Integrand size = 23, antiderivative size = 118 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\left (3 a^2-6 a b-b^2\right ) x}{8 (a+b)^3}+\frac {a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{(a+b)^3 d}-\frac {(5 a+b) \cosh (c+d x) \sinh (c+d x)}{8 (a+b)^2 d}+\frac {\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d} \]
1/8*(3*a^2-6*a*b-b^2)*x/(a+b)^3-1/8*(5*a+b)*cosh(d*x+c)*sinh(d*x+c)/(a+b)^ 2/d+1/4*cosh(d*x+c)^3*sinh(d*x+c)/(a+b)/d+a^(3/2)*arctan(b^(1/2)*tanh(d*x+ c)/a^(1/2))*b^(1/2)/(a+b)^3/d
Time = 0.89 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {4 \left (3 a^2-6 a b-b^2\right ) (c+d x)+32 a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )-8 a (a+b) \sinh (2 (c+d x))+(a+b)^2 \sinh (4 (c+d x))}{32 (a+b)^3 d} \]
(4*(3*a^2 - 6*a*b - b^2)*(c + d*x) + 32*a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Ta nh[c + d*x])/Sqrt[a]] - 8*a*(a + b)*Sinh[2*(c + d*x)] + (a + b)^2*Sinh[4*( c + d*x)])/(32*(a + b)^3*d)
Time = 0.36 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.28, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4146, 372, 402, 25, 397, 218, 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.
\(\displaystyle \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (i c+i d x)^4}{a-b \tan (i c+i d x)^2}dx\) |
\(\Big \downarrow \) 4146 |
\(\displaystyle \frac {\int \frac {\tanh ^4(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^3 \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 372 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\int \frac {(4 a+b) \tanh ^2(c+d x)+a}{\left (1-\tanh ^2(c+d x)\right )^2 \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{4 (a+b)}}{d}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {a (3 a-b)-b (5 a+b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 (a+b)}+\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}}{4 (a+b)}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\int \frac {a (3 a-b)-b (5 a+b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 (a+b)}}{4 (a+b)}}{d}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {\left (3 a^2-6 a b-b^2\right ) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {8 a^2 b \int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a+b}}{2 (a+b)}}{4 (a+b)}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {\left (3 a^2-6 a b-b^2\right ) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {8 a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a+b}}{2 (a+b)}}{4 (a+b)}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\tanh (c+d x)}{4 (a+b) \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(5 a+b) \tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {8 a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a+b}+\frac {\left (3 a^2-6 a b-b^2\right ) \text {arctanh}(\tanh (c+d x))}{a+b}}{2 (a+b)}}{4 (a+b)}}{d}\) |
(Tanh[c + d*x]/(4*(a + b)*(1 - Tanh[c + d*x]^2)^2) - (-1/2*((8*a^(3/2)*Sqr t[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(a + b) + ((3*a^2 - 6*a*b - b^2)*ArcTanh[Tanh[c + d*x]])/(a + b))/(a + b) + ((5*a + b)*Tanh[c + d*x])/ (2*(a + b)*(1 - Tanh[c + d*x]^2)))/(4*(a + b)))/d
3.1.25.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Leaf count of result is larger than twice the leaf count of optimal. \(214\) vs. \(2(104)=208\).
