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) - (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)

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Rubi [A]  time = 0.619208, antiderivative size = 147, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [A]  time = 0.0900951, 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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Maple [C]  time = 0.54, size = 531, normalized size = 3.6 \begin{align*} -{\frac{ax-1}{60\,{x}^{7}{a}^{12}} \left ( 15\,{a}^{10}{x}^{10} \left ({\frac{ax+1}{ax}} \right ) ^{7/2} \left ( -{\frac{ax-1}{ax}} \right ) ^{5/2}+30\,{a}^{8}{x}^{8} \left ({\frac{ax+1}{ax}} \right ) ^{7/2} \left ( -{\frac{ax-1}{ax}} \right ) ^{5/2}+30\, \left ({\frac{ax+1}{ax}} \right ) ^{7/2} \left ( -{\frac{ax-1}{ax}} \right ) ^{3/2}{x}^{8}{a}^{8}+30\,{x}^{6}\ln \left ({a}^{2}{x}^{2} \right ) \left ({\frac{ax+1}{ax}} \right ) ^{7/2} \left ( -{\frac{ax-1}{ax}} \right ) ^{5/2}{a}^{6}-30\,{a}^{7}{x}^{7} \left ({\frac{ax+1}{ax}} \right ) ^{7/2} \left ( -{\frac{ax-1}{ax}} \right ) ^{3/2}+60\, \left ({\frac{ax+1}{ax}} \right ) ^{7/2} \left ( -{\frac{ax-1}{ax}} \right ) ^{3/2}\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{ax+1}{ax}} \right ) ^{7/2} \left ( -{\frac{ax-1}{ax}} \right ) ^{3/2}{a}^{5}+30\, \left ({\frac{ax+1}{ax}} \right ) ^{7/2}\sqrt{-{\frac{ax-1}{ax}}}\ln \left ({a}^{2}{x}^{2} \right ){x}^{6}{a}^{6}+12\,{x}^{10}{a}^{10}-60\, \left ({\frac{ax+1}{ax}} \right ) ^{7/2}\sqrt{-{\frac{ax-1}{ax}}}\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{ax+1}{ax}} \right ) ^{7/2}\sqrt{-{\frac{ax-1}{ax}}}{a}^{4}-40\,{x}^{8}{a}^{8}+40\,{a}^{7}{x}^{7}+40\,{x}^{6}{a}^{6}-20\,{x}^{3}{a}^{3}-20\,{a}^{2}{x}^{2}+8\,ax+8 \right ) \left ({\frac{ax+1}{ax}} \right ) ^{-{\frac{7}{2}}} \left ( -{\frac{ax-1}{ax}} \right ) ^{-{\frac{7}{2}}}} \end{align*}

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)^(7/2)*(-(a*x-1)/a/x)^(3/2)*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)^(7/2)*(-(a*x-1)/a/x)
^(3/2)*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)^(7/2)*(-(a*x-1)/a/x)^(1/2)*ln(a^2*x^2)*x^6*a^6+12*x^10*a^10-60*((a*x+1)/a/x)^(7/2)*(-(a*x-1)/a/x)^(
1/2)*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*a^7*x^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., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}}\,{d x} \end{align*}

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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Fricas [A]  time = 2.03716, size = 135, normalized size = 0.92 \begin{align*} \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}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \int \frac{x^{5}}{a x \sqrt{-1 + \frac{1}{a x}} \sqrt{1 + \frac{1}{a x}} + 1}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}}\,{d x} \end{align*}

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)