Integrand size = 24, antiderivative size = 348 \[ \int \frac {\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=-\frac {\left (12 a-14 \sqrt {a} \sqrt {b}+5 b\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} \sqrt {b} d}+\frac {\left (12 a+14 \sqrt {a} \sqrt {b}+5 b\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} \sqrt {b} d}+\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}+\frac {\tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )} \] Output:
-1/64*(12*a-14*a^(1/2)*b^(1/2)+5*b)*arctanh((a^(1/2)-b^(1/2))^(1/2)*tanh(d *x+c)/a^(1/4))/a^(9/4)/(a^(1/2)-b^(1/2))^(5/2)/b^(1/2)/d+1/64*(12*a+14*a^( 1/2)*b^(1/2)+5*b)*arctanh((a^(1/2)+b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))/a^( 9/4)/(a^(1/2)+b^(1/2))^(5/2)/b^(1/2)/d+1/8*b*tanh(d*x+c)*(a*(a+3*b)-(a^2+6 *a*b+b^2)*tanh(d*x+c)^2)/a/(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)*(2*a*(5*a^2-9*a*b-4*b^2)/(a-b)^3-5*(2*a^2+3*a*b-b^ 2)*tanh(d*x+c)^2/(a-b)^2)/a^2/d/(a-2*a*tanh(d*x+c)^2+(a-b)*tanh(d*x+c)^4)
Time = 3.88 (sec) , antiderivative size = 343, normalized size of antiderivative = 0.99 \[ \int \frac {\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\frac {\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \left (12 a-14 \sqrt {a} \sqrt {b}+5 b\right ) \arctan \left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}} \sqrt {b}}+\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \left (12 a+14 \sqrt {a} \sqrt {b}+5 b\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}} \sqrt {b}}+\frac {4 \left (12 a^2+11 a b-5 b^2+b (-11 a+5 b) \cosh (2 (c+d x))\right ) \sinh (2 (c+d x))}{8 a-3 b+4 b \cosh (2 (c+d x))-b \cosh (4 (c+d x))}+\frac {128 a (a-b) (2 a+b-b \cosh (2 (c+d x))) \sinh (2 (c+d x))}{(-8 a+3 b-4 b \cosh (2 (c+d x))+b \cosh (4 (c+d x)))^2}}{64 a^2 (a-b)^2 d} \] Input:
Integrate[Sinh[c + d*x]^2/(a - b*Sinh[c + d*x]^4)^3,x]
Output:
(((Sqrt[a] + Sqrt[b])^2*(12*a - 14*Sqrt[a]*Sqrt[b] + 5*b)*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/(Sqrt[-a + Sqrt[a] *Sqrt[b]]*Sqrt[b]) + ((Sqrt[a] - Sqrt[b])^2*(12*a + 14*Sqrt[a]*Sqrt[b] + 5 *b)*ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]] )/(Sqrt[a + Sqrt[a]*Sqrt[b]]*Sqrt[b]) + (4*(12*a^2 + 11*a*b - 5*b^2 + b*(- 11*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)]) + (128*a*(a - b)*(2*a + b - 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)])^2)/(64*a^2*(a - b)^2*d)
Time = 0.94 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 25, 3696, 1672, 27, 2206, 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.
| \(\displaystyle \int \frac {\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin (i c+i d x)^2}{\left (a-b \sin (i c+i d x)^4\right )^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin (i c+i d x)^2}{\left (a-b \sin (i c+i d x)^4\right )^3}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\tanh ^2(c+d x) \left (1-\tanh ^2(c+d x)\right )^4}{\left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 1672 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\int \frac {2 \left (-\frac {8 a^2 b \tanh ^6(c+d x)}{a-b}+\frac {16 a^2 (a-2 b) b \tanh ^4(c+d x)}{(a-b)^2}-\frac {a b \left (8 a^3-29 b a^2+18 b^2 a-5 b^3\right ) \tanh ^2(c+d x)}{(a-b)^3}+\frac {a^2 b^2 (a+3 b)}{(a-b)^3}\right )}{\left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{16 a^2 b}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\int \frac {-\frac {8 a^2 b \tanh ^6(c+d x)}{a-b}+\frac {16 a^2 (a-2 b) b \tanh ^4(c+d x)}{(a-b)^2}-\frac {a b \left (8 a^3-29 b a^2+18 b^2 a-5 b^3\right ) \tanh ^2(c+d x)}{(a-b)^3}+\frac {a^2 b^2 (a+3 b)}{(a-b)^3}}{\left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{8 a^2 b}}{d}\) |
| \(\Big \downarrow \) 2206 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {-\frac {\int -\frac {2 a^2 b^2 \left (2 a (5 a-2 b)-\left (22 a^2-15 b a+5 b^2\right ) \tanh ^2(c+d x)\right )}{(a-b)^2 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{8 a^2 b}-\frac {b \tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}}{8 a^2 b}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {b \int \frac {2 a (5 a-2 b)-\left (22 a^2-15 b a+5 b^2\right ) \tanh ^2(c+d x)}{(a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a}d\tanh (c+d x)}{4 (a-b)^2}-\frac {b \tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}}{8 a^2 b}}{d}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {b \left (\frac {\left (\sqrt {a}-\sqrt {b}\right )^3 \left (14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \int \frac {1}{(a-b) \tanh ^2(c+d x)-\sqrt {a} \left (\sqrt {a}-\sqrt {b}\right )}d\tanh (c+d x)}{2 \sqrt {b}}-\frac {\left (\sqrt {a}+\sqrt {b}\right )^3 \left (-14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \int \frac {1}{(a-b) \tanh ^2(c+d x)-\sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )}d\tanh (c+d x)}{2 \sqrt {b}}\right )}{4 (a-b)^2}-\frac {b \tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}}{8 a^2 b}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {\frac {b \left (\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \left (-14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {b} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \left (14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \text {arctanh}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \sqrt {b} \sqrt {\sqrt {a}+\sqrt {b}}}\right )}{4 (a-b)^2}-\frac {b \tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{4 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}}{8 a^2 b}}{d}\) |
Input:
