Integrand size = 24, antiderivative size = 307 \[ \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=-\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} b^{3/2} d}+\frac {\left (2 \sqrt {a}+5 \sqrt {b}\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} b^{3/2} d}-\frac {a \tanh (c+d x) \left (3 a+b-4 (a+b) \tanh ^2(c+d x)\right )}{8 (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac {\tanh (c+d x) \left (\frac {(a-9 b) (a+b)}{(a-b)^3}-\frac {(a+19 b) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )} \] Output:
-1/64*(2*a^(1/2)-5*b^(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))^(5/2)/b^(3/2)/d+1/64*(2*a^(1/2)+5*b^(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))^(5/2)/b^(3/2)/d-1/8*a*tanh(d*x+c)*(3*a+b-4*(a+b)*tanh(d*x+c)^2)/(a- b)^3/d/(a-2*a*tanh(d*x+c)^2+(a-b)*tanh(d*x+c)^4)^2-1/32*tanh(d*x+c)*((a-9* b)*(a+b)/(a-b)^3-(a+19*b)*tanh(d*x+c)^2/(a-b)^2)/b/d/(a-2*a*tanh(d*x+c)^2+ (a-b)*tanh(d*x+c)^4)
Time = 12.37 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.08 \[ \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\frac {\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^2 \sqrt {b} \arctan \left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {\left (2 a^{3/2} \sqrt {b}+a b-8 \sqrt {a} b^{3/2}+5 b^2\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {8 b (5 a-14 b+(-2 a+5 b) \cosh (2 (c+d x))) \sinh (2 (c+d x))}{8 a-3 b+4 b \cosh (2 (c+d x))-b \cosh (4 (c+d x))}+\frac {64 a (a-b) b (-6 \sinh (2 (c+d x))+\sinh (4 (c+d x)))}{(-8 a+3 b-4 b \cosh (2 (c+d x))+b \cosh (4 (c+d x)))^2}}{64 (a-b)^2 b^2 d} \] Input:
Integrate[Sinh[c + d*x]^8/(a - b*Sinh[c + d*x]^4)^3,x]
Output:
(((2*Sqrt[a] - 5*Sqrt[b])*(Sqrt[a] + Sqrt[b])^2*Sqrt[b]*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[-a + S qrt[a]*Sqrt[b]]) + ((2*a^(3/2)*Sqrt[b] + a*b - 8*Sqrt[a]*b^(3/2) + 5*b^2)* ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(S qrt[a]*Sqrt[a + Sqrt[a]*Sqrt[b]]) + (8*b*(5*a - 14*b + (-2*a + 5*b)*Cosh[2 *(c + d*x)])*Sinh[2*(c + d*x)])/(8*a - 3*b + 4*b*Cosh[2*(c + d*x)] - b*Cos h[4*(c + d*x)]) + (64*a*(a - b)*b*(-6*Sinh[2*(c + d*x)] + Sinh[4*(c + d*x) ]))/(-8*a + 3*b - 4*b*Cosh[2*(c + d*x)] + b*Cosh[4*(c + d*x)])^2)/(64*(a - b)^2*b^2*d)
Time = 0.77 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.19, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.542, Rules used = {3042, 3696, 1598, 27, 1440, 27, 1602, 27, 1602, 25, 27, 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.
| \(\displaystyle \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (i c+i d x)^8}{\left (a-b \sin (i c+i d x)^4\right )^3}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\tanh ^8(c+d x) \left (1-\tanh ^2(c+d x)\right )}{\left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 1598 |
| \(\displaystyle \frac {\frac {\int -\frac {2 b \tanh ^8(c+d x)}{\left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{16 a b}+\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\int \frac {\tanh ^8(c+d x)}{\left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{8 a}}{d}\) |
| \(\Big \downarrow \) 1440 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {\tanh ^5(c+d x) \left (1-\tanh ^2(c+d x)\right )}{4 b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {\int \frac {2 a \tanh ^4(c+d x) \left (5-3 \tanh ^2(c+d x)\right )}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{8 a b}}{8 a}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {\tanh ^5(c+d x) \left (1-\tanh ^2(c+d x)\right )}{4 b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {\int \frac {\tanh ^4(c+d x) \left (5-3 \tanh ^2(c+d x)\right )}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{4 b}}{8 a}}{d}\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {\tanh ^5(c+d x) \left (1-\tanh ^2(c+d x)\right )}{4 b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {-\frac {\int -\frac {3 \tanh ^2(c+d x) \left (3 a-(a+5 b) \tanh ^2(c+d x)\right )}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{3 (a-b)}-\frac {\tanh ^3(c+d x)}{a-b}}{4 b}}{8 a}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {\tanh ^5(c+d x) \left (1-\tanh ^2(c+d x)\right )}{4 b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {\frac {\int \frac {\tanh ^2(c+d x) \left (3 a-(a+5 b) \tanh ^2(c+d x)\right )}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a-b}-\frac {\tanh ^3(c+d x)}{a-b}}{4 b}}{8 a}}{d}\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {\tanh ^5(c+d x) \left (1-\tanh ^2(c+d x)\right )}{4 b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {\frac {-\frac {\int -\frac {a \left ((a-13 b) \tanh ^2(c+d x)+a+5 b\right )}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a-b}-\frac {(a+5 b) \tanh (c+d x)}{a-b}}{a-b}-\frac {\tanh ^3(c+d x)}{a-b}}{4 b}}{8 a}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {\tanh ^5(c+d x) \left (1-\tanh ^2(c+d x)\right )}{4 b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {\frac {\frac {\int \frac {a \left ((a-13 b) \tanh ^2(c+d x)+a+5 b\right )}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a-b}-\frac {(a+5 b) \tanh (c+d x)}{a-b}}{a-b}-\frac {\tanh ^3(c+d x)}{a-b}}{4 b}}{8 a}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {\tanh ^5(c+d x) \left (1-\tanh ^2(c+d x)\right )}{4 b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {\frac {\frac {a \int \frac {(a-13 b) \tanh ^2(c+d x)+a+5 b}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a-b}-\frac {(a+5 b) \tanh (c+d x)}{a-b}}{a-b}-\frac {\tanh ^3(c+d x)}{a-b}}{4 b}}{8 a}}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {\tanh ^5(c+d x) \left (1-\tanh ^2(c+d x)\right )}{4 b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {\frac {\frac {a \left (\frac {1}{2} \left (-\frac {2 a^2-9 a b-5 b^2}{\sqrt {a} \sqrt {b}}+a-13 b\right ) \int \frac {1}{(a-b) \tanh ^2(c+d x)-\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tanh (c+d x)+\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^3 \int \frac {1}{(a-b) \tanh ^2(c+d x)-\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tanh (c+d x)}{2 \sqrt {a} \sqrt {b}}\right )}{a-b}-\frac {(a+5 b) \tanh (c+d x)}{a-b}}{a-b}-\frac {\tanh ^3(c+d x)}{a-b}}{4 b}}{8 a}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\tanh ^9(c+d x)}{8 a \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {\tanh ^5(c+d x) \left (1-\tanh ^2(c+d x)\right )}{4 b \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {\frac {\frac {a \left (-\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^2 \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {b} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\left (-\frac {2 a^2-9 a b-5 b^2}{\sqrt {a} \sqrt {b}}+a-13 b\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right ) \sqrt {\sqrt {a}+\sqrt {b}}}\right )}{a-b}-\frac {(a+5 b) \tanh (c+d x)}{a-b}}{a-b}-\frac {\tanh ^3(c+d x)}{a-b}}{4 b}}{8 a}}{d}\) |
Input:
Int[Sinh[c + d*x]^8/(a - b*Sinh[c + d*x]^4)^3,x]
Output:
(Tanh[c + d*x]^9/(8*a*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4)^ 2) - ((Tanh[c + d*x]^5*(1 - Tanh[c + d*x]^2))/(4*b*(a - 2*a*Tanh[c + d*x]^ 2 + (a - b)*Tanh[c + d*x]^4)) - (-(Tanh[c + d*x]^3/(a - b)) + ((a*(-1/2*(( 2*Sqrt[a] - 5*Sqrt[b])*(Sqrt[a] + Sqrt[b])^2*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[ b]]*Tanh[c + d*x])/a^(1/4)])/(a^(3/4)*Sqrt[Sqrt[a] - Sqrt[b]]*Sqrt[b]) - ( (a - 13*b - (2*a^2 - 9*a*b - 5*b^2)/(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]])))/(a - b) - ((a + 5*b)*Tanh[c + d*x])/(a - b))/(a - b ))/(4*b))/(8*a))/d
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_)^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)^(p_), x_Symbol] :> Simp[(-d^3)*(d*x)^(m - 3)*(2*a + b*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2* (p + 1)*(b^2 - 4*a*c))), x] + Simp[d^4/(2*(p + 1)*(b^2 - 4*a*c)) Int[(d*x )^(m - 4)*(2*a*(m - 3) + b*(m + 4*p + 3)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && Gt Q[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
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[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_.), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1) *((b*d - 2*a*e - (b*e - 2*c*d)*x^2)/(2*(p + 1)*(b^2 - 4*a*c))), x] - Simp[f ^2/(2*(p + 1)*(b^2 - 4*a*c)) Int[(f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1 )*Simp[(m - 1)*(b*d - 2*a*e) - (4*p + 4 + m + 1)*(b*e - 2*c*d)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3)) Int[(f*x)^(m - 2)* (a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c , 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | | IntegerQ[m])
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 = 11.69 (sec) , antiderivative size = 535, normalized size of antiderivative = 1.74
| method | result | size |
| derivativedivides | \(\frac {-\frac {512 \left (\frac {a \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a +49 b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \left (3 a^{2}+55 a b -48 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}+377 a b -784 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}+377 a b -784 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \left (3 a^{2}+55 a b -48 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a +49 b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {a \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a -4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a +6 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -16 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +a \right )^{2}}-\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 (\left (a +5 b \right ) \textit {\_R}^{6}+\left (5 a -47 b \right ) \textit {\_R}^{4}+\left (-5 a +47 b \right ) \textit {\_R}^{2}-a -5 b \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}}{128 b \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(535\) |
