Integrand size = 21, antiderivative size = 86 \[ \int \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {(a-b)^3 \arctan (\sinh (c+d x))}{d}+\frac {b \left (3 a^2-3 a b+b^2\right ) \sinh (c+d x)}{d}+\frac {(3 a-b) b^2 \sinh ^3(c+d x)}{3 d}+\frac {b^3 \sinh ^5(c+d x)}{5 d} \]
(a-b)^3*arctan(sinh(d*x+c))/d+b*(3*a^2-3*a*b+b^2)*sinh(d*x+c)/d+1/3*(3*a-b )*b^2*sinh(d*x+c)^3/d+1/5*b^3*sinh(d*x+c)^5/d
Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16 \[ \int \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {\sinh (c+d x) \left (\frac {15 (a-b)^3 \text {arctanh}\left (\sqrt {-\sinh ^2(c+d x)}\right )}{\sqrt {-\sinh ^2(c+d x)}}+b \left (45 a^2+15 a b \left (-3+\sinh ^2(c+d x)\right )+b^2 \left (15-5 \sinh ^2(c+d x)+3 \sinh ^4(c+d x)\right )\right )\right )}{15 d} \]
(Sinh[c + d*x]*((15*(a - b)^3*ArcTanh[Sqrt[-Sinh[c + d*x]^2]])/Sqrt[-Sinh[ c + d*x]^2] + b*(45*a^2 + 15*a*b*(-3 + Sinh[c + d*x]^2) + b^2*(15 - 5*Sinh [c + d*x]^2 + 3*Sinh[c + d*x]^4))))/(15*d)
Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3669, 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.
\(\displaystyle \int \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-b \sin (i c+i d x)^2\right )^3}{\cos (i c+i d x)}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {\left (b \sinh ^2(c+d x)+a\right )^3}{\sinh ^2(c+d x)+1}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (b^3 \sinh ^4(c+d x)+(3 a-b) b^2 \sinh ^2(c+d x)+b \left (3 a^2-3 b a+b^2\right )+\frac {(a-b)^3}{\sinh ^2(c+d x)+1}\right )d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (3 a^2-3 a b+b^2\right ) \sinh (c+d x)+(a-b)^3 \arctan (\sinh (c+d x))+\frac {1}{3} b^2 (3 a-b) \sinh ^3(c+d x)+\frac {1}{5} b^3 \sinh ^5(c+d x)}{d}\) |
((a - b)^3*ArcTan[Sinh[c + d*x]] + b*(3*a^2 - 3*a*b + b^2)*Sinh[c + d*x] + ((3*a - b)*b^2*Sinh[c + d*x]^3)/3 + (b^3*Sinh[c + d*x]^5)/5)/d
3.4.8.3.1 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[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.33
\[\frac {2 a^{3} \arctan \left ({\mathrm e}^{d x +c}\right )+3 a^{2} b \left (\sinh \left (d x +c \right )-2 \arctan \left ({\mathrm e}^{d x +c}\right )\right )+3 a \,b^{2} \left (\frac {\sinh \left (d x +c \right )^{3}}{3}-\sinh \left (d x +c \right )+2 \arctan \left ({\mathrm e}^{d x +c}\right )\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{5}}{5}-\frac {\sinh \left (d x +c \right )^{3}}{3}+\sinh \left (d x +c \right )-2 \arctan \left ({\mathrm e}^{d x +c}\right )\right )}{d}\]
1/d*(2*a^3*arctan(exp(d*x+c))+3*a^2*b*(sinh(d*x+c)-2*arctan(exp(d*x+c)))+3 *a*b^2*(1/3*sinh(d*x+c)^3-sinh(d*x+c)+2*arctan(exp(d*x+c)))+b^3*(1/5*sinh( d*x+c)^5-1/3*sinh(d*x+c)^3+sinh(d*x+c)-2*arctan(exp(d*x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 1114 vs. \(2 (82) = 164\).
Time = 0.29 (sec) , antiderivative size = 1114, normalized size of antiderivative = 12.95 \[ \int \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\text {Too large to display} \]
1/480*(3*b^3*cosh(d*x + c)^10 + 30*b^3*cosh(d*x + c)*sinh(d*x + c)^9 + 3*b ^3*sinh(d*x + c)^10 + 5*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^8 + 5*(27*b^3*cos h(d*x + c)^2 + 12*a*b^2 - 7*b^3)*sinh(d*x + c)^8 + 40*(9*b^3*cosh(d*x + c) ^3 + (12*a*b^2 - 7*b^3)*cosh(d*x + c))*sinh(d*x + c)^7 + 30*(24*a^2*b - 30 *a*b^2 + 11*b^3)*cosh(d*x + c)^6 + 10*(63*b^3*cosh(d*x + c)^4 + 72*a^2*b - 90*a*b^2 + 33*b^3 + 14*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^ 6 + 4*(189*b^3*cosh(d*x + c)^5 + 70*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^3 + 4 5*(24*a^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c))*sinh(d*x + c)^5 - 30*(24*a ^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c)^4 + 10*(63*b^3*cosh(d*x + c)^6 + 3 5*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^4 - 72*a^2*b + 90*a*b^2 - 33*b^3 + 45*( 24*a^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 40*(9*b^3 *cosh(d*x + c)^7 + 7*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^5 + 15*(24*a^2*b - 3 0*a*b^2 + 11*b^3)*cosh(d*x + c)^3 - 3*(24*a^2*b - 30*a*b^2 + 11*b^3)*cosh( d*x + c))*sinh(d*x + c)^3 - 3*b^3 - 5*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^2 + 5*(27*b^3*cosh(d*x + c)^8 + 28*(12*a*b^2 - 7*b^3)*cosh(d*x + c)^6 + 90*(2 4*a^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c)^4 - 12*a*b^2 + 7*b^3 - 36*(24*a ^2*b - 30*a*b^2 + 11*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 960*((a^3 - 3 *a^2*b + 3*a*b^2 - b^3)*cosh(d*x + c)^5 + 5*(a^3 - 3*a^2*b + 3*a*b^2 - b^3 )*cosh(d*x + c)^4*sinh(d*x + c) + 10*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cosh( d*x + c)^3*sinh(d*x + c)^2 + 10*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cosh(d*...
\[ \int \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\int \left (a + b \sinh ^{2}{\left (c + d x \right )}\right )^{3} \operatorname {sech}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (82) = 164\).
Time = 0.31 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.71 \[ \int \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=-\frac {1}{480} \, b^{3} {\left (\frac {{\left (35 \, e^{\left (-2 \, d x - 2 \, c\right )} - 330 \, e^{\left (-4 \, d x - 4 \, c\right )} - 3\right )} e^{\left (5 \, d x + 5 \, c\right )}}{d} + \frac {330 \, e^{\left (-d x - c\right )} - 35 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d} - \frac {960 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d}\right )} - \frac {1}{8} \, a b^{2} {\left (\frac {{\left (15 \, e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )} e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {15 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {48 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d}\right )} + \frac {3}{2} \, a^{2} b {\left (\frac {4 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac {e^{\left (d x + c\right )}}{d} - \frac {e^{\left (-d x - c\right )}}{d}\right )} + \frac {a^{3} \arctan \left (\sinh \left (d x + c\right )\right )}{d} \]
-1/480*b^3*((35*e^(-2*d*x - 2*c) - 330*e^(-4*d*x - 4*c) - 3)*e^(5*d*x + 5* c)/d + (330*e^(-d*x - c) - 35*e^(-3*d*x - 3*c) + 3*e^(-5*d*x - 5*c))/d - 9 60*arctan(e^(-d*x - c))/d) - 1/8*a*b^2*((15*e^(-2*d*x - 2*c) - 1)*e^(3*d*x + 3*c)/d - (15*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + 48*arctan(e^(-d*x - c ))/d) + 3/2*a^2*b*(4*arctan(e^(-d*x - c))/d + e^(d*x + c)/d - e^(-d*x - c) /d) + a^3*arctan(sinh(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (82) = 164\).