Int[Sinh[c + d*x]^2/(a - b*Sinh[c + d*x]^4)^3,x]
Output:
((b*Tanh[c + d*x]*(a*(a + 3*b) - (a^2 + 6*a*b + b^2)*Tanh[c + d*x]^2))/(8* a*(a - b)^3*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4)^2) - ((b*( ((Sqrt[a] + Sqrt[b])^2*(12*a - 14*Sqrt[a]*Sqrt[b] + 5*b)*ArcTanh[(Sqrt[Sqr t[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*a^(1/4)*Sqrt[Sqrt[a] - Sqrt[b] ]*Sqrt[b]) - ((Sqrt[a] - Sqrt[b])^2*(12*a + 14*Sqrt[a]*Sqrt[b] + 5*b)*ArcT anh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(2*a^(1/4)*Sqrt[Sqrt [a] + Sqrt[b]]*Sqrt[b])))/(4*(a - b)^2) - (b*Tanh[c + d*x]*((2*a*(5*a^2 - 9*a*b - 4*b^2))/(a - b)^3 - (5*(2*a^2 + 3*a*b - b^2)*Tanh[c + d*x]^2)/(a - b)^2))/(4*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4)))/(8*a^2*b) )/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_) + (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[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_), x_Symbol] :> With[{f = Coeff[PolynomialRemainder[x^m*(d + e*x^2)^q , a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[x^m*(d + e*x^ 2)^q, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a *b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x^2 + c*x^4)^(p + 1)* Simp[ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[x^m*(d + e*x^ 2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], x], x], x]] /; FreeQ[{a, b, c, d, e}, x ] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[q, 1] && IGtQ[m/2, 0]
Int[(Px_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[Poly nomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^ 4)^(p + 1)*((a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)/(2*a*(p + 1)*(b ^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x^2 + c *x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[Px, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4* p + 7)*(b*d - 2*a*e)*x^2, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Px, x ^2] && Expon[Px, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1]
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 = 10.68 (sec) , antiderivative size = 583, normalized size of antiderivative = 1.68
| method | result | size |
| derivativedivides | \(\frac {-\frac {8 \left (-\frac {\left (5 a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{2}+20 a b -18 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{64 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (15 a^{2}+8 a b -18 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{64 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{3}+2 a^{2} b -388 b^{2} a +160 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{3}+2 a^{2} b -388 b^{2} a +160 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (15 a^{2}+8 a b -18 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{64 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{2}+20 a b -18 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{64 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{64 \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 (a \left (-5 a +2 b \right ) \textit {\_R}^{6}+\left (39 a^{2}-28 a b +10 b^{2}\right ) \textit {\_R}^{4}+\left (-39 a^{2}+28 a b -10 b^{2}\right ) \textit {\_R}^{2}+5 a^{2}-2 a 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}}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(583\) |
| default | \(\frac {-\frac {8 \left (-\frac {\left (5 a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{2}+20 a b -18 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{64 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (15 a^{2}+8 a b -18 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{64 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{3}+2 a^{2} b -388 b^{2} a +160 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{3}+2 a^{2} b -388 b^{2} a +160 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (15 a^{2}+8 a b -18 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{64 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{2}+20 a b -18 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{64 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{64 \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 (a \left (-5 a +2 b \right ) \textit {\_R}^{6}+\left (39 a^{2}-28 a b +10 b^{2}\right ) \textit {\_R}^{4}+\left (-39 a^{2}+28 a b -10 b^{2}\right ) \textit {\_R}^{2}+5 a^{2}-2 a 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}}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}}{d}\) | \(583\) |
| risch | \(\text {Expression too large to display}\) | \(2536\) |
Input:
int(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-8*(-1/64*(5*a-2*b)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)+1/64*(25*a^2+ 20*a*b-18*b^2)/a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3-3/64/a*(15*a^2+8*a* b-18*b^2)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5+1/64*(25*a^3+2*a^2*b-388*a *b^2+160*b^3)/a^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7+1/64*(25*a^3+2*a^2 *b-388*a*b^2+160*b^3)/a^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9-3/64/a*(15 *a^2+8*a*b-18*b^2)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^11+1/64*(25*a^2+20* a*b-18*b^2)/a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^13-1/64*(5*a-2*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)^2-1/64/a^2/(a^2-2*a*b+b^2)*sum((a*(-5*a+2*b)*_R^6+(39*a ^2-28*a*b+10*b^2)*_R^4+(-39*a^2+28*a*b-10*b^2)*_R^2+5*a^2-2*a*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. 23355 vs. \(2 (296) = 592\).
