3.3.97 \(\int \frac {x^4 (d+e x)}{a^2-c^2 x^2} \, dx\) [297]

Optimal. Leaf size=95 \[ -\frac {a^2 d x}{c^4}-\frac {a^2 e x^2}{2 c^4}-\frac {d x^3}{3 c^2}-\frac {e x^4}{4 c^2}-\frac {a^3 (c d+a e) \log (a-c x)}{2 c^6}+\frac {a^3 (c d-a e) \log (a+c x)}{2 c^6} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 95, 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 = {815, 647, 31} \begin {gather*} -\frac {a^3 (a e+c d) \log (a-c x)}{2 c^6}+\frac {a^3 (c d-a e) \log (a+c x)}{2 c^6}-\frac {a^2 d x}{c^4}-\frac {a^2 e x^2}{2 c^4}-\frac {d x^3}{3 c^2}-\frac {e x^4}{4 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 815

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^4 (d+e x)}{a^2-c^2 x^2} \, dx &=\int \left (-\frac {a^2 d}{c^4}-\frac {a^2 e x}{c^4}-\frac {d x^2}{c^2}-\frac {e x^3}{c^2}+\frac {a^4 d+a^4 e x}{c^4 \left (a^2-c^2 x^2\right )}\right ) \, dx\\ &=-\frac {a^2 d x}{c^4}-\frac {a^2 e x^2}{2 c^4}-\frac {d x^3}{3 c^2}-\frac {e x^4}{4 c^2}+\frac {\int \frac {a^4 d+a^4 e x}{a^2-c^2 x^2} \, dx}{c^4}\\ &=-\frac {a^2 d x}{c^4}-\frac {a^2 e x^2}{2 c^4}-\frac {d x^3}{3 c^2}-\frac {e x^4}{4 c^2}-\frac {\left (a^3 (c d-a e)\right ) \int \frac {1}{-a c-c^2 x} \, dx}{2 c^4}+\frac {\left (a^3 (c d+a e)\right ) \int \frac {1}{a c-c^2 x} \, dx}{2 c^4}\\ &=-\frac {a^2 d x}{c^4}-\frac {a^2 e x^2}{2 c^4}-\frac {d x^3}{3 c^2}-\frac {e x^4}{4 c^2}-\frac {a^3 (c d+a e) \log (a-c x)}{2 c^6}+\frac {a^3 (c d-a e) \log (a+c x)}{2 c^6}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 86, normalized size = 0.91 \begin {gather*} -\frac {a^2 d x}{c^4}-\frac {a^2 e x^2}{2 c^4}-\frac {d x^3}{3 c^2}-\frac {e x^4}{4 c^2}+\frac {a^3 d \tanh ^{-1}\left (\frac {c x}{a}\right )}{c^5}-\frac {a^4 e \log \left (a^2-c^2 x^2\right )}{2 c^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*d*x)/c^4) - (a^2*e*x^2)/(2*c^4) - (d*x^3)/(3*c^2) - (e*x^4)/(4*c^2) + (a^3*d*ArcTanh[(c*x)/a])/c^5 - (a
^4*e*Log[a^2 - c^2*x^2])/(2*c^6)

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Maple [A]
time = 0.62, size = 85, normalized size = 0.89

method	result	size
default	\(-\frac {\frac {1}{4} c^{2} e \,x^{4}+\frac {1}{3} c^{2} d \,x^{3}+\frac {1}{2} a^{2} e \,x^{2}+a^{2} d x}{c^{4}}-\frac {a^{3} \left (a e +c d \right ) \ln \left (-c x +a \right )}{2 c^{6}}-\frac {a^{3} \left (a e -c d \right ) \ln \left (c x +a \right )}{2 c^{6}}\)	\(85\)
norman	\(-\frac {d \,x^{3}}{3 c^{2}}-\frac {e \,x^{4}}{4 c^{2}}-\frac {a^{2} d x}{c^{4}}-\frac {a^{2} e \,x^{2}}{2 c^{4}}-\frac {a^{3} \left (a e -c d \right ) \ln \left (c x +a \right )}{2 c^{6}}-\frac {a^{3} \left (a e +c d \right ) \ln \left (-c x +a \right )}{2 c^{6}}\)	\(86\)
risch	\(-\frac {d \,x^{3}}{3 c^{2}}-\frac {e \,x^{4}}{4 c^{2}}-\frac {a^{2} d x}{c^{4}}-\frac {a^{2} e \,x^{2}}{2 c^{4}}-\frac {a^{4} \ln \left (c x +a \right ) e}{2 c^{6}}+\frac {a^{3} \ln \left (c x +a \right ) d}{2 c^{5}}-\frac {a^{4} \ln \left (-c x +a \right ) e}{2 c^{6}}-\frac {a^{3} \ln \left (-c x +a \right ) d}{2 c^{5}}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(e*x+d)/(-c^2*x^2+a^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 94, normalized size = 0.99 \begin {gather*} -\frac {3 \, c^{2} x^{4} e + 4 \, c^{2} d x^{3} + 6 \, a^{2} x^{2} e + 12 \, a^{2} d x}{12 \, c^{4}} + \frac {{\left (a^{3} c d - a^{4} e\right )} \log \left (c x + a\right )}{2 \, c^{6}} - \frac {{\left (a^{3} c d + a^{4} e\right )} \log \left (c x - a\right )}{2 \, c^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(-c^2*x^2+a^2),x, algorithm="maxima")

