Integrand size = 12, antiderivative size = 57 \[ \int \left (b \tanh ^m(c+d x)\right )^n \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+m n),\frac {1}{2} (3+m n),\tanh ^2(c+d x)\right ) \tanh (c+d x) \left (b \tanh ^m(c+d x)\right )^n}{d (1+m n)} \] Output:
hypergeom([1, 1/2*m*n+1/2],[1/2*m*n+3/2],tanh(d*x+c)^2)*tanh(d*x+c)*(b*tan h(d*x+c)^m)^n/d/(m*n+1)
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \left (b \tanh ^m(c+d x)\right )^n \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+m n),\frac {1}{2} (3+m n),\tanh ^2(c+d x)\right ) \tanh (c+d x) \left (b \tanh ^m(c+d x)\right )^n}{d (1+m n)} \] Input:
Integrate[(b*Tanh[c + d*x]^m)^n,x]
Output:
(Hypergeometric2F1[1, (1 + m*n)/2, (3 + m*n)/2, Tanh[c + d*x]^2]*Tanh[c + d*x]*(b*Tanh[c + d*x]^m)^n)/(d*(1 + m*n))
Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4142, 3042, 3957, 25, 278}
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.
\(\displaystyle \int \left (b \tanh ^m(c+d x)\right )^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b (-i \tan (i c+i d x))^m\right )^ndx\) |
\(\Big \downarrow \) 4142 |
\(\displaystyle \tanh ^{-m n}(c+d x) \left (b \tanh ^m(c+d x)\right )^n \int \tanh ^{m n}(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \tanh ^{-m n}(c+d x) \left (b \tanh ^m(c+d x)\right )^n \int (-i \tan (i c+i d x))^{m n}dx\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle -\frac {\tanh ^{-m n}(c+d x) \left (b \tanh ^m(c+d x)\right )^n \int -\frac {\tanh ^{m n}(c+d x)}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tanh ^{-m n}(c+d x) \left (b \tanh ^m(c+d x)\right )^n \int \frac {\tanh ^{m n}(c+d x)}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {\tanh (c+d x) \left (b \tanh ^m(c+d x)\right )^n \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (m n+1),\frac {1}{2} (m n+3),\tanh ^2(c+d x)\right )}{d (m n+1)}\) |
Input:
Int[(b*Tanh[c + d*x]^m)^n,x]
Output:
(Hypergeometric2F1[1, (1 + m*n)/2, (3 + m*n)/2, Tanh[c + d*x]^2]*Tanh[c + d*x]*(b*Tanh[c + d*x]^m)^n)/(d*(1 + m*n))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Subst[Int [x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> S imp[b^IntPart[p]*((b*(c*Tan[e + f*x])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*Fr acPart[p])) Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; FreeQ[{ b, c, e, f, n, p}, x] && !IntegerQ[p] && !IntegerQ[n] && (EqQ[u, 1] || Ma tchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
\[\int \left (b \tanh \left (d x +c \right )^{m}\right )^{n}d x\]
Input:
int((b*tanh(d*x+c)^m)^n,x)
Output:
int((b*tanh(d*x+c)^m)^n,x)
\[ \int \left (b \tanh ^m(c+d x)\right )^n \, dx=\int { \left (b \tanh \left (d x + c\right )^{m}\right )^{n} \,d x } \] Input:
integrate((b*tanh(d*x+c)^m)^n,x, algorithm="fricas")
Output:
integral((b*tanh(d*x + c)^m)^n, x)
\[ \int \left (b \tanh ^m(c+d x)\right )^n \, dx=\int \left (b \tanh ^{m}{\left (c + d x \right )}\right )^{n}\, dx \] Input:
integrate((b*tanh(d*x+c)**m)**n,x)
Output:
Integral((b*tanh(c + d*x)**m)**n, x)
\[ \int \left (b \tanh ^m(c+d x)\right )^n \, dx=\int { \left (b \tanh \left (d x + c\right )^{m}\right )^{n} \,d x } \] Input:
integrate((b*tanh(d*x+c)^m)^n,x, algorithm="maxima")
Output:
integrate((b*tanh(d*x + c)^m)^n, x)
\[ \int \left (b \tanh ^m(c+d x)\right )^n \, dx=\int { \left (b \tanh \left (d x + c\right )^{m}\right )^{n} \,d x } \] Input:
integrate((b*tanh(d*x+c)^m)^n,x, algorithm="giac")
Output:
integrate((b*tanh(d*x + c)^m)^n, x)
Timed out. \[ \int \left (b \tanh ^m(c+d x)\right )^n \, dx=\int {\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^m\right )}^n \,d x \] Input:
int((b*tanh(c + d*x)^m)^n,x)
Output:
int((b*tanh(c + d*x)^m)^n, x)
\[ \int \left (b \tanh ^m(c+d x)\right )^n \, dx=b^{n} \left (\int \tanh \left (d x +c \right )^{m n}d x \right ) \] Input:
int((b*tanh(d*x+c)^m)^n,x)
Output:
b**n*int(tanh(c + d*x)**(m*n),x)