Integrand size = 14, antiderivative size = 74 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=-\frac {b \sqrt [3]{b \coth ^3(c+d x)}}{d}-\frac {b \coth ^2(c+d x) \sqrt [3]{b \coth ^3(c+d x)}}{3 d}+b x \sqrt [3]{b \coth ^3(c+d x)} \tanh (c+d x) \] Output:
-b*(b*coth(d*x+c)^3)^(1/3)/d-1/3*b*coth(d*x+c)^2*(b*coth(d*x+c)^3)^(1/3)/d +b*x*(b*coth(d*x+c)^3)^(1/3)*tanh(d*x+c)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.58 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=-\frac {\left (b \coth ^3(c+d x)\right )^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\tanh ^2(c+d x)\right ) \tanh (c+d x)}{3 d} \] Input:
Integrate[(b*Coth[c + d*x]^3)^(4/3),x]
Output:
-1/3*((b*Coth[c + d*x]^3)^(4/3)*Hypergeometric2F1[-3/2, 1, -1/2, Tanh[c + d*x]^2]*Tanh[c + d*x])/d
Time = 0.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4141, 3042, 3954, 25, 3042, 25, 3954, 24}
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.
\(\displaystyle \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (i b \tan \left (i c+i d x+\frac {\pi }{2}\right )^3\right )^{4/3}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle b \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \int \coth ^4(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \int \tan \left (i c+i d x+\frac {\pi }{2}\right )^4dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \left (-\int -\coth ^2(c+d x)dx-\frac {\coth ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \left (\int \coth ^2(c+d x)dx-\frac {\coth ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \left (-\frac {\coth ^3(c+d x)}{3 d}+\int -\tan \left (i c+i d x+\frac {\pi }{2}\right )^2dx\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \left (-\frac {\coth ^3(c+d x)}{3 d}-\int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^2dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \left (\int 1dx-\frac {\coth ^3(c+d x)}{3 d}-\frac {\coth (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle b \tanh (c+d x) \left (-\frac {\coth ^3(c+d x)}{3 d}-\frac {\coth (c+d x)}{d}+x\right ) \sqrt [3]{b \coth ^3(c+d x)}\) |
Input:
Int[(b*Coth[c + d*x]^3)^(4/3),x]
Output:
b*(b*Coth[c + d*x]^3)^(1/3)*(x - Coth[c + d*x]/d - Coth[c + d*x]^3/(3*d))* Tanh[c + d*x]
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[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Ta n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.76
method | result | size |
risch | \(-\frac {b {\left (\frac {b \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )}^{\frac {1}{3}} \left (-3 \,{\mathrm e}^{6 d x +6 c} d x +9 \,{\mathrm e}^{4 d x +4 c} d x -9 \,{\mathrm e}^{2 d x +2 c} d x +3 d x +12 \,{\mathrm e}^{4 d x +4 c}-12 \,{\mathrm e}^{2 d x +2 c}+8\right )}{3 \left ({\mathrm e}^{2 d x +2 c}+1\right ) \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} d}\) | \(130\) |
Input:
int((b*coth(d*x+c)^3)^(4/3),x,method=_RETURNVERBOSE)
Output:
-1/3*b*(b*(exp(2*d*x+2*c)+1)^3/(exp(2*d*x+2*c)-1)^3)^(1/3)*(-3*exp(6*d*x+6 *c)*d*x+9*exp(4*d*x+4*c)*d*x-9*exp(2*d*x+2*c)*d*x+3*d*x+12*exp(4*d*x+4*c)- 12*exp(2*d*x+2*c)+8)/(exp(2*d*x+2*c)+1)/(exp(2*d*x+2*c)-1)^2/d
Leaf count of result is larger than twice the leaf count of optimal. 1046 vs. \(2 (66) = 132\).
Time = 0.11 (sec) , antiderivative size = 1046, normalized size of antiderivative = 14.14 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\text {Too large to display} \] Input:
integrate((b*coth(d*x+c)^3)^(4/3),x, algorithm="fricas")
Output:
-1/3*(3*b*d*x*cosh(d*x + c)^6 - 3*(b*d*x*e^(2*d*x + 2*c) - b*d*x)*sinh(d*x + c)^6 - 18*(b*d*x*cosh(d*x + c)*e^(2*d*x + 2*c) - b*d*x*cosh(d*x + c))*s inh(d*x + c)^5 - 3*(3*b*d*x + 4*b)*cosh(d*x + c)^4 + 3*(15*b*d*x*cosh(d*x + c)^2 - 3*b*d*x - (15*b*d*x*cosh(d*x + c)^2 - 3*b*d*x - 4*b)*e^(2*d*x + 2 *c) - 4*b)*sinh(d*x + c)^4 + 12*(5*b*d*x*cosh(d*x + c)^3 - (3*b*d*x + 4*b) *cosh(d*x + c) - (5*b*d*x*cosh(d*x + c)^3 - (3*b*d*x + 4*b)*cosh(d*x + c)) *e^(2*d*x + 2*c))*sinh(d*x + c)^3 - 3*b*d*x + 3*(3*b*d*x + 4*b)*cosh(d*x + c)^2 + 3*(15*b*d*x*cosh(d*x + c)^4 + 3*b*d*x - 6*(3*b*d*x + 4*b)*cosh(d*x + c)^2 - (15*b*d*x*cosh(d*x + c)^4 + 3*b*d*x - 6*(3*b*d*x + 4*b)*cosh(d*x + c)^2 + 4*b)*e^(2*d*x + 2*c) + 4*b)*sinh(d*x + c)^2 - (3*b*d*x*cosh(d*x + c)^6 - 3*(3*b*d*x + 4*b)*cosh(d*x + c)^4 - 3*b*d*x + 3*(3*b*d*x + 4*b)*c osh(d*x + c)^2 - 8*b)*e^(2*d*x + 2*c) + 6*(3*b*d*x*cosh(d*x + c)^5 - 2*(3* b*d*x + 4*b)*cosh(d*x + c)^3 + (3*b*d*x + 4*b)*cosh(d*x + c) - (3*b*d*x*co sh(d*x + c)^5 - 2*(3*b*d*x + 4*b)*cosh(d*x + c)^3 + (3*b*d*x + 4*b)*cosh(d *x + c))*e^(2*d*x + 2*c))*sinh(d*x + c) - 8*b)*((b*e^(6*d*x + 6*c) + 3*b*e ^(4*d*x + 4*c) + 3*b*e^(2*d*x + 2*c) + b)/(e^(6*d*x + 6*c) - 3*e^(4*d*x + 4*c) + 3*e^(2*d*x + 2*c) - 1))^(1/3)/(d*cosh(d*x + c)^6 + (d*e^(2*d*x + 2* c) + d)*sinh(d*x + c)^6 + 6*(d*cosh(d*x + c)*e^(2*d*x + 2*c) + d*cosh(d*x + c))*sinh(d*x + c)^5 - 3*d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 + (5* d*cosh(d*x + c)^2 - d)*e^(2*d*x + 2*c) - d)*sinh(d*x + c)^4 + 4*(5*d*co...
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (66) = 132\).
Time = 65.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.19 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\begin {cases} x \left (b \coth ^{3}{\left (c \right )}\right )^{\frac {4}{3}} & \text {for}\: d = 0 \\- \frac {\left (b \coth ^{3}{\left (d x + \log {\left (- e^{- d x} \right )} \right )}\right )^{\frac {4}{3}} \log {\left (- e^{- d x} \right )}}{d} & \text {for}\: c = \log {\left (- e^{- d x} \right )} \\- \frac {\left (b \coth ^{3}{\left (d x + \log {\left (e^{- d x} \right )} \right )}\right )^{\frac {4}{3}} \log {\left (e^{- d x} \right )}}{d} & \text {for}\: c = \log {\left (e^{- d x} \right )} \\x \left (\frac {b}{\tanh ^{3}{\left (c + d x \right )}}\right )^{\frac {4}{3}} \tanh ^{4}{\left (c + d x \right )} - \frac {\left (\frac {b}{\tanh ^{3}{\left (c + d x \right )}}\right )^{\frac {4}{3}} \tanh ^{3}{\left (c + d x \right )}}{d} - \frac {\left (\frac {b}{\tanh ^{3}{\left (c + d x \right )}}\right )^{\frac {4}{3}} \tanh {\left (c + d x \right )}}{3 d} & \text {otherwise} \end {cases} \] Input:
integrate((b*coth(d*x+c)**3)**(4/3),x)
Output:
Piecewise((x*(b*coth(c)**3)**(4/3), Eq(d, 0)), (-(b*coth(d*x + log(-exp(-d *x)))**3)**(4/3)*log(-exp(-d*x))/d, Eq(c, log(-exp(-d*x)))), (-(b*coth(d*x + log(exp(-d*x)))**3)**(4/3)*log(exp(-d*x))/d, Eq(c, log(exp(-d*x)))), (x *(b/tanh(c + d*x)**3)**(4/3)*tanh(c + d*x)**4 - (b/tanh(c + d*x)**3)**(4/3 )*tanh(c + d*x)**3/d - (b/tanh(c + d*x)**3)**(4/3)*tanh(c + d*x)/(3*d), Tr ue))
Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\frac {{\left (d x + c\right )} b^{\frac {4}{3}}}{d} - \frac {4 \, {\left (3 \, b^{\frac {4}{3}} e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, b^{\frac {4}{3}} e^{\left (-4 \, d x - 4 \, c\right )} - 2 \, b^{\frac {4}{3}}\right )}}{3 \, d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} \] Input:
integrate((b*coth(d*x+c)^3)^(4/3),x, algorithm="maxima")
Output:
(d*x + c)*b^(4/3)/d - 4/3*(3*b^(4/3)*e^(-2*d*x - 2*c) - 3*b^(4/3)*e^(-4*d* x - 4*c) - 2*b^(4/3))/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6* d*x - 6*c) - 1))
\[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\int { \left (b \coth \left (d x + c\right )^{3}\right )^{\frac {4}{3}} \,d x } \] Input:
integrate((b*coth(d*x+c)^3)^(4/3),x, algorithm="giac")
Output:
integrate((b*coth(d*x + c)^3)^(4/3), x)
Timed out. \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\int {\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^3\right )}^{4/3} \,d x \] Input:
int((b*coth(c + d*x)^3)^(4/3),x)
Output:
int((b*coth(c + d*x)^3)^(4/3), x)
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.42 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\frac {b^{\frac {4}{3}} \left (-\coth \left (d x +c \right )^{3}-3 \coth \left (d x +c \right )+3 d x \right )}{3 d} \] Input:
int((b*coth(d*x+c)^3)^(4/3),x)
Output:
(b**(1/3)*b*( - coth(c + d*x)**3 - 3*coth(c + d*x) + 3*d*x))/(3*d)