Integrand size = 23, antiderivative size = 129 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {(4 a+3 b) \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a+b)^{3/2} d}+\frac {b \sinh (c+d x)}{4 a (a+b) d \left (a+(a+b) \sinh ^2(c+d x)\right )^2}+\frac {(4 a+3 b) \sinh (c+d x)}{8 a^2 (a+b) d \left (a+(a+b) \sinh ^2(c+d x)\right )} \] Output:
1/8*(4*a+3*b)*arctan((a+b)^(1/2)*sinh(d*x+c)/a^(1/2))/a^(5/2)/(a+b)^(3/2)/ d+1/4*b*sinh(d*x+c)/a/(a+b)/d/(a+(a+b)*sinh(d*x+c)^2)^2+1/8*(4*a+3*b)*sinh (d*x+c)/a^2/(a+b)/d/(a+(a+b)*sinh(d*x+c)^2)
Time = 0.51 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.95 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {-\frac {8 \sinh (c+d x)}{\left (a+(a+b) \sinh ^2(c+d x)\right )^2}+(4 a+3 b) \left (\frac {3 \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a+b}}+\frac {5 a \sinh (c+d x)+3 (a+b) \sinh ^3(c+d x)}{a^2 \left (a+(a+b) \sinh ^2(c+d x)\right )^2}\right )}{24 (a+b) d} \] Input:
Integrate[Sech[c + d*x]^3/(a + b*Tanh[c + d*x]^2)^3,x]
Output:
((-8*Sinh[c + d*x])/(a + (a + b)*Sinh[c + d*x]^2)^2 + (4*a + 3*b)*((3*ArcT an[(Sqrt[a + b]*Sinh[c + d*x])/Sqrt[a]])/(a^(5/2)*Sqrt[a + b]) + (5*a*Sinh [c + d*x] + 3*(a + b)*Sinh[c + d*x]^3)/(a^2*(a + (a + b)*Sinh[c + d*x]^2)^ 2)))/(24*(a + b)*d)
Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4159, 298, 215, 218}
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.
\(\displaystyle \int \frac {\text {sech}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (i c+i d x)^3}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 4159 |
\(\displaystyle \frac {\int \frac {\sinh ^2(c+d x)+1}{\left ((a+b) \sinh ^2(c+d x)+a\right )^3}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{a+b}+\frac {3}{a}\right ) \int \frac {1}{\left ((a+b) \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)+\frac {b \sinh (c+d x)}{4 a (a+b) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{a+b}+\frac {3}{a}\right ) \left (\frac {\int \frac {1}{(a+b) \sinh ^2(c+d x)+a}d\sinh (c+d x)}{2 a}+\frac {\sinh (c+d x)}{2 a \left ((a+b) \sinh ^2(c+d x)+a\right )}\right )+\frac {b \sinh (c+d x)}{4 a (a+b) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{a+b}+\frac {3}{a}\right ) \left (\frac {\arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {a+b}}+\frac {\sinh (c+d x)}{2 a \left ((a+b) \sinh ^2(c+d x)+a\right )}\right )+\frac {b \sinh (c+d x)}{4 a (a+b) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}}{d}\) |
Input:
Int[Sech[c + d*x]^3/(a + b*Tanh[c + d*x]^2)^3,x]
Output:
((b*Sinh[c + d*x])/(4*a*(a + b)*(a + (a + b)*Sinh[c + d*x]^2)^2) + ((3/a + (a + b)^(-1))*(ArcTan[(Sqrt[a + b]*Sinh[c + d*x])/Sqrt[a]]/(2*a^(3/2)*Sqr t[a + b]) + Sinh[c + d*x]/(2*a*(a + (a + b)*Sinh[c + d*x]^2))))/4)/d
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 *x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(341\) vs. \(2(115)=230\).
Time = 173.55 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.65
method | result | size |
derivativedivides | \(\frac {\frac {-\frac {\left (5 b +4 a \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a \left (a +b \right )}-\frac {\left (4 a^{2}+13 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 a^{2} \left (a +b \right )}+\frac {\left (4 a^{2}+13 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a^{2} \left (a +b \right )}+\frac {\left (5 b +4 a \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a \left (a +b \right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (4 a +3 b \right ) \left (\frac {\left (\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4 a \left (a +b \right )}}{d}\) | \(342\) |
default | \(\frac {\frac {-\frac {\left (5 b +4 a \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a \left (a +b \right )}-\frac {\left (4 a^{2}+13 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 a^{2} \left (a +b \right )}+\frac {\left (4 a^{2}+13 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a^{2} \left (a +b \right )}+\frac {\left (5 b +4 a \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a \left (a +b \right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (4 a +3 b \right ) \left (\frac {\left (\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4 a \left (a +b \right )}}{d}\) | \(342\) |
risch | \(\frac {{\mathrm e}^{d x +c} \left (4 \,{\mathrm e}^{6 d x +6 c} a^{2}+7 \,{\mathrm e}^{6 d x +6 c} a b +3 \,{\mathrm e}^{6 d x +6 c} b^{2}+4 \,{\mathrm e}^{4 d x +4 c} a^{2}-{\mathrm e}^{4 d x +4 c} a b -9 \,{\mathrm e}^{4 d x +4 c} b^{2}-4 \,{\mathrm e}^{2 d x +2 c} a^{2}+{\mathrm e}^{2 d x +2 c} b a +9 b^{2} {\mathrm e}^{2 d x +2 c}-4 a^{2}-7 a b -3 b^{2}\right )}{4 \left ({\mathrm e}^{4 d x +4 c} a +b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 \,{\mathrm e}^{2 d x +2 c} b +a +b \right )^{2} d \,a^{2} \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d a}-\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right ) b}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d a}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right ) b}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{2}}\) | \(443\) |
Input:
int(sech(d*x+c)^3/(a+tanh(d*x+c)^2*b)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(2*(-1/8*(5*b+4*a)/a/(a+b)*tanh(1/2*d*x+1/2*c)^7-1/8*(4*a^2+13*a*b+12* b^2)/a^2/(a+b)*tanh(1/2*d*x+1/2*c)^5+1/8*(4*a^2+13*a*b+12*b^2)/a^2/(a+b)*t anh(1/2*d*x+1/2*c)^3+1/8*(5*b+4*a)/a/(a+b)*tanh(1/2*d*x+1/2*c))/(tanh(1/2* d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*b*tanh(1/2*d*x+1/2*c)^2+a)^2+1/ 4/a*(4*a+3*b)/(a+b)*(1/2*(((a+b)*b)^(1/2)+b)/a/((a+b)*b)^(1/2)/((2*((a+b)* b)^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)+ a+2*b)*a)^(1/2))-1/2*(((a+b)*b)^(1/2)-b)/a/((a+b)*b)^(1/2)/((2*((a+b)*b)^( 1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)-a-2 *b)*a)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 3635 vs. \(2 (115) = 230\).
