Integrand size = 24, antiderivative size = 139 \[ \int \frac {\text {csch}^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{5/4} \sqrt {\sqrt {a}-\sqrt {b}} d}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{5/4} \sqrt {\sqrt {a}+\sqrt {b}} d}-\frac {\coth (c+d x)}{a d} \]
-coth(d*x+c)/a/d-1/2*arctanh((a^(1/2)-b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))* b^(1/2)/a^(5/4)/d/(a^(1/2)-b^(1/2))^(1/2)+1/2*arctanh((a^(1/2)+b^(1/2))^(1 /2)*tanh(d*x+c)/a^(1/4))*b^(1/2)/a^(5/4)/d/(a^(1/2)+b^(1/2))^(1/2)
Time = 1.33 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.03 \[ \int \frac {\text {csch}^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\frac {\frac {\sqrt {b} \arctan \left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}-2 \coth (c+d x)}{2 a d} \]
((Sqrt[b]*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a + Sqrt[a]*Sqr t[b]]])/Sqrt[-a + Sqrt[a]*Sqrt[b]] + (Sqrt[b]*ArcTanh[((Sqrt[a] + Sqrt[b]) *Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/Sqrt[a + Sqrt[a]*Sqrt[b]] - 2* Coth[c + d*x])/(2*a*d)
Time = 0.38 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 25, 3696, 1610, 2009}
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.
\(\displaystyle \int \frac {\text {csch}^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\sin (i c+i d x)^2 \left (a-b \sin (i c+i d x)^4\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\sin (i c+i d x)^2 \left (a-b \sin (i c+i d x)^4\right )}dx\) |
\(\Big \downarrow \) 3696 |
\(\displaystyle \frac {\int \frac {\coth ^2(c+d x) \left (1-\tanh ^2(c+d x)\right )^2}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 1610 |
\(\displaystyle \frac {\int \left (\frac {\coth ^2(c+d x)}{a}+\frac {b \tanh ^2(c+d x)}{a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{5/4} \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{5/4} \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {\coth (c+d x)}{a}}{d}\) |
(-1/2*(Sqrt[b]*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/( a^(5/4)*Sqrt[Sqrt[a] - Sqrt[b]]) + (Sqrt[b]*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b ]]*Tanh[c + d*x])/a^(1/4)])/(2*a^(5/4)*Sqrt[Sqrt[a] + Sqrt[b]]) - Coth[c + d*x]/a)/d
3.3.39.3.1 Defintions of rubi rules used
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*((d + e*x^2)^q/(a + b*x^2 + c*x^4)), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4 *a*c, 0] && IntegerQ[q] && IntegerQ[m]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2) ^(m/2 + 2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] & & IntegerQ[m/2] && IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.58 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}-\textit {\_R}^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}\right )}{a}}{d}\) | \(130\) |
default | \(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}-\textit {\_R}^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}\right )}{a}}{d}\) | \(130\) |
risch | \(-\frac {2}{a d \left ({\mathrm e}^{2 d x +2 c}-1\right )}+4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (65536 a^{6} d^{4}-65536 a^{5} b \,d^{4}\right ) \textit {\_Z}^{4}-512 a^{3} d^{2} \textit {\_Z}^{2} b +b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 d x +2 c}+\left (-\frac {8192 d^{3} a^{5}}{b^{2}}+\frac {8192 d^{3} a^{4}}{b}\right ) \textit {\_R}^{3}+\left (\frac {512 d^{2} a^{4}}{b^{2}}-\frac {512 a^{3} d^{2}}{b}\right ) \textit {\_R}^{2}+\frac {64 d \,a^{2} \textit {\_R}}{b}-\frac {2 a}{b}-1\right )\right )\) | \(151\) |
1/d*(-1/2/a*tanh(1/2*d*x+1/2*c)-1/2/a/tanh(1/2*d*x+1/2*c)-b/a*sum((_R^4-_R ^2)/(_R^7*a-3*_R^5*a+3*_R^3*a-8*_R^3*b-_R*a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R =RootOf(a*_Z^8-4*a*_Z^6+(6*a-16*b)*_Z^4-4*a*_Z^2+a)))
Leaf count of result is larger than twice the leaf count of optimal. 1305 vs. \(2 (99) = 198\).
