3.65 \(\int e^{2 \text{sech}^{-1}(a x)} x^4 \, dx\)

Optimal. Leaf size=203 \[ \frac{(1-a x) (a x+1)^4}{5 a^5}+\frac{\sqrt{\frac{1-a x}{a x+1}} \left (5-6 \sqrt{\frac{1-a x}{a x+1}}\right ) (a x+1)^4}{10 a^5}-\frac{\left (45 \sqrt{\frac{1-a x}{a x+1}}+4\right ) (a x+1)^3}{30 a^5}+\frac{5 \sqrt{\frac{1-a x}{a x+1}} (a x+1)^2}{4 a^5}+\frac{\left (4-\sqrt{\frac{1-a x}{a x+1}}\right ) (a x+1)}{4 a^5}-\frac{\tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{2 a^5} \]

[Out]

(5*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^2)/(4*a^5) + ((1 - a*x)*(1 + a*x)^4)/(5*a^5) + (Sqrt[(1 - a*x)/(1 + a*x
)]*(1 + a*x)^4*(5 - 6*Sqrt[(1 - a*x)/(1 + a*x)]))/(10*a^5) + ((1 + a*x)*(4 - Sqrt[(1 - a*x)/(1 + a*x)]))/(4*a^
5) - ((1 + a*x)^3*(4 + 45*Sqrt[(1 - a*x)/(1 + a*x)]))/(30*a^5) - ArcTan[Sqrt[(1 - a*x)/(1 + a*x)]]/(2*a^5)

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Rubi [A]  time = 0.699971, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6337, 1804, 1811, 1814, 639, 203} \[ \frac{(1-a x) (a x+1)^4}{5 a^5}+\frac{\sqrt{\frac{1-a x}{a x+1}} \left (5-6 \sqrt{\frac{1-a x}{a x+1}}\right ) (a x+1)^4}{10 a^5}-\frac{\left (45 \sqrt{\frac{1-a x}{a x+1}}+4\right ) (a x+1)^3}{30 a^5}+\frac{5 \sqrt{\frac{1-a x}{a x+1}} (a x+1)^2}{4 a^5}+\frac{\left (4-\sqrt{\frac{1-a x}{a x+1}}\right ) (a x+1)}{4 a^5}-\frac{\tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{2 a^5} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcSech[a*x])*x^4,x]

[Out]

(5*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^2)/(4*a^5) + ((1 - a*x)*(1 + a*x)^4)/(5*a^5) + (Sqrt[(1 - a*x)/(1 + a*x
)]*(1 + a*x)^4*(5 - 6*Sqrt[(1 - a*x)/(1 + a*x)]))/(10*a^5) + ((1 + a*x)*(4 - Sqrt[(1 - a*x)/(1 + a*x)]))/(4*a^
5) - ((1 + a*x)^3*(4 + 45*Sqrt[(1 - a*x)/(1 + a*x)]))/(30*a^5) - ArcTan[Sqrt[(1 - a*x)/(1 + a*x)]]/(2*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 1811

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[x*PolynomialQuotient[Pq, x, x]*(a + b*x^2)^p, x] /; Fre
eQ[{a, b, p}, x] && PolyQ[Pq, x] && EqQ[Coeff[Pq, x, 0], 0] &&  !MatchQ[Pq, x^(m_.)*(u_.) /; IntegerQ[m]]

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 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 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])

