3.45 \(\int e^{\text {sech}^{-1}(a x^2)} x^7 \, dx\)

Optimal. Leaf size=111 \[ \frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sin ^{-1}\left (a x^2\right )}{16 a^4}-\frac {x^2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sqrt {1-a^2 x^4}}{16 a^3}+\frac {x^6}{24 a}+\frac {1}{8} x^8 e^{\text {sech}^{-1}\left (a x^2\right )} \]

[Out]

1/24*x^6/a+1/8*(1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^8+1/16*arcsin(a*x^2)*(1/(a*x^2+1))^(1/2)*(a*x^2
+1)^(1/2)/a^4-1/16*x^2*(1/(a*x^2+1))^(1/2)*(a*x^2+1)^(1/2)*(-a^2*x^4+1)^(1/2)/a^3

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6335, 30, 259, 275, 321, 216} \[ -\frac {x^2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sqrt {1-a^2 x^4}}{16 a^3}+\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sin ^{-1}\left (a x^2\right )}{16 a^4}+\frac {x^6}{24 a}+\frac {1}{8} x^8 e^{\text {sech}^{-1}\left (a x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSech[a*x^2]*x^7,x]

[Out]

x^6/(24*a) + (E^ArcSech[a*x^2]*x^8)/8 - (x^2*Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*Sqrt[1 - a^2*x^4])/(16*a^3
) + (Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*ArcSin[a*x^2])/(16*a^4)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 259

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[(c*x)
^m*(a1*a2 + b1*b2*x^(2*n))^p, x] /; FreeQ[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (Intege
rQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 6335

Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*E^ArcSech[a*x^p])/(m + 1), x] + (Dist
[p/(a*(m + 1)), Int[x^(m - p), x], x] + Dist[(p*Sqrt[1 + a*x^p]*Sqrt[1/(1 + a*x^p)])/(a*(m + 1)), Int[x^(m - p
)/(Sqrt[1 + a*x^p]*Sqrt[1 - a*x^p]), x], x]) /; FreeQ[{a, m, p}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int e^{\text {sech}^{-1}\left (a x^2\right )} x^7 \, dx &=\frac {1}{8} e^{\text {sech}^{-1}\left (a x^2\right )} x^8+\frac {\int x^5 \, dx}{4 a}+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {x^5}{\sqrt {1-a x^2} \sqrt {1+a x^2}} \, dx}{4 a}\\ &=\frac {x^6}{24 a}+\frac {1}{8} e^{\text {sech}^{-1}\left (a x^2\right )} x^8+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {x^5}{\sqrt {1-a^2 x^4}} \, dx}{4 a}\\ &=\frac {x^6}{24 a}+\frac {1}{8} e^{\text {sech}^{-1}\left (a x^2\right )} x^8+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx,x,x^2\right )}{8 a}\\ &=\frac {x^6}{24 a}+\frac {1}{8} e^{\text {sech}^{-1}\left (a x^2\right )} x^8-\frac {x^2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \sqrt {1-a^2 x^4}}{16 a^3}+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx,x,x^2\right )}{16 a^3}\\ &=\frac {x^6}{24 a}+\frac {1}{8} e^{\text {sech}^{-1}\left (a x^2\right )} x^8-\frac {x^2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \sqrt {1-a^2 x^4}}{16 a^3}+\frac {\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \sin ^{-1}\left (a x^2\right )}{16 a^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.19, size = 111, normalized size = 1.00 \[ \frac {8 a^3 x^6-3 a \sqrt {\frac {1-a x^2}{a x^2+1}} \left (-2 a^3 x^8-2 a^2 x^6+a x^4+x^2\right )+3 i \log \left (2 \sqrt {\frac {1-a x^2}{a x^2+1}} \left (a x^2+1\right )-2 i a x^2\right )}{48 a^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSech[a*x^2]*x^7,x]

[Out]

(8*a^3*x^6 - 3*a*Sqrt[(1 - a*x^2)/(1 + a*x^2)]*(x^2 + a*x^4 - 2*a^2*x^6 - 2*a^3*x^8) + (3*I)*Log[(-2*I)*a*x^2
+ 2*Sqrt[(1 - a*x^2)/(1 + a*x^2)]*(1 + a*x^2)])/(48*a^4)

