Integrand size = 21, antiderivative size = 147 \[ \int \text {sech}(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {(2 a+b) \left (8 a^2+8 a b+5 b^2\right ) \arctan (\sinh (c+d x))}{16 d}+\frac {b \left (44 a^2+44 a b+15 b^2\right ) \text {sech}(c+d x) \tanh (c+d x)}{48 d}+\frac {5 b (2 a+b) \text {sech}^3(c+d x) \left (a+b+a \sinh ^2(c+d x)\right ) \tanh (c+d x)}{24 d}+\frac {b \text {sech}^5(c+d x) \left (a+b+a \sinh ^2(c+d x)\right )^2 \tanh (c+d x)}{6 d} \]
1/16*(2*a+b)*(8*a^2+8*a*b+5*b^2)*arctan(sinh(d*x+c))/d+1/48*b*(44*a^2+44*a *b+15*b^2)*sech(d*x+c)*tanh(d*x+c)/d+5/24*b*(2*a+b)*sech(d*x+c)^3*(a+b+a*s inh(d*x+c)^2)*tanh(d*x+c)/d+1/6*b*sech(d*x+c)^5*(a+b+a*sinh(d*x+c)^2)^2*ta nh(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.49 (sec) , antiderivative size = 1430, normalized size of antiderivative = 9.73 \[ \int \text {sech}(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx =\text {Too large to display} \]
(Coth[c + d*x]^6*Csch[c + d*x]*(a + b*Sech[c + d*x]^2)^3*(117228825*(a + b )^3*Sinh[c + d*x]^2 + 274542345*a*(a + b)^2*Sinh[c + d*x]^4 + 70189350*(a + b)^3*Sinh[c + d*x]^4 + 215549775*a^2*(a + b)*Sinh[c + d*x]^6 + 168951510 *a*(a + b)^2*Sinh[c + d*x]^6 + 4093425*(a + b)^3*Sinh[c + d*x]^6 + 5800945 5*a^3*Sinh[c + d*x]^8 + 135323370*a^2*(a + b)*Sinh[c + d*x]^8 + 9514449*a* (a + b)^2*Sinh[c + d*x]^8 + 36772890*a^3*Sinh[c + d*x]^10 + 7808535*a^2*(a + b)*Sinh[c + d*x]^10 - 75520*(a + b)^3*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^10 - 13824*(a + b)^3* HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 11/2}, -Sinh[c + d*x] ^2]*Sinh[c + d*x]^10 - 1024*(a + b)^3*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^10 + 2160711 *a^3*Sinh[c + d*x]^12 - 189696*a*(a + b)^2*HypergeometricPFQ[{3/2, 2, 2, 2 , 2}, {1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^12 - 38400*a*(a + b )^2*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^12 - 3072*a*(a + b)^2*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^12 - 1 58976*a^2*(a + b)*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, -S inh[c + d*x]^2]*Sinh[c + d*x]^14 - 35328*a^2*(a + b)*HypergeometricPFQ[{3/ 2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 11/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^14 - 3072*a^2*(a + b)*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2, 2}, {1, 1, 1,...
Time = 0.35 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4635, 315, 401, 25, 298, 216}
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.
