Integrand size = 23, antiderivative size = 384 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2} \, dx=\frac {\sqrt {2} a b \left (a^4-2 a^2 b^2-b^4\right ) \arctan \left (1-\sqrt {2} \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}-\frac {\sqrt {2} a b \left (a^4-2 a^2 b^2-b^4\right ) \arctan \left (1+\sqrt {2} \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}+\frac {a^2 \left (a^4-3 b^4\right ) \arctan (\sinh (c+d x))}{\left (a^4+b^4\right )^2 d}+\frac {b^2 \left (3 a^4-b^4\right ) \log (\cosh (c+d x))}{\left (a^4+b^4\right )^2 d}-\frac {2 b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2 d}-\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (1-\sqrt {2} \sqrt {\sinh (c+d x)}+\sinh (c+d x)\right )}{\sqrt {2} \left (a^4+b^4\right )^2 d}+\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (1+\sqrt {2} \sqrt {\sinh (c+d x)}+\sinh (c+d x)\right )}{\sqrt {2} \left (a^4+b^4\right )^2 d}+\frac {2 a b^2}{\left (a^4+b^4\right ) d \left (a+b \sqrt {\sinh (c+d x)}\right )} \]
a^2*(a^4-3*b^4)*arctan(sinh(d*x+c))/(a^4+b^4)^2/d+b^2*(3*a^4-b^4)*ln(cosh( d*x+c))/(a^4+b^4)^2/d-2*b^2*(3*a^4-b^4)*ln(a+b*sinh(d*x+c)^(1/2))/(a^4+b^4 )^2/d-1/2*a*b*(a^4+2*a^2*b^2-b^4)*ln(1+sinh(d*x+c)-2^(1/2)*sinh(d*x+c)^(1/ 2))/(a^4+b^4)^2/d*2^(1/2)+1/2*a*b*(a^4+2*a^2*b^2-b^4)*ln(1+sinh(d*x+c)+2^( 1/2)*sinh(d*x+c)^(1/2))/(a^4+b^4)^2/d*2^(1/2)-a*b*(a^4-2*a^2*b^2-b^4)*arct an(-1+2^(1/2)*sinh(d*x+c)^(1/2))*2^(1/2)/(a^4+b^4)^2/d-a*b*(a^4-2*a^2*b^2- b^4)*arctan(1+2^(1/2)*sinh(d*x+c)^(1/2))*2^(1/2)/(a^4+b^4)^2/d+2*a*b^2/(a^ 4+b^4)/d/(a+b*sinh(d*x+c)^(1/2))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.52 (sec) , antiderivative size = 300, normalized size of antiderivative = 0.78 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2} \, dx=\frac {-6 \sqrt {2} a^3 b^3 \arctan \left (1-\sqrt {2} \sqrt {\sinh (c+d x)}\right )+6 \sqrt {2} a^3 b^3 \arctan \left (1+\sqrt {2} \sqrt {\sinh (c+d x)}\right )+3 a^2 \left (a^4-3 b^4\right ) \arctan (\sinh (c+d x))-3 b^2 \left (-3 a^4+b^4\right ) \log (\cosh (c+d x))+6 b^2 \left (-3 a^4+b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )-3 \sqrt {2} a^3 b^3 \log \left (1-\sqrt {2} \sqrt {\sinh (c+d x)}+\sinh (c+d x)\right )+3 \sqrt {2} a^3 b^3 \log \left (1+\sqrt {2} \sqrt {\sinh (c+d x)}+\sinh (c+d x)\right )+\frac {6 a b^2 \left (a^4+b^4\right )}{a+b \sqrt {\sinh (c+d x)}}-4 a b \left (a^4-b^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\sinh ^2(c+d x)\right ) \sinh ^{\frac {3}{2}}(c+d x)}{3 \left (a^4+b^4\right )^2 d} \]
(-6*Sqrt[2]*a^3*b^3*ArcTan[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]]] + 6*Sqrt[2]*a^ 3*b^3*ArcTan[1 + Sqrt[2]*Sqrt[Sinh[c + d*x]]] + 3*a^2*(a^4 - 3*b^4)*ArcTan [Sinh[c + d*x]] - 3*b^2*(-3*a^4 + b^4)*Log[Cosh[c + d*x]] + 6*b^2*(-3*a^4 + b^4)*Log[a + b*Sqrt[Sinh[c + d*x]]] - 3*Sqrt[2]*a^3*b^3*Log[1 - Sqrt[2]* Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]] + 3*Sqrt[2]*a^3*b^3*Log[1 + Sqrt[2]*S qrt[Sinh[c + d*x]] + Sinh[c + d*x]] + (6*a*b^2*(a^4 + b^4))/(a + b*Sqrt[Si nh[c + d*x]]) - 4*a*b*(a^4 - b^4)*Hypergeometric2F1[3/4, 1, 7/4, -Sinh[c + d*x]^2]*Sinh[c + d*x]^(3/2))/(3*(a^4 + b^4)^2*d)
Time = 0.88 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3702, 7267, 7276, 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.