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 116, normalized size = 1.05 \[ \frac {8 \, a^{3} x^{6} + 3 \, {\left (2 \, a^{4} x^{8} - a^{2} x^{4}\right )} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} - 6 \, \arctan \left (\frac {a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} - 1}{a x^{2}}\right )}{48 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^7,x, algorithm="fricas")

[Out]

1/48*(8*a^3*x^6 + 3*(2*a^4*x^8 - a^2*x^4)*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) - 6*arctan((a*x
^2*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) - 1)/(a*x^2)))/a^4

________________________________________________________________________________________

giac [B]  time = 0.21, size = 205, normalized size = 1.85 \[ \frac {8 \, a^{2} x^{6} + 4 \, \sqrt {a^{2} x^{2} + a} \sqrt {-a^{2} x^{2} + a} {\left ({\left (a^{2} x^{2} + a\right )} {\left (\frac {2 \, {\left (a^{2} x^{2} + a\right )}}{a^{4}} - \frac {7}{a^{3}}\right )} + \frac {9}{a^{2}}\right )} + {\left (\sqrt {a^{2} x^{2} + a} \sqrt {-a^{2} x^{2} + a} {\left ({\left (a^{2} x^{2} + a\right )} {\left (2 \, {\left (a^{2} x^{2} + a\right )} {\left (\frac {3 \, {\left (a^{2} x^{2} + a\right )}}{a^{6}} - \frac {13}{a^{5}}\right )} + \frac {43}{a^{4}}\right )} - \frac {39}{a^{3}}\right )} - \frac {18 \, \arcsin \left (\frac {\sqrt {2} \sqrt {a^{2} x^{2} + a}}{2 \, \sqrt {a}}\right )}{a^{2}}\right )} a + \frac {24 \, \arcsin \left (\frac {\sqrt {2} \sqrt {a^{2} x^{2} + a}}{2 \, \sqrt {a}}\right )}{a}}{48 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^7,x, algorithm="giac")

[Out]

1/48*(8*a^2*x^6 + 4*sqrt(a^2*x^2 + a)*sqrt(-a^2*x^2 + a)*((a^2*x^2 + a)*(2*(a^2*x^2 + a)/a^4 - 7/a^3) + 9/a^2)
+ (sqrt(a^2*x^2 + a)*sqrt(-a^2*x^2 + a)*((a^2*x^2 + a)*(2*(a^2*x^2 + a)*(3*(a^2*x^2 + a)/a^6 - 13/a^5) + 43/a
^4) - 39/a^3) - 18*arcsin(1/2*sqrt(2)*sqrt(a^2*x^2 + a)/sqrt(a))/a^2)*a + 24*arcsin(1/2*sqrt(2)*sqrt(a^2*x^2 +
a)/sqrt(a))/a)/a^3

________________________________________________________________________________________

maple [A]  time = 0.22, size = 137, normalized size = 1.23 \[ \frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, x^{2} \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (2 x^{6} \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}\, a^{4}-x^{2} \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}\, a^{2}+\arctan \left (\frac {x^{2}}{\sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}}\right )\right )}{16 \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}\, a^{4}}+\frac {x^{6}}{6 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^7,x)

[Out]

1/16*(-(a*x^2-1)/a/x^2)^(1/2)*x^2*((a*x^2+1)/a/x^2)^(1/2)*(2*x^6*(-(a^2*x^4-1)/a^2)^(1/2)*a^4-x^2*(-(a^2*x^4-1
)/a^2)^(1/2)*a^2+arctan(x^2/(-(a^2*x^4-1)/a^2)^(1/2)))/(-(a^2*x^4-1)/a^2)^(1/2)/a^4+1/6*x^6/a