| \(\displaystyle \int \text {sech}(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (i c+i d x) \left (a+b \sec (i c+i d x)^2\right )^3dx\) |
| \(\Big \downarrow \) 4635 |
| \(\displaystyle \frac {\int \frac {\left (a \sinh ^2(c+d x)+a+b\right )^3}{\left (\sinh ^2(c+d x)+1\right )^4}d\sinh (c+d x)}{d}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {\left (a \sinh ^2(c+d x)+a+b\right ) \left (a (6 a+b) \sinh ^2(c+d x)+(a+b) (6 a+5 b)\right )}{\left (\sinh ^2(c+d x)+1\right )^3}d\sinh (c+d x)+\frac {b \sinh (c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^2}{6 \left (\sinh ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 401 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {5 b (2 a+b) \sinh (c+d x) \left (a \sinh ^2(c+d x)+a+b\right )}{4 \left (\sinh ^2(c+d x)+1\right )^2}-\frac {1}{4} \int -\frac {a \left (24 a^2+14 b a+5 b^2\right ) \sinh ^2(c+d x)+(a+b) \left (24 a^2+34 b a+15 b^2\right )}{\left (\sinh ^2(c+d x)+1\right )^2}d\sinh (c+d x)\right )+\frac {b \sinh (c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^2}{6 \left (\sinh ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {a \left (24 a^2+14 b a+5 b^2\right ) \sinh ^2(c+d x)+(a+b) \left (24 a^2+34 b a+15 b^2\right )}{\left (\sinh ^2(c+d x)+1\right )^2}d\sinh (c+d x)+\frac {5 b (2 a+b) \sinh (c+d x) \left (a \sinh ^2(c+d x)+a+b\right )}{4 \left (\sinh ^2(c+d x)+1\right )^2}\right )+\frac {b \sinh (c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^2}{6 \left (\sinh ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} (2 a+b) \left (8 a^2+8 a b+5 b^2\right ) \int \frac {1}{\sinh ^2(c+d x)+1}d\sinh (c+d x)+\frac {b \left (44 a^2+44 a b+15 b^2\right ) \sinh (c+d x)}{2 \left (\sinh ^2(c+d x)+1\right )}\right )+\frac {5 b (2 a+b) \sinh (c+d x) \left (a \sinh ^2(c+d x)+a+b\right )}{4 \left (\sinh ^2(c+d x)+1\right )^2}\right )+\frac {b \sinh (c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^2}{6 \left (\sinh ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} (2 a+b) \left (8 a^2+8 a b+5 b^2\right ) \arctan (\sinh (c+d x))+\frac {b \left (44 a^2+44 a b+15 b^2\right ) \sinh (c+d x)}{2 \left (\sinh ^2(c+d x)+1\right )}\right )+\frac {5 b (2 a+b) \sinh (c+d x) \left (a \sinh ^2(c+d x)+a+b\right )}{4 \left (\sinh ^2(c+d x)+1\right )^2}\right )+\frac {b \sinh (c+d x) \left (a \sinh ^2(c+d x)+a+b\right )^2}{6 \left (\sinh ^2(c+d x)+1\right )^3}}{d}\) |
((b*Sinh[c + d*x]*(a + b + a*Sinh[c + d*x]^2)^2)/(6*(1 + Sinh[c + d*x]^2)^ 3) + ((5*b*(2*a + b)*Sinh[c + d*x]*(a + b + a*Sinh[c + d*x]^2))/(4*(1 + Si nh[c + d*x]^2)^2) + ((3*(2*a + b)*(8*a^2 + 8*a*b + 5*b^2)*ArcTan[Sinh[c + d*x]])/2 + (b*(44*a^2 + 44*a*b + 15*b^2)*Sinh[c + d*x])/(2*(1 + Sinh[c + d *x]^2)))/4)/6)/d
3.1.69.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, 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/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1)) Int[(a + b*x^2)^(p + 1)*( c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + (b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L tQ[p, -1] && GtQ[q, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ ))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && In tegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 2.