3.63 \(\int \frac {1}{(a+b \tanh (c+d x))^3} \, dx\)

Optimal. Leaf size=129 \[ \frac {2 a b}{d \left (a^2-b^2\right )^2 (a+b \tanh (c+d x))}+\frac {b}{2 d \left (a^2-b^2\right ) (a+b \tanh (c+d x))^2}-\frac {b \left (3 a^2+b^2\right ) \log (a \cosh (c+d x)+b \sinh (c+d x))}{d \left (a^2-b^2\right )^3}+\frac {a x \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3} \]

[Out]

a*(a^2+3*b^2)*x/(a^2-b^2)^3-b*(3*a^2+b^2)*ln(a*cosh(d*x+c)+b*sinh(d*x+c))/(a^2-b^2)^3/d+1/2*b/(a^2-b^2)/d/(a+b
*tanh(d*x+c))^2+2*a*b/(a^2-b^2)^2/d/(a+b*tanh(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3483, 3529, 3531, 3530} \[ \frac {2 a b}{d \left (a^2-b^2\right )^2 (a+b \tanh (c+d x))}+\frac {b}{2 d \left (a^2-b^2\right ) (a+b \tanh (c+d x))^2}-\frac {b \left (3 a^2+b^2\right ) \log (a \cosh (c+d x)+b \sinh (c+d x))}{d \left (a^2-b^2\right )^3}+\frac {a x \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tanh[c + d*x])^(-3),x]

[Out]

(a*(a^2 + 3*b^2)*x)/(a^2 - b^2)^3 - (b*(3*a^2 + b^2)*Log[a*Cosh[c + d*x] + b*Sinh[c + d*x]])/((a^2 - b^2)^3*d)
+ b/(2*(a^2 - b^2)*d*(a + b*Tanh[c + d*x])^2) + (2*a*b)/((a^2 - b^2)^2*d*(a + b*Tanh[c + d*x]))

Rule 3483

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \tanh (c+d x))^3} \, dx &=\frac {b}{2 \left (a^2-b^2\right ) d (a+b \tanh (c+d x))^2}+\frac {\int \frac {a-b \tanh (c+d x)}{(a+b \tanh (c+d x))^2} \, dx}{a^2-b^2}\\ &=\frac {b}{2 \left (a^2-b^2\right ) d (a+b \tanh (c+d x))^2}+\frac {2 a b}{\left (a^2-b^2\right )^2 d (a+b \tanh (c+d x))}+\frac {\int \frac {a^2+b^2-2 a b \tanh (c+d x)}{a+b \tanh (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {a \left (a^2+3 b^2\right ) x}{\left (a^2-b^2\right )^3}+\frac {b}{2 \left (a^2-b^2\right ) d (a+b \tanh (c+d x))^2}+\frac {2 a b}{\left (a^2-b^2\right )^2 d (a+b \tanh (c+d x))}-\frac {\left (i b \left (3 a^2+b^2\right )\right ) \int \frac {-i b-i a \tanh (c+d x)}{a+b \tanh (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=\frac {a \left (a^2+3 b^2\right ) x}{\left (a^2-b^2\right )^3}-\frac {b \left (3 a^2+b^2\right ) \log (a \cosh (c+d x)+b \sinh (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {b}{2 \left (a^2-b^2\right ) d (a+b \tanh (c+d x))^2}+\frac {2 a b}{\left (a^2-b^2\right )^2 d (a+b \tanh (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.46, size = 122, normalized size = 0.95 \[ \frac {\frac {b \left (\frac {\left (a^2-b^2\right ) \left (5 a^2+4 a b \tanh (c+d x)-b^2\right )}{(a+b \tanh (c+d x))^2}-2 \left (3 a^2+b^2\right ) \log (a+b \tanh (c+d x))\right )}{\left (a^2-b^2\right )^3}-\frac {\log (1-\tanh (c+d x))}{(a+b)^3}+\frac {\log (\tanh (c+d x)+1)}{(a-b)^3}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tanh[c + d*x])^(-3),x]

[Out]

(-(Log[1 - Tanh[c + d*x]]/(a + b)^3) + Log[1 + Tanh[c + d*x]]/(a - b)^3 + (b*(-2*(3*a^2 + b^2)*Log[a + b*Tanh[
c + d*x]] + ((a^2 - b^2)*(5*a^2 - b^2 + 4*a*b*Tanh[c + d*x]))/(a + b*Tanh[c + d*x])^2))/(a^2 - b^2)^3)/(2*d)

________________________________________________________________________________________

fricas [B]  time = 0.65, size = 1427, normalized size = 11.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tanh(d*x+c))^3,x, algorithm="fricas")

