3.90 \(\int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx\)

Optimal. Leaf size=75 \[ \frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sin ^{-1}(c x)}{2 c^4}+\frac {\tanh ^{-1}(c x)}{c^4}-\frac {x \sqrt {1-c x}}{2 c^3 \sqrt {\frac {1}{c x+1}}}-\frac {x}{c^3} \]

[Out]

-x/c^3+arctanh(c*x)/c^4-1/2*x*(-c*x+1)^(1/2)/c^3/(1/(c*x+1))^(1/2)+1/2*arcsin(c*x)*(1/(c*x+1))^(1/2)*(c*x+1)^(
1/2)/c^4

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Rubi [A]  time = 0.17, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6341, 1956, 90, 41, 216, 321, 206} \[ -\frac {x \sqrt {1-c x}}{2 c^3 \sqrt {\frac {1}{c x+1}}}-\frac {x}{c^3}+\frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sin ^{-1}(c x)}{2 c^4}+\frac {\tanh ^{-1}(c x)}{c^4} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcSech[c*x]*x^3)/(1 - c^2*x^2),x]

[Out]

-(x/c^3) - (x*Sqrt[1 - c*x])/(2*c^3*Sqrt[(1 + c*x)^(-1)]) + (Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/(
2*c^4) + ArcTanh[c*x]/c^4

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 1956

Int[(x_)^(m_.)*((e_.)*((a_) + (b_.)*(x_)^(n_.))^(r_.))^(p_)*((f_.)*((c_) + (d_.)*(x_)^(n_.))^(s_))^(q_), x_Sym
bol] :> Dist[((e*(a + b*x^n)^r)^p*(f*(c + d*x^n)^s)^q)/((a + b*x^n)^(p*r)*(c + d*x^n)^(q*s)), Int[x^m*(a + b*x
^n)^(p*r)*(c + d*x^n)^(q*s), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r, s}, x]

Rule 6341

Int[(E^ArcSech[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d/(a*c), Int[((d*x)^(m
- 1)*Sqrt[1/(1 + c*x)])/Sqrt[1 - c*x], x], x] + Dist[d/c, Int[(d*x)^(m - 1)/(a + b*x^2), x], x] /; FreeQ[{a,
b, c, d, m}, x] && EqQ[b + a*c^2, 0]

Rubi steps

\begin {align*} \int \frac {e^{\text {sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx &=\frac {\int \frac {x^2 \sqrt {\frac {1}{1+c x}}}{\sqrt {1-c x}} \, dx}{c}+\frac {\int \frac {x^2}{1-c^2 x^2} \, dx}{c}\\ &=-\frac {x}{c^3}+\frac {\int \frac {1}{1-c^2 x^2} \, dx}{c^3}+\frac {\left (\sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x^2}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{c}\\ &=-\frac {x}{c^3}-\frac {x \sqrt {1-c x}}{2 c^3 \sqrt {\frac {1}{1+c x}}}+\frac {\tanh ^{-1}(c x)}{c^4}+\frac {\left (\sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{2 c^3}\\ &=-\frac {x}{c^3}-\frac {x \sqrt {1-c x}}{2 c^3 \sqrt {\frac {1}{1+c x}}}+\frac {\tanh ^{-1}(c x)}{c^4}+\frac {\left (\sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{2 c^3}\\ &=-\frac {x}{c^3}-\frac {x \sqrt {1-c x}}{2 c^3 \sqrt {\frac {1}{1+c x}}}+\frac {\sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sin ^{-1}(c x)}{2 c^4}+\frac {\tanh ^{-1}(c x)}{c^4}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 110, normalized size = 1.47 \[ -\frac {c^2 x^2 \sqrt {\frac {1-c x}{c x+1}}+2 c x+c x \sqrt {\frac {1-c x}{c x+1}}+\log (1-c x)-\log (c x+1)-i \log \left (2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)-2 i c x\right )}{2 c^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcSech[c*x]*x^3)/(1 - c^2*x^2),x]

[Out]

-1/2*(2*c*x + c*x*Sqrt[(1 - c*x)/(1 + c*x)] + c^2*x^2*Sqrt[(1 - c*x)/(1 + c*x)] + Log[1 - c*x] - Log[1 + c*x]
- I*Log[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)])/c^4

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fricas [B]  time = 1.57, size = 91, normalized size = 1.21 \[ -\frac {c^{2} x^{2} \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} + 2 \, c x + \arctan \left (\sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}}\right ) - \log \left (c x + 1\right ) + \log \left (c x - 1\right )}{2 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))*x^3/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

