Integrand size = 25, antiderivative size = 208 \[ \int \frac {\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (8 a^2-40 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{9/2} f}+\frac {8 a^2-40 a b+35 b^2}{24 a^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{8 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {\text {csch}^4(e+f x)}{4 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac {8 a^2-40 a b+35 b^2}{8 a^4 f \sqrt {a+b \sinh ^2(e+f x)}} \]
-1/8*(8*a^2-40*a*b+35*b^2)*arctanh((a+b*sinh(f*x+e)^2)^(1/2)/a^(1/2))/a^(9 /2)/f+1/24*(8*a^2-40*a*b+35*b^2)/a^3/f/(a+b*sinh(f*x+e)^2)^(3/2)-1/8*(8*a- 7*b)*csch(f*x+e)^2/a^2/f/(a+b*sinh(f*x+e)^2)^(3/2)-1/4*csch(f*x+e)^4/a/f/( a+b*sinh(f*x+e)^2)^(3/2)+1/8*(8*a^2-40*a*b+35*b^2)/a^4/f/(a+b*sinh(f*x+e)^ 2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.56 \[ \int \frac {\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {csch}^2(e+f x) \left (3 a \text {csch}^2(e+f x) \left (8 a-7 b+2 a \text {csch}^2(e+f x)\right )+\left (-8 a^2+40 a b-35 b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1+\frac {b \sinh ^2(e+f x)}{a}\right )\right )}{24 a^3 f \left (b+a \text {csch}^2(e+f x)\right ) \sqrt {a+b \sinh ^2(e+f x)}} \]
-1/24*(Csch[e + f*x]^2*(3*a*Csch[e + f*x]^2*(8*a - 7*b + 2*a*Csch[e + f*x] ^2) + (-8*a^2 + 40*a*b - 35*b^2)*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*S inh[e + f*x]^2)/a]))/(a^3*f*(b + a*Csch[e + f*x]^2)*Sqrt[a + b*Sinh[e + f* x]^2])
Time = 0.37 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.89, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 26, 3673, 100, 27, 87, 61, 61, 73, 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.
\(\displaystyle \int \frac {\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\tan (i e+i f x)^5 \left (a-b \sin (i e+i f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\left (a-b \sin (i e+i f x)^2\right )^{5/2} \tan (i e+i f x)^5}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\text {csch}^6(e+f x) \left (\sinh ^2(e+f x)+1\right )^2}{\left (b \sinh ^2(e+f x)+a\right )^{5/2}}d\sinh ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {\frac {\int \frac {\text {csch}^4(e+f x) \left (4 a \sinh ^2(e+f x)+8 a-7 b\right )}{2 \left (b \sinh ^2(e+f x)+a\right )^{5/2}}d\sinh ^2(e+f x)}{2 a}-\frac {\text {csch}^4(e+f x)}{2 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\text {csch}^4(e+f x) \left (4 a \sinh ^2(e+f x)+8 a-7 b\right )}{\left (b \sinh ^2(e+f x)+a\right )^{5/2}}d\sinh ^2(e+f x)}{4 a}-\frac {\text {csch}^4(e+f x)}{2 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\frac {\frac {\left (8 a^2-40 a b+35 b^2\right ) \int \frac {\text {csch}^2(e+f x)}{\left (b \sinh ^2(e+f x)+a\right )^{5/2}}d\sinh ^2(e+f x)}{2 a}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{4 a}-\frac {\text {csch}^4(e+f x)}{2 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {\frac {\left (8 a^2-40 a b+35 b^2\right ) \left (\frac {\int \frac {\text {csch}^2(e+f x)}{\left (b \sinh ^2(e+f x)+a\right )^{3/2}}d\sinh ^2(e+f x)}{a}+\frac {2}{3 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}\right )}{2 a}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{4 a}-\frac {\text {csch}^4(e+f x)}{2 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {\frac {\left (8 a^2-40 a b+35 b^2\right ) \left (\frac {\frac {\int \frac {\text {csch}^2(e+f x)}{\sqrt {b \sinh ^2(e+f x)+a}}d\sinh ^2(e+f x)}{a}+\frac {2}{a \sqrt {a+b \sinh ^2(e+f x)}}}{a}+\frac {2}{3 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}\right )}{2 a}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{4 a}-\frac {\text {csch}^4(e+f x)}{2 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {\frac {\left (8 a^2-40 a b+35 b^2\right ) \left (\frac {\frac {2 \int \frac {1}{\frac {\sinh ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sinh ^2(e+f x)+a}}{a b}+\frac {2}{a \sqrt {a+b \sinh ^2(e+f x)}}}{a}+\frac {2}{3 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}\right )}{2 a}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{4 a}-\frac {\text {csch}^4(e+f x)}{2 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\frac {\left (8 a^2-40 a b+35 b^2\right ) \left (\frac {\frac {2}{a \sqrt {a+b \sinh ^2(e+f x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}}{a}+\frac {2}{3 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}\right )}{2 a}-\frac {(8 a-7 b) \text {csch}^2(e+f x)}{a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{4 a}-\frac {\text {csch}^4(e+f x)}{2 a \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{2 f}\) |
(-1/2*Csch[e + f*x]^4/(a*(a + b*Sinh[e + f*x]^2)^(3/2)) + (-(((8*a - 7*b)* Csch[e + f*x]^2)/(a*(a + b*Sinh[e + f*x]^2)^(3/2))) + ((8*a^2 - 40*a*b + 3 5*b^2)*(2/(3*a*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((-2*ArcTanh[Sqrt[a + b*Si nh[e + f*x]^2]/Sqrt[a]])/a^(3/2) + 2/(a*Sqrt[a + b*Sinh[e + f*x]^2]))/a))/ (2*a))/(4*a))/(2*f)
3.6.6.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.69 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35
method | result | size |
default | \(\frac {\operatorname {`\,int/indef0`\,}\left (\frac {\cosh \left (f x +e \right )^{4}}{\left (b^{2} \sinh \left (f x +e \right )^{4}+2 \sinh \left (f x +e \right )^{2} a b +a^{2}\right ) \sinh \left (f x +e \right )^{5} \sqrt {a +b \sinh \left (f x +e \right )^{2}}}, \sinh \left (f x +e \right )\right )}{f}\) | \(73\) |
risch | \(\text {Expression too large to display}\) | \(2628058\) |
`int/indef0`(cosh(f*x+e)^4/(b^2*sinh(f*x+e)^4+2*sinh(f*x+e)^2*a*b+a^2)/sin h(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f
Leaf count of result is larger than twice the leaf count of optimal. 7450 vs. \(2 (184) = 368\).
Time = 1.29 (sec) , antiderivative size = 15102, normalized size of antiderivative = 72.61 \[ \int \frac {\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\coth ^{5}{\left (e + f x \right )}}{\left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\coth \left (f x + e\right )^{5}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {\coth ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Hanged} \]