Integrand size = 11, antiderivative size = 190 \[ \int \tanh ^p\left (a+\frac {\log (x)}{8}\right ) \, dx=\frac {1}{3} e^{-12 a} \left (-1+e^{2 a} \sqrt [4]{x}\right )^{1+p} \left (1+e^{2 a} \sqrt [4]{x}\right )^{1-p} \left (e^{4 a} \left (3+2 p^2\right )-2 e^{6 a} p \sqrt [4]{x}\right )+e^{-4 a} \left (-1+e^{2 a} \sqrt [4]{x}\right )^{1+p} \left (1+e^{2 a} \sqrt [4]{x}\right )^{1-p} \sqrt {x}-\frac {2^{2-p} e^{-8 a} p \left (2+p^2\right ) \left (-1+e^{2 a} \sqrt [4]{x}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (p,1+p,2+p,\frac {1}{2} \left (1-e^{2 a} \sqrt [4]{x}\right )\right )}{3 (1+p)} \]
1/3*(-1+exp(2*a)*x^(1/4))^(p+1)*(1+exp(2*a)*x^(1/4))^(1-p)*(exp(4*a)*(2*p^ 2+3)-2*exp(6*a)*p*x^(1/4))/exp(12*a)-1/3*2^(2-p)*p*(p^2+2)*(-1+exp(2*a)*x^ (1/4))^(p+1)*hypergeom([p, p+1],[2+p],1/2-1/2*exp(2*a)*x^(1/4))/exp(8*a)/( p+1)+(-1+exp(2*a)*x^(1/4))^(p+1)*(1+exp(2*a)*x^(1/4))^(1-p)*x^(1/2)/exp(4* a)
Time = 1.44 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.20 \[ \int \tanh ^p\left (a+\frac {\log (x)}{8}\right ) \, dx=\frac {e^{-8 a} \left (-1+e^{2 a} \sqrt [4]{x}\right ) \left (\frac {-1+e^{2 a} \sqrt [4]{x}}{2+2 e^{2 a} \sqrt [4]{x}}\right )^p \left (-8 p \left (1+e^{2 a} \sqrt [4]{x}\right )^p \operatorname {Hypergeometric2F1}\left (-2+p,1+p,2+p,\frac {1}{2}-\frac {1}{2} e^{2 a} \sqrt [4]{x}\right )+4 (1+2 p) \left (1+e^{2 a} \sqrt [4]{x}\right )^p \operatorname {Hypergeometric2F1}\left (-1+p,1+p,2+p,\frac {1}{2}-\frac {1}{2} e^{2 a} \sqrt [4]{x}\right )+(1+p) \left (2^p e^{4 a} \left (1+e^{2 a} \sqrt [4]{x}\right ) \sqrt {x}-2 \left (1+e^{2 a} \sqrt [4]{x}\right )^p \operatorname {Hypergeometric2F1}\left (p,1+p,2+p,\frac {1}{2}-\frac {1}{2} e^{2 a} \sqrt [4]{x}\right )\right )\right )}{1+p} \]
((-1 + E^(2*a)*x^(1/4))*((-1 + E^(2*a)*x^(1/4))/(2 + 2*E^(2*a)*x^(1/4)))^p *(-8*p*(1 + E^(2*a)*x^(1/4))^p*Hypergeometric2F1[-2 + p, 1 + p, 2 + p, 1/2 - (E^(2*a)*x^(1/4))/2] + 4*(1 + 2*p)*(1 + E^(2*a)*x^(1/4))^p*Hypergeometr ic2F1[-1 + p, 1 + p, 2 + p, 1/2 - (E^(2*a)*x^(1/4))/2] + (1 + p)*(2^p*E^(4 *a)*(1 + E^(2*a)*x^(1/4))*Sqrt[x] - 2*(1 + E^(2*a)*x^(1/4))^p*Hypergeometr ic2F1[p, 1 + p, 2 + p, 1/2 - (E^(2*a)*x^(1/4))/2])))/(E^(8*a)*(1 + p))
Time = 0.33 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {6067, 900, 111, 27, 164, 79}
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.
