Integrand size = 14, antiderivative size = 201 \[ \int \frac {1}{\left (a+b \tanh ^2(c+d x)\right )^4} \, dx=\frac {x}{(a+b)^4}+\frac {\sqrt {b} \left (35 a^3+35 a^2 b+21 a b^2+5 b^3\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{16 a^{7/2} (a+b)^4 d}+\frac {b \tanh (c+d x)}{6 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )^3}+\frac {b (11 a+5 b) \tanh (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {b \left (19 a^2+16 a b+5 b^2\right ) \tanh (c+d x)}{16 a^3 (a+b)^3 d \left (a+b \tanh ^2(c+d x)\right )} \] Output:
x/(a+b)^4+1/16*b^(1/2)*(35*a^3+35*a^2*b+21*a*b^2+5*b^3)*arctan(b^(1/2)*tan h(d*x+c)/a^(1/2))/a^(7/2)/(a+b)^4/d+1/6*b*tanh(d*x+c)/a/(a+b)/d/(a+b*tanh( d*x+c)^2)^3+1/24*b*(11*a+5*b)*tanh(d*x+c)/a^2/(a+b)^2/d/(a+b*tanh(d*x+c)^2 )^2+1/16*b*(19*a^2+16*a*b+5*b^2)*tanh(d*x+c)/a^3/(a+b)^3/d/(a+b*tanh(d*x+c )^2)
Time = 0.40 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (a+b \tanh ^2(c+d x)\right )^4} \, dx=\frac {\frac {3 \sqrt {b} \left (35 a^3+35 a^2 b+21 a b^2+5 b^3\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{7/2}}-24 \log (1-\tanh (c+d x))+24 \log (1+\tanh (c+d x))+\frac {8 b (a+b)^3 \tanh (c+d x)}{a \left (a+b \tanh ^2(c+d x)\right )^3}+\frac {2 b (a+b)^2 (11 a+5 b) \tanh (c+d x)}{a^2 \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {3 b (a+b) \left (19 a^2+16 a b+5 b^2\right ) \tanh (c+d x)}{a^3 \left (a+b \tanh ^2(c+d x)\right )}}{48 (a+b)^4 d} \] Input:
Integrate[(a + b*Tanh[c + d*x]^2)^(-4),x]
Output:
((3*Sqrt[b]*(35*a^3 + 35*a^2*b + 21*a*b^2 + 5*b^3)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/a^(7/2) - 24*Log[1 - Tanh[c + d*x]] + 24*Log[1 + Tanh[c + d*x]] + (8*b*(a + b)^3*Tanh[c + d*x])/(a*(a + b*Tanh[c + d*x]^2)^3) + (2 *b*(a + b)^2*(11*a + 5*b)*Tanh[c + d*x])/(a^2*(a + b*Tanh[c + d*x]^2)^2) + (3*b*(a + b)*(19*a^2 + 16*a*b + 5*b^2)*Tanh[c + d*x])/(a^3*(a + b*Tanh[c + d*x]^2)))/(48*(a + b)^4*d)
Time = 0.73 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.19, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4144, 316, 402, 27, 402, 25, 397, 218, 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.
| \(\displaystyle \int \frac {1}{\left (a+b \tanh ^2(c+d x)\right )^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a-b \tan (i c+i d x)^2\right )^4}dx\) |
| \(\Big \downarrow \) 4144 |
| \(\displaystyle \frac {\int \frac {1}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^4}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x)}{6 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^3}-\frac {\int \frac {5 b \tanh ^2(c+d x)+b-6 (a+b)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{6 a (a+b)}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x)}{6 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^3}-\frac {-\frac {\int \frac {3 \left (8 a^2+11 b a+5 b^2-b (11 a+5 b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{4 a (a+b)}-\frac {b (11 a+5 b) \tanh (c+d x)}{4 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{6 a (a+b)}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x)}{6 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^3}-\frac {-\frac {3 \int \frac {8 a^2+11 b a+5 b^2-b (11 a+5 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{4 a (a+b)}-\frac {b (11 a+5 b) \tanh (c+d x)}{4 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{6 a (a+b)}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x)}{6 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^3}-\frac {-\frac {3 \left (\frac {b \left (19 a^2+16 a b+5 b^2\right ) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac {\int -\frac {16 a^3+19 b a^2+16 b^2 a+5 b^3-b \left (19 a^2+16 b a+5 b^2\right ) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 a (a+b)}\right )}{4 a (a+b)}-\frac {b (11 a+5 b) \tanh (c+d x)}{4 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{6 a (a+b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x)}{6 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^3}-\frac {-\frac {3 \left (\frac {\int \frac {16 a^3+19 b a^2+16 b^2 a+5 b^3-b \left (19 a^2+16 b a+5 b^2\right ) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 a (a+b)}+\frac {b \left (19 a^2+16 a b+5 b^2\right ) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}\right )}{4 a (a+b)}-\frac {b (11 a+5 b) \tanh (c+d x)}{4 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{6 a (a+b)}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x)}{6 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^3}-\frac {-\frac {3 \left (\frac {\frac {16 a^3 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {b \left (35 a^3+35 a^2 b+21 a b^2+5 b^3\right ) \int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a+b}}{2 a (a+b)}+\frac {b \left (19 a^2+16 a b+5 b^2\right ) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}\right )}{4 a (a+b)}-\frac {b (11 a+5 b) \tanh (c+d x)}{4 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{6 a (a+b)}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x)}{6 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^3}-\frac {-\frac {3 \left (\frac {\frac {16 a^3 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {\sqrt {b} \left (35 a^3+35 a^2 b+21 a b^2+5 b^3\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{2 a (a+b)}+\frac {b \left (19 a^2+16 a b+5 b^2\right ) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}\right )}{4 a (a+b)}-\frac {b (11 a+5 b) \tanh (c+d x)}{4 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{6 a (a+b)}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x)}{6 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^3}-\frac {-\frac {3 \left (\frac {b \left (19 a^2+16 a b+5 b^2\right ) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac {\frac {16 a^3 \text {arctanh}(\tanh (c+d x))}{a+b}+\frac {\sqrt {b} \left (35 a^3+35 a^2 b+21 a b^2+5 b^3\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{2 a (a+b)}\right )}{4 a (a+b)}-\frac {b (11 a+5 b) \tanh (c+d x)}{4 a (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{6 a (a+b)}}{d}\) |
Input:
Int[(a + b*Tanh[c + d*x]^2)^(-4),x]
Output:
((b*Tanh[c + d*x])/(6*a*(a + b)*(a + b*Tanh[c + d*x]^2)^3) - (-1/4*(b*(11* a + 5*b)*Tanh[c + d*x])/(a*(a + b)*(a + b*Tanh[c + d*x]^2)^2) - (3*(((Sqrt [b]*(35*a^3 + 35*a^2*b + 21*a*b^2 + 5*b^3)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/ Sqrt[a]])/(Sqrt[a]*(a + b)) + (16*a^3*ArcTanh[Tanh[c + d*x]])/(a + b))/(2* a*(a + b)) + (b*(19*a^2 + 16*a*b + 5*b^2)*Tanh[c + d*x])/(2*a*(a + b)*(a + b*Tanh[c + d*x]^2))))/(4*a*(a + b)))/(6*a*(a + b)))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(a + b* (ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 0.30 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {\frac {b \left (\frac {\frac {b^{2} \left (19 a^{3}+35 a^{2} b +21 a \,b^{2}+5 b^{3}\right ) \tanh \left (d x +c \right )^{5}}{16 a^{3}}+\frac {b \left (17 a^{3}+33 a^{2} b +21 a \,b^{2}+5 b^{3}\right ) \tanh \left (d x +c \right )^{3}}{6 a^{2}}+\frac {\left (29 a^{3}+61 a^{2} b +43 a \,b^{2}+11 b^{3}\right ) \tanh \left (d x +c \right )}{16 a}}{\left (a +b \tanh \left (d x +c \right )^{2}\right )^{3}}+\frac {\left (35 a^{3}+35 a^{2} b +21 a \,b^{2}+5 b^{3}\right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{16 a^{3} \sqrt {a b}}\right )}{\left (a +b \right )^{4}}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{4}}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{4}}}{d}\) | \(219\) |
| default | \(\frac {\frac {b \left (\frac {\frac {b^{2} \left (19 a^{3}+35 a^{2} b +21 a \,b^{2}+5 b^{3}\right ) \tanh \left (d x +c \right )^{5}}{16 a^{3}}+\frac {b \left (17 a^{3}+33 a^{2} b +21 a \,b^{2}+5 b^{3}\right ) \tanh \left (d x +c \right )^{3}}{6 a^{2}}+\frac {\left (29 a^{3}+61 a^{2} b +43 a \,b^{2}+11 b^{3}\right ) \tanh \left (d x +c \right )}{16 a}}{\left (a +b \tanh \left (d x +c \right )^{2}\right )^{3}}+\frac {\left (35 a^{3}+35 a^{2} b +21 a \,b^{2}+5 b^{3}\right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{16 a^{3} \sqrt {a b}}\right )}{\left (a +b \right )^{4}}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{4}}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{4}}}{d}\) | \(219\) |
| risch | \(\text {Expression too large to display}\) | \(1031\) |
Input:
int(1/(a+b*tanh(d*x+c)^2)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(1/(a+b)^4*b*((1/16*b^2*(19*a^3+35*a^2*b+21*a*b^2+5*b^3)/a^3*tanh(d*x+ c)^5+1/6*b*(17*a^3+33*a^2*b+21*a*b^2+5*b^3)/a^2*tanh(d*x+c)^3+1/16*(29*a^3 +61*a^2*b+43*a*b^2+11*b^3)/a*tanh(d*x+c))/(a+b*tanh(d*x+c)^2)^3+1/16*(35*a ^3+35*a^2*b+21*a*b^2+5*b^3)/a^3/(a*b)^(1/2)*arctan(b*tanh(d*x+c)/(a*b)^(1/ 2)))+1/2/(a+b)^4*ln(1+tanh(d*x+c))-1/2/(a+b)^4*ln(-1+tanh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 9849 vs. \(2 (185) = 370\).