[Out]

((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*x*cosh(d*x + c)^4 + 4*(a^5 + 5*a^4*b + 10*a^3*b^2
+ 10*a^2*b^3 + 5*a*b^4 + b^5)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 +
5*a*b^4 + b^5)*d*x*sinh(d*x + c)^4 + 6*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 + (a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 +
a*b^4 + b^5)*d*x + 2*(3*a^3*b^2 - a^2*b^3 - 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4
- b^5)*d*x)*cosh(d*x + c)^2 + 2*(3*a^3*b^2 - a^2*b^3 - 3*a*b^4 + b^5 + 3*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*
b^3 + 5*a*b^4 + b^5)*d*x*cosh(d*x + c)^2 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*x)*sinh(d
*x + c)^2 - (3*a^4*b - 6*a^3*b^2 + 4*a^2*b^3 - 2*a*b^4 + b^5 + (3*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^
5)*cosh(d*x + c)^4 + 4*(3*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^5)*cosh(d*x + c)*sinh(d*x + c)^3 + (3*a^
4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^5)*sinh(d*x + c)^4 + 2*(3*a^4*b - 2*a^2*b^3 - b^5)*cosh(d*x + c)^2 +
2*(3*a^4*b - 2*a^2*b^3 - b^5 + 3*(3*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^5)*cosh(d*x + c)^2)*sinh(d*x
+ c)^2 + 4*((3*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^5)*cosh(d*x + c)^3 + (3*a^4*b - 2*a^2*b^3 - b^5)*co
sh(d*x + c))*sinh(d*x + c))*log(2*(a*cosh(d*x + c) + b*sinh(d*x + c))/(cosh(d*x + c) - sinh(d*x + c))) + 4*((a
^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*x*cosh(d*x + c)^3 + (3*a^3*b^2 - a^2*b^3 - 3*a*b^4 +
b^5 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^8 + 2*a^
7*b - 2*a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*d*cosh(d*x + c)^4 + 4*(a^8 + 2*a^7*b - 2*
a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^8 + 2*a^7*b
- 2*a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*d*sinh(d*x + c)^4 + 2*(a^8 - 4*a^6*b^2 + 6*a^
4*b^4 - 4*a^2*b^6 + b^8)*d*cosh(d*x + c)^2 + 2*(3*(a^8 + 2*a^7*b - 2*a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b
^6 - 2*a*b^7 - b^8)*d*cosh(d*x + c)^2 + (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*d)*sinh(d*x + c)^2 + (
a^8 - 2*a^7*b - 2*a^6*b^2 + 6*a^5*b^3 - 6*a^3*b^5 + 2*a^2*b^6 + 2*a*b^7 - b^8)*d + 4*((a^8 + 2*a^7*b - 2*a^6*b
^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*d*cosh(d*x + c)^3 + (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a
^2*b^6 + b^8)*d*cosh(d*x + c))*sinh(d*x + c))

________________________________________________________________________________________

giac [A]  time = 0.17, size = 205, normalized size = 1.59 \[ -\frac {\frac {{\left (3 \, a^{2} b + b^{3}\right )} \log \left ({\left | -a e^{\left (2 \, d x + 2 \, c\right )} - b e^{\left (2 \, d x + 2 \, c\right )} - a + b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {d x + c}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {2 \, {\left ({\left (3 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} e^{\left (2 \, d x + 2 \, c\right )} + \frac {3 \, {\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )}}{a + b}\right )}}{{\left (a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b\right )}^{2} {\left (a + b\right )}^{2} {\left (a - b\right )}^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tanh(d*x+c))^3,x, algorithm="giac")

[Out]

-((3*a^2*b + b^3)*log(abs(-a*e^(2*d*x + 2*c) - b*e^(2*d*x + 2*c) - a + b))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)
- (d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 2*((3*a^2*b^2 - 4*a*b^3 + b^4)*e^(2*d*x + 2*c) + 3*(a^3*b^2 - 2
*a^2*b^3 + a*b^4)/(a + b))/((a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)^2*(a + b)^2*(a - b)^3))/d

________________________________________________________________________________________

maple [A]  time = 0.11, size = 166, normalized size = 1.29 \[ -\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 d \left (a +b \right )^{3}}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 d \left (a -b \right )^{3}}+\frac {b}{2 d \left (a -b \right ) \left (a +b \right ) \left (a +b \tanh \left (d x +c \right )\right )^{2}}+\frac {2 a b}{d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \tanh \left (d x +c \right )\right )}-\frac {3 b \ln \left (a +b \tanh \left (d x +c \right )\right ) a^{2}}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {b^{3} \ln \left (a +b \tanh \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tanh(d*x+c))^3,x)

