Integrand size = 10, antiderivative size = 186 \[ \int e^{\frac {5}{2} \text {arctanh}(a x)} \, dx=\frac {5 (1-a x)^{3/4} \sqrt [4]{1+a x}}{a}+\frac {4 (1+a x)^{5/4}}{a \sqrt [4]{1-a x}}+\frac {5 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )}{\sqrt {2} a}-\frac {5 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )}{\sqrt {2} a}-\frac {5 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+a x}}{\sqrt [4]{1-a x} \left (1+\frac {\sqrt {1+a x}}{\sqrt {1-a x}}\right )}\right )}{\sqrt {2} a} \] Output:
5*(-a*x+1)^(3/4)*(a*x+1)^(1/4)/a+4*(a*x+1)^(5/4)/a/(-a*x+1)^(1/4)+5/2*arct an(1-2^(1/2)*(a*x+1)^(1/4)/(-a*x+1)^(1/4))*2^(1/2)/a-5/2*arctan(1+2^(1/2)* (a*x+1)^(1/4)/(-a*x+1)^(1/4))*2^(1/2)/a-5/2*arctanh(2^(1/2)*(a*x+1)^(1/4)/ (-a*x+1)^(1/4)/(1+(a*x+1)^(1/2)/(-a*x+1)^(1/2)))*2^(1/2)/a
Time = 0.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.94 \[ \int e^{\frac {5}{2} \text {arctanh}(a x)} \, dx=\frac {\frac {40 e^{\frac {1}{2} \text {arctanh}(a x)}}{1+e^{2 \text {arctanh}(a x)}}+\frac {32 e^{\frac {5}{2} \text {arctanh}(a x)}}{1+e^{2 \text {arctanh}(a x)}}+10 \sqrt {2} \arctan \left (1-\sqrt {2} e^{\frac {1}{2} \text {arctanh}(a x)}\right )-10 \sqrt {2} \arctan \left (1+\sqrt {2} e^{\frac {1}{2} \text {arctanh}(a x)}\right )+5 \sqrt {2} \log \left (1-\sqrt {2} e^{\frac {1}{2} \text {arctanh}(a x)}+e^{\text {arctanh}(a x)}\right )-5 \sqrt {2} \log \left (1+\sqrt {2} e^{\frac {1}{2} \text {arctanh}(a x)}+e^{\text {arctanh}(a x)}\right )}{4 a} \] Input:
Integrate[E^((5*ArcTanh[a*x])/2),x]
Output:
((40*E^(ArcTanh[a*x]/2))/(1 + E^(2*ArcTanh[a*x])) + (32*E^((5*ArcTanh[a*x] )/2))/(1 + E^(2*ArcTanh[a*x])) + 10*Sqrt[2]*ArcTan[1 - Sqrt[2]*E^(ArcTanh[ a*x]/2)] - 10*Sqrt[2]*ArcTan[1 + Sqrt[2]*E^(ArcTanh[a*x]/2)] + 5*Sqrt[2]*L og[1 - Sqrt[2]*E^(ArcTanh[a*x]/2) + E^ArcTanh[a*x]] - 5*Sqrt[2]*Log[1 + Sq rt[2]*E^(ArcTanh[a*x]/2) + E^ArcTanh[a*x]])/(4*a)
Time = 0.75 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.25, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.300, Rules used = {6675, 57, 60, 73, 854, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{\frac {5}{2} \text {arctanh}(a x)} \, dx\) |
\(\Big \downarrow \) 6675 |
\(\displaystyle \int \frac {(a x+1)^{5/4}}{(1-a x)^{5/4}}dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \int \frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (\frac {1}{2} \int \frac {1}{\sqrt [4]{1-a x} (a x+1)^{3/4}}dx-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \int \frac {\sqrt {1-a x}}{(a x+1)^{3/4}}d\sqrt [4]{1-a x}}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \int \frac {\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \left (\frac {1}{2} \int \frac {\sqrt {1-a x}+1}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}-\frac {1}{2} \int \frac {1-\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+\frac {1}{2} \int \frac {1}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {1}{2} \int \frac {1-\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\sqrt {1-a x}-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\sqrt {1-a x}-1}d\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {4 (a x+1)^{5/4}}{a \sqrt [4]{1-a x}}-5 \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2}}\right )\right )}{a}-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{a}\right )\) |
Input:
Int[E^((5*ArcTanh[a*x])/2),x]
Output:
(4*(1 + a*x)^(5/4))/(a*(1 - a*x)^(1/4)) - 5*(-(((1 - a*x)^(3/4)*(1 + a*x)^ (1/4))/a) - (2*((-(ArcTan[1 - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/S qrt[2]) + ArcTan[1 + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/Sqrt[2])/2 + (Log[1 + Sqrt[1 - a*x] - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/(2* Sqrt[2]) - Log[1 + Sqrt[1 - a*x] + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/ 4)]/(2*Sqrt[2]))/2))/a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n )^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 2^(-1)] && IntegersQ[m, p + (m + 1)/n]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_)), x_Symbol] :> Int[(1 + a*x)^(n/2)/(1 - a*x )^(n/2), x] /; FreeQ[{a, n}, x] && !IntegerQ[(n - 1)/2]
\[\int {\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}d x\]
Input:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x)
Output:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x)
Time = 0.08 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.27 \[ \int e^{\frac {5}{2} \text {arctanh}(a x)} \, dx=-\frac {4 \, {\left (a x - 9\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 10 \, \sqrt {2} \arctan \left (\sqrt {2} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + 10 \, \sqrt {2} \arctan \left (\sqrt {2} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) + 5 \, \sqrt {2} \log \left (\frac {a x + \sqrt {2} {\left (a x - 1\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {-a^{2} x^{2} + 1} - 1}{a x - 1}\right ) - 5 \, \sqrt {2} \log \left (\frac {a x - \sqrt {2} {\left (a x - 1\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {-a^{2} x^{2} + 1} - 1}{a x - 1}\right )}{4 \, a} \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x, algorithm="fricas")
Output:
-1/4*(4*(a*x - 9)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 10*sqrt(2)*arctan( sqrt(2)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 1) + 10*sqrt(2)*arctan(sqrt( 2)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - 1) + 5*sqrt(2)*log((a*x + sqrt(2) *(a*x - 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - sqrt(-a^2*x^2 + 1) - 1)/( a*x - 1)) - 5*sqrt(2)*log((a*x - sqrt(2)*(a*x - 1)*sqrt(-sqrt(-a^2*x^2 + 1 )/(a*x - 1)) - sqrt(-a^2*x^2 + 1) - 1)/(a*x - 1)))/a
\[ \int e^{\frac {5}{2} \text {arctanh}(a x)} \, dx=\int \left (\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}\, dx \] Input:
integrate(((a*x+1)/(-a**2*x**2+1)**(1/2))**(5/2),x)
Output:
Integral(((a*x + 1)/sqrt(-a**2*x**2 + 1))**(5/2), x)
\[ \int e^{\frac {5}{2} \text {arctanh}(a x)} \, dx=\int { \left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x, algorithm="maxima")
Output:
integrate(((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2), x)
\[ \int e^{\frac {5}{2} \text {arctanh}(a x)} \, dx=\int { \left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x, algorithm="giac")
Output:
integrate(((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2), x)
Timed out. \[ \int e^{\frac {5}{2} \text {arctanh}(a x)} \, dx=\int {\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2} \,d x \] Input:
int(((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2),x)
Output:
int(((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2), x)
\[ \int e^{\frac {5}{2} \text {arctanh}(a x)} \, dx=\frac {-2 \sqrt {a x +1}\, \left (-a^{2} x^{2}+1\right )^{\frac {3}{4}}+5 \left (\int \frac {\sqrt {a x +1}\, \left (-a^{2} x^{2}+1\right )^{\frac {3}{4}} x}{a^{3} x^{3}-a^{2} x^{2}-a x +1}d x \right ) a^{3} x -5 \left (\int \frac {\sqrt {a x +1}\, \left (-a^{2} x^{2}+1\right )^{\frac {3}{4}} x}{a^{3} x^{3}-a^{2} x^{2}-a x +1}d x \right ) a^{2}}{3 a \left (a x -1\right )} \] Input:
int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x)
Output:
( - 2*sqrt(a*x + 1)*( - a**2*x**2 + 1)**(3/4) + 5*int((sqrt(a*x + 1)*( - a **2*x**2 + 1)**(3/4)*x)/(a**3*x**3 - a**2*x**2 - a*x + 1),x)*a**3*x - 5*in t((sqrt(a*x + 1)*( - a**2*x**2 + 1)**(3/4)*x)/(a**3*x**3 - a**2*x**2 - a*x + 1),x)*a**2)/(3*a*(a*x - 1))