\(\int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\) [205]

Optimal result

Integrand size = 24, antiderivative size = 139 \[ \int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\frac {\sqrt {a} \arctan \left (\frac {\sqrt [4]{b} \cosh (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^{5/4} d}+\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt [4]{b} \cosh (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^{5/4} d}-\frac {\cosh (c+d x)}{b d} \] Output:

1/2*a^(1/2)*arctan(b^(1/4)*cosh(d*x+c)/(a^(1/2)-b^(1/2))^(1/2))/(a^(1/2)-b 
^(1/2))^(1/2)/b^(5/4)/d+1/2*a^(1/2)*arctanh(b^(1/4)*cosh(d*x+c)/(a^(1/2)+b 
^(1/2))^(1/2))/(a^(1/2)+b^(1/2))^(1/2)/b^(5/4)/d-cosh(d*x+c)/b/d
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 1.44 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.69 \[ \int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=-\frac {2 \cosh (c+d x)+a \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-c \text {$\#$1}-d x \text {$\#$1}-2 \log \left (-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}+c \text {$\#$1}^3+d x \text {$\#$1}^3+2 \log \left (-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}^3}{-b-8 a \text {$\#$1}^2+3 b \text {$\#$1}^2-3 b \text {$\#$1}^4+b \text {$\#$1}^6}\&\right ]}{2 b d} \] Input:

Integrate[Sinh[c + d*x]^5/(a - b*Sinh[c + d*x]^4),x]
 

Output:

-1/2*(2*Cosh[c + d*x] + a*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 4* 
b*#1^6 + b*#1^8 & , (-(c*#1) - d*x*#1 - 2*Log[-Cosh[(c + d*x)/2] - Sinh[(c 
 + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1 + c*#1^3 + d* 
x*#1^3 + 2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]* 
#1 - Sinh[(c + d*x)/2]*#1]*#1^3)/(-b - 8*a*#1^2 + 3*b*#1^2 - 3*b*#1^4 + b* 
#1^6) & ])/(b*d)
 

Rubi [A] (verified)

Time = 0.37 (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, 26, 3694, 1484, 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.

\(\displaystyle \int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \sin (i c+i d x)^5}{a-b \sin (i c+i d x)^4}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\sin (i c+i d x)^5}{a-b \sin (i c+i d x)^4}dx\)

\(\Big \downarrow \) 3694

\(\displaystyle \frac {\int \frac {\left (1-\cosh ^2(c+d x)\right )^2}{-b \cosh ^4(c+d x)+2 b \cosh ^2(c+d x)+a-b}d\cosh (c+d x)}{d}\)

\(\Big \downarrow \) 1484

\(\displaystyle \frac {\int \left (\frac {a}{b \left (-b \cosh ^4(c+d x)+2 b \cosh ^2(c+d x)+a-b\right )}-\frac {1}{b}\right )d\cosh (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\sqrt {a} \arctan \left (\frac {\sqrt [4]{b} \cosh (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{5/4} \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt [4]{b} \cosh (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 b^{5/4} \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {\cosh (c+d x)}{b}}{d}\)

Input:

Int[Sinh[c + d*x]^5/(a - b*Sinh[c + d*x]^4),x]
 

Output:

((Sqrt[a]*ArcTan[(b^(1/4)*Cosh[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqrt 
[Sqrt[a] - Sqrt[b]]*b^(5/4)) + (Sqrt[a]*ArcTanh[(b^(1/4)*Cosh[c + d*x])/Sq 
rt[Sqrt[a] + Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(5/4)) - Cosh[c + d*x 
]/b)/d
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 1484

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb 
ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 
 0] && IntegerQ[q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3694

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4)^p, 
 x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 
 1)/2]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.56 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.02

