3.1336 \(\int \frac{A+B x}{(d+e x)^2 (a+c x^2)} \, dx\)

Optimal. Leaf size=173 \[ \frac{\log \left (a+c x^2\right ) \left (-a B e^2-2 A c d e+B c d^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac{B d-A e}{(d+e x) \left (a e^2+c d^2\right )}-\frac{\log (d+e x) \left (-a B e^2-2 A c d e+B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (-a A e^2+2 a B d e+A c d^2\right )}{\sqrt{a} \left (a e^2+c d^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.169804, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {801, 635, 205, 260} \[ \frac{\log \left (a+c x^2\right ) \left (-a B e^2-2 A c d e+B c d^2\right )}{2 \left (a e^2+c d^2\right )^2}+\frac{B d-A e}{(d+e x) \left (a e^2+c d^2\right )}-\frac{\log (d+e x) \left (-a B e^2-2 A c d e+B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (-a A e^2+2 a B d e+A c d^2\right )}{\sqrt{a} \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.196738, size = 148, normalized size = 0.86 \[ \frac{\log \left (a+c x^2\right ) \left (-a B e^2-2 A c d e+B c d^2\right )+\frac{2 \left (a e^2+c d^2\right ) (B d-A e)}{d+e x}+\log (d+e x) \left (2 a B e^2+4 A c d e-2 B c d^2\right )+\frac{2 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (-a A e^2+2 a B d e+A c d^2\right )}{\sqrt{a}}}{2 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 312, normalized size = 1.8 \begin{align*} -{\frac{Ae}{ \left ( a{e}^{2}+c{d}^{2} \right ) \left ( ex+d \right ) }}+{\frac{Bd}{ \left ( a{e}^{2}+c{d}^{2} \right ) \left ( ex+d \right ) }}+2\,{\frac{\ln \left ( ex+d \right ) Acde}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}+{\frac{\ln \left ( ex+d \right ) aB{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}-{\frac{\ln \left ( ex+d \right ) Bc{d}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}-{\frac{c\ln \left ( c{x}^{2}+a \right ) Ade}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}-{\frac{\ln \left ( c{x}^{2}+a \right ) aB{e}^{2}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}+{\frac{c\ln \left ( c{x}^{2}+a \right ) B{d}^{2}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}-{\frac{aAc{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{A{c}^{2}{d}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+2\,{\frac{aBcde}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) } \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),x)

[Out]

-1/(a*e^2+c*d^2)/(e*x+d)*A*e+1/(a*e^2+c*d^2)/(e*x+d)*B*d+2/(a*e^2+c*d^2)^2*ln(e*x+d)*A*c*d*e+1/(a*e^2+c*d^2)^2
*ln(e*x+d)*a*B*e^2-1/(a*e^2+c*d^2)^2*ln(e*x+d)*B*c*d^2-1/(a*e^2+c*d^2)^2*c*ln(c*x^2+a)*A*d*e-1/2/(a*e^2+c*d^2)
^2*ln(c*x^2+a)*a*B*e^2+1/2/(a*e^2+c*d^2)^2*c*ln(c*x^2+a)*B*d^2-1/(a*e^2+c*d^2)^2*c/(a*c)^(1/2)*arctan(x*c/(a*c
)^(1/2))*A*a*e^2+1/(a*e^2+c*d^2)^2*c^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*d^2+2/(a*e^2+c*d^2)^2*c/(a*c)^(1/
2)*arctan(x*c/(a*c)^(1/2))*a*B*d*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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 49.3874, size = 1211, normalized size = 7. \begin{align*} \left [\frac{2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} -{\left (A c d^{3} + 2 \, B a d^{2} e - A a d e^{2} +{\left (A c d^{2} e + 2 \, B a d e^{2} - A a e^{3}\right )} x\right )} \sqrt{-\frac{c}{a}} \log \left (\frac{c x^{2} - 2 \, a x \sqrt{-\frac{c}{a}} - a}{c x^{2} + a}\right ) +{\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} +{\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (c x^{2} + a\right ) - 2 \,{\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} +{\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} +{\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}, \frac{2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} + 2 \,{\left (A c d^{3} + 2 \, B a d^{2} e - A a d e^{2} +{\left (A c d^{2} e + 2 \, B a d e^{2} - A a e^{3}\right )} x\right )} \sqrt{\frac{c}{a}} \arctan \left (x \sqrt{\frac{c}{a}}\right ) +{\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} +{\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (c x^{2} + a\right ) - 2 \,{\left (B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} +{\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} +{\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}\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),x, algorithm="fricas")

[Out]

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

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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),x)

[Out]

Timed out

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Giac [A]  time = 1.2111, size = 309, normalized size = 1.79 \begin{align*} \frac{{\left (A c^{2} d^{2} e^{2} + 2 \, B a c d e^{3} - A a c e^{4}\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 )}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt{a c}} + \frac{{\left (B c d^{2} - 2 \, A c d e - B a e^{2}\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^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} + \frac{\frac{B d e^{2}}{x e + d} - \frac{A e^{3}}{x e + d}}{c d^{2} e^{2} + a e^{4}} \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),x, algorithm="giac")

[Out]

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