Integrand size = 23, antiderivative size = 181 \[ \int \text {csch}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx=-\frac {3}{256} b \left (128 a^2+80 a b+21 b^2\right ) x-\frac {a^3 \coth (c+d x)}{d}+\frac {b \left (384 a^2+528 a b+193 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{256 d}-\frac {b^2 (208 a+149 b) \cosh ^3(c+d x) \sinh (c+d x)}{128 d}+\frac {b^2 (80 a+171 b) \cosh ^5(c+d x) \sinh (c+d x)}{160 d}-\frac {41 b^3 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac {b^3 \cosh ^9(c+d x) \sinh (c+d x)}{10 d} \]
-3/256*b*(128*a^2+80*a*b+21*b^2)*x-a^3*coth(d*x+c)/d+1/256*b*(384*a^2+528* a*b+193*b^2)*cosh(d*x+c)*sinh(d*x+c)/d-1/128*b^2*(208*a+149*b)*cosh(d*x+c) ^3*sinh(d*x+c)/d+1/160*b^2*(80*a+171*b)*cosh(d*x+c)^5*sinh(d*x+c)/d-41/80* b^3*cosh(d*x+c)^7*sinh(d*x+c)/d+1/10*b^3*cosh(d*x+c)^9*sinh(d*x+c)/d
Time = 6.53 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.74 \[ \int \text {csch}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx=\frac {-120 b \left (128 a^2+80 a b+21 b^2\right ) (c+d x)-10240 a^3 \coth (c+d x)+60 b \left (128 a^2+120 a b+35 b^2\right ) \sinh (2 (c+d x))-120 b^2 (12 a+5 b) \sinh (4 (c+d x))+10 b^2 (16 a+15 b) \sinh (6 (c+d x))-25 b^3 \sinh (8 (c+d x))+2 b^3 \sinh (10 (c+d x))}{10240 d} \]
(-120*b*(128*a^2 + 80*a*b + 21*b^2)*(c + d*x) - 10240*a^3*Coth[c + d*x] + 60*b*(128*a^2 + 120*a*b + 35*b^2)*Sinh[2*(c + d*x)] - 120*b^2*(12*a + 5*b) *Sinh[4*(c + d*x)] + 10*b^2*(16*a + 15*b)*Sinh[6*(c + d*x)] - 25*b^3*Sinh[ 8*(c + d*x)] + 2*b^3*Sinh[10*(c + d*x)])/(10240*d)
Time = 0.95 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.26, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 25, 3696, 1582, 25, 2336, 25, 2336, 27, 2336, 25, 1582, 25, 359, 219}
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.
| \(\displaystyle \int \text {csch}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\left (a+b \sin (i c+i d x)^4\right )^3}{\sin (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\left (b \sin (i c+i d x)^4+a\right )^3}{\sin (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\coth ^2(c+d x) \left ((a+b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^3}{\left (1-\tanh ^2(c+d x)\right )^6}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}-\frac {1}{10} \int -\frac {\coth ^2(c+d x) \left (-10 (a+b)^3 \tanh ^{10}(c+d x)+10 (5 a-b) (a+b)^2 \tanh ^8(c+d x)-10 \left (10 a^3+9 b a^2+b^3\right ) \tanh ^6(c+d x)+10 \left (10 a^3+3 b a^2-b^3\right ) \tanh ^4(c+d x)-\left (50 a^3+b^3\right ) \tanh ^2(c+d x)+10 a^3\right )}{\left (1-\tanh ^2(c+d x)\right )^5}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{10} \int \frac {\coth ^2(c+d x) \left (-10 (a+b)^3 \tanh ^{10}(c+d x)+10 (5 a-b) (a+b)^2 \tanh ^8(c+d x)-10 \left (10 a^3+9 b a^2+b^3\right ) \tanh ^6(c+d x)+10 \left (10 a^3+3 b a^2-b^3\right ) \tanh ^4(c+d x)-\left (50 a^3+b^3\right ) \tanh ^2(c+d x)+10 a^3\right )}{\left (1-\tanh ^2(c+d x)\right )^5}d\tanh (c+d x)+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {1}{10} \left (-\frac {1}{8} \int -\frac {\coth ^2(c+d x) \left (80 (a+b)^3 \tanh ^8(c+d x)-160 (2 a-b) (a+b)^2 \tanh ^6(c+d x)+240 \left (2 a^3+b a^2+b^3\right ) \tanh ^4(c+d x)-\left (320 a^3-33 b^3\right ) \tanh ^2(c+d x)+80 a^3\right )}{\left (1-\tanh ^2(c+d x)\right )^4}d\tanh (c+d x)-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \int \frac {\coth ^2(c+d x) \left (80 (a+b)^3 \tanh ^8(c+d x)-160 (2 a-b) (a+b)^2 \tanh ^6(c+d x)+240 \left (2 a^3+b a^2+b^3\right ) \tanh ^4(c+d x)-\left (320 a^3-33 b^3\right ) \tanh ^2(c+d x)+80 a^3\right )}{\left (1-\tanh ^2(c+d x)\right )^4}d\tanh (c+d x)-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {b^2 (80 a+171 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}-\frac {1}{6} \int -\frac {15 \coth ^2(c+d x) \left (-32 (a+b)^3 \tanh ^6(c+d x)+96 (a-b) (a+b)^2 \tanh ^4(c+d x)-\left (96 a^3+16 b^2 a+21 b^3\right ) \tanh ^2(c+d x)+32 a^3\right )}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)\right )-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} \int \frac {\coth ^2(c+d x) \left (-32 (a+b)^3 \tanh ^6(c+d x)+96 (a-b) (a+b)^2 \tanh ^4(c+d x)-\left (96 a^3+16 b^2 a+21 b^3\right ) \tanh ^2(c+d x)+32 a^3\right )}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)+\frac {b^2 (80 a+171 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} \left (-\frac {1}{4} \int -\frac {\coth ^2(c+d x) \left (128 (a+b)^3 \tanh ^4(c+d x)-\left (256 a^3-144 b^2 a-65 b^3\right ) \tanh ^2(c+d x)+128 a^3\right )}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)-\frac {b^2 (208 a+149 b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b^2 (80 a+171 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} \left (\frac {1}{4} \int \frac {\coth ^2(c+d x) \left (128 (a+b)^3 \tanh ^4(c+d x)-\left (256 a^3-144 b^2 a-65 b^3\right ) \tanh ^2(c+d x)+128 a^3\right )}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)-\frac {b^2 (208 a+149 b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b^2 (80 a+171 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} \left (\frac {1}{4} \left (\frac {b \left (384 a^2+528 a b+193 b^2\right ) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}-\frac {1}{2} \int -\frac {\coth ^2(c+d x) \left (256 a^3-\left (256 a^3+384 b a^2+240 b^2 a+63 b^3\right ) \tanh ^2(c+d x)\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)\right )-\frac {b^2 (208 a+149 b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b^2 (80 a+171 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {\coth ^2(c+d x) \left (256 a^3-\left (256 a^3+384 b a^2+240 b^2 a+63 b^3\right ) \tanh ^2(c+d x)\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {b \left (384 a^2+528 a b+193 b^2\right ) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )-\frac {b^2 (208 a+149 b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b^2 (80 a+171 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} \left (\frac {1}{4} \left (\frac {1}{2} \left (-3 b \left (128 a^2+80 a b+21 b^2\right ) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-256 a^3 \coth (c+d x)\right )+\frac {b \left (384 a^2+528 a b+193 b^2\right ) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )-\frac {b^2 (208 a+149 b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b^2 (80 a+171 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} \left (\frac {1}{4} \left (\frac {b \left (384 a^2+528 a b+193 b^2\right ) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}+\frac {1}{2} \left (-256 a^3 \coth (c+d x)-3 b \left (128 a^2+80 a b+21 b^2\right ) \text {arctanh}(\tanh (c+d x))\right )\right )-\frac {b^2 (208 a+149 b) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b^2 (80 a+171 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )-\frac {41 b^3 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b^3 \tanh (c+d x)}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
((b^3*Tanh[c + d*x])/(10*(1 - Tanh[c + d*x]^2)^5) + ((-41*b^3*Tanh[c + d*x ])/(8*(1 - Tanh[c + d*x]^2)^4) + ((b^2*(80*a + 171*b)*Tanh[c + d*x])/(2*(1 - Tanh[c + d*x]^2)^3) + (5*(-1/4*(b^2*(208*a + 149*b)*Tanh[c + d*x])/(1 - Tanh[c + d*x]^2)^2 + ((-3*b*(128*a^2 + 80*a*b + 21*b^2)*ArcTanh[Tanh[c + d*x]] - 256*a^3*Coth[c + d*x])/2 + (b*(384*a^2 + 528*a*b + 193*b^2)*Tanh[c + d*x])/(2*(1 - Tanh[c + d*x]^2)))/4))/2)/8)/10)/d
3.3.19.3.1 Defintions of rubi rules used
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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) ^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
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]
Time = 2.98 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.83
| method | result | size |
| parallelrisch | \(\frac {-2 a^{3} \coth \left (\frac {d x}{2}+\frac {c}{2}\right )+\operatorname {sech}\left (\frac {d x}{2}+\frac {c}{2}\right ) \operatorname {csch}\left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3}-3 \left (\frac {\left (-a^{2}-\frac {15}{16} a b -\frac {35}{128} b^{2}\right ) \sinh \left (2 d x +2 c \right )}{2}+\frac {3 \left (a +\frac {5 b}{12}\right ) b \sinh \left (4 d x +4 c \right )}{32}-\frac {b \left (a +\frac {15 b}{16}\right ) \sinh \left (6 d x +6 c \right )}{96}-\frac {b^{2} \sinh \left (10 d x +10 c \right )}{7680}+\frac {5 b^{2} \sinh \left (8 d x +8 c \right )}{3072}+d x \left (a^{2}+\frac {5}{8} a b +\frac {21}{128} b^{2}\right )\right ) b}{2 d}\) | \(151\) |
| derivativedivides | \(\frac {-a^{3} \coth \left (d x +c \right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+3 a \,b^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{5}}{6}-\frac {5 \sinh \left (d x +c \right )^{3}}{24}+\frac {5 \sinh \left (d x +c \right )}{16}\right ) \cosh \left (d x +c \right )-\frac {5 d x}{16}-\frac {5 c}{16}\right )+b^{3} \left (\left (\frac {\sinh \left (d x +c \right )^{9}}{10}-\frac {9 \sinh \left (d x +c \right )^{7}}{80}+\frac {21 \sinh \left (d x +c \right )^{5}}{160}-\frac {21 \sinh \left (d x +c \right )^{3}}{128}+\frac {63 \sinh \left (d x +c \right )}{256}\right ) \cosh \left (d x +c \right )-\frac {63 d x}{256}-\frac {63 c}{256}\right )}{d}\) | \(163\) |
| default | \(\frac {-a^{3} \coth \left (d x +c \right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+3 a \,b^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{5}}{6}-\frac {5 \sinh \left (d x +c \right )^{3}}{24}+\frac {5 \sinh \left (d x +c \right )}{16}\right ) \cosh \left (d x +c \right )-\frac {5 d x}{16}-\frac {5 c}{16}\right )+b^{3} \left (\left (\frac {\sinh \left (d x +c \right )^{9}}{10}-\frac {9 \sinh \left (d x +c \right )^{7}}{80}+\frac {21 \sinh \left (d x +c \right )^{5}}{160}-\frac {21 \sinh \left (d x +c \right )^{3}}{128}+\frac {63 \sinh \left (d x +c \right )}{256}\right ) \cosh \left (d x +c \right )-\frac {63 d x}{256}-\frac {63 c}{256}\right )}{d}\) | \(163\) |
