3.76 \(\int e^{-\text {sech}^{-1}(a x)} x^4 \, dx\)

Optimal. Leaf size=147 \[ -\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1)^5}{5 a^5}+\frac {\left (16 \sqrt {\frac {1-a x}{a x+1}}+5\right ) (a x+1)^4}{20 a^5}-\frac {\left (17 \sqrt {\frac {1-a x}{a x+1}}+15\right ) (a x+1)^3}{15 a^5}+\frac {\left (4 \sqrt {\frac {1-a x}{a x+1}}+9\right ) (a x+1)^2}{6 a^5}-\frac {x}{a^4} \]

[Out]

-x/a^4-1/5*(a*x+1)^5*((-a*x+1)/(a*x+1))^(1/2)/a^5+1/6*(a*x+1)^2*(9+4*((-a*x+1)/(a*x+1))^(1/2))/a^5+1/20*(a*x+1
)^4*(5+16*((-a*x+1)/(a*x+1))^(1/2))/a^5-1/15*(a*x+1)^3*(15+17*((-a*x+1)/(a*x+1))^(1/2))/a^5

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Rubi [A]  time = 0.62, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6337, 1804, 1814, 12, 261} \[ -\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1)^5}{5 a^5}+\frac {\left (16 \sqrt {\frac {1-a x}{a x+1}}+5\right ) (a x+1)^4}{20 a^5}-\frac {\left (17 \sqrt {\frac {1-a x}{a x+1}}+15\right ) (a x+1)^3}{15 a^5}+\frac {\left (4 \sqrt {\frac {1-a x}{a x+1}}+9\right ) (a x+1)^2}{6 a^5}-\frac {x}{a^4} \]

Antiderivative was successfully verified.

[In]

Int[x^4/E^ArcSech[a*x],x]

[Out]

-(x/a^4) - (Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^5)/(5*a^5) + ((1 + a*x)^2*(9 + 4*Sqrt[(1 - a*x)/(1 + a*x)]))/(
6*a^5) + ((1 + a*x)^4*(5 + 16*Sqrt[(1 - a*x)/(1 + a*x)]))/(20*a^5) - ((1 + a*x)^3*(15 + 17*Sqrt[(1 - a*x)/(1 +
a*x)]))/(15*a^5)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

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

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^4 \, dx &=\int \frac {x^4}{\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)^5 x (1+x)^3}{\left (1+x^2\right )^6} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{a^5}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^5}{5 a^5}-\frac {2 \operatorname {Subst}\left (\int \frac {-16+10 x+140 x^2-30 x^3-80 x^4+30 x^5+20 x^6-10 x^7}{\left (1+x^2\right )^5} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{5 a^5}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^5}{5 a^5}+\frac {(1+a x)^4 \left (5+16 \sqrt {\frac {1-a x}{1+a x}}\right )}{20 a^5}+\frac {\operatorname {Subst}\left (\int \frac {-128+560 x+800 x^2-320 x^3-160 x^4+80 x^5}{\left (1+x^2\right )^4} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{20 a^5}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^5}{5 a^5}+\frac {(1+a x)^4 \left (5+16 \sqrt {\frac {1-a x}{1+a x}}\right )}{20 a^5}-\frac {(1+a x)^3 \left (15+17 \sqrt {\frac {1-a x}{1+a x}}\right )}{15 a^5}-\frac {\operatorname {Subst}\left (\int \frac {-320+2400 x+960 x^2-480 x^3}{\left (1+x^2\right )^3} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{120 a^5}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^5}{5 a^5}+\frac {(1+a x)^2 \left (9+4 \sqrt {\frac {1-a x}{1+a x}}\right )}{6 a^5}+\frac {(1+a x)^4 \left (5+16 \sqrt {\frac {1-a x}{1+a x}}\right )}{20 a^5}-\frac {(1+a x)^3 \left (15+17 \sqrt {\frac {1-a x}{1+a x}}\right )}{15 a^5}+\frac {\operatorname {Subst}\left (\int \frac {1920 x}{\left (1+x^2\right )^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{480 a^5}\\ &=-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^5}{5 a^5}+\frac {(1+a x)^2 \left (9+4 \sqrt {\frac {1-a x}{1+a x}}\right )}{6 a^5}+\frac {(1+a x)^4 \left (5+16 \sqrt {\frac {1-a x}{1+a x}}\right )}{20 a^5}-\frac {(1+a x)^3 \left (15+17 \sqrt {\frac {1-a x}{1+a x}}\right )}{15 a^5}+\frac {4 \operatorname {Subst}\left (\int \frac {x}{\left (1+x^2\right )^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{a^5}\\ &=-\frac {x}{a^4}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)^5}{5 a^5}+\frac {(1+a x)^2 \left (9+4 \sqrt {\frac {1-a x}{1+a x}}\right )}{6 a^5}+\frac {(1+a x)^4 \left (5+16 \sqrt {\frac {1-a x}{1+a x}}\right )}{20 a^5}-\frac {(1+a x)^3 \left (15+17 \sqrt {\frac {1-a x}{1+a x}}\right )}{15 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 65, normalized size = 0.44 \[ \frac {15 a^4 x^4-4 \sqrt {\frac {1-a x}{a x+1}} (a x+1)^2 \left (3 a^3 x^3-3 a^2 x^2+2 a x-2\right )}{60 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/E^ArcSech[a*x],x]