Time = 0.37 (sec) , antiderivative size = 20020, normalized size of antiderivative = 99.60 \[ \int \frac {1}{\left (a+b \tanh ^2(c+d x)\right )^4} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*tanh(d*x+c)^2)^4,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{\left (a+b \tanh ^2(c+d x)\right )^4} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*tanh(d*x+c)**2)**4,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 925 vs. \(2 (185) = 370\).
Time = 0.42 (sec) , antiderivative size = 925, normalized size of antiderivative = 4.60 \[ \int \frac {1}{\left (a+b \tanh ^2(c+d x)\right )^4} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+b*tanh(d*x+c)^2)^4,x, algorithm="maxima")
Output:
-1/16*(35*a^3*b + 35*a^2*b^2 + 21*a*b^3 + 5*b^4)*arctan(1/2*((a + b)*e^(-2 *d*x - 2*c) + a - b)/sqrt(a*b))/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*sqrt(a*b)*d) + 1/24*(87*a^5*b + 319*a^4*b^2 + 450*a^3*b^3 + 306*a ^2*b^4 + 103*a*b^5 + 15*b^6 + 3*(145*a^5*b + 267*a^4*b^2 + 34*a^3*b^3 - 17 8*a^2*b^4 - 115*a*b^5 - 25*b^6)*e^(-2*d*x - 2*c) + 6*(145*a^5*b + 93*a^4*b ^2 - 6*a^3*b^3 + 106*a^2*b^4 + 85*a*b^5 + 25*b^6)*e^(-4*d*x - 4*c) + 2*(43 5*a^5*b + 29*a^4*b^2 + 162*a^3*b^3 - 306*a^2*b^4 - 245*a*b^5 - 75*b^6)*e^( -6*d*x - 6*c) + 3*(145*a^5*b + 17*a^4*b^2 - 58*a^3*b^3 + 150*a^2*b^4 + 105 *a*b^5 + 25*b^6)*e^(-8*d*x - 8*c) + 3*(29*a^5*b + 23*a^4*b^2 - 62*a^3*b^3 - 82*a^2*b^4 - 31*a*b^5 - 5*b^6)*e^(-10*d*x - 10*c))/((a^10 + 7*a^9*b + 21 *a^8*b^2 + 35*a^7*b^3 + 35*a^6*b^4 + 21*a^5*b^5 + 7*a^4*b^6 + a^3*b^7 + 6* (a^10 + 5*a^9*b + 9*a^8*b^2 + 5*a^7*b^3 - 5*a^6*b^4 - 9*a^5*b^5 - 5*a^4*b^ 6 - a^3*b^7)*e^(-2*d*x - 2*c) + 3*(5*a^10 + 19*a^9*b + 25*a^8*b^2 + 15*a^7 *b^3 + 15*a^6*b^4 + 25*a^5*b^5 + 19*a^4*b^6 + 5*a^3*b^7)*e^(-4*d*x - 4*c) + 4*(5*a^10 + 17*a^9*b + 21*a^8*b^2 + 9*a^7*b^3 - 9*a^6*b^4 - 21*a^5*b^5 - 17*a^4*b^6 - 5*a^3*b^7)*e^(-6*d*x - 6*c) + 3*(5*a^10 + 19*a^9*b + 25*a^8* b^2 + 15*a^7*b^3 + 15*a^6*b^4 + 25*a^5*b^5 + 19*a^4*b^6 + 5*a^3*b^7)*e^(-8 *d*x - 8*c) + 6*(a^10 + 5*a^9*b + 9*a^8*b^2 + 5*a^7*b^3 - 5*a^6*b^4 - 9*a^ 5*b^5 - 5*a^4*b^6 - a^3*b^7)*e^(-10*d*x - 10*c) + (a^10 + 7*a^9*b + 21*a^8 *b^2 + 35*a^7*b^3 + 35*a^6*b^4 + 21*a^5*b^5 + 7*a^4*b^6 + a^3*b^7)*e^(-...