[Out]

-1/2/d/(a+b)^3*ln(tanh(d*x+c)-1)+1/2/d/(a-b)^3*ln(1+tanh(d*x+c))+1/2/d*b/(a-b)/(a+b)/(a+b*tanh(d*x+c))^2+2/d*a
*b/(a+b)^2/(a-b)^2/(a+b*tanh(d*x+c))-3/d*b/(a+b)^3/(a-b)^3*ln(a+b*tanh(d*x+c))*a^2-1/d*b^3/(a+b)^3/(a-b)^3*ln(
a+b*tanh(d*x+c))

________________________________________________________________________________________

maxima [B]  time = 0.35, size = 325, normalized size = 2.52 \[ -\frac {{\left (3 \, a^{2} b + b^{3}\right )} \log \left (-{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} - a - b\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d} - \frac {2 \, {\left (3 \, a^{2} b^{2} + 3 \, a b^{3} + {\left (3 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )}}{{\left (a^{7} + a^{6} b - 3 \, a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6} - b^{7} + 2 \, {\left (a^{7} - a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - 3 \, a^{2} b^{5} - a b^{6} + b^{7}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{7} - 3 \, a^{6} b + a^{5} b^{2} + 5 \, a^{4} b^{3} - 5 \, a^{3} b^{4} - a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tanh(d*x+c))^3,x, algorithm="maxima")

[Out]

-(3*a^2*b + b^3)*log(-(a - b)*e^(-2*d*x - 2*c) - a - b)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d) - 2*(3*a^2*b^2
+ 3*a*b^3 + (3*a^2*b^2 - 2*a*b^3 - b^4)*e^(-2*d*x - 2*c))/((a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 +
3*a^2*b^5 - a*b^6 - b^7 + 2*(a^7 - a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7)*e^(-2
*d*x - 2*c) + (a^7 - 3*a^6*b + a^5*b^2 + 5*a^4*b^3 - 5*a^3*b^4 - a^2*b^5 + 3*a*b^6 - b^7)*e^(-4*d*x - 4*c))*d)
+ (d*x + c)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d)

________________________________________________________________________________________

mupad [B]  time = 2.11, size = 304, normalized size = 2.36 \[ \frac {\mathrm {tanh}\left (c+d\,x\right )\,\left (\frac {1}{a\,d}-\frac {a^4+a^2\,b^2}{a\,d\,\left (a^4-2\,a^2\,b^2+b^4\right )}\right )+\frac {a^2\,x}{\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}+\frac {b^2\,x\,{\mathrm {tanh}\left (c+d\,x\right )}^2}{a^3+3\,a^2\,b+3\,a\,b^2+b^3}+\frac {{\mathrm {tanh}\left (c+d\,x\right )}^2\,\left (\frac {b^5}{2}-\frac {5\,a^2\,b^3}{2}\right )}{a^2\,d\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,b\,x\,\mathrm {tanh}\left (c+d\,x\right )}{a^3+3\,a^2\,b+3\,a\,b^2+b^3}}{a^2+2\,a\,b\,\mathrm {tanh}\left (c+d\,x\right )+b^2\,{\mathrm {tanh}\left (c+d\,x\right )}^2}-\frac {\ln \left (a+b\,\mathrm {tanh}\left (c+d\,x\right )\right )\,\left (3\,a^2\,b+b^3\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )\,\left (3\,a^2\,b+b^3\right )}{d\,{\left (a^2-b^2\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tanh(c + d*x))^3,x)

[Out]

(tanh(c + d*x)*(1/(a*d) - (a^4 + a^2*b^2)/(a*d*(a^4 + b^4 - 2*a^2*b^2))) + (a^2*x)/((a + b)*(2*a*b + a^2 + b^2
)) + (b^2*x*tanh(c + d*x)^2)/(3*a*b^2 + 3*a^2*b + a^3 + b^3) + (tanh(c + d*x)^2*(b^5/2 - (5*a^2*b^3)/2))/(a^2*
d*(a^4 + b^4 - 2*a^2*b^2)) + (2*a*b*x*tanh(c + d*x))/(3*a*b^2 + 3*a^2*b + a^3 + b^3))/(a^2 + b^2*tanh(c + d*x)
^2 + 2*a*b*tanh(c + d*x)) - (log(a + b*tanh(c + d*x))*(3*a^2*b + b^3))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))
+ (log(tanh(c + d*x) + 1)*(3*a^2*b + b^3))/(d*(a^2 - b^2)^3)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tanh(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________