| \(\displaystyle \int \frac {\text {sech}(c+d x)}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (i c+i d x) \left (a+b \sqrt {-i \sin (i c+i d x)}\right )^2}dx\) |
| \(\Big \downarrow \) 3702 |
| \(\displaystyle \frac {\int \frac {1}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2 \left (\sinh ^2(c+d x)+1\right )}d\sinh (c+d x)}{d}\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {\sinh (c+d x)}}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2 \left (\sinh ^2(c+d x)+1\right )}d\sqrt {\sinh (c+d x)}}{d}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \frac {2 \int \left (-\frac {a b^3}{\left (a^4+b^4\right ) \left (a+b \sqrt {\sinh (c+d x)}\right )^2}+\frac {b^7-3 a^4 b^3}{\left (a^4+b^4\right )^2 \left (a+b \sqrt {\sinh (c+d x)}\right )}+\frac {4 a^3 b^3+\left (3 a^4-b^4\right ) \sinh ^{\frac {3}{2}}(c+d x) b^2-2 a \left (a^4-b^4\right ) \sinh (c+d x) b+a^2 \left (a^4-3 b^4\right ) \sqrt {\sinh (c+d x)}}{\left (a^4+b^4\right )^2 \left (\sinh ^2(c+d x)+1\right )}\right )d\sqrt {\sinh (c+d x)}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (\frac {a b^2}{\left (a^4+b^4\right ) \left (a+b \sqrt {\sinh (c+d x)}\right )}+\frac {b^2 \left (3 a^4-b^4\right ) \log \left (\sinh ^2(c+d x)+1\right )}{4 \left (a^4+b^4\right )^2}-\frac {b^2 \left (3 a^4-b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{\left (a^4+b^4\right )^2}+\frac {a^2 \left (a^4-3 b^4\right ) \arctan (\sinh (c+d x))}{2 \left (a^4+b^4\right )^2}+\frac {a b \left (a^4-2 a^2 b^2-b^4\right ) \arctan \left (1-\sqrt {2} \sqrt {\sinh (c+d x)}\right )}{\sqrt {2} \left (a^4+b^4\right )^2}-\frac {a b \left (a^4-2 a^2 b^2-b^4\right ) \arctan \left (\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )}{\sqrt {2} \left (a^4+b^4\right )^2}-\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (\sinh (c+d x)-\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )}{2 \sqrt {2} \left (a^4+b^4\right )^2}+\frac {a b \left (a^4+2 a^2 b^2-b^4\right ) \log \left (\sinh (c+d x)+\sqrt {2} \sqrt {\sinh (c+d x)}+1\right )}{2 \sqrt {2} \left (a^4+b^4\right )^2}\right )}{d}\) |
(2*((a*b*(a^4 - 2*a^2*b^2 - b^4)*ArcTan[1 - Sqrt[2]*Sqrt[Sinh[c + d*x]]])/ (Sqrt[2]*(a^4 + b^4)^2) - (a*b*(a^4 - 2*a^2*b^2 - b^4)*ArcTan[1 + Sqrt[2]* Sqrt[Sinh[c + d*x]]])/(Sqrt[2]*(a^4 + b^4)^2) + (a^2*(a^4 - 3*b^4)*ArcTan[ Sinh[c + d*x]])/(2*(a^4 + b^4)^2) - (b^2*(3*a^4 - b^4)*Log[a + b*Sqrt[Sinh [c + d*x]]])/(a^4 + b^4)^2 - (a*b*(a^4 + 2*a^2*b^2 - b^4)*Log[1 - Sqrt[2]* Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]])/(2*Sqrt[2]*(a^4 + b^4)^2) + (a*b*(a^ 4 + 2*a^2*b^2 - b^4)*Log[1 + Sqrt[2]*Sqrt[Sinh[c + d*x]] + Sinh[c + d*x]]) /(2*Sqrt[2]*(a^4 + b^4)^2) + (b^2*(3*a^4 - b^4)*Log[1 + Sinh[c + d*x]^2])/ (4*(a^4 + b^4)^2) + (a*b^2)/((a^4 + b^4)*(a + b*Sqrt[Sinh[c + d*x]]))))/d
3.5.18.3.1 Defintions of rubi rules used