Leaf count of result is larger than twice the leaf count of optimal. 750 vs. \(2 (185) = 370\).
Time = 0.19 (sec) , antiderivative size = 750, normalized size of antiderivative = 3.73 \[ \int \frac {1}{\left (a+b \tanh ^2(c+d x)\right )^4} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+b*tanh(d*x+c)^2)^4,x, algorithm="giac")
Output:
1/48*(3*(35*a^3*b + 35*a^2*b^2 + 21*a*b^3 + 5*b^4)*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*sqrt(a*b)) + 48*(d*x + c)/(a^4 + 4*a^3*b + 6*a^2*b ^2 + 4*a*b^3 + b^4) - 2*(87*a^5*b*e^(10*d*x + 10*c) + 69*a^4*b^2*e^(10*d*x + 10*c) - 186*a^3*b^3*e^(10*d*x + 10*c) - 246*a^2*b^4*e^(10*d*x + 10*c) - 93*a*b^5*e^(10*d*x + 10*c) - 15*b^6*e^(10*d*x + 10*c) + 435*a^5*b*e^(8*d* x + 8*c) + 51*a^4*b^2*e^(8*d*x + 8*c) - 174*a^3*b^3*e^(8*d*x + 8*c) + 450* a^2*b^4*e^(8*d*x + 8*c) + 315*a*b^5*e^(8*d*x + 8*c) + 75*b^6*e^(8*d*x + 8* c) + 870*a^5*b*e^(6*d*x + 6*c) + 58*a^4*b^2*e^(6*d*x + 6*c) + 324*a^3*b^3* e^(6*d*x + 6*c) - 612*a^2*b^4*e^(6*d*x + 6*c) - 490*a*b^5*e^(6*d*x + 6*c) - 150*b^6*e^(6*d*x + 6*c) + 870*a^5*b*e^(4*d*x + 4*c) + 558*a^4*b^2*e^(4*d *x + 4*c) - 36*a^3*b^3*e^(4*d*x + 4*c) + 636*a^2*b^4*e^(4*d*x + 4*c) + 510 *a*b^5*e^(4*d*x + 4*c) + 150*b^6*e^(4*d*x + 4*c) + 435*a^5*b*e^(2*d*x + 2* c) + 801*a^4*b^2*e^(2*d*x + 2*c) + 102*a^3*b^3*e^(2*d*x + 2*c) - 534*a^2*b ^4*e^(2*d*x + 2*c) - 345*a*b^5*e^(2*d*x + 2*c) - 75*b^6*e^(2*d*x + 2*c) + 87*a^5*b + 319*a^4*b^2 + 450*a^3*b^3 + 306*a^2*b^4 + 103*a*b^5 + 15*b^6)/( (a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*(a*e^(4*d*x + 4*c) + b*e ^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)^3))/d
Time = 1.29 (sec) , antiderivative size = 3685, normalized size of antiderivative = 18.33 \[ \int \frac {1}{\left (a+b \tanh ^2(c+d x)\right )^4} \, dx=\text {Too large to display} \] Input:
int(1/(a + b*tanh(c + d*x)^2)^4,x)
Output:
log(tanh(c + d*x) + 1)/(2*a^4*d + 2*b^4*d + 12*a^2*b^2*d + 8*a*b^3*d + 8*a ^3*b*d) + ((tanh(c + d*x)^3*(16*a*b^3 + 5*b^4 + 17*a^2*b^2))/(6*a^2*(3*a*b ^2 + 3*a^2*b + a^3 + b^3)) + (tanh(c + d*x)*(32*a*b^2 + 29*a^2*b + 11*b^3) )/(16*a*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) + (b^2*tanh(c + d*x)^5*(16*a*b^2 + 19*a^2*b + 5*b^3))/(16*a^2*(a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)))/(a^3*d + b^3*d*tanh(c + d*x)^6 + 3*a^2*b*d*tanh(c + d*x)^2 + 3*a*b^2*d*tanh(c + d *x)^4) - log(tanh(c + d*x) - 1)/(2*d*(a + b)^4) - (atan((((-a^7*b)^(1/2)*( (tanh(c + d*x)*(210*a*b^8 + 25*b^9 + 791*a^2*b^7 + 1820*a^3*b^6 + 2695*a^4 *b^5 + 2450*a^5*b^4 + 1481*a^6*b^3))/(128*(a^12*d^2 + 6*a^11*b*d^2 + a^6*b ^6*d^2 + 6*a^7*b^5*d^2 + 15*a^8*b^4*d^2 + 20*a^9*b^3*d^2 + 15*a^10*b^2*d^2 )) + ((((5*a^3*b^13*d^2)/4 + 14*a^4*b^12*d^2 + (287*a^5*b^11*d^2)/4 + 224* a^6*b^10*d^2 + (953*a^7*b^9*d^2)/2 + 728*a^8*b^8*d^2 + (1631*a^9*b^7*d^2)/ 2 + 668*a^10*b^6*d^2 + (1561*a^11*b^5*d^2)/4 + 154*a^12*b^4*d^2 + (147*a^1 3*b^3*d^2)/4 + 4*a^14*b^2*d^2)/(a^15*d^3 + 9*a^14*b*d^3 + a^6*b^9*d^3 + 9* a^7*b^8*d^3 + 36*a^8*b^7*d^3 + 84*a^9*b^6*d^3 + 126*a^10*b^5*d^3 + 126*a^1 1*b^4*d^3 + 84*a^12*b^3*d^3 + 36*a^13*b^2*d^3) - (tanh(c + d*x)*(-a^7*b)^( 1/2)*(21*a*b^2 + 35*a^2*b + 35*a^3 + 5*b^3)*(1024*a^6*b^11*d^2 + 7168*a^7* b^10*d^2 + 20480*a^8*b^9*d^2 + 28672*a^9*b^8*d^2 + 14336*a^10*b^7*d^2 - 14 336*a^11*b^6*d^2 - 28672*a^12*b^5*d^2 - 20480*a^13*b^4*d^2 - 7168*a^14*b^3 *d^2 - 1024*a^15*b^2*d^2))/(4096*(a^11*d + a^7*b^4*d + 4*a^8*b^3*d + 6*...
Time = 0.44 (sec) , antiderivative size = 1058, normalized size of antiderivative = 5.26 \[ \int \frac {1}{\left (a+b \tanh ^2(c+d x)\right )^4} \, dx =\text {Too large to display} \] Input:
int(1/(a+b*tanh(d*x+c)^2)^4,x)
Output:
(105*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tanh(c + d* x)**6*a**3*b**3 + 105*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)*b)/(sqrt(b)*sqrt (a)))*tanh(c + d*x)**6*a**2*b**4 + 63*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)* b)/(sqrt(b)*sqrt(a)))*tanh(c + d*x)**6*a*b**5 + 15*sqrt(b)*sqrt(a)*atan((t anh(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tanh(c + d*x)**6*b**6 + 315*sqrt(b)*sqr t(a)*atan((tanh(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tanh(c + d*x)**4*a**4*b**2 + 315*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tanh(c + d *x)**4*a**3*b**3 + 189*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)*b)/(sqrt(b)*sqr t(a)))*tanh(c + d*x)**4*a**2*b**4 + 45*sqrt(b)*sqrt(a)*atan((tanh(c + d*x) *b)/(sqrt(b)*sqrt(a)))*tanh(c + d*x)**4*a*b**5 + 315*sqrt(b)*sqrt(a)*atan( (tanh(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tanh(c + d*x)**2*a**5*b + 315*sqrt(b) *sqrt(a)*atan((tanh(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tanh(c + d*x)**2*a**4*b **2 + 189*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tanh(c + d*x)**2*a**3*b**3 + 45*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)*b)/(sqrt(b)* sqrt(a)))*tanh(c + d*x)**2*a**2*b**4 + 105*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)*b)/(sqrt(b)*sqrt(a)))*a**6 + 105*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)* b)/(sqrt(b)*sqrt(a)))*a**5*b + 63*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)*b)/( sqrt(b)*sqrt(a)))*a**4*b**2 + 15*sqrt(b)*sqrt(a)*atan((tanh(c + d*x)*b)/(s qrt(b)*sqrt(a)))*a**3*b**3 + 48*tanh(c + d*x)**6*a**4*b**3*d*x + 57*tanh(c + d*x)**5*a**4*b**3 + 105*tanh(c + d*x)**5*a**3*b**4 + 63*tanh(c + d*x...