Integrand size = 25, antiderivative size = 136 \[ \int \frac {\cosh ^3(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=-\frac {2 a \left (a^4+b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{b^6 d}+\frac {2 \left (a^4+b^4\right ) \sqrt {\sinh (c+d x)}}{b^5 d}-\frac {a^3 \sinh (c+d x)}{b^4 d}+\frac {2 a^2 \sinh ^{\frac {3}{2}}(c+d x)}{3 b^3 d}-\frac {a \sinh ^2(c+d x)}{2 b^2 d}+\frac {2 \sinh ^{\frac {5}{2}}(c+d x)}{5 b d} \]
-2*a*(a^4+b^4)*ln(a+b*sinh(d*x+c)^(1/2))/b^6/d-a^3*sinh(d*x+c)/b^4/d+2/3*a ^2*sinh(d*x+c)^(3/2)/b^3/d-1/2*a*sinh(d*x+c)^2/b^2/d+2/5*sinh(d*x+c)^(5/2) /b/d+2*(a^4+b^4)*sinh(d*x+c)^(1/2)/b^5/d
Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.86 \[ \int \frac {\cosh ^3(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\frac {-60 a \left (a^4+b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )+60 b \left (a^4+b^4\right ) \sqrt {\sinh (c+d x)}-30 a^3 b^2 \sinh (c+d x)+20 a^2 b^3 \sinh ^{\frac {3}{2}}(c+d x)-15 a b^4 \sinh ^2(c+d x)+12 b^5 \sinh ^{\frac {5}{2}}(c+d x)}{30 b^6 d} \]
(-60*a*(a^4 + b^4)*Log[a + b*Sqrt[Sinh[c + d*x]]] + 60*b*(a^4 + b^4)*Sqrt[ Sinh[c + d*x]] - 30*a^3*b^2*Sinh[c + d*x] + 20*a^2*b^3*Sinh[c + d*x]^(3/2) - 15*a*b^4*Sinh[c + d*x]^2 + 12*b^5*Sinh[c + d*x]^(5/2))/(30*b^6*d)
Time = 0.41 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3702, 2429, 2123, 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.
\(\displaystyle \int \frac {\cosh ^3(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i c+i d x)^3}{a+b \sqrt {-i \sin (i c+i d x)}}dx\) |
\(\Big \downarrow \) 3702 |
\(\displaystyle \frac {\int \frac {\sinh ^2(c+d x)+1}{a+b \sqrt {\sinh (c+d x)}}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 2429 |
\(\displaystyle \frac {2 \int \frac {\sqrt {\sinh (c+d x)} \left (\sinh ^2(c+d x)+1\right )}{a+b \sqrt {\sinh (c+d x)}}d\sqrt {\sinh (c+d x)}}{d}\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \frac {2 \int \left (-\frac {\sqrt {\sinh (c+d x)} a^3}{b^4}+\frac {\sinh (c+d x) a^2}{b^3}-\frac {\sinh ^{\frac {3}{2}}(c+d x) a}{b^2}-\frac {\left (a^4+b^4\right ) a}{b^5 \left (a+b \sqrt {\sinh (c+d x)}\right )}+\frac {\sinh ^2(c+d x)}{b}+\frac {a^4+b^4}{b^5}\right )d\sqrt {\sinh (c+d x)}}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-\frac {a \left (a^4+b^4\right ) \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{b^6}+\frac {\left (a^4+b^4\right ) \sqrt {\sinh (c+d x)}}{b^5}-\frac {a^3 \sinh (c+d x)}{2 b^4}+\frac {a^2 \sinh ^{\frac {3}{2}}(c+d x)}{3 b^3}-\frac {a \sinh ^2(c+d x)}{4 b^2}+\frac {\sinh ^{\frac {5}{2}}(c+d x)}{5 b}\right )}{d}\) |
(2*(-((a*(a^4 + b^4)*Log[a + b*Sqrt[Sinh[c + d*x]]])/b^6) + ((a^4 + b^4)*S qrt[Sinh[c + d*x]])/b^5 - (a^3*Sinh[c + d*x])/(2*b^4) + (a^2*Sinh[c + d*x] ^(3/2))/(3*b^3) - (a*Sinh[c + d*x]^2)/(4*b^2) + Sinh[c + d*x]^(5/2)/(5*b)) )/d
3.5.12.3.1 Defintions of rubi rules used
