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

Optimal. Leaf size=81 \[ -\frac {a^2 (a e+c d) \log (a-c x)}{2 c^5}-\frac {a^2 (c d-a e) \log (a+c x)}{2 c^5}-\frac {a^2 e x}{c^4}-\frac {d x^2}{2 c^2}-\frac {e x^3}{3 c^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {801, 633, 31} \[ -\frac {a^2 (a e+c d) \log (a-c x)}{2 c^5}-\frac {a^2 (c d-a e) \log (a+c x)}{2 c^5}-\frac {a^2 e x}{c^4}-\frac {d x^2}{2 c^2}-\frac {e x^3}{3 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 72, normalized size = 0.89 \[ \frac {a^3 e \tanh ^{-1}\left (\frac {c x}{a}\right )}{c^5}-\frac {a^2 e x}{c^4}-\frac {a^2 d \log \left (a^2-c^2 x^2\right )}{2 c^4}-\frac {d x^2}{2 c^2}-\frac {e x^3}{3 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.83, size = 75, normalized size = 0.93 \[ -\frac {2 \, c^{3} e x^{3} + 3 \, c^{3} d x^{2} + 6 \, a^{2} c e x + 3 \, {\left (a^{2} c d - a^{3} e\right )} \log \left (c x + a\right ) + 3 \, {\left (a^{2} c d + a^{3} e\right )} \log \left (c x - a\right )}{6 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 90, normalized size = 1.11 \[ -\frac {{\left (a^{2} c d - a^{3} e\right )} \log \left ({\left | c x + a \right |}\right )}{2 \, c^{5}} - \frac {{\left (a^{2} c d + a^{3} e\right )} \log \left ({\left | c x - a \right |}\right )}{2 \, c^{5}} - \frac {2 \, c^{4} x^{3} e + 3 \, c^{4} d x^{2} + 6 \, a^{2} c^{2} x e}{6 \, c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 94, normalized size = 1.16 \[ -\frac {e \,x^{3}}{3 c^{2}}-\frac {d \,x^{2}}{2 c^{2}}-\frac {a^{3} e \ln \left (c x -a \right )}{2 c^{5}}+\frac {a^{3} e \ln \left (c x +a \right )}{2 c^{5}}-\frac {a^{2} d \ln \left (c x -a \right )}{2 c^{4}}-\frac {a^{2} d \ln \left (c x +a \right )}{2 c^{4}}-\frac {a^{2} e x}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.55, size = 81, normalized size = 1.00 \[ -\frac {2 \, c^{2} e x^{3} + 3 \, c^{2} d x^{2} + 6 \, a^{2} e x}{6 \, c^{4}} - \frac {{\left (a^{2} c d - a^{3} e\right )} \log \left (c x + a\right )}{2 \, c^{5}} - \frac {{\left (a^{2} c d + a^{3} e\right )} \log \left (c x - a\right )}{2 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.09, size = 77, normalized size = 0.95 \[ \frac {\ln \left (a+c\,x\right )\,\left (a^3\,e-a^2\,c\,d\right )}{2\,c^5}-\frac {\ln \left (a-c\,x\right )\,\left (e\,a^3+c\,d\,a^2\right )}{2\,c^5}-\frac {d\,x^2}{2\,c^2}-\frac {e\,x^3}{3\,c^2}-\frac {a^2\,e\,x}{c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.44, size = 110, normalized size = 1.36 \[ - \frac {a^{2} e x}{c^{4}} + \frac {a^{2} \left (a e - c d\right ) \log {\left (x + \frac {a^{2} d + \frac {a^{2} \left (a e - c d\right )}{c}}{a^{2} e} \right )}}{2 c^{5}} - \frac {a^{2} \left (a e + c d\right ) \log {\left (x + \frac {a^{2} d - \frac {a^{2} \left (a e + c d\right )}{c}}{a^{2} e} \right )}}{2 c^{5}} - \frac {d x^{2}}{2 c^{2}} - \frac {e x^{3}}{3 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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