[Out]

(15*a^4*x^4 - 4*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^2*(-2 + 2*a*x - 3*a^2*x^2 + 3*a^3*x^3))/(60*a^5)

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fricas [A]  time = 1.97, size = 65, normalized size = 0.44 \[ \frac {15 \, a^{3} x^{4} - 4 \, {\left (3 \, a^{4} x^{5} - a^{2} x^{3} - 2 \, x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}}{60 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)),x, algorithm="fricas")

[Out]

1/60*(15*a^3*x^4 - 4*(3*a^4*x^5 - a^2*x^3 - 2*x)*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 \[ \int \frac {x^{4}}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.36, size = 531, normalized size = 3.61 \[ -\frac {\left (a x -1\right ) \left (15 a^{10} x^{10} \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {5}{2}}+30 a^{8} x^{8} \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {5}{2}}+30 \left (-\frac {a x -1}{a x}\right )^{\frac {3}{2}} \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} x^{8} a^{8}+30 x^{6} \ln \left (a^{2} x^{2}\right ) \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {5}{2}} a^{6}-30 a^{7} x^{7} \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {3}{2}}+60 \left (-\frac {a x -1}{a x}\right )^{\frac {3}{2}} \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \ln \left (a^{2} x^{2}\right ) x^{6} a^{6}+12 x^{11} a^{11}-60 x^{5} \ln \left (a^{2} x^{2}\right ) \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {3}{2}} a^{5}+30 \sqrt {-\frac {a x -1}{a x}}\, \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \ln \left (a^{2} x^{2}\right ) x^{6} a^{6}+12 x^{10} a^{10}-60 \sqrt {-\frac {a x -1}{a x}}\, \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \ln \left (a^{2} x^{2}\right ) x^{5} a^{5}-40 x^{9} a^{9}+30 x^{4} \ln \left (a^{2} x^{2}\right ) \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \sqrt {-\frac {a x -1}{a x}}\, a^{4}-40 x^{8} a^{8}+40 x^{7} a^{7}+40 x^{6} a^{6}-20 x^{3} a^{3}-20 a^{2} x^{2}+8 a x +8\right )}{60 x^{7} a^{12} \left (\frac {a x +1}{a x}\right )^{\frac {7}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)),x)

[Out]

-1/60*(a*x-1)/x^7*(15*a^10*x^10*((a*x+1)/a/x)^(7/2)*(-(a*x-1)/a/x)^(5/2)+30*a^8*x^8*((a*x+1)/a/x)^(7/2)*(-(a*x
-1)/a/x)^(5/2)+30*(-(a*x-1)/a/x)^(3/2)*((a*x+1)/a/x)^(7/2)*x^8*a^8+30*x^6*ln(a^2*x^2)*((a*x+1)/a/x)^(7/2)*(-(a
*x-1)/a/x)^(5/2)*a^6-30*a^7*x^7*((a*x+1)/a/x)^(7/2)*(-(a*x-1)/a/x)^(3/2)+60*(-(a*x-1)/a/x)^(3/2)*((a*x+1)/a/x)
^(7/2)*ln(a^2*x^2)*x^6*a^6+12*x^11*a^11-60*x^5*ln(a^2*x^2)*((a*x+1)/a/x)^(7/2)*(-(a*x-1)/a/x)^(3/2)*a^5+30*(-(
a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(7/2)*ln(a^2*x^2)*x^6*a^6+12*x^10*a^10-60*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(
7/2)*ln(a^2*x^2)*x^5*a^5-40*x^9*a^9+30*x^4*ln(a^2*x^2)*((a*x+1)/a/x)^(7/2)*(-(a*x-1)/a/x)^(1/2)*a^4-40*x^8*a^8
+40*x^7*a^7+40*x^6*a^6-20*x^3*a^3-20*a^2*x^2+8*a*x+8)/a^12/((a*x+1)/a/x)^(7/2)/(-(a*x-1)/a/x)^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.14, size = 73, normalized size = 0.50 \[ \frac {x^4}{4\,a}+\frac {\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {2\,x}{15\,a^4}+\frac {2}{15\,a^5}-\frac {x^5}{5}-\frac {x^4}{5\,a}+\frac {x^3}{15\,a^2}+\frac {x^2}{15\,a^3}\right )}{\sqrt {\frac {1}{a\,x}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x)),x)

[Out]

x^4/(4*a) + ((1/(a*x) - 1)^(1/2)*((2*x)/(15*a^4) + 2/(15*a^5) - x^5/5 - x^4/(5*a) + x^3/(15*a^2) + x^2/(15*a^3
)))/(1/(a*x) + 1)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \int \frac {x^{5}}{a x \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(1/a/x+(1/a/x-1)**(1/2)*(1+1/a/x)**(1/2)),x)

[Out]

a*Integral(x**5/(a*x*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)) + 1), x)

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