Integrand size = 24, antiderivative size = 125 \[ \int \frac {\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {\sqrt {a}-\sqrt {b}} \sqrt {b} d}+\frac {\text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {\sqrt {a}+\sqrt {b}} \sqrt {b} d} \] Output:
-1/2*arctanh((a^(1/2)-b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))/a^(1/4)/(a^(1/2) -b^(1/2))^(1/2)/b^(1/2)/d+1/2*arctanh((a^(1/2)+b^(1/2))^(1/2)*tanh(d*x+c)/ a^(1/4))/a^(1/4)/(a^(1/2)+b^(1/2))^(1/2)/b^(1/2)/d
Time = 4.41 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02 \[ \int \frac {\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\frac {\frac {\arctan \left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {\text {arctanh}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}}{2 \sqrt {b} d} \] Input:
Integrate[Sinh[c + d*x]^2/(a - b*Sinh[c + d*x]^4),x]
Output:
(ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]]/Sq rt[-a + Sqrt[a]*Sqrt[b]] + ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/Sqr t[a + Sqrt[a]*Sqrt[b]]]/Sqrt[a + Sqrt[a]*Sqrt[b]])/(2*Sqrt[b]*d)
Time = 0.34 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.34, 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, 1450, 221}
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.
\(\displaystyle \int \frac {\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin (i c+i d x)^2}{a-b \sin (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sin (i c+i d x)^2}{a-b \sin (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 3696 |
\(\displaystyle \frac {\int \frac {\tanh ^2(c+d x)}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 1450 |
\(\displaystyle \frac {\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {b}}\right ) \int \frac {1}{(a-b) \tanh ^2(c+d x)-\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tanh (c+d x)+\frac {1}{2} \left (\frac {\sqrt {a}}{\sqrt {b}}+1\right ) \int \frac {1}{(a-b) \tanh ^2(c+d x)-\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {-\frac {\left (\frac {\sqrt {a}}{\sqrt {b}}+1\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {\sqrt {a}-\sqrt {b}} \left (\sqrt {a}+\sqrt {b}\right )}-\frac {\left (1-\frac {\sqrt {a}}{\sqrt {b}}\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right ) \sqrt {\sqrt {a}+\sqrt {b}}}}{d}\) |
Input:
Int[Sinh[c + d*x]^2/(a - b*Sinh[c + d*x]^4),x]
Output:
(-1/2*((1 + Sqrt[a]/Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x ])/a^(1/4)])/(a^(1/4)*Sqrt[Sqrt[a] - Sqrt[b]]*(Sqrt[a] + Sqrt[b])) - ((1 - Sqrt[a]/Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)] )/(2*a^(1/4)*(Sqrt[a] - Sqrt[b])*Sqrt[Sqrt[a] + Sqrt[b]]))/d
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Wi th[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(d^2/2)*(b/q + 1) Int[(d*x)^(m - 2)/(b/ 2 + q/2 + c*x^2), x], x] - Simp[(d^2/2)*(b/q - 1) Int[(d*x)^(m - 2)/(b/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]
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 = 0.75 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}-\textit {\_R}^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}}{d}\) | \(94\) |
default | \(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}-\textit {\_R}^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}}{d}\) | \(94\) |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (256 a^{2} b^{2} d^{4}-256 a \,b^{3} d^{4}\right ) \textit {\_Z}^{4}-32 a \,d^{2} \textit {\_Z}^{2} b \right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 d x +2 c}+\left (-128 a^{2} b \,d^{3}+128 a \,b^{2} d^{3}\right ) \textit {\_R}^{3}+\left (32 d^{2} a^{2}-32 a \,d^{2} b \right ) \textit {\_R}^{2}+16 a d \textit {\_R} -\frac {2 a}{b}-1\right )\) | \(110\) |
Input:
int(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4),x,method=_RETURNVERBOSE)
Output:
-1/d*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. 975 vs. \(2 (85) = 170\).
Time = 0.12 (sec) , antiderivative size = 975, normalized size of antiderivative = 7.80 \[ \int \frac {\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4),x, algorithm="fricas")
Output:
-1/4*sqrt(((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1)/ ((a*b - b^2)*d^2))*log(2*(a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^ 3)*d^4)) + cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c) ^2 + 2*((a^2*b - a*b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - a* d)*sqrt(((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1)/(( a*b - b^2)*d^2)) - 1) + 1/4*sqrt(((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b ^2 + a*b^3)*d^4)) + 1)/((a*b - b^2)*d^2))*log(2*(a^2 - a*b)*d^2*sqrt(1/((a ^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d *x + c) + sinh(d*x + c)^2 - 2*((a^2*b - a*b^2)*d^3*sqrt(1/((a^3*b - 2*a^2* b^2 + a*b^3)*d^4)) - a*d)*sqrt(((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1)/((a*b - b^2)*d^2)) - 1) + 1/4*sqrt(-((a*b - b^2)*d^2* sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1)/((a*b - b^2)*d^2))*log(-2*( a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 2*((a^2*b - a*b^2)*d^ 3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + a*d)*sqrt(-((a*b - b^2)*d^2* sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1)/((a*b - b^2)*d^2)) - 1) - 1 /4*sqrt(-((a*b - b^2)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1)/( (a*b - b^2)*d^2))*log(-2*(a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^ 3)*d^4)) + cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c) ^2 - 2*((a^2*b - a*b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) +...
Timed out. \[ \int \frac {\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)**2/(a-b*sinh(d*x+c)**4),x)
Output:
Timed out
\[ \int \frac {\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{2}}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4),x, algorithm="maxima")
Output:
-integrate(sinh(d*x + c)^2/(b*sinh(d*x + c)^4 - a), x)
\[ \int \frac {\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{2}}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4),x, algorithm="giac")
Output:
sage0*x
Time = 13.66 (sec) , antiderivative size = 1859, normalized size of antiderivative = 14.87 \[ \int \frac {\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
int(sinh(c + d*x)^2/(a - b*sinh(c + d*x)^4),x)
Output:
log((((((262144*a^2*d^2*(102*a*b - 128*a^2 - 22*b^2 - 272*a^2*exp(2*c + 2* d*x) + 19*b^2*exp(2*c + 2*d*x) + 189*a*b*exp(2*c + 2*d*x)))/(b^6*(a - b)) - (131072*a^2*d^3*((a*b + (a*b^3)^(1/2))/(a*b^2*d^2*(a - b)))^(1/2)*(119*a *b^2 - 136*a^2*b + b^3 - 1024*a^3*exp(2*c + 2*d*x) + 9*b^3*exp(2*c + 2*d*x ) - 809*a*b^2*exp(2*c + 2*d*x) + 1808*a^2*b*exp(2*c + 2*d*x)))/(b^6*(a - b )))*((a*b + (a*b^3)^(1/2))/(a*b^2*d^2*(a - b)))^(1/2))/4 + (32768*a*d*(120 *a^2*b - 129*a*b^2 + b^3 - 1024*a^3*exp(2*c + 2*d*x) - b^3*exp(2*c + 2*d*x ) + 201*a*b^2*exp(2*c + 2*d*x) + 816*a^2*b*exp(2*c + 2*d*x)))/(b^7*(a - b) ))*((a*b + (a*b^3)^(1/2))/(a*b^2*d^2*(a - b)))^(1/2))/4 - (16384*a*(106*a* b - 128*a^2 - 2*b^2 + 240*a^2*exp(2*c + 2*d*x) + 3*b^2*exp(2*c + 2*d*x) - 275*a*b*exp(2*c + 2*d*x)))/(b^7*(a - b)))*(-(a*b + (a*b^3)^(1/2))/(16*(a*b ^3*d^2 - a^2*b^2*d^2)))^(1/2) - log((((((262144*a^2*d^2*(102*a*b - 128*a^2 - 22*b^2 - 272*a^2*exp(2*c + 2*d*x) + 19*b^2*exp(2*c + 2*d*x) + 189*a*b*e xp(2*c + 2*d*x)))/(b^6*(a - b)) + (131072*a^2*d^3*((a*b + (a*b^3)^(1/2))/( a*b^2*d^2*(a - b)))^(1/2)*(119*a*b^2 - 136*a^2*b + b^3 - 1024*a^3*exp(2*c + 2*d*x) + 9*b^3*exp(2*c + 2*d*x) - 809*a*b^2*exp(2*c + 2*d*x) + 1808*a^2* b*exp(2*c + 2*d*x)))/(b^6*(a - b)))*((a*b + (a*b^3)^(1/2))/(a*b^2*d^2*(a - b)))^(1/2))/4 - (32768*a*d*(120*a^2*b - 129*a*b^2 + b^3 - 1024*a^3*exp(2* c + 2*d*x) - b^3*exp(2*c + 2*d*x) + 201*a*b^2*exp(2*c + 2*d*x) + 816*a^2*b *exp(2*c + 2*d*x)))/(b^7*(a - b)))*((a*b + (a*b^3)^(1/2))/(a*b^2*d^2*(a...
