Integrand size = 23, antiderivative size = 153 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {\arctan (\sinh (c+d x))}{b^3 d}-\frac {\sqrt {a} \left (8 a^2+20 a b+15 b^2\right ) \arctan \left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {a+b}}\right )}{8 b^3 (a+b)^{5/2} d}-\frac {a \sinh (c+d x)}{4 b (a+b) d \left (a+b+a \sinh ^2(c+d x)\right )^2}-\frac {a (4 a+7 b) \sinh (c+d x)}{8 b^2 (a+b)^2 d \left (a+b+a \sinh ^2(c+d x)\right )} \] Output:
arctan(sinh(d*x+c))/b^3/d-1/8*a^(1/2)*(8*a^2+20*a*b+15*b^2)*arctan(a^(1/2) *sinh(d*x+c)/(a+b)^(1/2))/b^3/(a+b)^(5/2)/d-1/4*a*sinh(d*x+c)/b/(a+b)/d/(a +b+a*sinh(d*x+c)^2)^2-1/8*a*(4*a+7*b)*sinh(d*x+c)/b^2/(a+b)^2/d/(a+b+a*sin h(d*x+c)^2)
Time = 3.17 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.61 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^5(c+d x) \left (16 \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right ) (a+2 b+a \cosh (2 (c+d x)))^2 \text {sech}(c+d x)+\frac {\sqrt {a} \left (8 a^2+20 a b+15 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \text {csch}(c+d x) \sqrt {(\cosh (c)-\sinh (c))^2} (\cosh (c)+\sinh (c))}{\sqrt {a}}\right ) (a+2 b+a \cosh (2 (c+d x)))^2 \text {sech}(c+d x) (\cosh (c)-\sinh (c))}{(a+b)^{5/2} \sqrt {(\cosh (c)-\sinh (c))^2}}-\frac {8 a b^2 \tanh (c+d x)}{a+b}-\frac {2 a b (4 a+7 b) (a+2 b+a \cosh (2 (c+d x))) \tanh (c+d x)}{(a+b)^2}\right )}{64 b^3 d \left (a+b \text {sech}^2(c+d x)\right )^3} \] Input:
Integrate[Sech[c + d*x]^7/(a + b*Sech[c + d*x]^2)^3,x]
Output:
((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^5*(16*ArcTan[Tanh[(c + d*x) /2]]*(a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x] + (Sqrt[a]*(8*a^2 + 2 0*a*b + 15*b^2)*ArcTan[(Sqrt[a + b]*Csch[c + d*x]*Sqrt[(Cosh[c] - Sinh[c]) ^2]*(Cosh[c] + Sinh[c]))/Sqrt[a]]*(a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x]*(Cosh[c] - Sinh[c]))/((a + b)^(5/2)*Sqrt[(Cosh[c] - Sinh[c])^2]) - (8*a*b^2*Tanh[c + d*x])/(a + b) - (2*a*b*(4*a + 7*b)*(a + 2*b + a*Cosh[2* (c + d*x)])*Tanh[c + d*x])/(a + b)^2))/(64*b^3*d*(a + b*Sech[c + d*x]^2)^3 )
Time = 0.38 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4635, 316, 402, 397, 216, 218}
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.
