3.4.8 \(\int \frac {x^2 (d+e x)}{(a^2-c^2 x^2)^2} \, dx\) [308]

Optimal. Leaf size=77 \[ \frac {x (d+e x)}{2 c^2 \left (a^2-c^2 x^2\right )}+\frac {(c d+2 a e) \log (a-c x)}{4 a c^4}-\frac {(c d-2 a e) \log (a+c x)}{4 a c^4} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {833, 647, 31} \begin {gather*} \frac {x (d+e x)}{2 c^2 \left (a^2-c^2 x^2\right )}+\frac {(2 a e+c d) \log (a-c x)}{4 a c^4}-\frac {(c d-2 a e) \log (a+c x)}{4 a c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 833

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 64, normalized size = 0.83 \begin {gather*} \frac {\frac {a^2 e+c^2 d x}{a^2-c^2 x^2}-\frac {c d \tanh ^{-1}\left (\frac {c x}{a}\right )}{a}+e \log \left (a^2-c^2 x^2\right )}{2 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.56, size = 88, normalized size = 1.14

method	result	size
norman	\(\frac {\frac {a^{2} e}{2 c^{4}}+\frac {d x}{2 c^{2}}}{-c^{2} x^{2}+a^{2}}+\frac {\left (2 a e -c d \right ) \ln \left (c x +a \right )}{4 a \,c^{4}}+\frac {\left (2 a e +c d \right ) \ln \left (-c x +a \right )}{4 a \,c^{4}}\)	\(80\)
default	\(\frac {a e +c d}{4 c^{4} \left (-c x +a \right )}+\frac {\left (2 a e +c d \right ) \ln \left (-c x +a \right )}{4 a \,c^{4}}-\frac {-a e +c d}{4 c^{4} \left (c x +a \right )}+\frac {\left (2 a e -c d \right ) \ln \left (c x +a \right )}{4 a \,c^{4}}\)	\(88\)
risch	\(\frac {\frac {a^{2} e}{2 c^{4}}+\frac {d x}{2 c^{2}}}{-c^{2} x^{2}+a^{2}}+\frac {\ln \left (-c x -a \right ) e}{2 c^{4}}-\frac {\ln \left (-c x -a \right ) d}{4 a \,c^{3}}+\frac {\ln \left (c x -a \right ) e}{2 c^{4}}+\frac {\ln \left (c x -a \right ) d}{4 a \,c^{3}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 82, normalized size = 1.06 \begin {gather*} -\frac {c^{2} d x + a^{2} e}{2 \, {\left (c^{6} x^{2} - a^{2} c^{4}\right )}} - \frac {{\left (c d - 2 \, a e\right )} \log \left (c x + a\right )}{4 \, a c^{4}} + \frac {{\left (c d + 2 \, a e\right )} \log \left (c x - a\right )}{4 \, a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.25, size = 122, normalized size = 1.58 \begin {gather*} -\frac {2 \, a c^{2} d x + 2 \, a^{3} e + {\left (c^{3} d x^{2} - a^{2} c d - 2 \, {\left (a c^{2} x^{2} - a^{3}\right )} e\right )} \log \left (c x + a\right ) - {\left (c^{3} d x^{2} - a^{2} c d + 2 \, {\left (a c^{2} x^{2} - a^{3}\right )} e\right )} \log \left (c x - a\right )}{4 \, {\left (a c^{6} x^{2} - a^{3} c^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.33, size = 110, normalized size = 1.43 \begin {gather*} \frac {- a^{2} e - c^{2} d x}{- 2 a^{2} c^{4} + 2 c^{6} x^{2}} + \frac {\left (2 a e - c d\right ) \log {\left (x + \frac {2 a^{2} e - a \left (2 a e - c d\right )}{c^{2} d} \right )}}{4 a c^{4}} + \frac {\left (2 a e + c d\right ) \log {\left (x + \frac {2 a^{2} e - a \left (2 a e + c d\right )}{c^{2} d} \right )}}{4 a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.91, size = 85, normalized size = 1.10 \begin {gather*} -\frac {d x + \frac {a^{2} e}{c^{2}}}{2 \, {\left (c x + a\right )} {\left (c x - a\right )} c^{2}} - \frac {{\left (c d - 2 \, a e\right )} \log \left ({\left | c x + a \right |}\right )}{4 \, a c^{4}} + \frac {{\left (c d + 2 \, a e\right )} \log \left ({\left | c x - a \right |}\right )}{4 \, a c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.11, size = 103, normalized size = 1.34 \begin {gather*} \frac {a^2\,e}{2\,\left (a^2\,c^4-c^6\,x^2\right )}+\frac {d\,x}{2\,\left (a^2\,c^2-c^4\,x^2\right )}+\frac {e\,\ln \left (a+c\,x\right )}{2\,c^4}+\frac {e\,\ln \left (a-c\,x\right )}{2\,c^4}-\frac {d\,\ln \left (a+c\,x\right )}{4\,a\,c^3}+\frac {d\,\ln \left (a-c\,x\right )}{4\,a\,c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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