Integrand size = 12, antiderivative size = 97 \[ \int (b \tanh (c+d x))^{7/2} \, dx=\frac {b^{7/2} \arctan \left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{7/2} \text {arctanh}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b^3 \sqrt {b \tanh (c+d x)}}{d}-\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d} \]
b^(7/2)*arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/d+b^(7/2)*arctanh((b*tanh(d* x+c))^(1/2)/b^(1/2))/d-2*b^3*(b*tanh(d*x+c))^(1/2)/d-2/5*b*(b*tanh(d*x+c)) ^(5/2)/d
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int (b \tanh (c+d x))^{7/2} \, dx=-\frac {(b \tanh (c+d x))^{7/2} \left (-\arctan \left (\sqrt {\tanh (c+d x)}\right )-\text {arctanh}\left (\sqrt {\tanh (c+d x)}\right )+2 \sqrt {\tanh (c+d x)}+\frac {2}{5} \tanh ^{\frac {5}{2}}(c+d x)\right )}{d \tanh ^{\frac {7}{2}}(c+d x)} \]
-(((b*Tanh[c + d*x])^(7/2)*(-ArcTan[Sqrt[Tanh[c + d*x]]] - ArcTanh[Sqrt[Ta nh[c + d*x]]] + 2*Sqrt[Tanh[c + d*x]] + (2*Tanh[c + d*x]^(5/2))/5))/(d*Tan h[c + d*x]^(7/2)))
Time = 0.40 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3954, 3042, 3954, 3042, 3957, 25, 266, 756, 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.
\(\displaystyle \int (b \tanh (c+d x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (-i b \tan (i c+i d x))^{7/2}dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \int (b \tanh (c+d x))^{3/2}dx-\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}+b^2 \int (-i b \tan (i c+i d x))^{3/2}dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \left (b^2 \int \frac {1}{\sqrt {b \tanh (c+d x)}}dx-\frac {2 b \sqrt {b \tanh (c+d x)}}{d}\right )-\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}+b^2 \left (-\frac {2 b \sqrt {b \tanh (c+d x)}}{d}+b^2 \int \frac {1}{\sqrt {-i b \tan (i c+i d x)}}dx\right )\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle b^2 \left (-\frac {b^3 \int -\frac {1}{\sqrt {b \tanh (c+d x)} \left (b^2-b^2 \tanh ^2(c+d x)\right )}d(b \tanh (c+d x))}{d}-\frac {2 b \sqrt {b \tanh (c+d x)}}{d}\right )-\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b^2 \left (\frac {b^3 \int \frac {1}{\sqrt {b \tanh (c+d x)} \left (b^2-b^2 \tanh ^2(c+d x)\right )}d(b \tanh (c+d x))}{d}-\frac {2 b \sqrt {b \tanh (c+d x)}}{d}\right )-\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle b^2 \left (\frac {2 b^3 \int \frac {1}{b^2-b^4 \tanh ^4(c+d x)}d\sqrt {b \tanh (c+d x)}}{d}-\frac {2 b \sqrt {b \tanh (c+d x)}}{d}\right )-\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle b^2 \left (\frac {2 b^3 \left (\frac {\int \frac {1}{b-b^2 \tanh ^2(c+d x)}d\sqrt {b \tanh (c+d x)}}{2 b}+\frac {\int \frac {1}{b^2 \tanh ^2(c+d x)+b}d\sqrt {b \tanh (c+d x)}}{2 b}\right )}{d}-\frac {2 b \sqrt {b \tanh (c+d x)}}{d}\right )-\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle b^2 \left (\frac {2 b^3 \left (\frac {\int \frac {1}{b-b^2 \tanh ^2(c+d x)}d\sqrt {b \tanh (c+d x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \tanh (c+d x)\right )}{2 b^{3/2}}\right )}{d}-\frac {2 b \sqrt {b \tanh (c+d x)}}{d}\right )-\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle b^2 \left (\frac {2 b^3 \left (\frac {\arctan \left (\sqrt {b} \tanh (c+d x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \tanh (c+d x)\right )}{2 b^{3/2}}\right )}{d}-\frac {2 b \sqrt {b \tanh (c+d x)}}{d}\right )-\frac {2 b (b \tanh (c+d x))^{5/2}}{5 d}\) |
(-2*b*(b*Tanh[c + d*x])^(5/2))/(5*d) + b^2*((2*b^3*(ArcTan[Sqrt[b]*Tanh[c + d*x]]/(2*b^(3/2)) + ArcTanh[Sqrt[b]*Tanh[c + d*x]]/(2*b^(3/2))))/d - (2* b*Sqrt[b*Tanh[c + d*x]])/d)
3.1.13.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) 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 = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {b^{\frac {7}{2}} \arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d}+\frac {b^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d}-\frac {2 b^{3} \sqrt {b \tanh \left (d x +c \right )}}{d}-\frac {2 b \left (b \tanh \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d}\) | \(80\) |
default | \(\frac {b^{\frac {7}{2}} \arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d}+\frac {b^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{d}-\frac {2 b^{3} \sqrt {b \tanh \left (d x +c \right )}}{d}-\frac {2 b \left (b \tanh \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d}\) | \(80\) |
b^(7/2)*arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/d+b^(7/2)*arctanh((b*tanh(d* x+c))^(1/2)/b^(1/2))/d-2*b^3*(b*tanh(d*x+c))^(1/2)/d-2/5*b*(b*tanh(d*x+c)) ^(5/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 751 vs. \(2 (79) = 158\).