Rubi steps

\begin{align*} \int e^{2 \text{sech}^{-1}(a x)} x^4 \, dx &=\int x^4 \left (\frac{1}{a x}+\sqrt{\frac{1-a x}{1+a x}}+\frac{\sqrt{\frac{1-a x}{1+a x}}}{a x}\right )^2 \, dx\\ &=-\frac{4 \operatorname{Subst}\left (\int \frac{(-1+x)^2 x (1+x)^6}{\left (1+x^2\right )^6} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a^5}\\ &=\frac{(1-a x) (1+a x)^4}{5 a^5}+\frac{2 \operatorname{Subst}\left (\int \frac{-42 x-40 x^2+130 x^3+80 x^4-30 x^5-40 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{(1-a x) (1+a x)^4}{5 a^5}+\frac{2 \operatorname{Subst}\left (\int \frac{x \left (-42-40 x+130 x^2+80 x^3-30 x^4-40 x^5-10 x^6\right )}{\left (1+x^2\right )^5} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{5 a^5}\\ &=\frac{(1-a x) (1+a x)^4}{5 a^5}+\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4 \left (5-6 \sqrt{\frac{1-a x}{1+a x}}\right )}{10 a^5}-\frac{\operatorname{Subst}\left (\int \frac{160-48 x-960 x^2+160 x^3+320 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{(1-a x) (1+a x)^4}{5 a^5}+\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4 \left (5-6 \sqrt{\frac{1-a x}{1+a x}}\right )}{10 a^5}-\frac{(1+a x)^3 \left (4+45 \sqrt{\frac{1-a x}{1+a x}}\right )}{30 a^5}+\frac{\operatorname{Subst}\left (\int \frac{480-480 x-1920 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{5 \sqrt{\frac{1-a x}{1+a x}} (1+a x)^2}{4 a^5}+\frac{(1-a x) (1+a x)^4}{5 a^5}+\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4 \left (5-6 \sqrt{\frac{1-a x}{1+a x}}\right )}{10 a^5}-\frac{(1+a x)^3 \left (4+45 \sqrt{\frac{1-a x}{1+a x}}\right )}{30 a^5}-\frac{\operatorname{Subst}\left (\int \frac{480+1920 x}{\left (1+x^2\right )^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{480 a^5}\\ &=\frac{5 \sqrt{\frac{1-a x}{1+a x}} (1+a x)^2}{4 a^5}+\frac{(1-a x) (1+a x)^4}{5 a^5}+\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4 \left (5-6 \sqrt{\frac{1-a x}{1+a x}}\right )}{10 a^5}+\frac{(1+a x) \left (4-\sqrt{\frac{1-a x}{1+a x}}\right )}{4 a^5}-\frac{(1+a x)^3 \left (4+45 \sqrt{\frac{1-a x}{1+a x}}\right )}{30 a^5}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{2 a^5}\\ &=\frac{5 \sqrt{\frac{1-a x}{1+a x}} (1+a x)^2}{4 a^5}+\frac{(1-a x) (1+a x)^4}{5 a^5}+\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4 \left (5-6 \sqrt{\frac{1-a x}{1+a x}}\right )}{10 a^5}+\frac{(1+a x) \left (4-\sqrt{\frac{1-a x}{1+a x}}\right )}{4 a^5}-\frac{(1+a x)^3 \left (4+45 \sqrt{\frac{1-a x}{1+a x}}\right )}{30 a^5}-\frac{\tan ^{-1}\left (\sqrt{\frac{1-a x}{1+a x}}\right )}{2 a^5}\\ \end{align*}

Mathematica [C]  time = 0.141372, size = 105, normalized size = 0.52 \[ \frac{-12 a^5 x^5+40 a^3 x^3-15 a \sqrt{\frac{1-a x}{a x+1}} \left (-2 a^3 x^4-2 a^2 x^3+a x^2+x\right )+15 i \log \left (2 \sqrt{\frac{1-a x}{a x+1}} (a x+1)-2 i a x\right )}{60 a^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcSech[a*x])*x^4,x]

[Out]

(40*a^3*x^3 - 12*a^5*x^5 - 15*a*Sqrt[(1 - a*x)/(1 + a*x)]*(x + a*x^2 - 2*a^2*x^3 - 2*a^3*x^4) + (15*I)*Log[(-2
*I)*a*x + 2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)])/(60*a^5)

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Maple [C]  time = 0.206, size = 123, normalized size = 0.6 \begin{align*} -{\frac{{x}^{5}}{5}}+{\frac{2\,{x}^{3}}{3\,{a}^{2}}}+{\frac{x{\it csgn} \left ( a \right ) }{4\,{a}^{4}}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}} \left ( 2\,{\it csgn} \left ( a \right ){x}^{3}{a}^{3}\sqrt{-{a}^{2}{x}^{2}+1}-x\sqrt{-{a}^{2}{x}^{2}+1}{\it csgn} \left ( a \right ) a+\arctan \left ({x{\it csgn} \left ( a \right ) a{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, x^{3}}{3 \, a^{2}} + \frac{2 \,{\left (-\frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{4 \, a^{2}} + \frac{\sqrt{-a^{2} x^{2} + 1} x}{8 \, a^{2}} + \frac{\arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{8 \, \sqrt{a^{2}} a^{2}}\right )}}{a^{2}} - \int x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.09114, size = 225, normalized size = 1.11 \begin{align*} -\frac{12 \, a^{5} x^{5} - 40 \, a^{3} x^{3} - 15 \,{\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}} + 15 \, \arctan \left (\sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}}\right )}{60 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*a^5*x^5 - 40*a^3*x^3 - 15*(2*a^4*x^4 - a^2*x^2)*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 15*ar
ctan(sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x))))/a^5

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4}{\left (\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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