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \arctan \left ({\mathrm e}^{d x +c}\right )+3 a^{2} b \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+3 a \,b^{2} \left (\left (\frac {\operatorname {sech}\left (d x +c \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (d x +c \right )}{8}\right ) \tanh \left (d x +c \right )+\frac {3 \arctan \left ({\mathrm e}^{d x +c}\right )}{4}\right )+b^{3} \left (\left (\frac {\operatorname {sech}\left (d x +c \right )^{5}}{6}+\frac {5 \operatorname {sech}\left (d x +c \right )^{3}}{24}+\frac {5 \,\operatorname {sech}\left (d x +c \right )}{16}\right ) \tanh \left (d x +c \right )+\frac {5 \arctan \left ({\mathrm e}^{d x +c}\right )}{8}\right )}{d}\) | \(138\) |
| default | \(\frac {2 a^{3} \arctan \left ({\mathrm e}^{d x +c}\right )+3 a^{2} b \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+3 a \,b^{2} \left (\left (\frac {\operatorname {sech}\left (d x +c \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (d x +c \right )}{8}\right ) \tanh \left (d x +c \right )+\frac {3 \arctan \left ({\mathrm e}^{d x +c}\right )}{4}\right )+b^{3} \left (\left (\frac {\operatorname {sech}\left (d x +c \right )^{5}}{6}+\frac {5 \operatorname {sech}\left (d x +c \right )^{3}}{24}+\frac {5 \,\operatorname {sech}\left (d x +c \right )}{16}\right ) \tanh \left (d x +c \right )+\frac {5 \arctan \left ({\mathrm e}^{d x +c}\right )}{8}\right )}{d}\) | \(138\) |
| parts | \(\frac {a^{3} \arctan \left (\sinh \left (d x +c \right )\right )}{d}+\frac {b^{3} \left (\left (\frac {\operatorname {sech}\left (d x +c \right )^{5}}{6}+\frac {5 \operatorname {sech}\left (d x +c \right )^{3}}{24}+\frac {5 \,\operatorname {sech}\left (d x +c \right )}{16}\right ) \tanh \left (d x +c \right )+\frac {5 \arctan \left ({\mathrm e}^{d x +c}\right )}{8}\right )}{d}+\frac {3 a^{2} b \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}+\frac {3 a \,b^{2} \left (\left (\frac {\operatorname {sech}\left (d x +c \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (d x +c \right )}{8}\right ) \tanh \left (d x +c \right )+\frac {3 \arctan \left ({\mathrm e}^{d x +c}\right )}{4}\right )}{d}\) | \(145\) |
| parallelrisch | \(\frac {-15 i \left (\frac {b}{2}+a \right ) \left (\frac {2}{3}+\frac {\cosh \left (6 d x +6 c \right )}{15}+\frac {2 \cosh \left (4 d x +4 c \right )}{5}+\cosh \left (2 d x +2 c \right )\right ) \left (a^{2}+a b +\frac {5}{8} b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )+15 i \left (\frac {b}{2}+a \right ) \left (\frac {2}{3}+\frac {\cosh \left (6 d x +6 c \right )}{15}+\frac {2 \cosh \left (4 d x +4 c \right )}{5}+\cosh \left (2 d x +2 c \right )\right ) \left (a^{2}+a b +\frac {5}{8} b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )+9 \left (\left (a^{2}+\frac {17}{12} a b +\frac {85}{216} b^{2}\right ) \sinh \left (3 d x +3 c \right )+\left (\frac {1}{3} a^{2}+\frac {1}{4} a b +\frac {5}{72} b^{2}\right ) \sinh \left (5 d x +5 c \right )+\frac {2 \sinh \left (d x +c \right ) \left (a^{2}+\frac {7}{4} a b +\frac {11}{8} b^{2}\right )}{3}\right ) b}{d \left (10+\cosh \left (6 d x +6 c \right )+6 \cosh \left (4 d x +4 c \right )+15 \cosh \left (2 d x +2 c \right )\right )}\) | \(247\) |