\[ \int \frac {\sinh ^2(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\frac {64 e^{4 c} \left (\int \frac {e^{4 d x}}{e^{8 d x +8 c} b -4 e^{6 d x +6 c} b -16 e^{4 d x +4 c} a +6 e^{4 d x +4 c} b -4 e^{2 d x +2 c} b +b}d x \right ) a d -8 e^{4 c} \left (\int \frac {e^{4 d x}}{e^{8 d x +8 c} b -4 e^{6 d x +6 c} b -16 e^{4 d x +4 c} a +6 e^{4 d x +4 c} b -4 e^{2 d x +2 c} b +b}d x \right ) b d +16 e^{2 c} \left (\int \frac {e^{2 d x}}{e^{8 d x +8 c} b -4 e^{6 d x +6 c} b -16 e^{4 d x +4 c} a +6 e^{4 d x +4 c} b -4 e^{2 d x +2 c} b +b}d x \right ) b d -8 \left (\int \frac {1}{e^{8 d x +8 c} b -4 e^{6 d x +6 c} b -16 e^{4 d x +4 c} a +6 e^{4 d x +4 c} b -4 e^{2 d x +2 c} b +b}d x \right ) b d -\mathrm {log}\left (e^{8 d x +8 c} b -4 e^{6 d x +6 c} b -16 e^{4 d x +4 c} a +6 e^{4 d x +4 c} b -4 e^{2 d x +2 c} b +b \right )+8 d x}{2 b d} \] Input:
int(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4),x)
Output:
(64*e**(4*c)*int(e**(4*d*x)/(e**(8*c + 8*d*x)*b - 4*e**(6*c + 6*d*x)*b - 1 6*e**(4*c + 4*d*x)*a + 6*e**(4*c + 4*d*x)*b - 4*e**(2*c + 2*d*x)*b + b),x) *a*d - 8*e**(4*c)*int(e**(4*d*x)/(e**(8*c + 8*d*x)*b - 4*e**(6*c + 6*d*x)* b - 16*e**(4*c + 4*d*x)*a + 6*e**(4*c + 4*d*x)*b - 4*e**(2*c + 2*d*x)*b + b),x)*b*d + 16*e**(2*c)*int(e**(2*d*x)/(e**(8*c + 8*d*x)*b - 4*e**(6*c + 6 *d*x)*b - 16*e**(4*c + 4*d*x)*a + 6*e**(4*c + 4*d*x)*b - 4*e**(2*c + 2*d*x )*b + b),x)*b*d - 8*int(1/(e**(8*c + 8*d*x)*b - 4*e**(6*c + 6*d*x)*b - 16* e**(4*c + 4*d*x)*a + 6*e**(4*c + 4*d*x)*b - 4*e**(2*c + 2*d*x)*b + b),x)*b *d - log(e**(8*c + 8*d*x)*b - 4*e**(6*c + 6*d*x)*b - 16*e**(4*c + 4*d*x)*a + 6*e**(4*c + 4*d*x)*b - 4*e**(2*c + 2*d*x)*b + b) + 8*d*x)/(2*b*d)