3.307 \(\int \frac {x^3 (d+e x)}{(a^2-c^2 x^2)^2} \, dx\)

Optimal. Leaf size=84 \[ \frac {x^2 (d+e x)}{2 c^2 \left (a^2-c^2 x^2\right )}+\frac {(3 a e+2 c d) \log (a-c x)}{4 c^5}+\frac {(2 c d-3 a e) \log (a+c x)}{4 c^5}+\frac {3 e x}{2 c^4} \]

[Out]

3/2*e*x/c^4+1/2*x^2*(e*x+d)/c^2/(-c^2*x^2+a^2)+1/4*(3*a*e+2*c*d)*ln(-c*x+a)/c^5+1/4*(-3*a*e+2*c*d)*ln(c*x+a)/c
^5

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Rubi [A]  time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {819, 774, 633, 31} \[ \frac {x^2 (d+e x)}{2 c^2 \left (a^2-c^2 x^2\right )}+\frac {(3 a e+2 c d) \log (a-c x)}{4 c^5}+\frac {(2 c d-3 a e) \log (a+c x)}{4 c^5}+\frac {3 e x}{2 c^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x))/(a^2 - c^2*x^2)^2,x]

[Out]

(3*e*x)/(2*c^4) + (x^2*(d + e*x))/(2*c^2*(a^2 - c^2*x^2)) + ((2*c*d + 3*a*e)*Log[a - c*x])/(4*c^5) + ((2*c*d -
3*a*e)*Log[a + c*x])/(4*c^5)

Rule 31

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

Rule 633

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 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

Rule 819

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^3 (d+e x)}{\left (a^2-c^2 x^2\right )^2} \, dx &=\frac {x^2 (d+e x)}{2 c^2 \left (a^2-c^2 x^2\right )}-\frac {\int \frac {x \left (2 a^2 d+3 a^2 e x\right )}{a^2-c^2 x^2} \, dx}{2 a^2 c^2}\\ &=\frac {3 e x}{2 c^4}+\frac {x^2 (d+e x)}{2 c^2 \left (a^2-c^2 x^2\right )}+\frac {\int \frac {-3 a^4 e-2 a^2 c^2 d x}{a^2-c^2 x^2} \, dx}{2 a^2 c^4}\\ &=\frac {3 e x}{2 c^4}+\frac {x^2 (d+e x)}{2 c^2 \left (a^2-c^2 x^2\right )}-\frac {(2 c d-3 a e) \int \frac {1}{-a c-c^2 x} \, dx}{4 c^3}-\frac {(2 c d+3 a e) \int \frac {1}{a c-c^2 x} \, dx}{4 c^3}\\ &=\frac {3 e x}{2 c^4}+\frac {x^2 (d+e x)}{2 c^2 \left (a^2-c^2 x^2\right )}+\frac {(2 c d+3 a e) \log (a-c x)}{4 c^5}+\frac {(2 c d-3 a e) \log (a+c x)}{4 c^5}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 64, normalized size = 0.76 \[ \frac {\frac {a^2 c (d+e x)}{a^2-c^2 x^2}+c d \log \left (a^2-c^2 x^2\right )-3 a e \tanh ^{-1}\left (\frac {c x}{a}\right )+2 c e x}{2 c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x))/(a^2 - c^2*x^2)^2,x]

[Out]

(2*c*e*x + (a^2*c*(d + e*x))/(a^2 - c^2*x^2) - 3*a*e*ArcTanh[(c*x)/a] + c*d*Log[a^2 - c^2*x^2])/(2*c^5)

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fricas [A]  time = 0.83, size = 129, normalized size = 1.54 \[ \frac {4 \, c^{3} e x^{3} - 6 \, a^{2} c e x - 2 \, a^{2} c d - {\left (2 \, a^{2} c d - 3 \, a^{3} e - {\left (2 \, c^{3} d - 3 \, a c^{2} e\right )} x^{2}\right )} \log \left (c x + a\right ) - {\left (2 \, a^{2} c d + 3 \, a^{3} e - {\left (2 \, c^{3} d + 3 \, a c^{2} e\right )} x^{2}\right )} \log \left (c x - a\right )}{4 \, {\left (c^{7} x^{2} - a^{2} c^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)/(-c^2*x^2+a^2)^2,x, algorithm="fricas")

[Out]

1/4*(4*c^3*e*x^3 - 6*a^2*c*e*x - 2*a^2*c*d - (2*a^2*c*d - 3*a^3*e - (2*c^3*d - 3*a*c^2*e)*x^2)*log(c*x + a) -
(2*a^2*c*d + 3*a^3*e - (2*c^3*d + 3*a*c^2*e)*x^2)*log(c*x - a))/(c^7*x^2 - a^2*c^5)

