Integrand size = 23, antiderivative size = 77 \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\frac {(a+b) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2} d}-\frac {(a-b) \sinh (c+d x)}{2 a b d \left (a+b \sinh ^2(c+d x)\right )} \]
1/2*(a+b)*arctan(sinh(d*x+c)*b^(1/2)/a^(1/2))/a^(3/2)/b^(3/2)/d-1/2*(a-b)* sinh(d*x+c)/a/b/d/(a+b*sinh(d*x+c)^2)
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\frac {\frac {(a+b) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2}}-\frac {(a-b) \sinh (c+d x)}{2 a b \left (a+b \sinh ^2(c+d x)\right )}}{d} \]
(((a + b)*ArcTan[(Sqrt[b]*Sinh[c + d*x])/Sqrt[a]])/(2*a^(3/2)*b^(3/2)) - ( (a - b)*Sinh[c + d*x])/(2*a*b*(a + b*Sinh[c + d*x]^2)))/d
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3669, 298, 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.
\(\displaystyle \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i c+i d x)^3}{\left (a-b \sin (i c+i d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {\sinh ^2(c+d x)+1}{\left (b \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {(a+b) \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{2 a b}-\frac {(a-b) \sinh (c+d x)}{2 a b \left (a+b \sinh ^2(c+d x)\right )}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {(a+b) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2}}-\frac {(a-b) \sinh (c+d x)}{2 a b \left (a+b \sinh ^2(c+d x)\right )}}{d}\) |
(((a + b)*ArcTan[(Sqrt[b]*Sinh[c + d*x])/Sqrt[a]])/(2*a^(3/2)*b^(3/2)) - ( (a - b)*Sinh[c + d*x])/(2*a*b*(a + b*Sinh[c + d*x]^2)))/d
3.4.32.3.1 Defintions of rubi rules used
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), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
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 = 40.73 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {\left (a -b \right ) \sinh \left (d x +c \right )}{2 a b \left (a +b \sinh \left (d x +c \right )^{2}\right )}+\frac {\left (a +b \right ) \arctan \left (\frac {b \sinh \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a b \sqrt {a b}}}{d}\) | \(69\) |
default | \(\frac {-\frac {\left (a -b \right ) \sinh \left (d x +c \right )}{2 a b \left (a +b \sinh \left (d x +c \right )^{2}\right )}+\frac {\left (a +b \right ) \arctan \left (\frac {b \sinh \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a b \sqrt {a b}}}{d}\) | \(69\) |
risch | \(-\frac {{\mathrm e}^{d x +c} \left (a -b \right ) \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d a b \left (b \,{\mathrm e}^{4 d x +4 c}+4 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+b \right )}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, d b}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, d a}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, d b}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, d a}\) | \(238\) |
1/d*(-1/2*(a-b)/a/b*sinh(d*x+c)/(a+b*sinh(d*x+c)^2)+1/2*(a+b)/a/b/(a*b)^(1 /2)*arctan(b*sinh(d*x+c)/(a*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 803 vs. \(2 (65) = 130\).
Time = 0.28 (sec) , antiderivative size = 1615, normalized size of antiderivative = 20.97 \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
[-1/4*(4*(a^2*b - a*b^2)*cosh(d*x + c)^3 + 12*(a^2*b - a*b^2)*cosh(d*x + c )*sinh(d*x + c)^2 + 4*(a^2*b - a*b^2)*sinh(d*x + c)^3 + ((a*b + b^2)*cosh( d*x + c)^4 + 4*(a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a*b + b^2)*sin h(d*x + c)^4 + 2*(2*a^2 + a*b - b^2)*cosh(d*x + c)^2 + 2*(3*(a*b + b^2)*co sh(d*x + c)^2 + 2*a^2 + a*b - b^2)*sinh(d*x + c)^2 + a*b + b^2 + 4*((a*b + b^2)*cosh(d*x + c)^3 + (2*a^2 + a*b - b^2)*cosh(d*x + c))*sinh(d*x + c))* sqrt(-a*b)*log((b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b* sinh(d*x + c)^4 - 2*(2*a + b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 - 2 *a - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 - (2*a + b)*cosh(d*x + c))* sinh(d*x + c) - 4*(cosh(d*x + c)^3 + 3*cosh(d*x + c)*sinh(d*x + c)^2 + sin h(d*x + c)^3 + (3*cosh(d*x + c)^2 - 1)*sinh(d*x + c) - cosh(d*x + c))*sqrt (-a*b) + b)/(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sin h(d*x + c)^4 + 2*(2*a - b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 + (2*a - b)*cosh(d*x + c))*sin h(d*x + c) + b)) - 4*(a^2*b - a*b^2)*cosh(d*x + c) - 4*(a^2*b - a*b^2 - 3* (a^2*b - a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c))/(a^2*b^3*d*cosh(d*x + c)^4 + 4*a^2*b^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*b^3*d*sinh(d*x + c)^4 + a^2*b^3*d + 2*(2*a^3*b^2 - a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*a^2*b^3*d*co sh(d*x + c)^2 + (2*a^3*b^2 - a^2*b^3)*d)*sinh(d*x + c)^2 + 4*(a^2*b^3*d*co sh(d*x + c)^3 + (2*a^3*b^2 - a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c)), ...
Timed out. \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int { \frac {\cosh \left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \]
-((a*e^(3*c) - b*e^(3*c))*e^(3*d*x) - (a*e^c - b*e^c)*e^(d*x))/(a*b^2*d*e^ (4*d*x + 4*c) + a*b^2*d + 2*(2*a^2*b*d*e^(2*c) - a*b^2*d*e^(2*c))*e^(2*d*x )) + 1/8*integrate(8*((a*e^(3*c) + b*e^(3*c))*e^(3*d*x) + (a*e^c + b*e^c)* e^(d*x))/(a*b^2*e^(4*d*x + 4*c) + a*b^2 + 2*(2*a^2*b*e^(2*c) - a*b^2*e^(2* c))*e^(2*d*x)), x)
\[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int { \frac {\cosh \left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^3}{{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \]