Integrand size = 23, antiderivative size = 126 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{7/2} d}+\frac {\left (a^2-3 a b+3 b^2\right ) \tanh (c+d x)}{(a-b)^3 d}-\frac {(2 a-3 b) \tanh ^3(c+d x)}{3 (a-b)^2 d}+\frac {\tanh ^5(c+d x)}{5 (a-b) d} \]
-b^3*arctanh((a-b)^(1/2)*tanh(d*x+c)/a^(1/2))/(a-b)^(7/2)/d/a^(1/2)+(a^2-3 *a*b+3*b^2)*tanh(d*x+c)/(a-b)^3/d-1/3*(2*a-3*b)*tanh(d*x+c)^3/(a-b)^2/d+1/ 5*tanh(d*x+c)^5/(a-b)/d
Time = 0.82 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.94 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {-\frac {15 b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{7/2}}+\frac {\left (8 a^2-26 a b+33 b^2+\left (4 a^2-13 a b+9 b^2\right ) \text {sech}^2(c+d x)+3 (a-b)^2 \text {sech}^4(c+d x)\right ) \tanh (c+d x)}{(a-b)^3}}{15 d} \]
((-15*b^3*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^( 7/2)) + ((8*a^2 - 26*a*b + 33*b^2 + (4*a^2 - 13*a*b + 9*b^2)*Sech[c + d*x] ^2 + 3*(a - b)^2*Sech[c + d*x]^4)*Tanh[c + d*x])/(a - b)^3)/(15*d)
Time = 0.35 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3670, 300, 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.
\(\displaystyle \int \frac {\text {sech}^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (i c+i d x)^6 \left (a-b \sin (i c+i d x)^2\right )}dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \frac {\int \frac {\left (1-\tanh ^2(c+d x)\right )^3}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (\frac {\tanh ^4(c+d x)}{a-b}-\frac {(2 a-3 b) \tanh ^2(c+d x)}{(a-b)^2}+\frac {a^2-3 b a+3 b^2}{(a-b)^3}-\frac {b^3}{(a-b)^3 \left (a-(a-b) \tanh ^2(c+d x)\right )}\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\left (a^2-3 a b+3 b^2\right ) \tanh (c+d x)}{(a-b)^3}-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{7/2}}+\frac {\tanh ^5(c+d x)}{5 (a-b)}-\frac {(2 a-3 b) \tanh ^3(c+d x)}{3 (a-b)^2}}{d}\) |
(-((b^3*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^(7/ 2))) + ((a^2 - 3*a*b + 3*b^2)*Tanh[c + d*x])/(a - b)^3 - ((2*a - 3*b)*Tanh [c + d*x]^3)/(3*(a - b)^2) + Tanh[c + d*x]^5/(5*(a - b)))/d
3.4.28.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(345\) vs. \(2(114)=228\).
Time = 0.16 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.75
\[\frac {-\frac {2 \left (\left (-a^{2}+3 a b -3 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+\left (-\frac {4}{3} a^{2}+\frac {16}{3} a b -8 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (-\frac {58}{15} a^{2}+\frac {166}{15} a b -\frac {66}{5} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {4}{3} a^{2}+\frac {16}{3} a b -8 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{2}+3 a b -3 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{5}}+\frac {2 b^{3} a \left (\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{\left (a -b \right )^{3}}}{d}\]
1/d*(-2/(a-b)^3*((-a^2+3*a*b-3*b^2)*tanh(1/2*d*x+1/2*c)^9+(-4/3*a^2+16/3*a *b-8*b^2)*tanh(1/2*d*x+1/2*c)^7+(-58/15*a^2+166/15*a*b-66/5*b^2)*tanh(1/2* d*x+1/2*c)^5+(-4/3*a^2+16/3*a*b-8*b^2)*tanh(1/2*d*x+1/2*c)^3+(-a^2+3*a*b-3 *b^2)*tanh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^2+1)^5+2*b^3/(a-b)^3*a*(1/ 2*((-b*(a-b))^(1/2)+b)/a/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^( 1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))-1/ 2*((-b*(a-b))^(1/2)-b)/a/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^( 1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 2895 vs. \(2 (114) = 228\).