| risch | \(\frac {b \,{\mathrm e}^{d x +c} \left (72 a^{2} {\mathrm e}^{10 d x +10 c}+54 a b \,{\mathrm e}^{10 d x +10 c}+15 b^{2} {\mathrm e}^{10 d x +10 c}+216 a^{2} {\mathrm e}^{8 d x +8 c}+306 a b \,{\mathrm e}^{8 d x +8 c}+85 b^{2} {\mathrm e}^{8 d x +8 c}+144 a^{2} {\mathrm e}^{6 d x +6 c}+252 a b \,{\mathrm e}^{6 d x +6 c}+198 b^{2} {\mathrm e}^{6 d x +6 c}-144 a^{2} {\mathrm e}^{4 d x +4 c}-252 a b \,{\mathrm e}^{4 d x +4 c}-198 \,{\mathrm e}^{4 d x +4 c} b^{2}-216 a^{2} {\mathrm e}^{2 d x +2 c}-306 a b \,{\mathrm e}^{2 d x +2 c}-85 \,{\mathrm e}^{2 d x +2 c} b^{2}-72 a^{2}-54 a b -15 b^{2}\right )}{24 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{6}}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a^{3}}{d}+\frac {3 i b \ln \left ({\mathrm e}^{d x +c}+i\right ) a^{2}}{2 d}+\frac {9 i b^{2} \ln \left ({\mathrm e}^{d x +c}+i\right ) a}{8 d}+\frac {5 i b^{3} \ln \left ({\mathrm e}^{d x +c}+i\right )}{16 d}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a^{3}}{d}-\frac {3 i b \ln \left ({\mathrm e}^{d x +c}-i\right ) a^{2}}{2 d}-\frac {9 i b^{2} \ln \left ({\mathrm e}^{d x +c}-i\right ) a}{8 d}-\frac {5 i b^{3} \ln \left ({\mathrm e}^{d x +c}-i\right )}{16 d}\) | \(403\) |
1/d*(2*a^3*arctan(exp(d*x+c))+3*a^2*b*(1/2*sech(d*x+c)*tanh(d*x+c)+arctan( exp(d*x+c)))+3*a*b^2*((1/4*sech(d*x+c)^3+3/8*sech(d*x+c))*tanh(d*x+c)+3/4* arctan(exp(d*x+c)))+b^3*((1/6*sech(d*x+c)^5+5/24*sech(d*x+c)^3+5/16*sech(d *x+c))*tanh(d*x+c)+5/8*arctan(exp(d*x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 3465 vs. \(2 (139) = 278\).
Time = 0.29 (sec) , antiderivative size = 3465, normalized size of antiderivative = 23.57 \[ \int \text {sech}(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\text {Too large to display} \]
1/24*(3*(24*a^2*b + 18*a*b^2 + 5*b^3)*cosh(d*x + c)^11 + 33*(24*a^2*b + 18 *a*b^2 + 5*b^3)*cosh(d*x + c)*sinh(d*x + c)^10 + 3*(24*a^2*b + 18*a*b^2 + 5*b^3)*sinh(d*x + c)^11 + (216*a^2*b + 306*a*b^2 + 85*b^3)*cosh(d*x + c)^9 + (216*a^2*b + 306*a*b^2 + 85*b^3 + 165*(24*a^2*b + 18*a*b^2 + 5*b^3)*cos h(d*x + c)^2)*sinh(d*x + c)^9 + 9*(55*(24*a^2*b + 18*a*b^2 + 5*b^3)*cosh(d *x + c)^3 + (216*a^2*b + 306*a*b^2 + 85*b^3)*cosh(d*x + c))*sinh(d*x + c)^ 8 + 18*(8*a^2*b + 14*a*b^2 + 11*b^3)*cosh(d*x + c)^7 + 18*(55*(24*a^2*b + 18*a*b^2 + 5*b^3)*cosh(d*x + c)^4 + 8*a^2*b + 14*a*b^2 + 11*b^3 + 2*(216*a ^2*b + 306*a*b^2 + 85*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^7 + 42*(33*(24*a ^2*b + 18*a*b^2 + 5*b^3)*cosh(d*x + c)^5 + 2*(216*a^2*b + 306*a*b^2 + 85*b ^3)*cosh(d*x + c)^3 + 3*(8*a^2*b + 14*a*b^2 + 11*b^3)*cosh(d*x + c))*sinh( d*x + c)^6 - 18*(8*a^2*b + 14*a*b^2 + 11*b^3)*cosh(d*x + c)^5 + 18*(77*(24 *a^2*b + 18*a*b^2 + 5*b^3)*cosh(d*x + c)^6 + 7*(216*a^2*b + 306*a*b^2 + 85 *b^3)*cosh(d*x + c)^4 - 8*a^2*b - 14*a*b^2 - 11*b^3 + 21*(8*a^2*b + 14*a*b ^2 + 11*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 18*(55*(24*a^2*b + 18*a*b^ 2 + 5*b^3)*cosh(d*x + c)^7 + 7*(216*a^2*b + 306*a*b^2 + 85*b^3)*cosh(d*x + c)^5 + 35*(8*a^2*b + 14*a*b^2 + 11*b^3)*cosh(d*x + c)^3 - 5*(8*a^2*b + 14 *a*b^2 + 11*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 - (216*a^2*b + 306*a*b^2 + 85*b^3)*cosh(d*x + c)^3 + (495*(24*a^2*b + 18*a*b^2 + 5*b^3)*cosh(d*x + c )^8 + 84*(216*a^2*b + 306*a*b^2 + 85*b^3)*cosh(d*x + c)^6 + 630*(8*a^2*...