Time = 1.20 (sec) , antiderivative size = 23355, normalized size of antiderivative = 67.11 \[ \int \frac {\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)**2/(a-b*sinh(d*x+c)**4)**3,x)
Output:
Timed out
\[ \int \frac {\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{2}}{{\left (b \sinh \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \] Input:
integrate(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4)^3,x, algorithm="maxima")
Output:
-1/16*(11*a*b^2 - 5*b^3 + (12*a^2*b*e^(14*c) - 11*a*b^2*e^(14*c) + 5*b^3*e ^(14*c))*e^(14*d*x) - (104*a^2*b*e^(12*c) - 85*a*b^2*e^(12*c) + 35*b^3*e^( 12*c))*e^(12*d*x) - (320*a^3*e^(10*c) - 652*a^2*b*e^(10*c) + 407*a*b^2*e^( 10*c) - 105*b^3*e^(10*c))*e^(10*d*x) + (1408*a^3*e^(8*c) - 1696*a^2*b*e^(8 *c) + 865*a*b^2*e^(8*c) - 175*b^3*e^(8*c))*e^(8*d*x) + (320*a^3*e^(6*c) + 756*a^2*b*e^(6*c) - 849*a*b^2*e^(6*c) + 175*b^3*e^(6*c))*e^(6*d*x) - (248* a^2*b*e^(4*c) - 383*a*b^2*e^(4*c) + 105*b^3*e^(4*c))*e^(4*d*x) - (12*a^2*b *e^(2*c) + 77*a*b^2*e^(2*c) - 35*b^3*e^(2*c))*e^(2*d*x))/(a^4*b^2*d - 2*a^ 3*b^3*d + a^2*b^4*d + (a^4*b^2*d*e^(16*c) - 2*a^3*b^3*d*e^(16*c) + a^2*b^4 *d*e^(16*c))*e^(16*d*x) - 8*(a^4*b^2*d*e^(14*c) - 2*a^3*b^3*d*e^(14*c) + a ^2*b^4*d*e^(14*c))*e^(14*d*x) - 4*(8*a^5*b*d*e^(12*c) - 23*a^4*b^2*d*e^(12 *c) + 22*a^3*b^3*d*e^(12*c) - 7*a^2*b^4*d*e^(12*c))*e^(12*d*x) + 8*(16*a^5 *b*d*e^(10*c) - 39*a^4*b^2*d*e^(10*c) + 30*a^3*b^3*d*e^(10*c) - 7*a^2*b^4* d*e^(10*c))*e^(10*d*x) + 2*(128*a^6*d*e^(8*c) - 352*a^5*b*d*e^(8*c) + 355* a^4*b^2*d*e^(8*c) - 166*a^3*b^3*d*e^(8*c) + 35*a^2*b^4*d*e^(8*c))*e^(8*d*x ) + 8*(16*a^5*b*d*e^(6*c) - 39*a^4*b^2*d*e^(6*c) + 30*a^3*b^3*d*e^(6*c) - 7*a^2*b^4*d*e^(6*c))*e^(6*d*x) - 4*(8*a^5*b*d*e^(4*c) - 23*a^4*b^2*d*e^(4* c) + 22*a^3*b^3*d*e^(4*c) - 7*a^2*b^4*d*e^(4*c))*e^(4*d*x) - 8*(a^4*b^2*d* e^(2*c) - 2*a^3*b^3*d*e^(2*c) + a^2*b^4*d*e^(2*c))*e^(2*d*x)) - 1/4*integr ate(1/2*((12*a^2*e^(6*c) - 11*a*b*e^(6*c) + 5*b^2*e^(6*c))*e^(6*d*x) - ...
\[ \int \frac {\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\int { -\frac {\sinh \left (d x + c\right )^{2}}{{\left (b \sinh \left (d x + c\right )^{4} - a\right )}^{3}} \,d x } \] Input:
integrate(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4)^3,x, algorithm="giac")
Output:
sage0*x
Timed out. \[ \int \frac {\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {sinh}\left (c+d\,x\right )}^2}{{\left (a-b\,{\mathrm {sinh}\left (c+d\,x\right )}^4\right )}^3} \,d x \] Input:
int(sinh(c + d*x)^2/(a - b*sinh(c + d*x)^4)^3,x)
Output:
int(sinh(c + d*x)^2/(a - b*sinh(c + d*x)^4)^3, x)
\[ \int \frac {\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx=\text {too large to display} \] Input:
int(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4)^3,x)
Output:
(64*(11083077207982080*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 + 2 0*d*x)*a**3*b**4 + 1069056*e**(20*c + 20*d*x)*a**2*b**5 - 8016*e**(20*c + 20*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 + 8139 264*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 + 2642411 520*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)...