| \(\displaystyle \int \frac {\text {sech}^7(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (i c+i d x)^7}{\left (a+b \sec (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4635 |
| \(\displaystyle \frac {\int \frac {1}{\left (\sinh ^2(c+d x)+1\right ) \left (a \sinh ^2(c+d x)+a+b\right )^3}d\sinh (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int \frac {-3 a \sinh ^2(c+d x)+a+4 b}{\left (\sinh ^2(c+d x)+1\right ) \left (a \sinh ^2(c+d x)+a+b\right )^2}d\sinh (c+d x)}{4 b (a+b)}-\frac {a \sinh (c+d x)}{4 b (a+b) \left (a \sinh ^2(c+d x)+a+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {4 a^2-(4 a+7 b) \sinh ^2(c+d x) a+9 b a+8 b^2}{\left (\sinh ^2(c+d x)+1\right ) \left (a \sinh ^2(c+d x)+a+b\right )}d\sinh (c+d x)}{2 b (a+b)}-\frac {a (4 a+7 b) \sinh (c+d x)}{2 b (a+b) \left (a \sinh ^2(c+d x)+a+b\right )}}{4 b (a+b)}-\frac {a \sinh (c+d x)}{4 b (a+b) \left (a \sinh ^2(c+d x)+a+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 (a+b)^2 \int \frac {1}{\sinh ^2(c+d x)+1}d\sinh (c+d x)}{b}-\frac {a \left (8 a^2+20 a b+15 b^2\right ) \int \frac {1}{a \sinh ^2(c+d x)+a+b}d\sinh (c+d x)}{b}}{2 b (a+b)}-\frac {a (4 a+7 b) \sinh (c+d x)}{2 b (a+b) \left (a \sinh ^2(c+d x)+a+b\right )}}{4 b (a+b)}-\frac {a \sinh (c+d x)}{4 b (a+b) \left (a \sinh ^2(c+d x)+a+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 (a+b)^2 \arctan (\sinh (c+d x))}{b}-\frac {a \left (8 a^2+20 a b+15 b^2\right ) \int \frac {1}{a \sinh ^2(c+d x)+a+b}d\sinh (c+d x)}{b}}{2 b (a+b)}-\frac {a (4 a+7 b) \sinh (c+d x)}{2 b (a+b) \left (a \sinh ^2(c+d x)+a+b\right )}}{4 b (a+b)}-\frac {a \sinh (c+d x)}{4 b (a+b) \left (a \sinh ^2(c+d x)+a+b\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 (a+b)^2 \arctan (\sinh (c+d x))}{b}-\frac {\sqrt {a} \left (8 a^2+20 a b+15 b^2\right ) \arctan \left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {a+b}}\right )}{b \sqrt {a+b}}}{2 b (a+b)}-\frac {a (4 a+7 b) \sinh (c+d x)}{2 b (a+b) \left (a \sinh ^2(c+d x)+a+b\right )}}{4 b (a+b)}-\frac {a \sinh (c+d x)}{4 b (a+b) \left (a \sinh ^2(c+d x)+a+b\right )^2}}{d}\) |
Input:
Int[Sech[c + d*x]^7/(a + b*Sech[c + d*x]^2)^3,x]
Output:
(-1/4*(a*Sinh[c + d*x])/(b*(a + b)*(a + b + a*Sinh[c + d*x]^2)^2) + (((8*( a + b)^2*ArcTan[Sinh[c + d*x]])/b - (Sqrt[a]*(8*a^2 + 20*a*b + 15*b^2)*Arc Tan[(Sqrt[a]*Sinh[c + d*x])/Sqrt[a + b]])/(b*Sqrt[a + b]))/(2*b*(a + b)) - (a*(4*a + 7*b)*Sinh[c + d*x])/(2*b*(a + b)*(a + b + a*Sinh[c + d*x]^2)))/ (4*b*(a + b)))/d
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)^(-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)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 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[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]
Leaf count of result is larger than twice the leaf count of optimal. \(313\) vs. \(2(139)=278\).
Time = 2.93 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.05
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 a \left (\frac {-\frac {b \left (4 a +9 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 \left (a +b \right )}-\frac {b \left (4 a^{2}-11 a b -27 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 \left (a +b \right )^{2}}+\frac {b \left (4 a^{2}-11 a b -27 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 \left (a +b \right )^{2}}+\frac {b \left (4 a +9 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a +8 b}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (8 a^{2}+20 a b +15 b^{2}\right ) \left (\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sqrt {b}}{2 \sqrt {a}}\right )}{2 \sqrt {a +b}\, \sqrt {a}}+\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {b}}{2 \sqrt {a}}\right )}{2 \sqrt {a +b}\, \sqrt {a}}\right )}{8 a^{2}+16 a b +8 b^{2}}\right )}{b^{3}}+\frac {2 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\) | \(314\) |