\[ \int \text {sech}(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3} \operatorname {sech}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 365 vs. \(2 (139) = 278\).
Time = 0.29 (sec) , antiderivative size = 365, normalized size of antiderivative = 2.48 \[ \int \text {sech}(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=-\frac {1}{24} \, b^{3} {\left (\frac {15 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {15 \, e^{\left (-d x - c\right )} + 85 \, e^{\left (-3 \, d x - 3 \, c\right )} + 198 \, e^{\left (-5 \, d x - 5 \, c\right )} - 198 \, e^{\left (-7 \, d x - 7 \, c\right )} - 85 \, e^{\left (-9 \, d x - 9 \, c\right )} - 15 \, e^{\left (-11 \, d x - 11 \, c\right )}}{d {\left (6 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 20 \, e^{\left (-6 \, d x - 6 \, c\right )} + 15 \, e^{\left (-8 \, d x - 8 \, c\right )} + 6 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}}\right )} - \frac {3}{4} \, a b^{2} {\left (\frac {3 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {3 \, e^{\left (-d x - c\right )} + 11 \, e^{\left (-3 \, d x - 3 \, c\right )} - 11 \, e^{\left (-5 \, d x - 5 \, c\right )} - 3 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} - 3 \, a^{2} b {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac {a^{3} \arctan \left (\sinh \left (d x + c\right )\right )}{d} \]
-1/24*b^3*(15*arctan(e^(-d*x - c))/d - (15*e^(-d*x - c) + 85*e^(-3*d*x - 3 *c) + 198*e^(-5*d*x - 5*c) - 198*e^(-7*d*x - 7*c) - 85*e^(-9*d*x - 9*c) - 15*e^(-11*d*x - 11*c))/(d*(6*e^(-2*d*x - 2*c) + 15*e^(-4*d*x - 4*c) + 20*e ^(-6*d*x - 6*c) + 15*e^(-8*d*x - 8*c) + 6*e^(-10*d*x - 10*c) + e^(-12*d*x - 12*c) + 1))) - 3/4*a*b^2*(3*arctan(e^(-d*x - c))/d - (3*e^(-d*x - c) + 1 1*e^(-3*d*x - 3*c) - 11*e^(-5*d*x - 5*c) - 3*e^(-7*d*x - 7*c))/(d*(4*e^(-2 *d*x - 2*c) + 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) + 1))) - 3*a^2*b*(arctan(e^(-d*x - c))/d - (e^(-d*x - c) - e^(-3*d*x - 3*c) )/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1))) + a^3*arctan(sinh(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (139) = 278\).