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x^{6}}{6 \, a} + \frac {-\frac {{\left (-a^{2} x^{4} + 1\right )}^{\frac {3}{2}} x^{2}}{8 \, a^{2}} + \frac {\sqrt {-a^{2} x^{4} + 1} x^{2}}{16 \, a^{2}} + \frac {\arcsin \left (a x^{2}\right )}{16 \, a^{3}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^7,x, algorithm="maxima")

[Out]

1/6*x^6/a + integrate(sqrt(a*x^2 + 1)*sqrt(-a*x^2 + 1)*x^5, x)/a

________________________________________________________________________________________

mupad [B]  time = 14.45, size = 521, normalized size = 4.69 \[ \frac {\ln \left (\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}}{16\,a^4}-\frac {\frac {1{}\mathrm {i}}{2048\,a^4}+\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^2}+\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^4\,11{}\mathrm {i}}{1024\,a^4\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^6\,7{}\mathrm {i}}{512\,a^4\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^6}-\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^8\,239{}\mathrm {i}}{2048\,a^4\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^8}+\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^{10}\,1{}\mathrm {i}}{512\,a^4\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^{10}}}{\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^4}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^6}+\frac {6\,{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^8}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^{10}}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^{10}}+\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^{12}}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^{12}}}-\frac {\ln \left (\frac {\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x^2}+1}-1}\right )\,1{}\mathrm {i}}{16\,a^4}+\frac {x^6}{6\,a}-\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{512\,a^4\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^4\,1{}\mathrm {i}}{2048\,a^4\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*((1/(a*x^2) - 1)^(1/2)*(1/(a*x^2) + 1)^(1/2) + 1/(a*x^2)),x)

[Out]

(log(((1/(a*x^2) - 1)^(1/2) - 1i)^2/((1/(a*x^2) + 1)^(1/2) - 1)^2 + 1)*1i)/(16*a^4) - (1i/(2048*a^4) + (((1/(a
*x^2) - 1)^(1/2) - 1i)^2*1i)/(256*a^4*((1/(a*x^2) + 1)^(1/2) - 1)^2) + (((1/(a*x^2) - 1)^(1/2) - 1i)^4*11i)/(1
024*a^4*((1/(a*x^2) + 1)^(1/2) - 1)^4) + (((1/(a*x^2) - 1)^(1/2) - 1i)^6*7i)/(512*a^4*((1/(a*x^2) + 1)^(1/2) -
1)^6) - (((1/(a*x^2) - 1)^(1/2) - 1i)^8*239i)/(2048*a^4*((1/(a*x^2) + 1)^(1/2) - 1)^8) + (((1/(a*x^2) - 1)^(1
/2) - 1i)^10*1i)/(512*a^4*((1/(a*x^2) + 1)^(1/2) - 1)^10))/(((1/(a*x^2) - 1)^(1/2) - 1i)^4/((1/(a*x^2) + 1)^(1
/2) - 1)^4 + (4*((1/(a*x^2) - 1)^(1/2) - 1i)^6)/((1/(a*x^2) + 1)^(1/2) - 1)^6 + (6*((1/(a*x^2) - 1)^(1/2) - 1i
)^8)/((1/(a*x^2) + 1)^(1/2) - 1)^8 + (4*((1/(a*x^2) - 1)^(1/2) - 1i)^10)/((1/(a*x^2) + 1)^(1/2) - 1)^10 + ((1/
(a*x^2) - 1)^(1/2) - 1i)^12/((1/(a*x^2) + 1)^(1/2) - 1)^12) - (log(((1/(a*x^2) - 1)^(1/2) - 1i)/((1/(a*x^2) +
1)^(1/2) - 1))*1i)/(16*a^4) + x^6/(6*a) - (((1/(a*x^2) - 1)^(1/2) - 1i)^2*1i)/(512*a^4*((1/(a*x^2) + 1)^(1/2)
- 1)^2) - (((1/(a*x^2) - 1)^(1/2) - 1i)^4*1i)/(2048*a^4*((1/(a*x^2) + 1)^(1/2) - 1)^4)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x**2+(1/a/x**2-1)**(1/2)*(1/a/x**2+1)**(1/2))*x**7,x)

[Out]

Timed out

________________________________________________________________________________________