Integrand size = 15, antiderivative size = 115 \[ \int \frac {1}{a-b \sinh ^4(c+d x)} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}-\sqrt {b}} d}+\frac {\text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}+\sqrt {b}} d} \] Output:
1/2*arctanh((a^(1/2)-b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))/a^(3/4)/(a^(1/2)- b^(1/2))^(1/2)/d+1/2*arctanh((a^(1/2)+b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))/ a^(3/4)/(a^(1/2)+b^(1/2))^(1/2)/d
Time = 1.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11 \[ \int \frac {1}{a-b \sinh ^4(c+d x)} \, dx=\frac {-\frac {\arctan \left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {\text {arctanh}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}}{2 \sqrt {a} d} \] Input:
Integrate[(a - b*Sinh[c + d*x]^4)^(-1),x]
Output:
(-(ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]]/ Sqrt[-a + Sqrt[a]*Sqrt[b]]) + ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/ Sqrt[a + Sqrt[a]*Sqrt[b]]]/Sqrt[a + Sqrt[a]*Sqrt[b]])/(2*Sqrt[a]*d)
Time = 0.34 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.46, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3688, 1480, 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.
\(\displaystyle \int \frac {1}{a-b \sinh ^4(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{a-b \sin (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 3688 |
\(\displaystyle \frac {\int \frac {1-\tanh ^2(c+d x)}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {-\frac {1}{2} \left (1-\frac {\sqrt {b}}{\sqrt {a}}\right ) \int \frac {1}{(a-b) \tanh ^2(c+d x)-\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tanh (c+d x)-\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a}}+1\right ) \int \frac {1}{(a-b) \tanh ^2(c+d x)-\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\left (\frac {\sqrt {b}}{\sqrt {a}}+1\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {\sqrt {a}-\sqrt {b}} \left (\sqrt {a}+\sqrt {b}\right )}+\frac {\left (1-\frac {\sqrt {b}}{\sqrt {a}}\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right ) \sqrt {\sqrt {a}+\sqrt {b}}}}{d}\) |
Input:
Int[(a - b*Sinh[c + d*x]^4)^(-1),x]
Output:
(((1 + Sqrt[b]/Sqrt[a])*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^ (1/4)])/(2*a^(1/4)*Sqrt[Sqrt[a] - Sqrt[b]]*(Sqrt[a] + Sqrt[b])) + ((1 - Sq rt[b]/Sqrt[a])*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_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + 2*a*ff^2*x^2 + ( a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x]] / ; FreeQ[{a, b, e, f}, x] && IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.68 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (-\textit {\_R}^{6}+3 \textit {\_R}^{4}-3 \textit {\_R}^{2}+1\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}}{4 d}\) | \(102\) |
default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (-\textit {\_R}^{6}+3 \textit {\_R}^{4}-3 \textit {\_R}^{2}+1\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}}{4 d}\) | \(102\) |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (256 a^{4} d^{4}-256 b \,d^{4} a^{3}\right ) \textit {\_Z}^{4}-32 a^{2} d^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 d x +2 c}+\left (-\frac {128 d^{3} a^{4}}{b}+128 a^{3} d^{3}\right ) \textit {\_R}^{3}+\left (\frac {32 a^{3} d^{2}}{b}-32 d^{2} a^{2}\right ) \textit {\_R}^{2}+\left (\frac {8 a^{2} d}{b}+8 a d \right ) \textit {\_R} -\frac {2 a}{b}-1\right )\) | \(124\) |
Input:
int(1/(a-b*sinh(d*x+c)^4),x,method=_RETURNVERBOSE)
Output:
1/4/d*sum((-_R^6+3*_R^4-3*_R^2+1)/(_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. \(2 (79) = 158\).
Time = 0.17 (sec) , antiderivative size = 975, normalized size of antiderivative = 8.48 \[ \int \frac {1}{a-b \sinh ^4(c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(1/(a-b*sinh(d*x+c)^4),x, algorithm="fricas")
Output:
-1/4*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/(( a^2 - a*b)*d^2))*log(2*(a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2) *d^4)) + b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - a* b*d)*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/(( a^2 - a*b)*d^2)) - b) + 1/4*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2))*log(2*(a^3 - a^2*b)*d^2*sqrt(b/((a ^5 - 2*a^4*b + a^3*b^2)*d^4)) + b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh (d*x + c) + b*sinh(d*x + c)^2 - 2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4* b + a^3*b^2)*d^4)) - a*b*d)*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2)) - b) + 1/4*sqrt(-((a^2 - a*b)*d^2* sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-2*(a^ 3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2*((a^4 - a^3*b)* d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + a*b*d)*sqrt(-((a^2 - a*b)*d^ 2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^2)) - b) - 1 /4*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a ^2 - a*b)*d^2))*log(-2*(a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2) *d^4)) + b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - 2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) +...