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x _)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si mp[ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.44 (sec) , antiderivative size = 376, normalized size of antiderivative = 0.98
| method | result | size |
| default | \(\frac {-\frac {2 b^{2} \left (\frac {2 \left (a^{4}+b^{4}\right ) b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}-a^{2}}+\frac {\left (3 a^{4}-b^{4}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}-a^{2}\right )}{2}\right )}{\left (a^{4}+b^{4}\right )^{2}}+\frac {\left (3 a^{4} b^{2}-b^{6}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+2 \left (a^{6}-3 a^{2} b^{4}\right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{8}+2 a^{4} b^{4}+b^{8}}}{d}+\frac {\operatorname {`\,int/indef0`\,}\left (\frac {2 a b \sqrt {\sinh \left (d x +c \right )}\, \left (b^{4} \sinh \left (d x +c \right )^{2}-2 a^{2} b^{2} \sinh \left (d x +c \right )+a^{4}\right )}{-b^{8} \cosh \left (d x +c \right )^{6}+4 a^{2} b^{6} \cosh \left (d x +c \right )^{4} \sinh \left (d x +c \right )+\left (-6 a^{4} b^{4}+2 b^{8}\right ) \cosh \left (d x +c \right )^{4}+\left (4 a^{6} b^{2}-4 a^{2} b^{6}\right ) \cosh \left (d x +c \right )^{2} \sinh \left (d x +c \right )+\left (-a^{8}+6 a^{4} b^{4}-b^{8}\right ) \cosh \left (d x +c \right )^{2}}, \sinh \left (d x +c \right )\right )}{d}\) | \(376\) |
1/d*(-2*b^2/(a^4+b^4)^2*(2*(a^4+b^4)*b^2*tanh(1/2*d*x+1/2*c)/(tanh(1/2*d*x +1/2*c)^2*a^2+2*tanh(1/2*d*x+1/2*c)*b^2-a^2)+1/2*(3*a^4-b^4)*ln(tanh(1/2*d *x+1/2*c)^2*a^2+2*tanh(1/2*d*x+1/2*c)*b^2-a^2))+2/(a^8+2*a^4*b^4+b^8)*(1/2 *(3*a^4*b^2-b^6)*ln(tanh(1/2*d*x+1/2*c)^2+1)+(a^6-3*a^2*b^4)*arctan(tanh(1 /2*d*x+1/2*c))))+`int/indef0`(2*a*b*sinh(d*x+c)^(1/2)*(b^4*sinh(d*x+c)^2-2 *a^2*b^2*sinh(d*x+c)+a^4)/(-b^8*cosh(d*x+c)^6+4*a^2*b^6*cosh(d*x+c)^4*sinh (d*x+c)+(-6*a^4*b^4+2*b^8)*cosh(d*x+c)^4+(4*a^6*b^2-4*a^2*b^6)*cosh(d*x+c) ^2*sinh(d*x+c)+(-a^8+6*a^4*b^4-b^8)*cosh(d*x+c)^2),sinh(d*x+c))/d
Leaf count of result is larger than twice the leaf count of optimal. 10240 vs. \(2 (357) = 714\).
Time = 5.15 (sec) , antiderivative size = 10240, normalized size of antiderivative = 26.67 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2} \, dx=\text {Too large to display} \]
\[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2} \, dx=\int \frac {\operatorname {sech}{\left (c + d x \right )}}{\left (a + b \sqrt {\sinh {\left (c + d x \right )}}\right )^{2}}\, dx \]
\[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )}{{\left (b \sqrt {\sinh \left (d x + c\right )} + a\right )}^{2}} \,d x } \]
\[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )}{{\left (b \sqrt {\sinh \left (d x + c\right )} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \sqrt {\sinh (c+d x)}\right )^2} \, dx=\int \frac {1}{\mathrm {cosh}\left (c+d\,x\right )\,{\left (a+b\,\sqrt {\mathrm {sinh}\left (c+d\,x\right )}\right )}^2} \,d x \]