Integrand size = 23, antiderivative size = 154 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {5 \sqrt {b} (3 a+7 b) \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{8 a^{9/2} d}-\frac {(a+3 b) \cosh (c+d x)}{a^4 d}+\frac {\cosh ^3(c+d x)}{3 a^3 d}+\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 d \left (b+a \cosh ^2(c+d x)\right )^2}-\frac {b (9 a+13 b) \cosh (c+d x)}{8 a^4 d \left (b+a \cosh ^2(c+d x)\right )} \]
-(a+3*b)*cosh(d*x+c)/a^4/d+1/3*cosh(d*x+c)^3/a^3/d+1/4*b^2*(a+b)*cosh(d*x+ c)/a^4/d/(b+a*cosh(d*x+c)^2)^2-1/8*b*(9*a+13*b)*cosh(d*x+c)/a^4/d/(b+a*cos h(d*x+c)^2)+5/8*(3*a+7*b)*arctan(cosh(d*x+c)*a^(1/2)/b^(1/2))*b^(1/2)/a^(9 /2)/d
Result contains complex when optimal does not.
Time = 13.81 (sec) , antiderivative size = 1364, normalized size of antiderivative = 8.86 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx =\text {Too large to display} \]
(-3*((3*(ArcTan[(Sqrt[a] - I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]] + Arc Tan[(Sqrt[a] + I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]]))/Sqrt[a] + (2*Sq rt[b]*Cosh[c + d*x]*(3*a + 10*b + 3*a*Cosh[2*(c + d*x)]))/(a + 2*b + a*Cos h[2*(c + d*x)])^2)*(a + 2*b + a*Cosh[2*c + 2*d*x])^3*Sech[c + d*x]^6)/(819 2*b^(5/2)*d*(a + b*Sech[c + d*x]^2)^3) - ((-((3*a - 4*b)*(ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh [c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/S qrt[b]] + ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Si nh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sin h[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]])) - (2*Sqrt[a]*Sqrt[b]*Cosh[c + d*x]*(3* a^2 + 6*a*b + 8*b^2 + a*(3*a - 4*b)*Cosh[2*(c + d*x)]))/(a + 2*b + a*Cosh[ 2*(c + d*x)])^2)*(a + 2*b + a*Cosh[2*c + 2*d*x])^3*Sech[c + d*x]^6)/(2048* a^(3/2)*b^(5/2)*d*(a + b*Sech[c + d*x]^2)^3) + ((3*(3*a^4 - 40*a^3*b + 720 *a^2*b^2 + 6720*a*b^3 + 8960*b^4)*ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(C osh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] + 3*(3*a^4 - 40* a^3*b + 720*a^2*b^2 + 6720*a*b^3 + 8960*b^4)*ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] + (2* Sqrt[a]*Sqrt[b]*Cosh[c + d*x]*(9*a^5 - 90*a^4*b - 10144*a^3*b^2 - 48672...
Time = 0.48 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 26, 4621, 360, 25, 2345, 1467, 2009}
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.
| \(\displaystyle \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {i \sin (i c+i d x)^3}{\left (a+b \sec (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \frac {\sin (i c+i d x)^3}{\left (b \sec (i c+i d x)^2+a\right )^3}dx\) |
| \(\Big \downarrow \) 4621 |
| \(\displaystyle -\frac {\int \frac {\cosh ^6(c+d x) \left (1-\cosh ^2(c+d x)\right )}{\left (a \cosh ^2(c+d x)+b\right )^3}d\cosh (c+d x)}{d}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle -\frac {-\frac {\int -\frac {-4 a^3 \cosh ^6(c+d x)+4 a^2 (a+b) \cosh ^4(c+d x)-4 a b (a+b) \cosh ^2(c+d x)+b^2 (a+b)}{\left (a \cosh ^2(c+d x)+b\right )^2}d\cosh (c+d x)}{4 a^4}-\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {\int \frac {-4 a^3 \cosh ^6(c+d x)+4 a^2 (a+b) \cosh ^4(c+d x)-4 a b (a+b) \cosh ^2(c+d x)+b^2 (a+b)}{\left (a \cosh ^2(c+d x)+b\right )^2}d\cosh (c+d x)}{4 a^4}-\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle -\frac {\frac {\frac {b (9 a+13 b) \cosh (c+d x)}{2 \left (a \cosh ^2(c+d x)+b\right )}-\frac {\int \frac {8 a^2 b \cosh ^4(c+d x)-8 a b (a+2 b) \cosh ^2(c+d x)+b^2 (7 a+11 b)}{a \cosh ^2(c+d x)+b}d\cosh (c+d x)}{2 b}}{4 a^4}-\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 1467 |