| risch | \(-\frac {3 a^{2} b x}{2}-\frac {15 a \,b^{2} x}{16}-\frac {63 b^{3} x}{256}+\frac {b^{3} {\mathrm e}^{10 d x +10 c}}{10240 d}-\frac {5 b^{3} {\mathrm e}^{8 d x +8 c}}{4096 d}+\frac {b^{2} {\mathrm e}^{6 d x +6 c} a}{128 d}+\frac {15 b^{3} {\mathrm e}^{6 d x +6 c}}{2048 d}-\frac {9 \,{\mathrm e}^{4 d x +4 c} a \,b^{2}}{128 d}-\frac {15 \,{\mathrm e}^{4 d x +4 c} b^{3}}{512 d}+\frac {3 \,{\mathrm e}^{2 d x +2 c} a^{2} b}{8 d}+\frac {45 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}}{128 d}+\frac {105 \,{\mathrm e}^{2 d x +2 c} b^{3}}{1024 d}-\frac {3 \,{\mathrm e}^{-2 d x -2 c} a^{2} b}{8 d}-\frac {45 \,{\mathrm e}^{-2 d x -2 c} a \,b^{2}}{128 d}-\frac {105 \,{\mathrm e}^{-2 d x -2 c} b^{3}}{1024 d}+\frac {9 \,{\mathrm e}^{-4 d x -4 c} a \,b^{2}}{128 d}+\frac {15 \,{\mathrm e}^{-4 d x -4 c} b^{3}}{512 d}-\frac {b^{2} {\mathrm e}^{-6 d x -6 c} a}{128 d}-\frac {15 b^{3} {\mathrm e}^{-6 d x -6 c}}{2048 d}+\frac {5 b^{3} {\mathrm e}^{-8 d x -8 c}}{4096 d}-\frac {b^{3} {\mathrm e}^{-10 d x -10 c}}{10240 d}-\frac {2 a^{3}}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )}\) | \(357\) |
1/2*(-2*a^3*coth(1/2*d*x+1/2*c)+sech(1/2*d*x+1/2*c)*csch(1/2*d*x+1/2*c)*a^ 3-3*(1/2*(-a^2-15/16*a*b-35/128*b^2)*sinh(2*d*x+2*c)+3/32*(a+5/12*b)*b*sin h(4*d*x+4*c)-1/96*b*(a+15/16*b)*sinh(6*d*x+6*c)-1/7680*b^2*sinh(10*d*x+10* c)+5/3072*b^2*sinh(8*d*x+8*c)+d*x*(a^2+5/8*a*b+21/128*b^2))*b)/d
Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (169) = 338\).
Time = 0.26 (sec) , antiderivative size = 474, normalized size of antiderivative = 2.62 \[ \int \text {csch}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx=\frac {2 \, b^{3} \cosh \left (d x + c\right )^{11} + 22 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{10} - 27 \, b^{3} \cosh \left (d x + c\right )^{9} + 3 \, {\left (110 \, b^{3} \cosh \left (d x + c\right )^{3} - 81 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{8} + 5 \, {\left (32 \, a b^{2} + 35 \, b^{3}\right )} \cosh \left (d x + c\right )^{7} + 7 \, {\left (132 \, b^{3} \cosh \left (d x + c\right )^{5} - 324 \, b^{3} \cosh \left (d x + c\right )^{3} + 5 \, {\left (32 \, a b^{2} + 35 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{6} - 50 \, {\left (32 \, a b^{2} + 15 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + {\left (660 \, b^{3} \cosh \left (d x + c\right )^{7} - 3402 \, b^{3} \cosh \left (d x + c\right )^{5} + 175 \, {\left (32 \, a b^{2} + 35 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - 250 \, {\left (32 \, a b^{2} + 15 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 60 \, {\left (128 \, a^{2} b + 144 \, a b^{2} + 45 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + {\left (110 \, b^{3} \cosh \left (d x + c\right )^{9} - 972 \, b^{3} \cosh \left (d x + c\right )^{7} + 105 \, {\left (32 \, a b^{2} + 35 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} - 500 \, {\left (32 \, a b^{2} + 15 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 180 \, {\left (128 \, a^{2} b + 144 \, a b^{2} + 45 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 20 \, {\left (1024 \, a^{3} + 384 \, a^{2} b + 360 \, a b^{2} + 105 \, b^{3}\right )} \cosh \left (d x + c\right ) + 80 \, {\left (256 \, a^{3} - 3 \, {\left (128 \, a^{2} b + 80 \, a b^{2} + 21 \, b^{3}\right )} d x\right )} \sinh \left (d x + c\right )}{20480 \, d \sinh \left (d x + c\right )} \]
1/20480*(2*b^3*cosh(d*x + c)^11 + 22*b^3*cosh(d*x + c)*sinh(d*x + c)^10 - 27*b^3*cosh(d*x + c)^9 + 3*(110*b^3*cosh(d*x + c)^3 - 81*b^3*cosh(d*x + c) )*sinh(d*x + c)^8 + 5*(32*a*b^2 + 35*b^3)*cosh(d*x + c)^7 + 7*(132*b^3*cos h(d*x + c)^5 - 324*b^3*cosh(d*x + c)^3 + 5*(32*a*b^2 + 35*b^3)*cosh(d*x + c))*sinh(d*x + c)^6 - 50*(32*a*b^2 + 15*b^3)*cosh(d*x + c)^5 + (660*b^3*co sh(d*x + c)^7 - 3402*b^3*cosh(d*x + c)^5 + 175*(32*a*b^2 + 35*b^3)*cosh(d* x + c)^3 - 250*(32*a*b^2 + 15*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + 60*(12 8*a^2*b + 144*a*b^2 + 45*b^3)*cosh(d*x + c)^3 + (110*b^3*cosh(d*x + c)^9 - 972*b^3*cosh(d*x + c)^7 + 105*(32*a*b^2 + 35*b^3)*cosh(d*x + c)^5 - 500*( 32*a*b^2 + 15*b^3)*cosh(d*x + c)^3 + 180*(128*a^2*b + 144*a*b^2 + 45*b^3)* cosh(d*x + c))*sinh(d*x + c)^2 - 20*(1024*a^3 + 384*a^2*b + 360*a*b^2 + 10 5*b^3)*cosh(d*x + c) + 80*(256*a^3 - 3*(128*a^2*b + 80*a*b^2 + 21*b^3)*d*x )*sinh(d*x + c))/(d*sinh(d*x + c))
Timed out. \[ \int \text {csch}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.57 \[ \int \text {csch}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx=-\frac {3}{8} \, a^{2} b {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{20480} \, b^{3} {\left (\frac {{\left (25 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 \, e^{\left (-4 \, d x - 4 \, c\right )} + 600 \, e^{\left (-6 \, d x - 6 \, c\right )} - 2100 \, e^{\left (-8 \, d x - 8 \, c\right )} - 2\right )} e^{\left (10 \, d x + 10 \, c\right )}}{d} + \frac {5040 \, {\left (d x + c\right )}}{d} + \frac {2100 \, e^{\left (-2 \, d x - 2 \, c\right )} - 600 \, e^{\left (-4 \, d x - 4 \, c\right )} + 150 \, e^{\left (-6 \, d x - 6 \, c\right )} - 25 \, e^{\left (-8 \, d x - 8 \, c\right )} + 2 \, e^{\left (-10 \, d x - 10 \, c\right )}}{d}\right )} - \frac {1}{128} \, a b^{2} {\left (\frac {{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac {120 \, {\left (d x + c\right )}}{d} + \frac {45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} + \frac {2 \, a^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]
-3/8*a^2*b*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/20480*b^3*(( 25*e^(-2*d*x - 2*c) - 150*e^(-4*d*x - 4*c) + 600*e^(-6*d*x - 6*c) - 2100*e ^(-8*d*x - 8*c) - 2)*e^(10*d*x + 10*c)/d + 5040*(d*x + c)/d + (2100*e^(-2* d*x - 2*c) - 600*e^(-4*d*x - 4*c) + 150*e^(-6*d*x - 6*c) - 25*e^(-8*d*x - 8*c) + 2*e^(-10*d*x - 10*c))/d) - 1/128*a*b^2*((9*e^(-2*d*x - 2*c) - 45*e^ (-4*d*x - 4*c) - 1)*e^(6*d*x + 6*c)/d + 120*(d*x + c)/d + (45*e^(-2*d*x - 2*c) - 9*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/d) + 2*a^3/(d*(e^(-2*d*x - 2 *c) - 1))
Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (169) = 338\).