Time = 0.19 (sec) , antiderivative size = 6622, normalized size of antiderivative = 51.33 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(sech(d*x+c)^3/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{3}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(sech(d*x+c)**3/(a+b*tanh(d*x+c)**2)**3,x)
Output:
Integral(sech(c + d*x)**3/(a + b*tanh(c + d*x)**2)**3, x)
\[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{3}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \] Input:
integrate(sech(d*x+c)^3/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
Output:
1/4*((4*a^2*e^(7*c) + 7*a*b*e^(7*c) + 3*b^2*e^(7*c))*e^(7*d*x) + (4*a^2*e^ (5*c) - a*b*e^(5*c) - 9*b^2*e^(5*c))*e^(5*d*x) - (4*a^2*e^(3*c) - a*b*e^(3 *c) - 9*b^2*e^(3*c))*e^(3*d*x) - (4*a^2*e^c + 7*a*b*e^c + 3*b^2*e^c)*e^(d* x))/(a^5*d + 3*a^4*b*d + 3*a^3*b^2*d + a^2*b^3*d + (a^5*d*e^(8*c) + 3*a^4* b*d*e^(8*c) + 3*a^3*b^2*d*e^(8*c) + a^2*b^3*d*e^(8*c))*e^(8*d*x) + 4*(a^5* d*e^(6*c) + a^4*b*d*e^(6*c) - a^3*b^2*d*e^(6*c) - a^2*b^3*d*e^(6*c))*e^(6* d*x) + 2*(3*a^5*d*e^(4*c) + a^4*b*d*e^(4*c) + a^3*b^2*d*e^(4*c) + 3*a^2*b^ 3*d*e^(4*c))*e^(4*d*x) + 4*(a^5*d*e^(2*c) + a^4*b*d*e^(2*c) - a^3*b^2*d*e^ (2*c) - a^2*b^3*d*e^(2*c))*e^(2*d*x)) + 8*integrate(1/32*((4*a*e^(3*c) + 3 *b*e^(3*c))*e^(3*d*x) + (4*a*e^c + 3*b*e^c)*e^(d*x))/(a^4 + 2*a^3*b + a^2* b^2 + (a^4*e^(4*c) + 2*a^3*b*e^(4*c) + a^2*b^2*e^(4*c))*e^(4*d*x) + 2*(a^4 *e^(2*c) - a^2*b^2*e^(2*c))*e^(2*d*x)), x)
Exception generated. \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sech(d*x+c)^3/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Timed out. \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^3\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \] Input:
int(1/(cosh(c + d*x)^3*(a + b*tanh(c + d*x)^2)^3),x)
Output:
int(1/(cosh(c + d*x)^3*(a + b*tanh(c + d*x)^2)^3), x)
Time = 0.29 (sec) , antiderivative size = 2228, normalized size of antiderivative = 17.27 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:
int(sech(d*x+c)^3/(a+b*tanh(d*x+c)^2)^3,x)
Output:
(4*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - s qrt(b))/sqrt(a))*a**3 + 11*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**( c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b + 10*e**(8*c + 8*d*x)*sqrt (a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*b**2 + 3*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**3 + 16*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a + b)*atan((e** (c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3 + 12*e**(6*c + 6*d*x)*sqrt( a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b - 16*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*b**2 - 12*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a + b)*atan((e **(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**3 + 24*e**(4*c + 4*d*x)*sqr t(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3 + 2*e**(4*c + 4*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - s qrt(b))/sqrt(a))*a**2*b + 12*e**(4*c + 4*d*x)*sqrt(a)*sqrt(a + b)*atan((e* *(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*b**2 + 18*e**(4*c + 4*d*x)*sq rt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**3 + 16*e**(2*c + 2*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3 + 12*e**(2*c + 2*d*x)*sqrt(a)*sqrt(a + b)*atan((e* *(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b - 16*e**(2*c + 2*d*x)*sq rt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*...