Time = 0.30 (sec) , antiderivative size = 1556, normalized size of antiderivative = 16.04 \[ \int (b \tanh (c+d x))^{7/2} \, dx=\text {Too large to display} \]
[-1/20*(10*(b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^ 3*sinh(d*x + c)^4 + 2*b^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)^2 + b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 + b^3*cosh(d*x + c))*sinh (d*x + c))*sqrt(-b)*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c ) + sinh(d*x + c)^2)*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 - b)) - 5 *(b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 + 2*b^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)^2 + b^3)*si nh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 + b^3*cosh(d*x + c))*sinh(d*x + c)) *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) + sin h(d*x + c)^2 + 1)*sqrt(-b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)) - 2*b)/(cos h(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)) + 16*(3*b^3 *cosh(d*x + c)^4 + 12*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + 3*b^3*sinh(d*x + c)^4 + 4*b^3*cosh(d*x + c)^2 + 3*b^3 + 2*(9*b^3*cosh(d*x + c)^2 + 2*b^3)* sinh(d*x + c)^2 + 4*(3*b^3*cosh(d*x + c)^3 + 2*b^3*cosh(d*x + c))*sinh(d*x + c))*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/(d*cosh(d*x + c)^4 + 4*d*cosh( d*x + c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 + 2*d*cosh(d*x + c)^2 + 2*...
\[ \int (b \tanh (c+d x))^{7/2} \, dx=\int \left (b \tanh {\left (c + d x \right )}\right )^{\frac {7}{2}}\, dx \]
\[ \int (b \tanh (c+d x))^{7/2} \, dx=\int { \left (b \tanh \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (79) = 158\).
Time = 0.42 (sec) , antiderivative size = 293, normalized size of antiderivative = 3.02 \[ \int (b \tanh (c+d x))^{7/2} \, dx=\frac {10 \, b^{\frac {7}{2}} \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right ) - 5 \, b^{\frac {7}{2}} \log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right ) - \frac {16 \, {\left (5 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{4} b^{4} + 10 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{3} b^{\frac {9}{2}} + 20 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{2} b^{5} + 10 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )} b^{\frac {11}{2}} + 3 \, b^{6}\right )}}{{\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} + \sqrt {b}\right )}^{5}}}{10 \, d} \]
1/10*(10*b^(7/2)*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))/sqrt(b)) - 5*b^(7/2)*log(abs(-sqrt(b)*e^(2*d*x + 2*c) + sqrt(b*e^(4 *d*x + 4*c) - b))) - 16*(5*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4* c) - b))^4*b^4 + 10*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b) )^3*b^(9/2) + 20*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))^2 *b^5 + 10*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))*b^(11/2) + 3*b^6)/(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b) + sqrt(b) )^5)/d
Time = 2.16 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86 \[ \int (b \tanh (c+d x))^{7/2} \, dx=\frac {b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d}-\frac {2\,b^3\,\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{d}-\frac {2\,b\,{\left (b\,\mathrm {tanh}\left (c+d\,x\right )\right )}^{5/2}}{5\,d}-\frac {b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}\,1{}\mathrm {i}}{\sqrt {b}}\right )\,1{}\mathrm {i}}{d} \]