Time = 0.32 (sec) , antiderivative size = 6046, normalized size of antiderivative = 47.98 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {\text {sech}^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\int \frac {\operatorname {sech}^{6}{\left (c + d x \right )}}{a + b \sinh ^{2}{\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\text {sech}^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
\[ \int \frac {\text {sech}^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{6}}{b \sinh \left (d x + c\right )^{2} + a} \,d x } \]
Time = 3.76 (sec) , antiderivative size = 1152, normalized size of antiderivative = 9.14 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {16}{\left (a\,d-b\,d\right )\,\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}{5\,\left (a\,d-b\,d\right )\,\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 {\mathrm {atan}\left (\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {4\,b}{d\,{\left (a-b\right )}^3\,\sqrt {b^6}\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {\left (2\,a-b\right )\,\left (2\,a^4\,d\,\sqrt {b^6}+b^4\,d\,\sqrt {b^6}-5\,a\,b^3\,d\,\sqrt {b^6}-7\,a^3\,b\,d\,\sqrt {b^6}+9\,a^2\,b^2\,d\,\sqrt {b^6}\right )}{b^5\,\sqrt {-a\,d^2\,{\left (a-b\right )}^7}\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\,\sqrt {-a^8\,d^2+7\,a^7\,b\,d^2-21\,a^6\,b^2\,d^2+35\,a^5\,b^3\,d^2-35\,a^4\,b^4\,d^2+21\,a^3\,b^5\,d^2-7\,a^2\,b^6\,d^2+a\,b^7\,d^2}}\right )-\frac {\left (2\,a-b\right )\,\left (b^4\,d\,\sqrt {b^6}-3\,a\,b^3\,d\,\sqrt {b^6}-a^3\,b\,d\,\sqrt {b^6}+3\,a^2\,b^2\,d\,\sqrt {b^6}\right )}{b^5\,\sqrt {-a\,d^2\,{\left (a-b\right )}^7}\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\,\sqrt {-a^8\,d^2+7\,a^7\,b\,d^2-21\,a^6\,b^2\,d^2+35\,a^5\,b^3\,d^2-35\,a^4\,b^4\,d^2+21\,a^3\,b^5\,d^2-7\,a^2\,b^6\,d^2+a\,b^7\,d^2}}\right )\,\left (\frac {b^4\,\sqrt {-a^8\,d^2+7\,a^7\,b\,d^2-21\,a^6\,b^2\,d^2+35\,a^5\,b^3\,d^2-35\,a^4\,b^4\,d^2+21\,a^3\,b^5\,d^2-7\,a^2\,b^6\,d^2+a\,b^7\,d^2}}{2}+\frac {3\,a^2\,b^2\,\sqrt {-a^8\,d^2+7\,a^7\,b\,d^2-21\,a^6\,b^2\,d^2+35\,a^5\,b^3\,d^2-35\,a^4\,b^4\,d^2+21\,a^3\,b^5\,d^2-7\,a^2\,b^6\,d^2+a\,b^7\,d^2}}{2}-\frac {3\,a\,b^3\,\sqrt {-a^8\,d^2+7\,a^7\,b\,d^2-21\,a^6\,b^2\,d^2+35\,a^5\,b^3\,d^2-35\,a^4\,b^4\,d^2+21\,a^3\,b^5\,d^2-7\,a^2\,b^6\,d^2+a\,b^7\,d^2}}{2}-\frac {a^3\,b\,\sqrt {-a^8\,d^2+7\,a^7\,b\,d^2-21\,a^6\,b^2\,d^2+35\,a^5\,b^3\,d^2-35\,a^4\,b^4\,d^2+21\,a^3\,b^5\,d^2-7\,a^2\,b^6\,d^2+a\,b^7\,d^2}}{2}\right )\right )\,\sqrt {b^6}}{\sqrt {-a^8\,d^2+7\,a^7\,b\,d^2-21\,a^6\,b^2\,d^2+35\,a^5\,b^3\,d^2-35\,a^4\,b^4\,d^2+21\,a^3\,b^5\,d^2-7\,a^2\,b^6\,d^2+a\,b^7\,d^2}}-\frac {2\,b^2}{\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )\,{\left (a-b\right )}^2\,\left (a\,d-b\,d\right )}+\frac {4\,\left (a\,b-b^2\right )}{{\left (a-b\right )}^2\,\left (a\,d-b\,d\right )\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {8\,\left (4\,a-3\,b\right )}{3\,\left (a-b\right )\,\left (a\,d-b\,d\right )\,\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 )} \]
16/((a*d - b*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/(5*(a*d - b*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 + 1 0*d*x) + 1)) + (atan((exp(2*c)*exp(2*d*x)*((4*b)/(d*(a - b)^3*(b^6)^(1/2)* (3*a*b^2 - 3*a^2*b + a^3 - b^3)) + ((2*a - b)*(2*a^4*d*(b^6)^(1/2) + b^4*d *(b^6)^(1/2) - 5*a*b^3*d*(b^6)^(1/2) - 7*a^3*b*d*(b^6)^(1/2) + 9*a^2*b^2*d *(b^6)^(1/2)))/(b^5*(-a*d^2*(a - b)^7)^(1/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^ 3)*(a*b^7*d^2 - a^8*d^2 + 7*a^7*b*d^2 - 7*a^2*b^6*d^2 + 21*a^3*b^5*d^2 - 3 5*a^4*b^4*d^2 + 35*a^5*b^3*d^2 - 21*a^6*b^2*d^2)^(1/2))) - ((2*a - b)*(b^4 *d*(b^6)^(1/2) - 3*a*b^3*d*(b^6)^(1/2) - a^3*b*d*(b^6)^(1/2) + 3*a^2*b^2*d *(b^6)^(1/2)))/(b^5*(-a*d^2*(a - b)^7)^(1/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^ 3)*(a*b^7*d^2 - a^8*d^2 + 7*a^7*b*d^2 - 7*a^2*b^6*d^2 + 21*a^3*b^5*d^2 - 3 5*a^4*b^4*d^2 + 35*a^5*b^3*d^2 - 21*a^6*b^2*d^2)^(1/2)))*((b^4*(a*b^7*d^2 - a^8*d^2 + 7*a^7*b*d^2 - 7*a^2*b^6*d^2 + 21*a^3*b^5*d^2 - 35*a^4*b^4*d^2 + 35*a^5*b^3*d^2 - 21*a^6*b^2*d^2)^(1/2))/2 + (3*a^2*b^2*(a*b^7*d^2 - a^8* d^2 + 7*a^7*b*d^2 - 7*a^2*b^6*d^2 + 21*a^3*b^5*d^2 - 35*a^4*b^4*d^2 + 35*a ^5*b^3*d^2 - 21*a^6*b^2*d^2)^(1/2))/2 - (3*a*b^3*(a*b^7*d^2 - a^8*d^2 + 7* a^7*b*d^2 - 7*a^2*b^6*d^2 + 21*a^3*b^5*d^2 - 35*a^4*b^4*d^2 + 35*a^5*b^3*d ^2 - 21*a^6*b^2*d^2)^(1/2))/2 - (a^3*b*(a*b^7*d^2 - a^8*d^2 + 7*a^7*b*d^2 - 7*a^2*b^6*d^2 + 21*a^3*b^5*d^2 - 35*a^4*b^4*d^2 + 35*a^5*b^3*d^2 - 21...