| default | \(\frac {-\frac {2 a \left (\frac {-\frac {b \left (4 a +9 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 \left (a +b \right )}-\frac {b \left (4 a^{2}-11 a b -27 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 \left (a +b \right )^{2}}+\frac {b \left (4 a^{2}-11 a b -27 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 \left (a +b \right )^{2}}+\frac {b \left (4 a +9 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a +8 b}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (8 a^{2}+20 a b +15 b^{2}\right ) \left (\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sqrt {b}}{2 \sqrt {a}}\right )}{2 \sqrt {a +b}\, \sqrt {a}}+\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {b}}{2 \sqrt {a}}\right )}{2 \sqrt {a +b}\, \sqrt {a}}\right )}{8 a^{2}+16 a b +8 b^{2}}\right )}{b^{3}}+\frac {2 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\) | \(314\) |
| risch | \(-\frac {{\mathrm e}^{d x +c} a \left (4 a^{2} {\mathrm e}^{6 d x +6 c}+7 a b \,{\mathrm e}^{6 d x +6 c}+4 a^{2} {\mathrm e}^{4 d x +4 c}+31 a b \,{\mathrm e}^{4 d x +4 c}+36 b^{2} {\mathrm e}^{4 d x +4 c}-4 a^{2} {\mathrm e}^{2 d x +2 c}-31 a b \,{\mathrm e}^{2 d x +2 c}-36 b^{2} {\mathrm e}^{2 d x +2 c}-4 a^{2}-7 a b \right )}{4 d \,b^{2} \left (a +b \right )^{2} \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2}}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right )}{d \,b^{3}}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right )}{d \,b^{3}}+\frac {\sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-\left (a +b \right ) a}\, {\mathrm e}^{d x +c}}{a}-1\right ) a^{2}}{2 \left (a +b \right )^{3} d \,b^{3}}+\frac {5 \sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-\left (a +b \right ) a}\, {\mathrm e}^{d x +c}}{a}-1\right ) a}{4 \left (a +b \right )^{3} d \,b^{2}}+\frac {15 \sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-\left (a +b \right ) a}\, {\mathrm e}^{d x +c}}{a}-1\right )}{16 \left (a +b \right )^{3} d b}-\frac {\sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-\left (a +b \right ) a}\, {\mathrm e}^{d x +c}}{a}-1\right ) a^{2}}{2 \left (a +b \right )^{3} d \,b^{3}}-\frac {5 \sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-\left (a +b \right ) a}\, {\mathrm e}^{d x +c}}{a}-1\right ) a}{4 \left (a +b \right )^{3} d \,b^{2}}-\frac {15 \sqrt {-\left (a +b \right ) a}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-\left (a +b \right ) a}\, {\mathrm e}^{d x +c}}{a}-1\right )}{16 \left (a +b \right )^{3} d b}\) | \(538\) |
Input:
int(sech(d*x+c)^7/(a+b*sech(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-2*a/b^3*((-1/8*b*(4*a+9*b)/(a+b)*tanh(1/2*d*x+1/2*c)^7-1/8*b*(4*a^2- 11*a*b-27*b^2)/(a+b)^2*tanh(1/2*d*x+1/2*c)^5+1/8*b*(4*a^2-11*a*b-27*b^2)/( a+b)^2*tanh(1/2*d*x+1/2*c)^3+1/8*b*(4*a+9*b)/(a+b)*tanh(1/2*d*x+1/2*c))/(t anh(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+1/8*(8*a^2+20*a*b+15*b^2)/(a^2+2*a*b+b^2)* (1/2/(a+b)^(1/2)/a^(1/2)*arctan(1/2*(2*(a+b)^(1/2)*tanh(1/2*d*x+1/2*c)-2*b ^(1/2))/a^(1/2))+1/2/(a+b)^(1/2)/a^(1/2)*arctan(1/2*(2*(a+b)^(1/2)*tanh(1/ 2*d*x+1/2*c)+2*b^(1/2))/a^(1/2))))+2/b^3*arctan(tanh(1/2*d*x+1/2*c)))
Leaf count of result is larger than twice the leaf count of optimal. 4335 vs. \(2 (139) = 278\).
