Integrand size = 10, antiderivative size = 435 \[ \int \frac {1}{a+b \sinh ^5(x)} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt [5]{b}-\sqrt [5]{a} \tanh \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}+b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+b^{2/5}}}+\frac {2 (-1)^{9/10} \text {arctanh}\left (\frac {(-1)^{9/10} \left (\sqrt [5]{-1} \sqrt [5]{b}+\sqrt [5]{a} \tanh \left (\frac {x}{2}\right )\right )}{\sqrt {-(-1)^{4/5} a^{2/5}+\sqrt [5]{-1} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {-(-1)^{4/5} a^{2/5}+\sqrt [5]{-1} b^{2/5}}}+\frac {2 \sqrt [5]{-1} \text {arctanh}\left (\frac {\sqrt [5]{b}+\sqrt [5]{-1} \sqrt [5]{a} \tanh \left (\frac {x}{2}\right )}{\sqrt {(-1)^{2/5} a^{2/5}+b^{2/5}}}\right )}{5 a^{4/5} \sqrt {(-1)^{2/5} a^{2/5}+b^{2/5}}}+\frac {2 (-1)^{9/10} \text {arctanh}\left (\frac {(-1)^{3/10} \left (\sqrt [5]{b}+(-1)^{3/5} \sqrt [5]{a} \tanh \left (\frac {x}{2}\right )\right )}{\sqrt {-(-1)^{4/5} a^{2/5}+(-1)^{3/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {-(-1)^{4/5} a^{2/5}+(-1)^{3/5} b^{2/5}}}-\frac {2 (-1)^{9/10} \text {arctanh}\left (\frac {i \sqrt [5]{b}-(-1)^{9/10} \sqrt [5]{a} \tanh \left (\frac {x}{2}\right )}{\sqrt {-(-1)^{4/5} a^{2/5}-b^{2/5}}}\right )}{5 a^{4/5} \sqrt {-(-1)^{4/5} a^{2/5}-b^{2/5}}} \] Output:
-2/5*arctanh((b^(1/5)-a^(1/5)*tanh(1/2*x))/(a^(2/5)+b^(2/5))^(1/2))/a^(4/5 )/(a^(2/5)+b^(2/5))^(1/2)+2/5*(-1)^(9/10)*arctanh((-1)^(9/10)*((-1)^(1/5)* b^(1/5)+a^(1/5)*tanh(1/2*x))/(-(-1)^(4/5)*a^(2/5)+(-1)^(1/5)*b^(2/5))^(1/2 ))/a^(4/5)/(-(-1)^(4/5)*a^(2/5)+(-1)^(1/5)*b^(2/5))^(1/2)+2/5*(-1)^(1/5)*a rctanh((b^(1/5)+(-1)^(1/5)*a^(1/5)*tanh(1/2*x))/((-1)^(2/5)*a^(2/5)+b^(2/5 ))^(1/2))/a^(4/5)/((-1)^(2/5)*a^(2/5)+b^(2/5))^(1/2)+2/5*(-1)^(9/10)*arcta nh((-1)^(3/10)*(b^(1/5)+(-1)^(3/5)*a^(1/5)*tanh(1/2*x))/(-(-1)^(4/5)*a^(2/ 5)+(-1)^(3/5)*b^(2/5))^(1/2))/a^(4/5)/(-(-1)^(4/5)*a^(2/5)+(-1)^(3/5)*b^(2 /5))^(1/2)-2/5*(-1)^(9/10)*arctanh((I*b^(1/5)-(-1)^(9/10)*a^(1/5)*tanh(1/2 *x))/(-(-1)^(4/5)*a^(2/5)-b^(2/5))^(1/2))/a^(4/5)/(-(-1)^(4/5)*a^(2/5)-b^( 2/5))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 5.05 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.32 \[ \int \frac {1}{a+b \sinh ^5(x)} \, dx=\frac {8}{5} \text {RootSum}\left [-b+5 b \text {$\#$1}^2-10 b \text {$\#$1}^4+32 a \text {$\#$1}^5+10 b \text {$\#$1}^6-5 b \text {$\#$1}^8+b \text {$\#$1}^{10}\&,\frac {x \text {$\#$1}^3+2 \log \left (-\cosh \left (\frac {x}{2}\right )-\sinh \left (\frac {x}{2}\right )+\cosh \left (\frac {x}{2}\right ) \text {$\#$1}-\sinh \left (\frac {x}{2}\right ) \text {$\#$1}\right ) \text {$\#$1}^3}{b-4 b \text {$\#$1}^2+16 a \text {$\#$1}^3+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8}\&\right ] \] Input:
Integrate[(a + b*Sinh[x]^5)^(-1),x]
Output:
(8*RootSum[-b + 5*b*#1^2 - 10*b*#1^4 + 32*a*#1^5 + 10*b*#1^6 - 5*b*#1^8 + b*#1^10 & , (x*#1^3 + 2*Log[-Cosh[x/2] - Sinh[x/2] + Cosh[x/2]*#1 - Sinh[x /2]*#1]*#1^3)/(b - 4*b*#1^2 + 16*a*#1^3 + 6*b*#1^4 - 4*b*#1^6 + b*#1^8) & ])/5
Time = 1.21 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3692, 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.
