Integrand size = 16, antiderivative size = 183 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {a+b-b \tanh ^2(c+d x)}}\right )}{a^{7/2} d}-\frac {b \tanh (c+d x)}{5 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^{5/2}}-\frac {b (9 a+5 b) \tanh (c+d x)}{15 a^2 (a+b)^2 d \left (a+b-b \tanh ^2(c+d x)\right )^{3/2}}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{15 a^3 (a+b)^3 d \sqrt {a+b-b \tanh ^2(c+d x)}} \]
arctanh(a^(1/2)*tanh(d*x+c)/(a+b-b*tanh(d*x+c)^2)^(1/2))/a^(7/2)/d-1/15*b* (33*a^2+40*a*b+15*b^2)*tanh(d*x+c)/a^3/(a+b)^3/d/(a+b-b*tanh(d*x+c)^2)^(1/ 2)-1/5*b*tanh(d*x+c)/a/(a+b)/d/(a+b-b*tanh(d*x+c)^2)^(5/2)-1/15*b*(9*a+5*b )*tanh(d*x+c)/a^2/(a+b)^2/d/(a+b-b*tanh(d*x+c)^2)^(3/2)
Time = 1.86 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \, dx=\frac {\text {sech}^7(c+d x) \left (\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {a+b+a \sinh ^2(c+d x)}}\right ) (a+2 b+a \cosh (2 (c+d x)))^{7/2}}{a^{7/2} d}-\frac {b (a+2 b+a \cosh (2 (c+d x))) \left (135 a^4+480 a^3 b+709 a^2 b^2+460 a b^3+120 b^4+4 a \left (45 a^3+135 a^2 b+117 a b^2+35 b^3\right ) \cosh (2 (c+d x))+a^2 \left (45 a^2+60 a b+23 b^2\right ) \cosh (4 (c+d x))\right ) \sinh (c+d x)}{15 a^3 (a+b)^3 d}\right )}{16 \left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \]
(Sech[c + d*x]^7*((Sqrt[2]*ArcTanh[(Sqrt[a]*Sinh[c + d*x])/Sqrt[a + b + a* Sinh[c + d*x]^2]]*(a + 2*b + a*Cosh[2*(c + d*x)])^(7/2))/(a^(7/2)*d) - (b* (a + 2*b + a*Cosh[2*(c + d*x)])*(135*a^4 + 480*a^3*b + 709*a^2*b^2 + 460*a *b^3 + 120*b^4 + 4*a*(45*a^3 + 135*a^2*b + 117*a*b^2 + 35*b^3)*Cosh[2*(c + d*x)] + a^2*(45*a^2 + 60*a*b + 23*b^2)*Cosh[4*(c + d*x)])*Sinh[c + d*x])/ (15*a^3*(a + b)^3*d)))/(16*(a + b*Sech[c + d*x]^2)^(7/2))
Time = 0.38 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 4616, 316, 25, 402, 25, 402, 27, 291, 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.
\(\displaystyle \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a+b \sec (i c+i d x)^2\right )^{7/2}}dx\) |
\(\Big \downarrow \) 4616 |
\(\displaystyle \frac {\int \frac {1}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^{7/2}}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {-\frac {\int -\frac {4 b \tanh ^2(c+d x)+5 a+b}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^{5/2}}d\tanh (c+d x)}{5 a (a+b)}-\frac {b \tanh (c+d x)}{5 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {4 b \tanh ^2(c+d x)+5 a+b}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^{5/2}}d\tanh (c+d x)}{5 a (a+b)}-\frac {b \tanh (c+d x)}{5 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}}}{d}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {-\frac {\int -\frac {15 a^2+12 b a+5 b^2+2 b (9 a+5 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^{3/2}}d\tanh (c+d x)}{3 a (a+b)}-\frac {b (9 a+5 b) \tanh (c+d x)}{3 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tanh (c+d x)}{5 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\frac {\int \frac {15 a^2+12 b a+5 b^2+2 b (9 a+5 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^{3/2}}d\tanh (c+d x)}{3 a (a+b)}-\frac {b (9 a+5 b) \tanh (c+d x)}{3 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tanh (c+d x)}{5 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}}}{d}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {\frac {-\frac {\int -\frac {15 (a+b)^3}{\left (1-\tanh ^2(c+d x)\right ) \sqrt {-b \tanh ^2(c+d x)+a+b}}d\tanh (c+d x)}{a (a+b)}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{a (a+b) \sqrt {a-b \tanh ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \tanh (c+d x)}{3 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tanh (c+d x)}{5 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\frac {15 (a+b)^2 \int \frac {1}{\left (1-\tanh ^2(c+d x)\right ) \sqrt {-b \tanh ^2(c+d x)+a+b}}d\tanh (c+d x)}{a}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{a (a+b) \sqrt {a-b \tanh ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \tanh (c+d x)}{3 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tanh (c+d x)}{5 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}}}{d}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {\frac {\frac {15 (a+b)^2 \int \frac {1}{1-\frac {a \tanh ^2(c+d x)}{-b \tanh ^2(c+d x)+a+b}}d\frac {\tanh (c+d x)}{\sqrt {-b \tanh ^2(c+d x)+a+b}}}{a}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{a (a+b) \sqrt {a-b \tanh ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \tanh (c+d x)}{3 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tanh (c+d x)}{5 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {\frac {15 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {a-b \tanh ^2(c+d x)+b}}\right )}{a^{3/2}}-\frac {b \left (33 a^2+40 a b+15 b^2\right ) \tanh (c+d x)}{a (a+b) \sqrt {a-b \tanh ^2(c+d x)+b}}}{3 a (a+b)}-\frac {b (9 a+5 b) \tanh (c+d x)}{3 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{3/2}}}{5 a (a+b)}-\frac {b \tanh (c+d x)}{5 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^{5/2}}}{d}\) |
(-1/5*(b*Tanh[c + d*x])/(a*(a + b)*(a + b - b*Tanh[c + d*x]^2)^(5/2)) + (- 1/3*(b*(9*a + 5*b)*Tanh[c + d*x])/(a*(a + b)*(a + b - b*Tanh[c + d*x]^2)^( 3/2)) + ((15*(a + b)^2*ArcTanh[(Sqrt[a]*Tanh[c + d*x])/Sqrt[a + b - b*Tanh [c + d*x]^2]])/a^(3/2) - (b*(33*a^2 + 40*a*b + 15*b^2)*Tanh[c + d*x])/(a*( a + b)*Sqrt[a + b - b*Tanh[c + d*x]^2]))/(3*a*(a + b)))/(5*a*(a + b)))/d
3.3.20.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
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[((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[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
\[\int \frac {1}{\left (a +b \operatorname {sech}\left (d x +c \right )^{2}\right )^{\frac {7}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 8940 vs. \(2 (165) = 330\).
Time = 1.83 (sec) , antiderivative size = 18565, normalized size of antiderivative = 101.45 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
\[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{\frac {7}{2}}} \,d x } \]
Exception generated. \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^{7/2}} \,d x \]