[Out]

-1/12*(3*c^2*x^4*e + 4*c^2*d*x^3 + 6*a^2*x^2*e + 12*a^2*d*x)/c^4 + 1/2*(a^3*c*d - a^4*e)*log(c*x + a)/c^6 - 1/
2*(a^3*c*d + a^4*e)*log(c*x - a)/c^6

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Fricas [A]
time = 3.58, size = 93, normalized size = 0.98 \begin {gather*} -\frac {4 \, c^{4} d x^{3} + 12 \, a^{2} c^{2} d x + 3 \, {\left (c^{4} x^{4} + 2 \, a^{2} c^{2} x^{2}\right )} e - 6 \, {\left (a^{3} c d - a^{4} e\right )} \log \left (c x + a\right ) + 6 \, {\left (a^{3} c d + a^{4} e\right )} \log \left (c x - a\right )}{12 \, c^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(-c^2*x^2+a^2),x, algorithm="fricas")

[Out]

-1/12*(4*c^4*d*x^3 + 12*a^2*c^2*d*x + 3*(c^4*x^4 + 2*a^2*c^2*x^2)*e - 6*(a^3*c*d - a^4*e)*log(c*x + a) + 6*(a^
3*c*d + a^4*e)*log(c*x - a))/c^6

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Sympy [A]
time = 0.21, size = 129, normalized size = 1.36 \begin {gather*} - \frac {a^{3} \left (a e - c d\right ) \log {\left (x + \frac {a^{4} e - a^{3} \left (a e - c d\right )}{a^{2} c^{2} d} \right )}}{2 c^{6}} - \frac {a^{3} \left (a e + c d\right ) \log {\left (x + \frac {a^{4} e - a^{3} \left (a e + c d\right )}{a^{2} c^{2} d} \right )}}{2 c^{6}} - \frac {a^{2} d x}{c^{4}} - \frac {a^{2} e x^{2}}{2 c^{4}} - \frac {d x^{3}}{3 c^{2}} - \frac {e x^{4}}{4 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(e*x+d)/(-c**2*x**2+a**2),x)

[Out]

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

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Giac [A]
time = 0.64, size = 102, normalized size = 1.07 \begin {gather*} \frac {{\left (a^{3} c d - a^{4} e\right )} \log \left ({\left | c x + a \right |}\right )}{2 \, c^{6}} - \frac {{\left (a^{3} c d + a^{4} e\right )} \log \left ({\left | c x - a \right |}\right )}{2 \, c^{6}} - \frac {3 \, c^{6} x^{4} e + 4 \, c^{6} d x^{3} + 6 \, a^{2} c^{4} x^{2} e + 12 \, a^{2} c^{4} d x}{12 \, c^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x+d)/(-c^2*x^2+a^2),x, algorithm="giac")

[Out]

1/2*(a^3*c*d - a^4*e)*log(abs(c*x + a))/c^6 - 1/2*(a^3*c*d + a^4*e)*log(abs(c*x - a))/c^6 - 1/12*(3*c^6*x^4*e
+ 4*c^6*d*x^3 + 6*a^2*c^4*x^2*e + 12*a^2*c^4*d*x)/c^8

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Mupad [B]
time = 0.10, size = 89, normalized size = 0.94 \begin {gather*} -\frac {\ln \left (a+c\,x\right )\,\left (a^4\,e-a^3\,c\,d\right )}{2\,c^6}-\frac {\ln \left (a-c\,x\right )\,\left (e\,a^4+c\,d\,a^3\right )}{2\,c^6}-\frac {d\,x^3}{3\,c^2}-\frac {e\,x^4}{4\,c^2}-\frac {a^2\,e\,x^2}{2\,c^4}-\frac {a^2\,d\,x}{c^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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