\(\displaystyle \int \frac {1}{a+b \sinh ^5(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{a-i b \sin (i x)^5}dx\) |
\(\Big \downarrow \) 3692 |
\(\displaystyle \int \left (-\frac {(-1)^{9/10}}{5 a^{4/5} \left (-(-1)^{9/10} \sqrt [5]{a}-i \sqrt [5]{b} \sinh (x)\right )}-\frac {(-1)^{9/10}}{5 a^{4/5} \left (-(-1)^{9/10} \sqrt [5]{a}-\sqrt [10]{-1} \sqrt [5]{b} \sinh (x)\right )}-\frac {(-1)^{9/10}}{5 a^{4/5} \left ((-1)^{3/10} \sqrt [5]{b} \sinh (x)-(-1)^{9/10} \sqrt [5]{a}\right )}-\frac {(-1)^{9/10}}{5 a^{4/5} \left ((-1)^{7/10} \sqrt [5]{b} \sinh (x)-(-1)^{9/10} \sqrt [5]{a}\right )}-\frac {(-1)^{9/10}}{5 a^{4/5} \left (-(-1)^{9/10} \sqrt [5]{a}-(-1)^{9/10} \sqrt [5]{b} \sinh (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \text {arctanh}\left (\frac {\sqrt [5]{b}-\sqrt [5]{a} \tanh \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}+b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+b^{2/5}}}+\frac {2 (-1)^{9/10} \text {arctanh}\left (\frac {(-1)^{9/10} \left (\sqrt [5]{a} \tanh \left (\frac {x}{2}\right )+\sqrt [5]{-1} \sqrt [5]{b}\right )}{\sqrt {\sqrt [5]{-1} b^{2/5}-(-1)^{4/5} a^{2/5}}}\right )}{5 a^{4/5} \sqrt {\sqrt [5]{-1} b^{2/5}-(-1)^{4/5} a^{2/5}}}+\frac {2 \sqrt [5]{-1} \text {arctanh}\left (\frac {\sqrt [5]{-1} \sqrt [5]{a} \tanh \left (\frac {x}{2}\right )+\sqrt [5]{b}}{\sqrt {(-1)^{2/5} a^{2/5}+b^{2/5}}}\right )}{5 a^{4/5} \sqrt {(-1)^{2/5} a^{2/5}+b^{2/5}}}+\frac {2 (-1)^{9/10} \text {arctanh}\left (\frac {(-1)^{3/10} \left ((-1)^{3/5} \sqrt [5]{a} \tanh \left (\frac {x}{2}\right )+\sqrt [5]{b}\right )}{\sqrt {(-1)^{3/5} b^{2/5}-(-1)^{4/5} a^{2/5}}}\right )}{5 a^{4/5} \sqrt {(-1)^{3/5} b^{2/5}-(-1)^{4/5} a^{2/5}}}-\frac {2 (-1)^{9/10} \text {arctanh}\left (\frac {-(-1)^{9/10} \sqrt [5]{a} \tanh \left (\frac {x}{2}\right )+i \sqrt [5]{b}}{\sqrt {-(-1)^{4/5} a^{2/5}-b^{2/5}}}\right )}{5 a^{4/5} \sqrt {-(-1)^{4/5} a^{2/5}-b^{2/5}}}\) |
Input:
Int[(a + b*Sinh[x]^5)^(-1),x]
Output:
(-2*ArcTanh[(b^(1/5) - a^(1/5)*Tanh[x/2])/Sqrt[a^(2/5) + b^(2/5)]])/(5*a^( 4/5)*Sqrt[a^(2/5) + b^(2/5)]) + (2*(-1)^(9/10)*ArcTanh[((-1)^(9/10)*((-1)^ (1/5)*b^(1/5) + a^(1/5)*Tanh[x/2]))/Sqrt[-((-1)^(4/5)*a^(2/5)) + (-1)^(1/5 )*b^(2/5)]])/(5*a^(4/5)*Sqrt[-((-1)^(4/5)*a^(2/5)) + (-1)^(1/5)*b^(2/5)]) + (2*(-1)^(1/5)*ArcTanh[(b^(1/5) + (-1)^(1/5)*a^(1/5)*Tanh[x/2])/Sqrt[(-1) ^(2/5)*a^(2/5) + b^(2/5)]])/(5*a^(4/5)*Sqrt[(-1)^(2/5)*a^(2/5) + b^(2/5)]) + (2*(-1)^(9/10)*ArcTanh[((-1)^(3/10)*(b^(1/5) + (-1)^(3/5)*a^(1/5)*Tanh[ x/2]))/Sqrt[-((-1)^(4/5)*a^(2/5)) + (-1)^(3/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[ -((-1)^(4/5)*a^(2/5)) + (-1)^(3/5)*b^(2/5)]) - (2*(-1)^(9/10)*ArcTanh[(I*b ^(1/5) - (-1)^(9/10)*a^(1/5)*Tanh[x/2])/Sqrt[-((-1)^(4/5)*a^(2/5)) - b^(2/ 5)]])/(5*a^(4/5)*Sqrt[-((-1)^(4/5)*a^(2/5)) - b^(2/5)])
Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f , n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.26
method | result | size |
default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{10}-5 a \,\textit {\_Z}^{8}+10 a \,\textit {\_Z}^{6}-32 b \,\textit {\_Z}^{5}-10 a \,\textit {\_Z}^{4}+5 a \,\textit {\_Z}^{2}-a \right )}{\sum }\frac {\left (-\textit {\_R}^{8}+4 \textit {\_R}^{6}-6 \textit {\_R}^{4}+4 \textit {\_R}^{2}-1\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{9} a -4 \textit {\_R}^{7} a +6 \textit {\_R}^{5} a -16 \textit {\_R}^{4} b -4 \textit {\_R}^{3} a +\textit {\_R} a}\right )}{5}\) | \(113\) |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (-1+\left (9765625 a^{10}+9765625 a^{8} b^{2}\right ) \textit {\_Z}^{10}-1953125 a^{8} \textit {\_Z}^{8}+156250 a^{6} \textit {\_Z}^{6}-6250 a^{4} \textit {\_Z}^{4}+125 a^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+\left (-\frac {11718750 a^{10}}{b}-11718750 a^{8} b \right ) \textit {\_R}^{9}+\left (\frac {1171875 a^{9}}{b}+1171875 a^{7} b \right ) \textit {\_R}^{8}+\left (\frac {2109375 a^{8}}{b}-234375 a^{6} b \right ) \textit {\_R}^{7}+\left (-\frac {218750 a^{7}}{b}+15625 a^{5} b \right ) \textit {\_R}^{6}+\left (-\frac {143750 a^{6}}{b}-3125 a^{4} b \right ) \textit {\_R}^{5}+\frac {15625 a^{5} \textit {\_R}^{4}}{b}+\frac {4375 a^{4} \textit {\_R}^{3}}{b}-\frac {500 a^{3} \textit {\_R}^{2}}{b}-\frac {50 a^{2} \textit {\_R}}{b}+\frac {6 a}{b}\right )\) | \(206\) |
Input:
int(1/(a+b*sinh(x)^5),x,method=_RETURNVERBOSE)
Output:
1/5*sum((-_R^8+4*_R^6-6*_R^4+4*_R^2-1)/(_R^9*a-4*_R^7*a+6*_R^5*a-16*_R^4*b -4*_R^3*a+_R*a)*ln(tanh(1/2*x)-_R),_R=RootOf(_Z^10*a-5*_Z^8*a+10*_Z^6*a-32 *_Z^5*b-10*_Z^4*a+5*_Z^2*a-a))
Exception generated. \[ \int \frac {1}{a+b \sinh ^5(x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(a+b*sinh(x)^5),x, algorithm="fricas")
Output:
Exception raised: RuntimeError >> no explicit roots found
\[ \int \frac {1}{a+b \sinh ^5(x)} \, dx=\int \frac {1}{a + b \sinh ^{5}{\left (x \right )}}\, dx \] Input:
integrate(1/(a+b*sinh(x)**5),x)
Output:
Integral(1/(a + b*sinh(x)**5), x)
\[ \int \frac {1}{a+b \sinh ^5(x)} \, dx=\int { \frac {1}{b \sinh \left (x\right )^{5} + a} \,d x } \] Input:
integrate(1/(a+b*sinh(x)^5),x, algorithm="maxima")
Output:
integrate(1/(b*sinh(x)^5 + a), x)
\[ \int \frac {1}{a+b \sinh ^5(x)} \, dx=\int { \frac {1}{b \sinh \left (x\right )^{5} + a} \,d x } \] Input:
integrate(1/(a+b*sinh(x)^5),x, algorithm="giac")
Output:
integrate(1/(b*sinh(x)^5 + a), x)
Timed out. \[ \int \frac {1}{a+b \sinh ^5(x)} \, dx=\text {Hanged} \] Input:
int(1/(a + b*sinh(x)^5),x)
Output:
\text{Hanged}
\[ \int \frac {1}{a+b \sinh ^5(x)} \, dx=\int \frac {1}{\sinh \left (x \right )^{5} b +a}d x \] Input:
int(1/(a+b*sinh(x)^5),x)
Output:
int(1/(sinh(x)**5*b + a),x)