\(\displaystyle \int \tanh ^p\left (a+\frac {\log (x)}{8}\right ) \, dx\) |
\(\Big \downarrow \) 6067 |
\(\displaystyle \int \left (e^{2 a} \sqrt [4]{x}-1\right )^p \left (e^{2 a} \sqrt [4]{x}+1\right )^{-p}dx\) |
\(\Big \downarrow \) 900 |
\(\displaystyle 4 \int \left (e^{2 a} \sqrt [4]{x}-1\right )^p \left (e^{2 a} \sqrt [4]{x}+1\right )^{-p} x^{3/4}d\sqrt [4]{x}\) |
\(\Big \downarrow \) 111 |
\(\displaystyle 4 \left (\frac {1}{4} e^{-4 a} \int 2 \left (e^{2 a} \sqrt [4]{x}-1\right )^p \left (e^{2 a} \sqrt [4]{x}+1\right )^{-p} \left (1-e^{2 a} p \sqrt [4]{x}\right ) \sqrt [4]{x}d\sqrt [4]{x}+\frac {1}{4} e^{-4 a} \sqrt {x} \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \left (e^{2 a} \sqrt [4]{x}+1\right )^{1-p}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \left (\frac {1}{2} e^{-4 a} \int \left (e^{2 a} \sqrt [4]{x}-1\right )^p \left (e^{2 a} \sqrt [4]{x}+1\right )^{-p} \left (1-e^{2 a} p \sqrt [4]{x}\right ) \sqrt [4]{x}d\sqrt [4]{x}+\frac {1}{4} e^{-4 a} \sqrt {x} \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \left (e^{2 a} \sqrt [4]{x}+1\right )^{1-p}\right )\) |
\(\Big \downarrow \) 164 |
\(\displaystyle 4 \left (\frac {1}{2} e^{-4 a} \left (\frac {1}{6} e^{-8 a} \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \left (e^{2 a} \sqrt [4]{x}+1\right )^{1-p} \left (e^{4 a} \left (2 p^2+3\right )-2 e^{6 a} p \sqrt [4]{x}\right )-\frac {2}{3} e^{-2 a} p \left (p^2+2\right ) \int \left (e^{2 a} \sqrt [4]{x}-1\right )^p \left (e^{2 a} \sqrt [4]{x}+1\right )^{-p}d\sqrt [4]{x}\right )+\frac {1}{4} e^{-4 a} \sqrt {x} \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \left (e^{2 a} \sqrt [4]{x}+1\right )^{1-p}\right )\) |
\(\Big \downarrow \) 79 |
\(\displaystyle 4 \left (\frac {1}{2} e^{-4 a} \left (\frac {1}{6} e^{-8 a} \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \left (e^{2 a} \sqrt [4]{x}+1\right )^{1-p} \left (e^{4 a} \left (2 p^2+3\right )-2 e^{6 a} p \sqrt [4]{x}\right )-\frac {e^{-4 a} 2^{1-p} p \left (p^2+2\right ) \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \operatorname {Hypergeometric2F1}\left (p,p+1,p+2,\frac {1}{2} \left (1-e^{2 a} \sqrt [4]{x}\right )\right )}{3 (p+1)}\right )+\frac {1}{4} e^{-4 a} \sqrt {x} \left (e^{2 a} \sqrt [4]{x}-1\right )^{p+1} \left (e^{2 a} \sqrt [4]{x}+1\right )^{1-p}\right )\) |
4*(((-1 + E^(2*a)*x^(1/4))^(1 + p)*(1 + E^(2*a)*x^(1/4))^(1 - p)*Sqrt[x])/ (4*E^(4*a)) + (((-1 + E^(2*a)*x^(1/4))^(1 + p)*(1 + E^(2*a)*x^(1/4))^(1 - p)*(E^(4*a)*(3 + 2*p^2) - 2*E^(6*a)*p*x^(1/4)))/(6*E^(8*a)) - (2^(1 - p)*p *(2 + p^2)*(-1 + E^(2*a)*x^(1/4))^(1 + p)*Hypergeometric2F1[p, 1 + p, 2 + p, (1 - E^(2*a)*x^(1/4))/2])/(3*E^(4*a)*(1 + p)))/(2*E^(4*a)))
3.2.68.3.1 Defintions of rubi rules used
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)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol ] :> With[{g = Denominator[n]}, Simp[g Subst[Int[x^(g - 1)*(a + b*x^(g*n) )^p*(c + d*x^(g*n))^q, x], x, x^(1/g)], x]] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[n]
Int[Tanh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(-1 + E^(2* a*d)*x^(2*b*d))^p/(1 + E^(2*a*d)*x^(2*b*d))^p, x] /; FreeQ[{a, b, d, p}, x]
\[\int \tanh \left (a +\frac {\ln \left (x \right )}{8}\right )^{p}d x\]
\[ \int \tanh ^p\left (a+\frac {\log (x)}{8}\right ) \, dx=\int { \tanh \left (a + \frac {1}{8} \, \log \left (x\right )\right )^{p} \,d x } \]
\[ \int \tanh ^p\left (a+\frac {\log (x)}{8}\right ) \, dx=\int \tanh ^{p}{\left (a + \frac {\log {\left (x \right )}}{8} \right )}\, dx \]
\[ \int \tanh ^p\left (a+\frac {\log (x)}{8}\right ) \, dx=\int { \tanh \left (a + \frac {1}{8} \, \log \left (x\right )\right )^{p} \,d x } \]
\[ \int \tanh ^p\left (a+\frac {\log (x)}{8}\right ) \, dx=\int { \tanh \left (a + \frac {1}{8} \, \log \left (x\right )\right )^{p} \,d x } \]
Timed out. \[ \int \tanh ^p\left (a+\frac {\log (x)}{8}\right ) \, dx=\int {\mathrm {tanh}\left (a+\frac {\ln \left (x\right )}{8}\right )}^p \,d x \]