Time = 0.45 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.96 \[ \int \text {csch}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx=\frac {2 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} - 25 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 160 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 150 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 1440 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 600 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 7680 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 7200 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2100 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - 240 \, {\left (128 \, a^{2} b + 80 \, a b^{2} + 21 \, b^{3}\right )} {\left (d x + c\right )} - \frac {40960 \, a^{3}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} + {\left (35072 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} + 21920 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 5754 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} - 7680 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 7200 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 2100 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 1440 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 600 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 160 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 150 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 25 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b^{3}\right )} e^{\left (-10 \, d x - 10 \, c\right )}}{20480 \, d} \]
1/20480*(2*b^3*e^(10*d*x + 10*c) - 25*b^3*e^(8*d*x + 8*c) + 160*a*b^2*e^(6 *d*x + 6*c) + 150*b^3*e^(6*d*x + 6*c) - 1440*a*b^2*e^(4*d*x + 4*c) - 600*b ^3*e^(4*d*x + 4*c) + 7680*a^2*b*e^(2*d*x + 2*c) + 7200*a*b^2*e^(2*d*x + 2* c) + 2100*b^3*e^(2*d*x + 2*c) - 240*(128*a^2*b + 80*a*b^2 + 21*b^3)*(d*x + c) - 40960*a^3/(e^(2*d*x + 2*c) - 1) + (35072*a^2*b*e^(10*d*x + 10*c) + 2 1920*a*b^2*e^(10*d*x + 10*c) + 5754*b^3*e^(10*d*x + 10*c) - 7680*a^2*b*e^( 8*d*x + 8*c) - 7200*a*b^2*e^(8*d*x + 8*c) - 2100*b^3*e^(8*d*x + 8*c) + 144 0*a*b^2*e^(6*d*x + 6*c) + 600*b^3*e^(6*d*x + 6*c) - 160*a*b^2*e^(4*d*x + 4 *c) - 150*b^3*e^(4*d*x + 4*c) + 25*b^3*e^(2*d*x + 2*c) - 2*b^3)*e^(-10*d*x - 10*c))/d
Time = 1.85 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.46 \[ \int \text {csch}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx=\frac {5\,b^3\,{\mathrm {e}}^{-8\,c-8\,d\,x}}{4096\,d}-\frac {2\,a^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {5\,b^3\,{\mathrm {e}}^{8\,c+8\,d\,x}}{4096\,d}-\frac {b^3\,{\mathrm {e}}^{-10\,c-10\,d\,x}}{10240\,d}+\frac {b^3\,{\mathrm {e}}^{10\,c+10\,d\,x}}{10240\,d}-\frac {3\,b\,x\,\left (128\,a^2+80\,a\,b+21\,b^2\right )}{256}-\frac {3\,b\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (128\,a^2+120\,a\,b+35\,b^2\right )}{1024\,d}+\frac {3\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (128\,a^2+120\,a\,b+35\,b^2\right )}{1024\,d}+\frac {3\,b^2\,{\mathrm {e}}^{-4\,c-4\,d\,x}\,\left (12\,a+5\,b\right )}{512\,d}-\frac {3\,b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (12\,a+5\,b\right )}{512\,d}-\frac {b^2\,{\mathrm {e}}^{-6\,c-6\,d\,x}\,\left (16\,a+15\,b\right )}{2048\,d}+\frac {b^2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (16\,a+15\,b\right )}{2048\,d} \]
(5*b^3*exp(- 8*c - 8*d*x))/(4096*d) - (2*a^3)/(d*(exp(2*c + 2*d*x) - 1)) - (5*b^3*exp(8*c + 8*d*x))/(4096*d) - (b^3*exp(- 10*c - 10*d*x))/(10240*d) + (b^3*exp(10*c + 10*d*x))/(10240*d) - (3*b*x*(80*a*b + 128*a^2 + 21*b^2)) /256 - (3*b*exp(- 2*c - 2*d*x)*(120*a*b + 128*a^2 + 35*b^2))/(1024*d) + (3 *b*exp(2*c + 2*d*x)*(120*a*b + 128*a^2 + 35*b^2))/(1024*d) + (3*b^2*exp(- 4*c - 4*d*x)*(12*a + 5*b))/(512*d) - (3*b^2*exp(4*c + 4*d*x)*(12*a + 5*b)) /(512*d) - (b^2*exp(- 6*c - 6*d*x)*(16*a + 15*b))/(2048*d) + (b^2*exp(6*c + 6*d*x)*(16*a + 15*b))/(2048*d)