3.77 \(\int e^{-\text {sech}^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=163 \[ -\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1)^4}{4 a^4}+\frac {\left (9 \sqrt {\frac {1-a x}{a x+1}}+4\right ) (a x+1)^3}{12 a^4}-\frac {\left (5 \sqrt {\frac {1-a x}{a x+1}}+8\right ) (a x+1)^2}{8 a^4}+\frac {\left (\sqrt {\frac {1-a x}{a x+1}}+8\right ) (a x+1)}{8 a^4}+\frac {\tan ^{-1}\left (\sqrt {\frac {1-a x}{a x+1}}\right )}{4 a^4} \]

[Out]

1/4*arctan(((-a*x+1)/(a*x+1))^(1/2))/a^4-1/4*(a*x+1)^4*((-a*x+1)/(a*x+1))^(1/2)/a^4+1/8*(a*x+1)*(8+((-a*x+1)/(
a*x+1))^(1/2))/a^4-1/8*(a*x+1)^2*(8+5*((-a*x+1)/(a*x+1))^(1/2))/a^4+1/12*(a*x+1)^3*(4+9*((-a*x+1)/(a*x+1))^(1/
2))/a^4

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Rubi [A]  time = 0.57, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6337, 1804, 1814, 639, 203} \[ -\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1)^4}{4 a^4}+\frac {\left (9 \sqrt {\frac {1-a x}{a x+1}}+4\right ) (a x+1)^3}{12 a^4}-\frac {\left (5 \sqrt {\frac {1-a x}{a x+1}}+8\right ) (a x+1)^2}{8 a^4}+\frac {\left (\sqrt {\frac {1-a x}{a x+1}}+8\right ) (a x+1)}{8 a^4}+\frac {\tan ^{-1}\left (\sqrt {\frac {1-a x}{a x+1}}\right )}{4 a^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3/E^ArcSech[a*x],x]

[Out]

-(Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^4)/(4*a^4) + ((1 + a*x)*(8 + Sqrt[(1 - a*x)/(1 + a*x)]))/(8*a^4) - ((1 +
a*x)^2*(8 + 5*Sqrt[(1 - a*x)/(1 + a*x)]))/(8*a^4) + ((1 + a*x)^3*(4 + 9*Sqrt[(1 - a*x)/(1 + a*x)]))/(12*a^4)
+ ArcTan[Sqrt[(1 - a*x)/(1 + a*x)]]/(4*a^4)

Rule 203

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

Rule 639

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

Rule 1804

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[((c*x)^m*(a + b*x^2)^(p + 1)*(a*g - b*f*x))/(2*a*b*(p + 1)), x] + Dist[c/(2*a*b*(p + 1)), Int[
(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 6337

Int[E^(ArcSech[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[(1 - u)/(1 + u)] + (1*Sqrt[(1 - u)/(1 +
u)])/u)^n, x] /; FreeQ[m, x] && IntegerQ[n]

Rubi steps

\begin {align*} \int e^{-\text {sech}^{-1}(a x)} x^3 \, dx &=\int \frac {x^3}{\frac {1}{a x}+\sqrt {\frac {1-a x}{1+a x}}+\frac {\sqrt {\frac {1-a x}{1+a x}}}{a x}} \, dx\\ &=-\frac {4 \operatorname {Subst}\left (\int \frac {(-1+x)^4 x (1+x)^2}{\left (1+x^2\right )^5} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{a^4}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac {\operatorname {Subst}\left (\int \frac {8-8 x-48 x^2+16 x^3+16 x^4-8 x^5}{\left (1+x^2\right )^4} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{2 a^4}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac {(1+a x)^3 \left (4+9 \sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}-\frac {\operatorname {Subst}\left (\int \frac {24-144 x-96 x^2+48 x^3}{\left (1+x^2\right )^3} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}-\frac {(1+a x)^2 \left (8+5 \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}+\frac {(1+a x)^3 \left (4+9 \sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}+\frac {\operatorname {Subst}\left (\int \frac {24-192 x}{\left (1+x^2\right )^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{48 a^4}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac {(1+a x) \left (8+\sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}-\frac {(1+a x)^2 \left (8+5 \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}+\frac {(1+a x)^3 \left (4+9 \sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{4 a^4}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac {(1+a x) \left (8+\sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}-\frac {(1+a x)^2 \left (8+5 \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a^4}+\frac {(1+a x)^3 \left (4+9 \sqrt {\frac {1-a x}{1+a x}}\right )}{12 a^4}+\frac {\tan ^{-1}\left (\sqrt {\frac {1-a x}{1+a x}}\right )}{4 a^4}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 97, normalized size = 0.60 \[ \frac {8 a^3 x^3+3 a \sqrt {\frac {1-a x}{a x+1}} \left (-2 a^3 x^4-2 a^2 x^3+a x^2+x\right )-3 i \log \left (2 \sqrt {\frac {1-a x}{a x+1}} (a x+1)-2 i a x\right )}{24 a^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3/E^ArcSech[a*x],x]

[Out]

(8*a^3*x^3 + 3*a*Sqrt[(1 - a*x)/(1 + a*x)]*(x + a*x^2 - 2*a^2*x^3 - 2*a^3*x^4) - (3*I)*Log[(-2*I)*a*x + 2*Sqrt
[(1 - a*x)/(1 + a*x)]*(1 + a*x)])/(24*a^4)

