\(\int (b \coth ^3(c+d x))^{4/3} \, dx\) [33]

Optimal result

Integrand size = 14, antiderivative size = 74 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=-\frac {b \sqrt [3]{b \coth ^3(c+d x)}}{d}-\frac {b \coth ^2(c+d x) \sqrt [3]{b \coth ^3(c+d x)}}{3 d}+b x \sqrt [3]{b \coth ^3(c+d x)} \tanh (c+d x) \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=-\frac {b \sqrt [3]{b \coth ^3(c+d x)}}{d}-\frac {b \coth ^2(c+d x) \sqrt [3]{b \coth ^3(c+d x)}}{3 d}+b x \tanh (c+d x) \sqrt [3]{b \coth ^3(c+d x)} \]

[In]

Int[(b*Coth[c + d*x]^3)^(4/3),x]

[Out]

-((b*(b*Coth[c + d*x]^3)^(1/3))/d) - (b*Coth[c + d*x]^2*(b*Coth[c + d*x]^3)^(1/3))/(3*d) + b*x*(b*Coth[c + d*x
]^3)^(1/3)*Tanh[c + d*x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[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
]])

Rubi steps \begin{align*} \text {integral}& = \left (b \sqrt [3]{b \coth ^3(c+d x)} \tanh (c+d x)\right ) \int \coth ^4(c+d x) \, dx \\ & = -\frac {b \coth ^2(c+d x) \sqrt [3]{b \coth ^3(c+d x)}}{3 d}+\left (b \sqrt [3]{b \coth ^3(c+d x)} \tanh (c+d x)\right ) \int \coth ^2(c+d x) \, 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}+\left (b \sqrt [3]{b \coth ^3(c+d x)} \tanh (c+d x)\right ) \int 1 \, 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) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.58 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=-\frac {\left (b \coth ^3(c+d x)\right )^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\tanh ^2(c+d x)\right ) \tanh (c+d x)}{3 d} \]

[In]

Integrate[(b*Coth[c + d*x]^3)^(4/3),x]

[Out]

-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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.76

method	result	size
risch	\(-\frac {b {\left (\frac {b \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\right )}^{\frac {1}{3}} \left (-3 \,{\mathrm e}^{6 d x +6 c} d x +9 \,{\mathrm e}^{4 d x +4 c} d x -9 \,{\mathrm e}^{2 d x +2 c} d x +3 d x +12 \,{\mathrm e}^{4 d x +4 c}-12 \,{\mathrm e}^{2 d x +2 c}+8\right )}{3 \left ({\mathrm e}^{2 d x +2 c}+1\right ) \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} d}\)	\(130\)

[In]

int((b*coth(d*x+c)^3)^(4/3),x,method=_RETURNVERBOSE)

[Out]

-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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1046 vs. \(2 (66) = 132\).

Time = 0.30 (sec) , antiderivative size = 1046, normalized size of antiderivative = 14.14 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\text {Too large to display} \]

[In]

integrate((b*coth(d*x+c)^3)^(4/3),x, algorithm="fricas")

[Out]

-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))*sinh(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)*co
sh(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)*cos
h(d*x + c)^4 - 3*b*d*x + 3*(3*b*d*x + 4*b)*cosh(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*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))*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*cosh(d*x + c)^3 - 3*d*cosh(d*x + c) + (5*d*cosh(d*x + c)^3 - 3*d*
cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c)^3 + 3*d*cosh(d*x + c)^2 + 3*(5*d*cosh(d*x + c)^4 - 6*d*cosh(d*x
+ c)^2 + (5*d*cosh(d*x + c)^4 - 6*d*cosh(d*x + c)^2 + d)*e^(2*d*x + 2*c) + d)*sinh(d*x + c)^2 + (d*cosh(d*x +
c)^6 - 3*d*cosh(d*x + c)^4 + 3*d*cosh(d*x + c)^2 - d)*e^(2*d*x + 2*c) + 6*(d*cosh(d*x + c)^5 - 2*d*cosh(d*x +
c)^3 + d*cosh(d*x + c) + (d*cosh(d*x + c)^5 - 2*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x
+ c) - d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (66) = 132\).

Time = 68.06 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.04 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\begin {cases} x \left (b \coth ^{3}{\left (c \right )}\right )^{\frac {4}{3}} & \text {for}\: d = 0 \\- \frac {\left (b \coth ^{3}{\left (d x + \log {\left (- e^{- d x} \right )} \right )}\right )^{\frac {4}{3}} \log {\left (- e^{- d x} \right )}}{d} & \text {for}\: c = \log {\left (- e^{- d x} \right )} \\x \left (b \coth ^{3}{\left (d x + \log {\left (e^{- d x} \right )} \right )}\right )^{\frac {4}{3}} & \text {for}\: c = \log {\left (e^{- d x} \right )} \\x \left (\frac {b}{\tanh ^{3}{\left (c + d x \right )}}\right )^{\frac {4}{3}} \tanh ^{4}{\left (c + d x \right )} - \frac {\left (\frac {b}{\tanh ^{3}{\left (c + d x \right )}}\right )^{\frac {4}{3}} \tanh ^{3}{\left (c + d x \right )}}{d} - \frac {\left (\frac {b}{\tanh ^{3}{\left (c + d x \right )}}\right )^{\frac {4}{3}} \tanh {\left (c + d x \right )}}{3 d} & \text {otherwise} \end {cases} \]

[In]

integrate((b*coth(d*x+c)**3)**(4/3),x)

[Out]

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)))), (x*(b*coth(d*x + log(exp(-d*x)))**3)**(4/3), 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), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18 \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\frac {{\left (d x + c\right )} b^{\frac {4}{3}}}{d} - \frac {4 \, {\left (3 \, b^{\frac {4}{3}} e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, b^{\frac {4}{3}} e^{\left (-4 \, d x - 4 \, c\right )} - 2 \, b^{\frac {4}{3}}\right )}}{3 \, d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} \]

[In]

integrate((b*coth(d*x+c)^3)^(4/3),x, algorithm="maxima")

[Out]

(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))

Giac [F]

\[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\int { \left (b \coth \left (d x + c\right )^{3}\right )^{\frac {4}{3}} \,d x } \]

[In]

integrate((b*coth(d*x+c)^3)^(4/3),x, algorithm="giac")

[Out]

integrate((b*coth(d*x + c)^3)^(4/3), x)

Mupad [F(-1)]

Timed out. \[ \int \left (b \coth ^3(c+d x)\right )^{4/3} \, dx=\int {\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^3\right )}^{4/3} \,d x \]

[In]

int((b*coth(c + d*x)^3)^(4/3),x)

[Out]

int((b*coth(c + d*x)^3)^(4/3), x)