Time = 0.32 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.37 \[ \int \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {3 \, b^{3} e^{\left (5 \, d x + 5 \, c\right )} + 60 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} - 35 \, b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 720 \, a^{2} b e^{\left (d x + c\right )} - 900 \, a b^{2} e^{\left (d x + c\right )} + 330 \, b^{3} e^{\left (d x + c\right )} + 960 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) - {\left (720 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 900 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 330 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 35 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{3}\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{480 \, d} \]
1/480*(3*b^3*e^(5*d*x + 5*c) + 60*a*b^2*e^(3*d*x + 3*c) - 35*b^3*e^(3*d*x + 3*c) + 720*a^2*b*e^(d*x + c) - 900*a*b^2*e^(d*x + c) + 330*b^3*e^(d*x + c) + 960*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*arctan(e^(d*x + c)) - (720*a^2*b* e^(4*d*x + 4*c) - 900*a*b^2*e^(4*d*x + 4*c) + 330*b^3*e^(4*d*x + 4*c) + 60 *a*b^2*e^(2*d*x + 2*c) - 35*b^3*e^(2*d*x + 2*c) + 3*b^3)*e^(-5*d*x - 5*c)) /d
Time = 1.81 (sec) , antiderivative size = 294, normalized size of antiderivative = 3.42 \[ \int \text {sech}(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {{\mathrm {e}}^{c+d\,x}\,\left (24\,a^2\,b-30\,a\,b^2+11\,b^3\right )}{16\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (24\,a^2\,b-30\,a\,b^2+11\,b^3\right )}{16\,d}+\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (a^3\,\sqrt {d^2}-b^3\,\sqrt {d^2}+3\,a\,b^2\,\sqrt {d^2}-3\,a^2\,b\,\sqrt {d^2}\right )}{d\,\sqrt {a^6-6\,a^5\,b+15\,a^4\,b^2-20\,a^3\,b^3+15\,a^2\,b^4-6\,a\,b^5+b^6}}\right )\,\sqrt {a^6-6\,a^5\,b+15\,a^4\,b^2-20\,a^3\,b^3+15\,a^2\,b^4-6\,a\,b^5+b^6}}{\sqrt {d^2}}-\frac {b^3\,{\mathrm {e}}^{-5\,c-5\,d\,x}}{160\,d}+\frac {b^3\,{\mathrm {e}}^{5\,c+5\,d\,x}}{160\,d}-\frac {b^2\,{\mathrm {e}}^{-3\,c-3\,d\,x}\,\left (12\,a-7\,b\right )}{96\,d}+\frac {b^2\,{\mathrm {e}}^{3\,c+3\,d\,x}\,\left (12\,a-7\,b\right )}{96\,d} \]
(exp(c + d*x)*(24*a^2*b - 30*a*b^2 + 11*b^3))/(16*d) - (exp(- c - d*x)*(24 *a^2*b - 30*a*b^2 + 11*b^3))/(16*d) + (2*atan((exp(d*x)*exp(c)*(a^3*(d^2)^ (1/2) - b^3*(d^2)^(1/2) + 3*a*b^2*(d^2)^(1/2) - 3*a^2*b*(d^2)^(1/2)))/(d*( a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)^(1/2 )))*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2) ^(1/2))/(d^2)^(1/2) - (b^3*exp(- 5*c - 5*d*x))/(160*d) + (b^3*exp(5*c + 5* d*x))/(160*d) - (b^2*exp(- 3*c - 3*d*x)*(12*a - 7*b))/(96*d) + (b^2*exp(3* c + 3*d*x)*(12*a - 7*b))/(96*d)