A*B*a^6*c^6*d^9*e^3*x - 808*A*B*a^7*c^5*
d^7*e^5*x - 1320*A*B*a^8*c^4*d^5*e^7*x - 276*A*B*a^9*c^3*d^3*e^9*x + 20*A*B*a^2*c^4*d^9*e^3*(-a^3*c)^(3/2))*(c
*(a^3*((3*B*d^2*e^2)/2 - 2*A*d*e^3) + a*((3*A*d^2*e^2*(-a^3*c)^(1/2))/2 - (B*d^3*e*(-a^3*c)^(1/2))/2)) - a^2*(
(3*A*e^4*(-a^3*c)^(1/2))/4 - (3*B*d*e^3*(-a^3*c)^(1/2))/2) - (B*a^4*e^4)/2 + (A*c^2*d^4*(-a^3*c)^(1/2))/4))/(a
^6*e^6 + a^3*c^3*d^6 + 3*a^5*c*d^2*e^4 + 3*a^4*c^2*d^4*e^2) - (log(d + e*x)*(c*(3*B*d^2*e^2 - 4*A*d*e^3) - B*a
*e^4))/(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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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