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fricas [A]  time = 0.78, size = 95, normalized size = 0.58 \[ \frac {8 \, a^{3} x^{3} - 3 \, {\left (2 \, a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 3 \, \arctan \left (\sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}\right )}{24 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*a^3*x^3 - 3*(2*a^4*x^4 - a^2*x^2)*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 3*arctan(sqrt((a*x +
1)/(a*x))*sqrt(-(a*x - 1)/(a*x))))/a^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 19.83, size = 795, normalized size = 4.88 \[ \frac {\ln \left (\frac {a\,\sqrt {\frac {1}{a\,x}+1}-\frac {1}{x}+a\,\sqrt {\frac {1}{a\,x}-1}\,1{}\mathrm {i}}{2\,a-2\,a\,\sqrt {\frac {1}{a\,x}+1}+\frac {1}{x}}\right )\,3{}\mathrm {i}}{8\,a^4}+\frac {\frac {1{}\mathrm {i}}{1024\,a^4}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,3{}\mathrm {i}}{128\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,53{}\mathrm {i}}{512\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6\,87{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8\,657{}\mathrm {i}}{1024\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{10}\,121{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{10}}}{\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}+\frac {6\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{10}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{10}}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{12}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{12}}}+\frac {\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\,1{}\mathrm {i}}{8\,a^4}+\frac {\frac {1{}\mathrm {i}}{32\,a^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{32\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}}{\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}}-\frac {\ln \left (\frac {2\,a\,\sqrt {\frac {a+\frac {1}{x}}{a}}-\frac {2}{x}+a\,\sqrt {-\frac {a-\frac {1}{x}}{a}}\,2{}\mathrm {i}}{2\,a+\frac {1}{x}-2\,a\,\sqrt {\frac {a+\frac {1}{x}}{a}}}\right )\,1{}\mathrm {i}}{2\,a^4}+\frac {x^3}{3\,a}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,1{}\mathrm {i}}{1024\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((a*(1/(a*x) - 1)^(1/2)*1i + a*(1/(a*x) + 1)^(1/2) - 1/x)/(2*a - 2*a*(1/(a*x) + 1)^(1/2) + 1/x))*3i)/(8*a^
4) + (1i/(1024*a^4) - (((1/(a*x) - 1)^(1/2) - 1i)^2*3i)/(128*a^4*((1/(a*x) + 1)^(1/2) - 1)^2) - (((1/(a*x) - 1
)^(1/2) - 1i)^4*53i)/(512*a^4*((1/(a*x) + 1)^(1/2) - 1)^4) + (((1/(a*x) - 1)^(1/2) - 1i)^6*87i)/(256*a^4*((1/(
a*x) + 1)^(1/2) - 1)^6) + (((1/(a*x) - 1)^(1/2) - 1i)^8*657i)/(1024*a^4*((1/(a*x) + 1)^(1/2) - 1)^8) + (((1/(a
*x) - 1)^(1/2) - 1i)^10*121i)/(256*a^4*((1/(a*x) + 1)^(1/2) - 1)^10))/(((1/(a*x) - 1)^(1/2) - 1i)^4/((1/(a*x)
+ 1)^(1/2) - 1)^4 + (4*((1/(a*x) - 1)^(1/2) - 1i)^6)/((1/(a*x) + 1)^(1/2) - 1)^6 + (6*((1/(a*x) - 1)^(1/2) - 1
i)^8)/((1/(a*x) + 1)^(1/2) - 1)^8 + (4*((1/(a*x) - 1)^(1/2) - 1i)^10)/((1/(a*x) + 1)^(1/2) - 1)^10 + ((1/(a*x)
- 1)^(1/2) - 1i)^12/((1/(a*x) + 1)^(1/2) - 1)^12) + (log(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1)
)*1i)/(8*a^4) + (1i/(32*a^4) + (((1/(a*x) - 1)^(1/2) - 1i)^2*1i)/(16*a^4*((1/(a*x) + 1)^(1/2) - 1)^2) - (((1/(
a*x) - 1)^(1/2) - 1i)^4*15i)/(32*a^4*((1/(a*x) + 1)^(1/2) - 1)^4))/(((1/(a*x) - 1)^(1/2) - 1i)^2/((1/(a*x) + 1
)^(1/2) - 1)^2 + (2*((1/(a*x) - 1)^(1/2) - 1i)^4)/((1/(a*x) + 1)^(1/2) - 1)^4 + ((1/(a*x) - 1)^(1/2) - 1i)^6/(
(1/(a*x) + 1)^(1/2) - 1)^6) - (log((a*(-(a - 1/x)/a)^(1/2)*2i - 2/x + 2*a*((a + 1/x)/a)^(1/2))/(2*a + 1/x - 2*
a*((a + 1/x)/a)^(1/2)))*1i)/(2*a^4) + x^3/(3*a) + (((1/(a*x) - 1)^(1/2) - 1i)^2*1i)/(256*a^4*((1/(a*x) + 1)^(1
/2) - 1)^2) + (((1/(a*x) - 1)^(1/2) - 1i)^4*1i)/(1024*a^4*((1/(a*x) + 1)^(1/2) - 1)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \int \frac {x^{4}}{a x \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Integral(x**4/(a*x*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)) + 1), x)

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