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giac [A]  time = 0.16, size = 88, normalized size = 1.05 \[ \frac {x e}{c^{4}} + \frac {{\left (2 \, c d - 3 \, a e\right )} \log \left ({\left | c x + a \right |}\right )}{4 \, c^{5}} + \frac {{\left (2 \, c d + 3 \, a e\right )} \log \left ({\left | c x - a \right |}\right )}{4 \, c^{5}} - \frac {a^{2} x e + a^{2} d}{2 \, {\left (c x + a\right )} {\left (c x - a\right )} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)/(-c^2*x^2+a^2)^2,x, algorithm="giac")

[Out]

x*e/c^4 + 1/4*(2*c*d - 3*a*e)*log(abs(c*x + a))/c^5 + 1/4*(2*c*d + 3*a*e)*log(abs(c*x - a))/c^5 - 1/2*(a^2*x*e
+ a^2*d)/((c*x + a)*(c*x - a)*c^4)

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maple [A]  time = 0.06, size = 126, normalized size = 1.50 \[ -\frac {a^{2} e}{4 \left (c x +a \right ) c^{5}}-\frac {a^{2} e}{4 \left (c x -a \right ) c^{5}}+\frac {a d}{4 \left (c x +a \right ) c^{4}}-\frac {a d}{4 \left (c x -a \right ) c^{4}}+\frac {3 a e \ln \left (c x -a \right )}{4 c^{5}}-\frac {3 a e \ln \left (c x +a \right )}{4 c^{5}}+\frac {d \ln \left (c x -a \right )}{2 c^{4}}+\frac {d \ln \left (c x +a \right )}{2 c^{4}}+\frac {e x}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x+d)/(-c^2*x^2+a^2)^2,x)

[Out]

e*x/c^4-3/4/c^5*ln(c*x+a)*a*e+1/2/c^4*ln(c*x+a)*d-1/4/c^5*a^2/(c*x+a)*e+1/4/c^4*a/(c*x+a)*d+3/4/c^5*ln(c*x-a)*
a*e+1/2/c^4*ln(c*x-a)*d-1/4/c^5*a^2/(c*x-a)*e-1/4/c^4*a/(c*x-a)*d

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maxima [A]  time = 0.69, size = 81, normalized size = 0.96 \[ -\frac {a^{2} e x + a^{2} d}{2 \, {\left (c^{6} x^{2} - a^{2} c^{4}\right )}} + \frac {e x}{c^{4}} + \frac {{\left (2 \, c d - 3 \, a e\right )} \log \left (c x + a\right )}{4 \, c^{5}} + \frac {{\left (2 \, c d + 3 \, a e\right )} \log \left (c x - a\right )}{4 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)/(-c^2*x^2+a^2)^2,x, algorithm="maxima")

[Out]

-1/2*(a^2*e*x + a^2*d)/(c^6*x^2 - a^2*c^4) + e*x/c^4 + 1/4*(2*c*d - 3*a*e)*log(c*x + a)/c^5 + 1/4*(2*c*d + 3*a
*e)*log(c*x - a)/c^5

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mupad [B]  time = 1.08, size = 81, normalized size = 0.96 \[ \frac {\frac {a^2\,d}{2}+\frac {a^2\,e\,x}{2}}{a^2\,c^4-c^6\,x^2}-\frac {\ln \left (a+c\,x\right )\,\left (3\,a\,e-2\,c\,d\right )}{4\,c^5}+\frac {\ln \left (a-c\,x\right )\,\left (3\,a\,e+2\,c\,d\right )}{4\,c^5}+\frac {e\,x}{c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d + e*x))/(a^2 - c^2*x^2)^2,x)

[Out]

((a^2*d)/2 + (a^2*e*x)/2)/(a^2*c^4 - c^6*x^2) - (log(a + c*x)*(3*a*e - 2*c*d))/(4*c^5) + (log(a - c*x)*(3*a*e
+ 2*c*d))/(4*c^5) + (e*x)/c^4

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sympy [A]  time = 0.74, size = 110, normalized size = 1.31 \[ \frac {- a^{2} d - a^{2} e x}{- 2 a^{2} c^{4} + 2 c^{6} x^{2}} + \frac {e x}{c^{4}} - \frac {\left (3 a e - 2 c d\right ) \log {\left (x + \frac {2 d + \frac {3 a e - 2 c d}{c}}{3 e} \right )}}{4 c^{5}} + \frac {\left (3 a e + 2 c d\right ) \log {\left (x + \frac {2 d - \frac {3 a e + 2 c d}{c}}{3 e} \right )}}{4 c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x+d)/(-c**2*x**2+a**2)**2,x)

[Out]

(-a**2*d - a**2*e*x)/(-2*a**2*c**4 + 2*c**6*x**2) + e*x/c**4 - (3*a*e - 2*c*d)*log(x + (2*d + (3*a*e - 2*c*d)/
c)/(3*e))/(4*c**5) + (3*a*e + 2*c*d)*log(x + (2*d - (3*a*e + 2*c*d)/c)/(3*e))/(4*c**5)

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