| default | \(\frac {-\frac {512 \left (\frac {a \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a +49 b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \left (3 a^{2}+55 a b -48 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}+377 a b -784 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}+377 a b -784 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \left (3 a^{2}+55 a b -48 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a +49 b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {a \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{8192 b \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a -4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a +6 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -16 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +a \right )^{2}}-\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 (\left (a +5 b \right ) \textit {\_R}^{6}+\left (5 a -47 b \right ) \textit {\_R}^{4}+\left (-5 a +47 b \right ) \textit {\_R}^{2}-a -5 b \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}}{128 b \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(535\) |
| risch | \(\text {Expression too large to display}\) | \(1574\) |
Input:
int(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-512*(1/8192*a*(a+5*b)/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)-1/8192*( 5*a+49*b)*a/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3+3/8192/b*(3*a^2+55*a*b -48*b^2)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5-1/8192*(5*a^2+377*a*b-784*b ^2)/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7-1/8192*(5*a^2+377*a*b-784*b^2) /b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9+3/8192/b*(3*a^2+55*a*b-48*b^2)/(a ^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^11-1/8192*(5*a+49*b)*a/b/(a^2-2*a*b+b^2) *tanh(1/2*d*x+1/2*c)^13+1/8192*a*(a+5*b)/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/ 2*c)^15)/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x +1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2-1/12 8/b/(a^2-2*a*b+b^2)*sum(((a+5*b)*_R^6+(5*a-47*b)*_R^4+(-5*a+47*b)*_R^2-a-5 *b)/(_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. 20486 vs. \(2 (257) = 514\).
Time = 0.73 (sec) , antiderivative size = 20486, normalized size of antiderivative = 66.73 \[ \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)**8/(a-b*sinh(d*x+c)**4)**3,x)
Output:
Timed out
\[ \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{8}}{{\left (b \sinh \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \] Input:
integrate(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x, algorithm="maxima")
Output:
-1/8*(2*a*b^2 - 5*b^3 + (a*b^2*e^(14*c) - 4*b^3*e^(14*c))*e^(14*d*x) - (32 *a^2*b*e^(12*c) - 58*a*b^2*e^(12*c) - b^3*e^(12*c))*e^(12*d*x) + 3*(48*a^2 *b*e^(10*c) - 73*a*b^2*e^(10*c) + 20*b^3*e^(10*c))*e^(10*d*x) + (256*a^3*e ^(8*c) - 832*a^2*b*e^(8*c) + 550*a*b^2*e^(8*c) - 175*b^3*e^(8*c))*e^(8*d*x ) + (112*a^2*b*e^(6*c) - 533*a*b^2*e^(6*c) + 220*b^3*e^(6*c))*e^(6*d*x) - (32*a^2*b*e^(4*c) - 158*a*b^2*e^(4*c) + 141*b^3*e^(4*c))*e^(4*d*x) - (17*a *b^2*e^(2*c) - 44*b^3*e^(2*c))*e^(2*d*x))/(a^2*b^4*d - 2*a*b^5*d + b^6*d + (a^2*b^4*d*e^(16*c) - 2*a*b^5*d*e^(16*c) + b^6*d*e^(16*c))*e^(16*d*x) - 8 *(a^2*b^4*d*e^(14*c) - 2*a*b^5*d*e^(14*c) + b^6*d*e^(14*c))*e^(14*d*x) - 4 *(8*a^3*b^3*d*e^(12*c) - 23*a^2*b^4*d*e^(12*c) + 22*a*b^5*d*e^(12*c) - 7*b ^6*d*e^(12*c))*e^(12*d*x) + 8*(16*a^3*b^3*d*e^(10*c) - 39*a^2*b^4*d*e^(10* c) + 30*a*b^5*d*e^(10*c) - 7*b^6*d*e^(10*c))*e^(10*d*x) + 2*(128*a^4*b^2*d *e^(8*c) - 352*a^3*b^3*d*e^(8*c) + 355*a^2*b^4*d*e^(8*c) - 166*a*b^5*d*e^( 8*c) + 35*b^6*d*e^(8*c))*e^(8*d*x) + 8*(16*a^3*b^3*d*e^(6*c) - 39*a^2*b^4* d*e^(6*c) + 30*a*b^5*d*e^(6*c) - 7*b^6*d*e^(6*c))*e^(6*d*x) - 4*(8*a^3*b^3 *d*e^(4*c) - 23*a^2*b^4*d*e^(4*c) + 22*a*b^5*d*e^(4*c) - 7*b^6*d*e^(4*c))* e^(4*d*x) - 8*(a^2*b^4*d*e^(2*c) - 2*a*b^5*d*e^(2*c) + b^6*d*e^(2*c))*e^(2 *d*x)) - 1/256*integrate(64*((a*e^(6*c) - 4*b*e^(6*c))*e^(6*d*x) + (a*e^(2 *c) - 4*b*e^(2*c))*e^(2*d*x) + 18*b*e^(4*d*x + 4*c))/(a^2*b^2 - 2*a*b^3 + b^4 + (a^2*b^2*e^(8*c) - 2*a*b^3*e^(8*c) + b^4*e^(8*c))*e^(8*d*x) - 4*(...