method	result	size
risch	\(-\frac {{\mathrm e}^{d x +c}}{2 b d}-\frac {{\mathrm e}^{-d x -c}}{2 b d}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (256 a \,b^{5} d^{4}-256 b^{6} d^{4}\right ) \textit {\_Z}^{4}+32 a \,d^{2} \textit {\_Z}^{2} b^{3}-a^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 d x +2 c}+\left (\left (\frac {128 b^{4} d^{3}}{a}-\frac {128 b^{5} d^{3}}{a^{2}}\right ) \textit {\_R}^{3}+\left (8 b d +\frac {8 b^{2} d}{a}\right ) \textit {\_R} \right ) {\mathrm e}^{d x +c}+1\right )\right )\)	\(142\)
derivativedivides	\(\frac {\frac {2 a^{2} \left (\frac {\arctan \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \sqrt {a b}-2 a}{4 \sqrt {-a b +\sqrt {a b}\, a}}\right )}{4 a \sqrt {-a b +\sqrt {a b}\, a}}-\frac {\arctan \left (\frac {-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \sqrt {a b}+2 a}{4 \sqrt {-a b -\sqrt {a b}\, a}}\right )}{4 a \sqrt {-a b -\sqrt {a b}\, a}}\right )}{b}-\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(174\)
default	\(\frac {\frac {2 a^{2} \left (\frac {\arctan \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \sqrt {a b}-2 a}{4 \sqrt {-a b +\sqrt {a b}\, a}}\right )}{4 a \sqrt {-a b +\sqrt {a b}\, a}}-\frac {\arctan \left (\frac {-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \sqrt {a b}+2 a}{4 \sqrt {-a b -\sqrt {a b}\, a}}\right )}{4 a \sqrt {-a b -\sqrt {a b}\, a}}\right )}{b}-\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {1}{b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(174\)

Input:

int(sinh(d*x+c)^5/(a-b*sinh(d*x+c)^4),x,method=_RETURNVERBOSE)
 

Output:

-1/2/b/d*exp(d*x+c)-1/2/b/d*exp(-d*x-c)+sum(_R*ln(exp(2*d*x+2*c)+((128/a*b 
^4*d^3-128/a^2*b^5*d^3)*_R^3+(8*b*d+8/a*b^2*d)*_R)*exp(d*x+c)+1),_R=RootOf 
((256*a*b^5*d^4-256*b^6*d^4)*_Z^4+32*a*d^2*_Z^2*b^3-a^2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1247 vs. \(2 (99) = 198\).

Time = 0.16 (sec) , antiderivative size = 1247, normalized size of antiderivative = 8.97 \[ \int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(sinh(d*x+c)^5/(a-b*sinh(d*x+c)^4),x, algorithm="fricas")
 

Output:

1/4*((b*d*cosh(d*x + c) + b*d*sinh(d*x + c))*sqrt(-((a*b^2 - b^3)*d^2*sqrt 
(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))*log(a^2*co 
sh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 + 
a^2 + 2*(a^2*b*d*cosh(d*x + c) + a^2*b*d*sinh(d*x + c) - ((a*b^4 - b^5)*d^ 
3*cosh(d*x + c) + (a*b^4 - b^5)*d^3*sinh(d*x + c))*sqrt(a^3/((a^2*b^5 - 2* 
a*b^6 + b^7)*d^4)))*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 
+ b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2))) - (b*d*cosh(d*x + c) + b*d*sinh(d* 
x + c))*sqrt(-((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) 
 + a)/((a*b^2 - b^3)*d^2))*log(a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*s 
inh(d*x + c) + a^2*sinh(d*x + c)^2 + a^2 - 2*(a^2*b*d*cosh(d*x + c) + a^2* 
b*d*sinh(d*x + c) - ((a*b^4 - b^5)*d^3*cosh(d*x + c) + (a*b^4 - b^5)*d^3*s 
inh(d*x + c))*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)))*sqrt(-((a*b^2 - b 
^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) + a)/((a*b^2 - b^3)*d^2) 
)) + (b*d*cosh(d*x + c) + b*d*sinh(d*x + c))*sqrt(((a*b^2 - b^3)*d^2*sqrt( 
a^3/((a^2*b^5 - 2*a*b^6 + b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))*log(a^2*cos 
h(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c) + a^2*sinh(d*x + c)^2 + a 
^2 + 2*(a^2*b*d*cosh(d*x + c) + a^2*b*d*sinh(d*x + c) + ((a*b^4 - b^5)*d^3 
*cosh(d*x + c) + (a*b^4 - b^5)*d^3*sinh(d*x + c))*sqrt(a^3/((a^2*b^5 - 2*a 
*b^6 + b^7)*d^4)))*sqrt(((a*b^2 - b^3)*d^2*sqrt(a^3/((a^2*b^5 - 2*a*b^6 + 
b^7)*d^4)) - a)/((a*b^2 - b^3)*d^2))) - (b*d*cosh(d*x + c) + b*d*sinh(d...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Timed out} \] Input:

integrate(sinh(d*x+c)**5/(a-b*sinh(d*x+c)**4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{5}}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:

integrate(sinh(d*x+c)^5/(a-b*sinh(d*x+c)^4),x, algorithm="maxima")
 

Output:

-1/2*(e^(2*d*x + 2*c) + 1)*e^(-d*x - c)/(b*d) - 1/32*integrate(256*(a*e^(5 
*d*x + 5*c) - a*e^(3*d*x + 3*c))/(b^2*e^(8*d*x + 8*c) - 4*b^2*e^(6*d*x + 6 
*c) - 4*b^2*e^(2*d*x + 2*c) + b^2 - 2*(8*a*b*e^(4*c) - 3*b^2*e^(4*c))*e^(4 
*d*x)), x)
 

Giac [F]

\[ \int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{5}}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:

integrate(sinh(d*x+c)^5/(a-b*sinh(d*x+c)^4),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 8.88 (sec) , antiderivative size = 1046, normalized size of antiderivative = 7.53 \[ \int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx =\text {Too large to display} \] Input:

int(sinh(c + d*x)^5/(a - b*sinh(c + d*x)^4),x)
 

Output:

log((((((4194304*a^6*d^2*(exp(2*c + 2*d*x) + 1)*(3*a + b))/(b^9*(a - b)^2) 
 + (16777216*a^6*d^3*exp(c + d*x)*(-((a^3*b^5)^(1/2) + a*b^3)/(b^5*d^2*(a 
- b)))^(1/2))/(b^8*(a - b)))*(-((a^3*b^5)^(1/2) + a*b^3)/(b^5*d^2*(a - b)) 
)^(1/2))/4 - (2097152*a^7*d*exp(c + d*x))/(b^11*(a - b)))*(-((a^3*b^5)^(1/ 
2) + a*b^3)/(b^5*d^2*(a - b)))^(1/2))/4 - (262144*a^7*(exp(2*c + 2*d*x) + 
1)*(a + b))/(b^12*(a - b)^2))*(((a^3*b^5)^(1/2) + a*b^3)/(16*(b^6*d^2 - a* 
b^5*d^2)))^(1/2) - log((((((4194304*a^6*d^2*(exp(2*c + 2*d*x) + 1)*(3*a + 
b))/(b^9*(a - b)^2) - (16777216*a^6*d^3*exp(c + d*x)*(-((a^3*b^5)^(1/2) + 
a*b^3)/(b^5*d^2*(a - b)))^(1/2))/(b^8*(a - b)))*(-((a^3*b^5)^(1/2) + a*b^3 
)/(b^5*d^2*(a - b)))^(1/2))/4 + (2097152*a^7*d*exp(c + d*x))/(b^11*(a - b) 
))*(-((a^3*b^5)^(1/2) + a*b^3)/(b^5*d^2*(a - b)))^(1/2))/4 - (262144*a^7*( 
exp(2*c + 2*d*x) + 1)*(a + b))/(b^12*(a - b)^2))*(((a^3*b^5)^(1/2) + a*b^3 
)/(16*(b^6*d^2 - a*b^5*d^2)))^(1/2) - log((((((4194304*a^6*d^2*(exp(2*c + 
2*d*x) + 1)*(3*a + b))/(b^9*(a - b)^2) - (16777216*a^6*d^3*exp(c + d*x)*(( 
(a^3*b^5)^(1/2) - a*b^3)/(b^5*d^2*(a - b)))^(1/2))/(b^8*(a - b)))*(((a^3*b 
^5)^(1/2) - a*b^3)/(b^5*d^2*(a - b)))^(1/2))/4 + (2097152*a^7*d*exp(c + d* 
x))/(b^11*(a - b)))*(((a^3*b^5)^(1/2) - a*b^3)/(b^5*d^2*(a - b)))^(1/2))/4 
 - (262144*a^7*(exp(2*c + 2*d*x) + 1)*(a + b))/(b^12*(a - b)^2))*(-((a^3*b 
^5)^(1/2) - a*b^3)/(16*(b^6*d^2 - a*b^5*d^2)))^(1/2) + log((((((4194304*a^ 
6*d^2*(exp(2*c + 2*d*x) + 1)*(3*a + b))/(b^9*(a - b)^2) + (16777216*a^6...
 

Reduce [F]

\[ \int \frac {\sinh ^5(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\frac {-\cosh \left (d x +c \right )-\left (\int \frac {\sinh \left (d x +c \right )}{\sinh \left (d x +c \right )^{4} b -a}d x \right ) a d}{b d} \] Input:

int(sinh(d*x+c)^5/(a-b*sinh(d*x+c)^4),x)
 

Output:

( - (cosh(c + d*x) + int(sinh(c + d*x)/(sinh(c + d*x)**4*b - a),x)*a*d))/( 
b*d)