3.1345 \(\int \frac{A+B x}{(d+e x)^2 (a+c x^2)^2} \, dx\)

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

[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)

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Rubi [A]  time = 0.410702, antiderivative size = 290, normalized size of antiderivative = 1., 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} \[ -\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )\right )}{2 a^{3/2} \left (a e^2+c d^2\right )^3}-\frac{a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}+\frac{e^2 \log \left (a+c x^2\right ) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{2 \left (a e^2+c d^2\right )^3}+\frac{e \left (-3 a A e^2+4 a B d e+A c d^2\right )}{2 a (d+e x) \left (a e^2+c d^2\right )^2}-\frac{e^2 \log (d+e x) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{\left (a e^2+c d^2\right )^3} \]

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 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])

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 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 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]

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*}

Mathematica [A]  time = 0.389556, size = 251, normalized size = 0.87 \[ \frac{\frac{\left (a e^2+c d^2\right ) \left (a^2 B e^2-a c (A e (e x-2 d)+B d (d-2 e x))+A c^2 d^2 x\right )}{a \left (a+c x^2\right )}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+2 a B d e \left (3 a e^2-c d^2\right )\right )}{a^{3/2}}-e^2 \log \left (a+c x^2\right ) \left (a B e^2+4 A c d e-3 B c d^2\right )-\frac{2 e^2 \left (a e^2+c d^2\right ) (A e-B d)}{d+e x}+2 e^2 \log (d+e x) \left (a B e^2+4 A c d e-3 B c d^2\right )}{2 \left (a e^2+c d^2\right )^3} \]

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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Maple [B]  time = 0.017, size = 661, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

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]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.16611, size = 667, normalized size = 2.3 \begin{align*} \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{align*}

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))