3.4.11 \(\int \frac {d+e x}{x (a^2-c^2 x^2)^2} \, dx\)

Optimal. Leaf size=84 \[ -\frac {(a e+2 c d) \log (a-c x)}{4 a^4 c}-\frac {(2 c d-a e) \log (a+c x)}{4 a^4 c}+\frac {d \log (x)}{a^4}+\frac {d+e x}{2 a^2 \left (a^2-c^2 x^2\right )} \]

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Rubi [A]  time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {823, 801} \begin {gather*} \frac {d+e x}{2 a^2 \left (a^2-c^2 x^2\right )}-\frac {(a e+2 c d) \log (a-c x)}{4 a^4 c}-\frac {(2 c d-a e) \log (a+c x)}{4 a^4 c}+\frac {d \log (x)}{a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x)/(2*a^2*(a^2 - c^2*x^2)) + (d*Log[x])/a^4 - ((2*c*d + a*e)*Log[a - c*x])/(4*a^4*c) - ((2*c*d - a*e)*L
og[a + c*x])/(4*a^4*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 {d+e x}{x \left (a^2-c^2 x^2\right )^2} \, dx &=\frac {d+e x}{2 a^2 \left (a^2-c^2 x^2\right )}+\frac {\int \frac {2 a^2 c^2 d+a^2 c^2 e x}{x \left (a^2-c^2 x^2\right )} \, dx}{2 a^4 c^2}\\ &=\frac {d+e x}{2 a^2 \left (a^2-c^2 x^2\right )}+\frac {\int \left (\frac {2 c^2 d}{x}+\frac {c^2 (2 c d+a e)}{2 (a-c x)}-\frac {c^2 (2 c d-a e)}{2 (a+c x)}\right ) \, dx}{2 a^4 c^2}\\ &=\frac {d+e x}{2 a^2 \left (a^2-c^2 x^2\right )}+\frac {d \log (x)}{a^4}-\frac {(2 c d+a e) \log (a-c x)}{4 a^4 c}-\frac {(2 c d-a e) \log (a+c x)}{4 a^4 c}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 65, normalized size = 0.77 \begin {gather*} \frac {\frac {a^2 (d+e x)}{a^2-c^2 x^2}-d \log \left (a^2-c^2 x^2\right )+\frac {a e \tanh ^{-1}\left (\frac {c x}{a}\right )}{c}+2 d \log (x)}{2 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2*(d + e*x))/(a^2 - c^2*x^2) + (a*e*ArcTanh[(c*x)/a])/c + 2*d*Log[x] - d*Log[a^2 - c^2*x^2])/(2*a^4)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 139, normalized size = 1.65 \begin {gather*} -\frac {2 \, a^{2} c e x + 2 \, a^{2} c d - {\left (2 \, a^{2} c d - a^{3} e - {\left (2 \, c^{3} d - a c^{2} e\right )} x^{2}\right )} \log \left (c x + a\right ) - {\left (2 \, a^{2} c d + a^{3} e - {\left (2 \, c^{3} d + a c^{2} e\right )} x^{2}\right )} \log \left (c x - a\right ) - 4 \, {\left (c^{3} d x^{2} - a^{2} c d\right )} \log \relax (x)}{4 \, {\left (a^{4} c^{3} x^{2} - a^{6} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a^2*c*e*x + 2*a^2*c*d - (2*a^2*c*d - a^3*e - (2*c^3*d - a*c^2*e)*x^2)*log(c*x + a) - (2*a^2*c*d + a^3*
e - (2*c^3*d + a*c^2*e)*x^2)*log(c*x - a) - 4*(c^3*d*x^2 - a^2*c*d)*log(x))/(a^4*c^3*x^2 - a^6*c)

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giac [A]  time = 0.16, size = 94, normalized size = 1.12 \begin {gather*} \frac {d \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {{\left (2 \, c d - a e\right )} \log \left ({\left | c x + a \right |}\right )}{4 \, a^{4} c} - \frac {{\left (2 \, c d + a e\right )} \log \left ({\left | c x - a \right |}\right )}{4 \, a^{4} c} - \frac {a^{2} x e + a^{2} d}{2 \, {\left (c x + a\right )} {\left (c x - a\right )} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*log(abs(x))/a^4 - 1/4*(2*c*d - a*e)*log(abs(c*x + a))/(a^4*c) - 1/4*(2*c*d + a*e)*log(abs(c*x - a))/(a^4*c)
- 1/2*(a^2*x*e + a^2*d)/((c*x + a)*(c*x - a)*a^4)