Time = 0.31 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.11 \[ \int \text {sech}(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {3 \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} {\left (16 \, a^{3} + 24 \, a^{2} b + 18 \, a b^{2} + 5 \, b^{3}\right )} + \frac {4 \, {\left (72 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} + 54 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} + 15 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} + 576 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 576 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 160 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 1152 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 1440 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 528 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}^{3}}}{96 \, d} \]
1/96*(3*(pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(16*a^3 + 24*a^2*b + 18*a*b^2 + 5*b^3) + 4*(72*a^2*b*(e^(d*x + c) - e^(-d*x - c))^5 + 54*a*b^2*(e^(d*x + c) - e^(-d*x - c))^5 + 15*b^3*(e^(d*x + c) - e^(-d*x - c))^5 + 576*a^2*b*(e^(d*x + c) - e^(-d*x - c))^3 + 576*a*b^2*(e^(d*x + c ) - e^(-d*x - c))^3 + 160*b^3*(e^(d*x + c) - e^(-d*x - c))^3 + 1152*a^2*b* (e^(d*x + c) - e^(-d*x - c)) + 1440*a*b^2*(e^(d*x + c) - e^(-d*x - c)) + 5 28*b^3*(e^(d*x + c) - e^(-d*x - c)))/((e^(d*x + c) - e^(-d*x - c))^2 + 4)^ 3)/d
Time = 2.15 (sec) , antiderivative size = 535, normalized size of antiderivative = 3.64 \[ \int \text {sech}(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (16\,a^3\,\sqrt {d^2}+5\,b^3\,\sqrt {d^2}+18\,a\,b^2\,\sqrt {d^2}+24\,a^2\,b\,\sqrt {d^2}\right )}{d\,\sqrt {256\,a^6+768\,a^5\,b+1152\,a^4\,b^2+1024\,a^3\,b^3+564\,a^2\,b^4+180\,a\,b^5+25\,b^6}}\right )\,\sqrt {256\,a^6+768\,a^5\,b+1152\,a^4\,b^2+1024\,a^3\,b^3+564\,a^2\,b^4+180\,a\,b^5+25\,b^6}}{8\,\sqrt {d^2}}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (54\,a\,b^2-b^3\right )}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {80\,b^3\,{\mathrm {e}}^{c+d\,x}}{3\,d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )}+\frac {6\,{\mathrm {e}}^{c+d\,x}\,\left (2\,a\,b^2-3\,b^3\right )}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {32\,b^3\,{\mathrm {e}}^{c+d\,x}}{3\,d\,\left (6\,{\mathrm {e}}^{2\,c+2\,d\,x}+15\,{\mathrm {e}}^{4\,c+4\,d\,x}+20\,{\mathrm {e}}^{6\,c+6\,d\,x}+15\,{\mathrm {e}}^{8\,c+8\,d\,x}+6\,{\mathrm {e}}^{10\,c+10\,d\,x}+{\mathrm {e}}^{12\,c+12\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (24\,a^2\,b+18\,a\,b^2+5\,b^3\right )}{8\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (-72\,a^2\,b+18\,a\,b^2+5\,b^3\right )}{12\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]
(atan((exp(d*x)*exp(c)*(16*a^3*(d^2)^(1/2) + 5*b^3*(d^2)^(1/2) + 18*a*b^2* (d^2)^(1/2) + 24*a^2*b*(d^2)^(1/2)))/(d*(180*a*b^5 + 768*a^5*b + 256*a^6 + 25*b^6 + 564*a^2*b^4 + 1024*a^3*b^3 + 1152*a^4*b^2)^(1/2)))*(180*a*b^5 + 768*a^5*b + 256*a^6 + 25*b^6 + 564*a^2*b^4 + 1024*a^3*b^3 + 1152*a^4*b^2)^ (1/2))/(8*(d^2)^(1/2)) - (exp(c + d*x)*(54*a*b^2 - b^3))/(3*d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) + (80*b^3*exp(c + d* x))/(3*d*(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*x) + 10*exp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1)) + (6*exp(c + d*x)*(2*a*b^2 - 3*b^3))/(d*(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1)) - (32*b^3*exp(c + d*x))/(3*d*(6*exp(2*c + 2*d*x) + 15*exp(4*c + 4*d*x) + 20*exp(6*c + 6*d*x) + 15*exp(8*c + 8*d*x) + 6*exp (10*c + 10*d*x) + exp(12*c + 12*d*x) + 1)) + (exp(c + d*x)*(18*a*b^2 + 24* a^2*b + 5*b^3))/(8*d*(exp(2*c + 2*d*x) + 1)) + (exp(c + d*x)*(18*a*b^2 - 7 2*a^2*b + 5*b^3))/(12*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1))