\[ \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{8}}{{\left (b \sinh \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \] Input:
integrate(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x, algorithm="giac")
Output:
sage0*x
Timed out. \[ \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {sinh}\left (c+d\,x\right )}^8}{{\left (a-b\,{\mathrm {sinh}\left (c+d\,x\right )}^4\right )}^3} \,d x \] Input:
int(sinh(c + d*x)^8/(a - b*sinh(c + d*x)^4)^3,x)
Output:
int(sinh(c + d*x)^8/(a - b*sinh(c + d*x)^4)^3, x)
\[ \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\text {too large to display} \] Input:
int(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x)
Output:
(4*(29554872554618880*e**(20*c + 16*d*x)*int(e**(4*d*x)/(3440640*e**(24*c + 24*d*x)*a**4*b**3 + 2150400*e**(24*c + 24*d*x)*a**3*b**4 + 16128*e**(24* c + 24*d*x)*a**2*b**5 - 96*e**(24*c + 24*d*x)*a*b**6 + 35*e**(24*c + 24*d* x)*b**7 - 41287680*e**(22*c + 22*d*x)*a**4*b**3 - 25804800*e**(22*c + 22*d *x)*a**3*b**4 - 193536*e**(22*c + 22*d*x)*a**2*b**5 + 1152*e**(22*c + 22*d *x)*a*b**6 - 420*e**(22*c + 22*d*x)*b**7 - 165150720*e**(20*c + 20*d*x)*a* *5*b**2 + 123863040*e**(20*c + 20*d*x)*a**4*b**3 + 141152256*e**(20*c + 20 *d*x)*a**3*b**4 + 1069056*e**(20*c + 20*d*x)*a**2*b**5 - 8016*e**(20*c + 2 0*d*x)*a*b**6 + 2310*e**(20*c + 20*d*x)*b**7 + 1321205760*e**(18*c + 18*d* x)*a**5*b**2 + 68812800*e**(18*c + 18*d*x)*a**4*b**3 - 466894848*e**(18*c + 18*d*x)*a**3*b**4 - 3585024*e**(18*c + 18*d*x)*a**2*b**5 + 34560*e**(18* c + 18*d*x)*a*b**6 - 7700*e**(18*c + 18*d*x)*b**7 + 2642411520*e**(16*c + 16*d*x)*a**6*b - 2972712960*e**(16*c + 16*d*x)*a**5*b**2 - 1174634496*e**( 16*c + 16*d*x)*a**4*b**3 + 1042698240*e**(16*c + 16*d*x)*a**3*b**4 + 81392 64*e**(16*c + 16*d*x)*a**2*b**5 - 94560*e**(16*c + 16*d*x)*a*b**6 + 17325* e**(16*c + 16*d*x)*b**7 - 10569646080*e**(14*c + 14*d*x)*a**6*b + 26424115 20*e**(14*c + 14*d*x)*a**5*b**2 + 3005743104*e**(14*c + 14*d*x)*a**4*b**3 - 1659469824*e**(14*c + 14*d*x)*a**3*b**4 - 13138944*e**(14*c + 14*d*x)*a* *2*b**5 + 170112*e**(14*c + 14*d*x)*a*b**6 - 27720*e**(14*c + 14*d*x)*b**7 - 14092861440*e**(12*c + 12*d*x)*a**7 + 7046430720*e**(12*c + 12*d*x)*...