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maple [A]  time = 0.06, size = 129, normalized size = 1.54 \begin {gather*} -\frac {e}{4 \left (c x +a \right ) a^{2} c}-\frac {e}{4 \left (c x -a \right ) a^{2} c}-\frac {e \ln \left (c x -a \right )}{4 a^{3} c}+\frac {e \ln \left (c x +a \right )}{4 a^{3} c}+\frac {d}{4 \left (c x +a \right ) a^{3}}-\frac {d}{4 \left (c x -a \right ) a^{3}}+\frac {d \ln \relax (x )}{a^{4}}-\frac {d \ln \left (c x -a \right )}{2 a^{4}}-\frac {d \ln \left (c x +a \right )}{2 a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.69, size = 80, normalized size = 0.95 \begin {gather*} -\frac {e x + d}{2 \, {\left (a^{2} c^{2} x^{2} - a^{4}\right )}} + \frac {d \log \relax (x)}{a^{4}} - \frac {{\left (2 \, c d - a e\right )} \log \left (c x + a\right )}{4 \, a^{4} c} - \frac {{\left (2 \, c d + a e\right )} \log \left (c x - a\right )}{4 \, a^{4} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(e*x + d)/(a^2*c^2*x^2 - a^4) + d*log(x)/a^4 - 1/4*(2*c*d - a*e)*log(c*x + a)/(a^4*c) - 1/4*(2*c*d + a*e)
*log(c*x - a)/(a^4*c)

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mupad [B]  time = 0.11, size = 82, normalized size = 0.98 \begin {gather*} \frac {\frac {d}{2\,a^2}+\frac {e\,x}{2\,a^2}}{a^2-c^2\,x^2}+\frac {d\,\ln \relax (x)}{a^4}+\frac {\ln \left (a+c\,x\right )\,\left (a\,e-2\,c\,d\right )}{4\,a^4\,c}-\frac {\ln \left (a-c\,x\right )\,\left (a\,e+2\,c\,d\right )}{4\,a^4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d/(2*a^2) + (e*x)/(2*a^2))/(a^2 - c^2*x^2) + (d*log(x))/a^4 + (log(a + c*x)*(a*e - 2*c*d))/(4*a^4*c) - (log(a
- c*x)*(a*e + 2*c*d))/(4*a^4*c)

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sympy [B]  time = 1.83, size = 231, normalized size = 2.75 \begin {gather*} \frac {- d - e x}{- 2 a^{4} + 2 a^{2} c^{2} x^{2}} + \frac {d \log {\relax (x )}}{a^{4}} + \frac {\left (a e - 2 c d\right ) \log {\left (x + \frac {- 4 a^{2} d e^{2} + \frac {a^{2} e^{2} \left (a e - 2 c d\right )}{c} - 48 c^{2} d^{3} - 12 c d^{2} \left (a e - 2 c d\right ) + 6 d \left (a e - 2 c d\right )^{2}}{a^{2} e^{3} - 36 c^{2} d^{2} e} \right )}}{4 a^{4} c} - \frac {\left (a e + 2 c d\right ) \log {\left (x + \frac {- 4 a^{2} d e^{2} - \frac {a^{2} e^{2} \left (a e + 2 c d\right )}{c} - 48 c^{2} d^{3} + 12 c d^{2} \left (a e + 2 c d\right ) + 6 d \left (a e + 2 c d\right )^{2}}{a^{2} e^{3} - 36 c^{2} d^{2} e} \right )}}{4 a^{4} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-d - e*x)/(-2*a**4 + 2*a**2*c**2*x**2) + d*log(x)/a**4 + (a*e - 2*c*d)*log(x + (-4*a**2*d*e**2 + a**2*e**2*(a
*e - 2*c*d)/c - 48*c**2*d**3 - 12*c*d**2*(a*e - 2*c*d) + 6*d*(a*e - 2*c*d)**2)/(a**2*e**3 - 36*c**2*d**2*e))/(
4*a**4*c) - (a*e + 2*c*d)*log(x + (-4*a**2*d*e**2 - a**2*e**2*(a*e + 2*c*d)/c - 48*c**2*d**3 + 12*c*d**2*(a*e
+ 2*c*d) + 6*d*(a*e + 2*c*d)**2)/(a**2*e**3 - 36*c**2*d**2*e))/(4*a**4*c)

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