| \(\displaystyle -\frac {\frac {\frac {b (9 a+13 b) \cosh (c+d x)}{2 \left (a \cosh ^2(c+d x)+b\right )}-\frac {\int \left (8 a b \cosh ^2(c+d x)-8 b (a+3 b)+\frac {5 \left (7 b^3+3 a b^2\right )}{a \cosh ^2(c+d x)+b}\right )d\cosh (c+d x)}{2 b}}{4 a^4}-\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {\frac {b (9 a+13 b) \cosh (c+d x)}{2 \left (a \cosh ^2(c+d x)+b\right )}-\frac {\frac {5 b^{3/2} (3 a+7 b) \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{\sqrt {a}}+\frac {8}{3} a b \cosh ^3(c+d x)-8 b (a+3 b) \cosh (c+d x)}{2 b}}{4 a^4}-\frac {b^2 (a+b) \cosh (c+d x)}{4 a^4 \left (a \cosh ^2(c+d x)+b\right )^2}}{d}\) |
-((-1/4*(b^2*(a + b)*Cosh[c + d*x])/(a^4*(b + a*Cosh[c + d*x]^2)^2) + ((b* (9*a + 13*b)*Cosh[c + d*x])/(2*(b + a*Cosh[c + d*x]^2)) - ((5*b^(3/2)*(3*a + 7*b)*ArcTan[(Sqrt[a]*Cosh[c + d*x])/Sqrt[b]])/Sqrt[a] - 8*b*(a + 3*b)*C osh[c + d*x] + (8*a*b*Cosh[c + d*x]^3)/3)/(2*b))/(4*a^4))/d)
3.1.42.3.1 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[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[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)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 2] && IntegerQ[n] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(347\) vs. \(2(138)=276\).
Time = 0.21 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.26
\[\frac {\frac {1}{3 a^{3} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {1}{2 a^{3} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {a +6 b}{2 a^{4} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {1}{3 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -6 b}{2 a^{4} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b \left (\frac {\left (-\frac {9}{8} a^{2}+\frac {1}{4} a b +\frac {11}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {\left (27 a^{3}+15 a^{2} b +5 a \,b^{2}+33 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 \left (a +b \right )}+\left (-\frac {27}{8} a^{2}-\frac {5}{4} a b +\frac {33}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {9 a^{2}}{8}-\frac {5 a b}{2}-\frac {11 b^{2}}{8}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {5 \left (3 a +7 b \right ) \arctan \left (\frac {2 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 a -2 b}{4 \sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{a^{4}}}{d}\]
1/d*(1/3/a^3/(1+tanh(1/2*d*x+1/2*c))^3-1/2/a^3/(1+tanh(1/2*d*x+1/2*c))^2-1 /2*(a+6*b)/a^4/(1+tanh(1/2*d*x+1/2*c))-1/3/a^3/(tanh(1/2*d*x+1/2*c)-1)^3-1 /2/a^3/(tanh(1/2*d*x+1/2*c)-1)^2-1/2/a^4*(-a-6*b)/(tanh(1/2*d*x+1/2*c)-1)+ 2*b/a^4*(((-9/8*a^2+1/4*a*b+11/8*b^2)*tanh(1/2*d*x+1/2*c)^6-1/8*(27*a^3+15 *a^2*b+5*a*b^2+33*b^3)/(a+b)*tanh(1/2*d*x+1/2*c)^4+(-27/8*a^2-5/4*a*b+33/8 *b^2)*tanh(1/2*d*x+1/2*c)^2-9/8*a^2-5/2*a*b-11/8*b^2)/(tanh(1/2*d*x+1/2*c) ^4*a+tanh(1/2*d*x+1/2*c)^4*b+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2* c)^2*b+a+b)^2+5/16*(3*a+7*b)/(a*b)^(1/2)*arctan(1/4*(2*(a+b)*tanh(1/2*d*x+ 1/2*c)^2+2*a-2*b)/(a*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 4793 vs. \(2 (138) = 276\).
Time = 0.35 (sec) , antiderivative size = 8667, normalized size of antiderivative = 56.28 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\sinh \left (d x + c\right )^{3}}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^6\,{\mathrm {sinh}\left (c+d\,x\right )}^3}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^3} \,d x \]