Time = 11.14 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.82
method | result | size |
risch | \(\frac {3 a^{2} x}{8 \left (a +b \right )^{3}}-\frac {3 a x b}{4 \left (a +b \right )^{3}}-\frac {x \,b^{2}}{8 \left (a +b \right )^{3}}+\frac {{\mathrm e}^{4 d x +4 c}}{64 d \left (a +b \right )}-\frac {a \,{\mathrm e}^{2 d x +2 c}}{8 \left (a +b \right )^{2} d}+\frac {a \,{\mathrm e}^{-2 d x -2 c}}{8 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {{\mathrm e}^{-4 d x -4 c}}{64 d \left (a +b \right )}+\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right )}{2 \left (a +b \right )^{3} d}-\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right )}{2 \left (a +b \right )^{3} d}\) | \(215\) |
derivativedivides | \(\frac {\frac {8}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a -3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a -b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-3 a^{2}+6 a b +b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}-\frac {8}{\left (32 a +32 b \right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {-a +3 b}{8 \left (a +b \right )^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {3 a -b}{8 \left (a +b \right )^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (3 a^{2}-6 a b -b^{2}\right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 \left (a +b \right )^{3}}-\frac {2 a^{3} b \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{\left (a +b \right )^{3}}}{d}\) | \(434\) |
default | \(\frac {\frac {8}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a -3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a -b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-3 a^{2}+6 a b +b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}-\frac {8}{\left (32 a +32 b \right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {-a +3 b}{8 \left (a +b \right )^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {3 a -b}{8 \left (a +b \right )^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (3 a^{2}-6 a b -b^{2}\right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 \left (a +b \right )^{3}}-\frac {2 a^{3} b \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{\left (a +b \right )^{3}}}{d}\) | \(434\) |
3/8*a^2*x/(a+b)^3-3/4*a*x/(a+b)^3*b-1/8*x/(a+b)^3*b^2+1/64/d/(a+b)*exp(4*d *x+4*c)-1/8*a/(a+b)^2/d*exp(2*d*x+2*c)+1/8*a/(a^2+2*a*b+b^2)/d*exp(-2*d*x- 2*c)-1/64/d/(a+b)*exp(-4*d*x-4*c)+1/2*(-a*b)^(1/2)*a/(a+b)^3/d*ln(exp(2*d* x+2*c)+(2*(-a*b)^(1/2)+a-b)/(a+b))-1/2*(-a*b)^(1/2)*a/(a+b)^3/d*ln(exp(2*d *x+2*c)-(2*(-a*b)^(1/2)-a+b)/(a+b))
Leaf count of result is larger than twice the leaf count of optimal. 860 vs. \(2 (104) = 208\).
Time = 0.30 (sec) , antiderivative size = 2024, normalized size of antiderivative = 17.15 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \]
[1/64*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(d* x + c)*sinh(d*x + c)^7 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^8 + 8*(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c)^4 - 8*(a^2 + a*b)*cosh(d*x + c)^6 + 4*(7*(a ^2 + 2*a*b + b^2)*cosh(d*x + c)^2 - 2*a^2 - 2*a*b)*sinh(d*x + c)^6 + 8*(7* (a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 - 6*(a^2 + a*b)*cosh(d*x + c))*sinh(d* x + c)^5 + 2*(35*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(3*a^2 - 6*a*b - b^2)*d*x - 60*(a^2 + a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(a^2 + 2 *a*b + b^2)*cosh(d*x + c)^5 + 4*(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c) - 20*(a^2 + a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a^2 + a*b)*cosh(d*x + c)^2 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 12*(3*a^2 - 6*a*b - b^2 )*d*x*cosh(d*x + c)^2 - 30*(a^2 + a*b)*cosh(d*x + c)^4 + 2*a^2 + 2*a*b)*si nh(d*x + c)^2 + 32*(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)^3*sinh(d*x + c) + 6*a*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4)*sqrt(-a*b)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2) *sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2 )*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((a ^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c) + 4*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(-a*b))/((a + b)*cosh(d*x + c)^4 +...
\[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int \frac {\sinh ^{4}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 514 vs. \(2 (104) = 208\).
Time = 0.34 (sec) , antiderivative size = 514, normalized size of antiderivative = 4.36 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=-\frac {{\left (a b - b^{2}\right )} {\left (d x + c\right )}}{2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} + \frac {{\left (8 \, b e^{\left (-2 \, d x - 2 \, c\right )} + a + b\right )} e^{\left (4 \, d x + 4 \, c\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} - \frac {b \log \left ({\left (a + b\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a - b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {b \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b} d} - \frac {{\left (a^{2} b - 6 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b} d} - \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b} d} - \frac {3 \, b \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{8 \, \sqrt {a b} {\left (a + b\right )} d} - \frac {8 \, b e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {3 \, {\left (d x + c\right )}}{8 \, {\left (a + b\right )} d} - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, {\left (a + b\right )} d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, {\left (a + b\right )} d} \]
-1/2*(a*b - b^2)*(d*x + c)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d) + 1/64*(8*b *e^(-2*d*x - 2*c) + a + b)*e^(4*d*x + 4*c)/((a^2 + 2*a*b + b^2)*d) - 1/4*b *log((a + b)*e^(4*d*x + 4*c) + 2*(a - b)*e^(2*d*x + 2*c) + a + b)/((a^2 + 2*a*b + b^2)*d) + 1/4*b*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^2 + 2*a*b + b^2)*d) + 1/4*(a*b - b^2)*arctan(1/2*((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)*d) - 1/8*(a^2*b - 6*a*b^2 + b^3)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)*d) - 1/4*(a*b - b ^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/((a^2 + 2*a*b + b^2)*sqrt(a*b)*d) - 3/8*b*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b) /sqrt(a*b))/(sqrt(a*b)*(a + b)*d) - 1/64*(8*b*e^(-2*d*x - 2*c) + (a + b)*e ^(-4*d*x - 4*c))/((a^2 + 2*a*b + b^2)*d) + 3/8*(d*x + c)/((a + b)*d) - 1/8 *e^(2*d*x + 2*c)/((a + b)*d) + 1/8*e^(-2*d*x - 2*c)/((a + b)*d)
\[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{4}}{b \tanh \left (d x + c\right )^{2} + a} \,d x } \]
Time = 2.32 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.12 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d\,\left (a+b\right )}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d\,\left (a+b\right )}-\frac {x\,\left (-3\,a^2+6\,a\,b+b^2\right )}{8\,{\left (a+b\right )}^3}+\frac {a\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d\,{\left (a+b\right )}^2}-\frac {a\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d\,{\left (a+b\right )}^2}+\frac {{\left (-a\right )}^{3/2}\,\sqrt {b}\,\ln \left ({\left (-a\right )}^{3/2}\,b^{3/2}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )-2\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\left (-a\right )}^{5/2}\,\sqrt {b}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )\right )}{2\,d\,{\left (a+b\right )}^3}-\frac {{\left (-a\right )}^{3/2}\,\sqrt {b}\,\ln \left (2\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\left (-a\right )}^{3/2}\,b^{3/2}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )+{\left (-a\right )}^{5/2}\,\sqrt {b}\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )\right )}{2\,d\,{\left (a+b\right )}^3} \]
exp(4*c + 4*d*x)/(64*d*(a + b)) - exp(- 4*c - 4*d*x)/(64*d*(a + b)) - (x*( 6*a*b - 3*a^2 + b^2))/(8*(a + b)^3) + (a*exp(- 2*c - 2*d*x))/(8*d*(a + b)^ 2) - (a*exp(2*c + 2*d*x))/(8*d*(a + b)^2) + ((-a)^(3/2)*b^(1/2)*log((-a)^( 3/2)*b^(3/2)*(exp(2*c + 2*d*x) - 1) - 2*a^2*b*exp(2*c + 2*d*x) + (-a)^(5/2 )*b^(1/2)*(exp(2*c + 2*d*x) + 1)))/(2*d*(a + b)^3) - ((-a)^(3/2)*b^(1/2)*l og(2*a^2*b*exp(2*c + 2*d*x) + (-a)^(3/2)*b^(3/2)*(exp(2*c + 2*d*x) - 1) + (-a)^(5/2)*b^(1/2)*(exp(2*c + 2*d*x) + 1)))/(2*d*(a + b)^3)