Time = 0.35 (sec) , antiderivative size = 1305, normalized size of antiderivative = 9.39 \[ \int \frac {\text {csch}^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Too large to display} \]
-1/4*((a*d*cosh(d*x + c)^2 + 2*a*d*cosh(d*x + c)*sinh(d*x + c) + a*d*sinh( d*x + c)^2 - a*d)*sqrt(((a^3 - a^2*b)*d^2*sqrt(b^3/((a^7 - 2*a^6*b + a^5*b ^2)*d^4)) + b)/((a^3 - a^2*b)*d^2))*log(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d *x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 + 2*(a^4 - a^3*b)*d^2*sqrt(b^3 /((a^7 - 2*a^6*b + a^5*b^2)*d^4)) - b^2 + 2*((a^5 - a^4*b)*d^3*sqrt(b^3/(( a^7 - 2*a^6*b + a^5*b^2)*d^4)) - a^2*b*d)*sqrt(((a^3 - a^2*b)*d^2*sqrt(b^3 /((a^7 - 2*a^6*b + a^5*b^2)*d^4)) + b)/((a^3 - a^2*b)*d^2))) - (a*d*cosh(d *x + c)^2 + 2*a*d*cosh(d*x + c)*sinh(d*x + c) + a*d*sinh(d*x + c)^2 - a*d) *sqrt(((a^3 - a^2*b)*d^2*sqrt(b^3/((a^7 - 2*a^6*b + a^5*b^2)*d^4)) + b)/(( a^3 - a^2*b)*d^2))*log(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 + 2*(a^4 - a^3*b)*d^2*sqrt(b^3/((a^7 - 2*a^6*b + a^5*b^2)*d^4)) - b^2 - 2*((a^5 - a^4*b)*d^3*sqrt(b^3/((a^7 - 2*a^6*b + a ^5*b^2)*d^4)) - a^2*b*d)*sqrt(((a^3 - a^2*b)*d^2*sqrt(b^3/((a^7 - 2*a^6*b + a^5*b^2)*d^4)) + b)/((a^3 - a^2*b)*d^2))) - (a*d*cosh(d*x + c)^2 + 2*a*d *cosh(d*x + c)*sinh(d*x + c) + a*d*sinh(d*x + c)^2 - a*d)*sqrt(-((a^3 - a^ 2*b)*d^2*sqrt(b^3/((a^7 - 2*a^6*b + a^5*b^2)*d^4)) - b)/((a^3 - a^2*b)*d^2 ))*log(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh( d*x + c)^2 - 2*(a^4 - a^3*b)*d^2*sqrt(b^3/((a^7 - 2*a^6*b + a^5*b^2)*d^4)) - b^2 + 2*((a^5 - a^4*b)*d^3*sqrt(b^3/((a^7 - 2*a^6*b + a^5*b^2)*d^4)) + a^2*b*d)*sqrt(-((a^3 - a^2*b)*d^2*sqrt(b^3/((a^7 - 2*a^6*b + a^5*b^2)*d...
Timed out. \[ \int \frac {\text {csch}^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\text {csch}^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\operatorname {csch}\left (d x + c\right )^{2}}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \]
-2/(a*d*e^(2*d*x + 2*c) - a*d) - 4*integrate((b*e^(6*d*x + 6*c) - 2*b*e^(4 *d*x + 4*c) + b*e^(2*d*x + 2*c))/(a*b*e^(8*d*x + 8*c) - 4*a*b*e^(6*d*x + 6 *c) - 4*a*b*e^(2*d*x + 2*c) + a*b - 2*(8*a^2*e^(4*c) - 3*a*b*e^(4*c))*e^(4 *d*x)), x)
\[ \int \frac {\text {csch}^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\operatorname {csch}\left (d x + c\right )^{2}}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \]
Time = 12.31 (sec) , antiderivative size = 2128, normalized size of antiderivative = 15.31 \[ \int \frac {\text {csch}^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Too large to display} \]
log((((((4194304*d^2*(512*a^4 - 1184*a^3*b - 253*a*b^3 - b^4 + 930*a^2*b^2 + b^4*exp(2*c + 2*d*x) + 627*a*b^3*exp(2*c + 2*d*x) + 768*a^3*b*exp(2*c + 2*d*x) - 1392*a^2*b^2*exp(2*c + 2*d*x)))/(a^2*b^4*(a - b)^2) - (16777216* d^3*(((a^5*b^3)^(1/2) + a^3*b)/(a^5*d^2*(a - b)))^(1/2)*(40*a*b^2 - 35*b^3 + 512*a^3*exp(2*c + 2*d*x) + 64*b^3*exp(2*c + 2*d*x) + 326*a*b^2*exp(2*c + 2*d*x) - 896*a^2*b*exp(2*c + 2*d*x)))/(b^5*(a - b)))*(((a^5*b^3)^(1/2) + a^3*b)/(a^5*d^2*(a - b)))^(1/2))/4 - (2097152*d*(256*a^2*b - 256*a*b^2 - 5*b^3 - 1024*a^3*exp(2*c + 2*d*x) + 6*b^3*exp(2*c + 2*d*x) + 756*a*b^2*exp (2*c + 2*d*x) + 256*a^2*b*exp(2*c + 2*d*x)))/(a^3*b^4*(a - b)))*(((a^5*b^3 )^(1/2) + a^3*b)/(a^5*d^2*(a - b)))^(1/2))/4 - (524288*(185*a*b^2 - 464*a^ 2*b + 256*a^3 + 24*b^3 - 1024*a^3*exp(2*c + 2*d*x) - 35*b^3*exp(2*c + 2*d* x) - 988*a*b^2*exp(2*c + 2*d*x) + 2048*a^2*b*exp(2*c + 2*d*x)))/(a^4*b^3*( a - b)^2))*(((a^5*b^3)^(1/2) + a^3*b)/(16*(a^6*d^2 - a^5*b*d^2)))^(1/2) - log((((((4194304*d^2*(512*a^4 - 1184*a^3*b - 253*a*b^3 - b^4 + 930*a^2*b^2 + b^4*exp(2*c + 2*d*x) + 627*a*b^3*exp(2*c + 2*d*x) + 768*a^3*b*exp(2*c + 2*d*x) - 1392*a^2*b^2*exp(2*c + 2*d*x)))/(a^2*b^4*(a - b)^2) + (16777216* d^3*(((a^5*b^3)^(1/2) + a^3*b)/(a^5*d^2*(a - b)))^(1/2)*(40*a*b^2 - 35*b^3 + 512*a^3*exp(2*c + 2*d*x) + 64*b^3*exp(2*c + 2*d*x) + 326*a*b^2*exp(2*c + 2*d*x) - 896*a^2*b*exp(2*c + 2*d*x)))/(b^5*(a - b)))*(((a^5*b^3)^(1/2) + a^3*b)/(a^5*d^2*(a - b)))^(1/2))/4 + (2097152*d*(256*a^2*b - 256*a*b^2...