-1/2*(c^2*x^2*sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1)/(c*x)) + 2*c*x + arctan(sqrt((c*x + 1)/(c*x))*sqrt(-(c*x -
1)/(c*x))) - log(c*x + 1) + log(c*x - 1))/c^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{3} {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))*x^3/(-c^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-x^3*(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))/(c^2*x^2 - 1), x)

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maple [C]  time = 0.08, size = 117, normalized size = 1.56 \[ -\frac {\sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}\, \left (x \sqrt {-c^{2} x^{2}+1}\, \mathrm {csgn}\relax (c ) c -\arctan \left (\frac {\mathrm {csgn}\relax (c ) c x}{\sqrt {-c^{2} x^{2}+1}}\right )\right ) \mathrm {csgn}\relax (c )}{2 c^{3} \sqrt {-c^{2} x^{2}+1}}-\frac {x}{c^{3}}+\frac {\ln \left (c x +1\right )}{2 c^{4}}-\frac {\ln \left (c x -1\right )}{2 c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))*x^3/(-c^2*x^2+1),x)

[Out]

-1/2*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)/c^3*(x*(-c^2*x^2+1)^(1/2)*csgn(c)*c-arctan(csgn(c)*c*x/(-c^2*x
^2+1)^(1/2)))/(-c^2*x^2+1)^(1/2)*csgn(c)-x/c^3+1/2/c^4*ln(c*x+1)-1/2/c^4*ln(c*x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {x}{c^{3}} + \frac {\log \left (c x + 1\right )}{2 \, c^{4}} - \frac {\log \left (c x - 1\right )}{2 \, c^{4}} - \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} x^{2}}{c^{3} x^{2} - c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))*x^3/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

-x/c^3 + 1/2*log(c*x + 1)/c^4 - 1/2*log(c*x - 1)/c^4 - integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^2/(c^3*x^2 - c
), x)

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mupad [B]  time = 8.08, size = 340, normalized size = 4.53 \[ \frac {\mathrm {atanh}\left (c\,x\right )-c\,x}{c^4}-\frac {\ln \left (\frac {\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{c\,x}+1}-1}\right )\,1{}\mathrm {i}}{2\,c^4}-\frac {\frac {1{}\mathrm {i}}{32\,c^4}+\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,c^4\,{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{32\,c^4\,{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^4}}{\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^6}}+\frac {\ln \left (\frac {2\,c\,\sqrt {\frac {c+\frac {1}{x}}{c}}-\frac {2}{x}+c\,\sqrt {-\frac {c-\frac {1}{x}}{c}}\,2{}\mathrm {i}}{2\,c+\frac {1}{x}-2\,c\,\sqrt {\frac {c+\frac {1}{x}}{c}}}\right )\,1{}\mathrm {i}}{2\,c^4}-\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{32\,c^4\,{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3*((1/(c*x) - 1)^(1/2)*(1/(c*x) + 1)^(1/2) + 1/(c*x)))/(c^2*x^2 - 1),x)

[Out]

(atanh(c*x) - c*x)/c^4 - (log(((1/(c*x) - 1)^(1/2) - 1i)/((1/(c*x) + 1)^(1/2) - 1))*1i)/(2*c^4) - (1i/(32*c^4)
+ (((1/(c*x) - 1)^(1/2) - 1i)^2*1i)/(16*c^4*((1/(c*x) + 1)^(1/2) - 1)^2) - (((1/(c*x) - 1)^(1/2) - 1i)^4*15i)
/(32*c^4*((1/(c*x) + 1)^(1/2) - 1)^4))/(((1/(c*x) - 1)^(1/2) - 1i)^2/((1/(c*x) + 1)^(1/2) - 1)^2 + (2*((1/(c*x
) - 1)^(1/2) - 1i)^4)/((1/(c*x) + 1)^(1/2) - 1)^4 + ((1/(c*x) - 1)^(1/2) - 1i)^6/((1/(c*x) + 1)^(1/2) - 1)^6)
+ (log((c*(-(c - 1/x)/c)^(1/2)*2i - 2/x + 2*c*((c + 1/x)/c)^(1/2))/(2*c + 1/x - 2*c*((c + 1/x)/c)^(1/2)))*1i)/
(2*c^4) - (((1/(c*x) - 1)^(1/2) - 1i)^2*1i)/(32*c^4*((1/(c*x) + 1)^(1/2) - 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {x^{2}}{c^{2} x^{2} - 1}\, dx + \int \frac {c x^{3} \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{2} - 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(-1+1/c/x)**(1/2)*(1+1/c/x)**(1/2))*x**3/(-c**2*x**2+1),x)

[Out]

-(Integral(x**2/(c**2*x**2 - 1), x) + Integral(c*x**3*sqrt(-1 + 1/(c*x))*sqrt(1 + 1/(c*x))/(c**2*x**2 - 1), x)
)/c

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