Time = 0.45 (sec) , antiderivative size = 7993, normalized size of antiderivative = 52.24 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(sech(d*x+c)^7/(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{7}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(sech(d*x+c)**7/(a+b*sech(d*x+c)**2)**3,x)
Output:
Integral(sech(c + d*x)**7/(a + b*sech(c + d*x)**2)**3, x)
\[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{7}}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \] Input:
integrate(sech(d*x+c)^7/(a+b*sech(d*x+c)^2)^3,x, algorithm="maxima")
Output:
-1/4*((4*a^3*e^(7*c) + 7*a^2*b*e^(7*c))*e^(7*d*x) + (4*a^3*e^(5*c) + 31*a^ 2*b*e^(5*c) + 36*a*b^2*e^(5*c))*e^(5*d*x) - (4*a^3*e^(3*c) + 31*a^2*b*e^(3 *c) + 36*a*b^2*e^(3*c))*e^(3*d*x) - (4*a^3*e^c + 7*a^2*b*e^c)*e^(d*x))/(a^ 4*b^2*d + 2*a^3*b^3*d + a^2*b^4*d + (a^4*b^2*d*e^(8*c) + 2*a^3*b^3*d*e^(8* c) + a^2*b^4*d*e^(8*c))*e^(8*d*x) + 4*(a^4*b^2*d*e^(6*c) + 4*a^3*b^3*d*e^( 6*c) + 5*a^2*b^4*d*e^(6*c) + 2*a*b^5*d*e^(6*c))*e^(6*d*x) + 2*(3*a^4*b^2*d *e^(4*c) + 14*a^3*b^3*d*e^(4*c) + 27*a^2*b^4*d*e^(4*c) + 24*a*b^5*d*e^(4*c ) + 8*b^6*d*e^(4*c))*e^(4*d*x) + 4*(a^4*b^2*d*e^(2*c) + 4*a^3*b^3*d*e^(2*c ) + 5*a^2*b^4*d*e^(2*c) + 2*a*b^5*d*e^(2*c))*e^(2*d*x)) + 2*arctan(e^(d*x + c))/(b^3*d) - 128*integrate(1/512*((8*a^3*e^(3*c) + 20*a^2*b*e^(3*c) + 1 5*a*b^2*e^(3*c))*e^(3*d*x) + (8*a^3*e^c + 20*a^2*b*e^c + 15*a*b^2*e^c)*e^( d*x))/(a^3*b^3 + 2*a^2*b^4 + a*b^5 + (a^3*b^3*e^(4*c) + 2*a^2*b^4*e^(4*c) + a*b^5*e^(4*c))*e^(4*d*x) + 2*(a^3*b^3*e^(2*c) + 4*a^2*b^4*e^(2*c) + 5*a* b^5*e^(2*c) + 2*b^6*e^(2*c))*e^(2*d*x)), x)
Exception generated. \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sech(d*x+c)^7/(a+b*sech(d*x+c)^2)^3,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Timed out. \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^7\,{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^3} \,d x \] Input:
int(1/(cosh(c + d*x)^7*(a + b/cosh(c + d*x)^2)^3),x)
Output:
int(1/(cosh(c + d*x)^7*(a + b/cosh(c + d*x)^2)^3), x)
Time = 0.44 (sec) , antiderivative size = 9430, normalized size of antiderivative = 61.63 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:
int(sech(d*x+c)^7/(a+b*sech(d*x+c)^2)^3,x)
Output:
(32*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**6 + 96*e**(8*c + 8*d*x)*atan(e* *(c + d*x))*a**5*b + 96*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**4*b**2 + 32 *e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**3*b**3 + 128*e**(6*c + 6*d*x)*atan (e**(c + d*x))*a**6 + 640*e**(6*c + 6*d*x)*atan(e**(c + d*x))*a**5*b + 115 2*e**(6*c + 6*d*x)*atan(e**(c + d*x))*a**4*b**2 + 896*e**(6*c + 6*d*x)*ata n(e**(c + d*x))*a**3*b**3 + 256*e**(6*c + 6*d*x)*atan(e**(c + d*x))*a**2*b **4 + 192*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a**6 + 1088*e**(4*c + 4*d*x) *atan(e**(c + d*x))*a**5*b + 2624*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a**4 *b**2 + 3264*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a**3*b**3 + 2048*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a**2*b**4 + 512*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a*b**5 + 128*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a**6 + 640*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a**5*b + 1152*e**(2*c + 2*d*x)*atan(e**(c + d* x))*a**4*b**2 + 896*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a**3*b**3 + 256*e* *(2*c + 2*d*x)*atan(e**(c + d*x))*a**2*b**4 + 32*atan(e**(c + d*x))*a**6 + 96*atan(e**(c + d*x))*a**5*b + 96*atan(e**(c + d*x))*a**4*b**2 + 32*atan( e**(c + d*x))*a**3*b**3 + 16*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)* sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt( 2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**4 + 40*e**(8*c + 8*d*x)*sqrt(b)*sqrt (a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a )/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a**3*b + 30*e**(8*c ...