Timed out. \[ \int \frac {1}{a-b \sinh ^4(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(1/(a-b*sinh(d*x+c)**4),x)
Output:
Timed out
\[ \int \frac {1}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {1}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(1/(a-b*sinh(d*x+c)^4),x, algorithm="maxima")
Output:
-integrate(1/(b*sinh(d*x + c)^4 - a), x)
\[ \int \frac {1}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {1}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(1/(a-b*sinh(d*x+c)^4),x, algorithm="giac")
Output:
sage0*x
Time = 10.99 (sec) , antiderivative size = 1787, normalized size of antiderivative = 15.54 \[ \int \frac {1}{a-b \sinh ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(a - b*sinh(c + d*x)^4),x)
Output:
log((((((524288*d^2*(31*a*b^2 - 128*a^2*b + 128*a^3 - b^3 + 256*a^3*exp(2* c + 2*d*x) + b^3*exp(2*c + 2*d*x) + 21*a*b^2*exp(2*c + 2*d*x) - 240*a^2*b* exp(2*c + 2*d*x)))/(b^5*(a - b)) + (1048576*a*d^3*((a^2 - (a^3*b)^(1/2))/( a^3*d^2*(a - b)))^(1/2)*(45*a*b^2 - 104*a^2*b + 64*a^3 - 3*b^3 + 4*b^3*exp (2*c + 2*d*x) - 50*a*b^2*exp(2*c + 2*d*x) + 48*a^2*b*exp(2*c + 2*d*x)))/(b ^5*(a - b)))*((a^2 - (a^3*b)^(1/2))/(a^3*d^2*(a - b)))^(1/2))/4 + (262144* d*(72*a*b - 64*a^2 - 9*b^2 + 256*a^2*exp(2*c + 2*d*x) + 31*b^2*exp(2*c + 2 *d*x) - 288*a*b*exp(2*c + 2*d*x)))/(b^5*(a - b)))*((a^2 - (a^3*b)^(1/2))/( a^3*d^2*(a - b)))^(1/2))/4 + (32768*(128*a*b - 128*a^2 - 15*b^2 + 256*a^2* exp(2*c + 2*d*x) + 29*b^2*exp(2*c + 2*d*x) - 304*a*b*exp(2*c + 2*d*x)))/(a *b^5*(a - b)))*((a^2 - (a^3*b)^(1/2))/(16*(a^4*d^2 - a^3*b*d^2)))^(1/2) - log((((((524288*d^2*(31*a*b^2 - 128*a^2*b + 128*a^3 - b^3 + 256*a^3*exp(2* c + 2*d*x) + b^3*exp(2*c + 2*d*x) + 21*a*b^2*exp(2*c + 2*d*x) - 240*a^2*b* exp(2*c + 2*d*x)))/(b^5*(a - b)) - (1048576*a*d^3*((a^2 - (a^3*b)^(1/2))/( a^3*d^2*(a - b)))^(1/2)*(45*a*b^2 - 104*a^2*b + 64*a^3 - 3*b^3 + 4*b^3*exp (2*c + 2*d*x) - 50*a*b^2*exp(2*c + 2*d*x) + 48*a^2*b*exp(2*c + 2*d*x)))/(b ^5*(a - b)))*((a^2 - (a^3*b)^(1/2))/(a^3*d^2*(a - b)))^(1/2))/4 - (262144* d*(72*a*b - 64*a^2 - 9*b^2 + 256*a^2*exp(2*c + 2*d*x) + 31*b^2*exp(2*c + 2 *d*x) - 288*a*b*exp(2*c + 2*d*x)))/(b^5*(a - b)))*((a^2 - (a^3*b)^(1/2))/( a^3*d^2*(a - b)))^(1/2))/4 + (32768*(128*a*b - 128*a^2 - 15*b^2 + 256*a...
\[ \int \frac {1}{a-b \sinh ^4(c+d x)} \, dx=-\left (\int \frac {1}{\sinh \left (d x +c \right )^{4} b -a}d x \right ) \] Input:
int(1/(a-b*sinh(d*x+c)^4),x)
Output:
- int(1/(sinh(c + d*x)**4*b - a),x)