Integrand size = 12, antiderivative size = 78 \[ \int (b \tanh (c+d x))^{5/2} \, dx=-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{5/2} \text {arctanh}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d} \] Output:
-b^(5/2)*arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/d+b^(5/2)*arctanh((b*tanh(d *x+c))^(1/2)/b^(1/2))/d-2/3*b*(b*tanh(d*x+c))^(3/2)/d
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int (b \tanh (c+d x))^{5/2} \, dx=-\frac {(b \tanh (c+d x))^{5/2} \left (\arctan \left (\sqrt {\tanh (c+d x)}\right )-\text {arctanh}\left (\sqrt {\tanh (c+d x)}\right )+\frac {2}{3} \tanh ^{\frac {3}{2}}(c+d x)\right )}{d \tanh ^{\frac {5}{2}}(c+d x)} \] Input:
Integrate[(b*Tanh[c + d*x])^(5/2),x]
Output:
-(((b*Tanh[c + d*x])^(5/2)*(ArcTan[Sqrt[Tanh[c + d*x]]] - ArcTanh[Sqrt[Tan h[c + d*x]]] + (2*Tanh[c + d*x]^(3/2))/3))/(d*Tanh[c + d*x]^(5/2)))
Time = 0.36 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3954, 3042, 3957, 25, 266, 827, 216, 219}
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.
\(\displaystyle \int (b \tanh (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (-i b \tan (i c+i d x))^{5/2}dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \int \sqrt {b \tanh (c+d x)}dx-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d}+b^2 \int \sqrt {-i b \tan (i c+i d x)}dx\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle -\frac {b^3 \int -\frac {\sqrt {b \tanh (c+d x)}}{b^2-b^2 \tanh ^2(c+d x)}d(b \tanh (c+d x))}{d}-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b^3 \int \frac {\sqrt {b \tanh (c+d x)}}{b^2-b^2 \tanh ^2(c+d x)}d(b \tanh (c+d x))}{d}-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 b^3 \int \frac {b^2 \tanh ^2(c+d x)}{b^2-b^4 \tanh ^4(c+d x)}d\sqrt {b \tanh (c+d x)}}{d}-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {2 b^3 \left (\frac {1}{2} \int \frac {1}{b-b^2 \tanh ^2(c+d x)}d\sqrt {b \tanh (c+d x)}-\frac {1}{2} \int \frac {1}{b^2 \tanh ^2(c+d x)+b}d\sqrt {b \tanh (c+d x)}\right )}{d}-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2 b^3 \left (\frac {1}{2} \int \frac {1}{b-b^2 \tanh ^2(c+d x)}d\sqrt {b \tanh (c+d x)}-\frac {\arctan \left (\sqrt {b} \tanh (c+d x)\right )}{2 \sqrt {b}}\right )}{d}-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 b^3 \left (\frac {\text {arctanh}\left (\sqrt {b} \tanh (c+d x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \tanh (c+d x)\right )}{2 \sqrt {b}}\right )}{d}-\frac {2 b (b \tanh (c+d x))^{3/2}}{3 d}\) |
Input:
Int[(b*Tanh[c + d*x])^(5/2),x]
Output:
(2*b^3*(-1/2*ArcTan[Sqrt[b]*Tanh[c + d*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Tanh[ c + d*x]]/(2*Sqrt[b])))/d - (2*b*(b*Tanh[c + d*x])^(3/2))/(3*d)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
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]
Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(-\frac {b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d}+\frac {b^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d}-\frac {2 b \left (b \tanh \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d}\) | \(63\) |
default | \(-\frac {b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d}+\frac {b^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d}-\frac {2 b \left (b \tanh \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d}\) | \(63\) |
Input:
int((b*tanh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-b^(5/2)*arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/d+b^(5/2)*arctanh((b*tanh(d *x+c))^(1/2)/b^(1/2))/d-2/3*b*(b*tanh(d*x+c))^(3/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (62) = 124\).
Time = 0.13 (sec) , antiderivative size = 969, normalized size of antiderivative = 12.42 \[ \int (b \tanh (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:
integrate((b*tanh(d*x+c))^(5/2),x, algorithm="fricas")
Output:
[-1/12*(6*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*s inh(d*x + c)^2 + b^2)*sqrt(-b)*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*s inh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(-b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2)) - 3*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2 *sinh(d*x + c)^2 + b^2)*sqrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(-b)*sqrt(b*sinh(d*x + c)/cos h(d*x + c)) - 2*b)/(cosh(d*x + c)^4 + 4*cosh(d*x + c)^3*sinh(d*x + c) + 6* cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d *x + c)^4)) + 8*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 - b^2)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/(d*cosh(d *x + c)^2 + 2*d*cosh(d*x + c)*sinh(d*x + c) + d*sinh(d*x + c)^2 + d), -1/1 2*(6*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d *x + c)^2 + b^2)*sqrt(b)*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d* x + c) + sinh(d*x + c)^2 + 1)*sqrt(b*sinh(d*x + c)/cosh(d*x + c))/sqrt(b)) - 3*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d *x + c)^2 + b^2)*sqrt(b)*log(2*b*cosh(d*x + c)^4 + 8*b*cosh(d*x + c)^3*sin h(d*x + c) + 12*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*cosh(d*x + c)*s...
\[ \int (b \tanh (c+d x))^{5/2} \, dx=\int \left (b \tanh {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((b*tanh(d*x+c))**(5/2),x)
Output:
Integral((b*tanh(c + d*x))**(5/2), x)
\[ \int (b \tanh (c+d x))^{5/2} \, dx=\int { \left (b \tanh \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((b*tanh(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((b*tanh(d*x + c))^(5/2), x)
Exception generated. \[ \int (b \tanh (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((b*tanh(d*x+c))^(5/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Time = 2.35 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.79 \[ \int (b \tanh (c+d x))^{5/2} \, dx=\frac {b^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d}-\frac {b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d}-\frac {2\,b\,{\left (b\,\mathrm {tanh}\left (c+d\,x\right )\right )}^{3/2}}{3\,d} \] Input:
int((b*tanh(c + d*x))^(5/2),x)
Output:
(b^(5/2)*atanh((b*tanh(c + d*x))^(1/2)/b^(1/2)))/d - (b^(5/2)*atan((b*tanh (c + d*x))^(1/2)/b^(1/2)))/d - (2*b*(b*tanh(c + d*x))^(3/2))/(3*d)
\[ \int (b \tanh (c+d x))^{5/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\tanh \left (d x +c \right )}\, \tanh \left (d x +c \right )^{2}d x \right ) b^{2} \] Input:
int((b*tanh(d*x+c))^(5/2),x)
Output:
sqrt(b)*int(sqrt(tanh(c + d*x))*tanh(c + d*x)**2,x)*b**2