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{g = Denominator [n]}, Simp[g Subst[Int[x^(g - 1)*(Pq /. x -> x^g)*(a + b*x^(g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && FractionQ[n]
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])
Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {\frac {\frac {2 \sinh \left (d x +c \right )^{\frac {5}{2}} b^{4}}{5}-\frac {a \sinh \left (d x +c \right )^{2} b^{3}}{2}+\frac {2 a^{2} \sinh \left (d x +c \right )^{\frac {3}{2}} b^{2}}{3}-a^{3} \sinh \left (d x +c \right ) b +2 a^{4} \sqrt {\sinh \left (d x +c \right )}+2 b^{4} \sqrt {\sinh \left (d x +c \right )}}{b^{5}}-\frac {2 a \left (a^{4}+b^{4}\right ) \ln \left (a +b \sqrt {\sinh \left (d x +c \right )}\right )}{b^{6}}}{d}\) | \(117\) |
default | \(\frac {\frac {\frac {2 \sinh \left (d x +c \right )^{\frac {5}{2}} b^{4}}{5}-\frac {a \sinh \left (d x +c \right )^{2} b^{3}}{2}+\frac {2 a^{2} \sinh \left (d x +c \right )^{\frac {3}{2}} b^{2}}{3}-a^{3} \sinh \left (d x +c \right ) b +2 a^{4} \sqrt {\sinh \left (d x +c \right )}+2 b^{4} \sqrt {\sinh \left (d x +c \right )}}{b^{5}}-\frac {2 a \left (a^{4}+b^{4}\right ) \ln \left (a +b \sqrt {\sinh \left (d x +c \right )}\right )}{b^{6}}}{d}\) | \(117\) |
1/d*(2/b^5*(1/5*sinh(d*x+c)^(5/2)*b^4-1/4*a*sinh(d*x+c)^2*b^3+1/3*a^2*sinh (d*x+c)^(3/2)*b^2-1/2*a^3*sinh(d*x+c)*b+a^4*sinh(d*x+c)^(1/2)+b^4*sinh(d*x +c)^(1/2))-2*a*(a^4+b^4)/b^6*ln(a+b*sinh(d*x+c)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 879 vs. \(2 (122) = 244\).
Time = 0.77 (sec) , antiderivative size = 879, normalized size of antiderivative = 6.46 \[ \int \frac {\cosh ^3(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\text {Too large to display} \]
-1/120*(15*a*b^4*cosh(d*x + c)^4 + 15*a*b^4*sinh(d*x + c)^4 + 60*a^3*b^2*c osh(d*x + c)^3 - 60*a^3*b^2*cosh(d*x + c) + 15*a*b^4 + 60*(a*b^4*cosh(d*x + c) + a^3*b^2)*sinh(d*x + c)^3 - 120*((a^5 + a*b^4)*d*x + (a^5 + a*b^4)*c )*cosh(d*x + c)^2 + 30*(3*a*b^4*cosh(d*x + c)^2 + 6*a^3*b^2*cosh(d*x + c) - 4*(a^5 + a*b^4)*d*x - 4*(a^5 + a*b^4)*c)*sinh(d*x + c)^2 - 120*((a^5 + a *b^4)*cosh(d*x + c)^2 + 2*(a^5 + a*b^4)*cosh(d*x + c)*sinh(d*x + c) + (a^5 + a*b^4)*sinh(d*x + c)^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a^2*cosh(d*x + c) - b^2 + 2*(b^2*cosh(d*x + c) + a^2)*sinh(d*x + c) - 4*(a*b*cosh(d*x + c) + a*b*sinh(d*x + c))*sqrt(sinh(d*x + c)))/(b^2*cosh(d *x + c)^2 + b^2*sinh(d*x + c)^2 - 2*a^2*cosh(d*x + c) - b^2 + 2*(b^2*cosh( d*x + c) - a^2)*sinh(d*x + c))) + 120*((a^5 + a*b^4)*cosh(d*x + c)^2 + 2*( a^5 + a*b^4)*cosh(d*x + c)*sinh(d*x + c) + (a^5 + a*b^4)*sinh(d*x + c)^2)* log(2*(b^2*sinh(d*x + c) - a^2)/(cosh(d*x + c) - sinh(d*x + c))) + 60*(a*b ^4*cosh(d*x + c)^3 + 3*a^3*b^2*cosh(d*x + c)^2 - a^3*b^2 - 4*((a^5 + a*b^4 )*d*x + (a^5 + a*b^4)*c)*cosh(d*x + c))*sinh(d*x + c) - 4*(3*b^5*cosh(d*x + c)^4 + 3*b^5*sinh(d*x + c)^4 + 10*a^2*b^3*cosh(d*x + c)^3 - 10*a^2*b^3*c osh(d*x + c) + 3*b^5 + 2*(6*b^5*cosh(d*x + c) + 5*a^2*b^3)*sinh(d*x + c)^3 + 6*(10*a^4*b + 9*b^5)*cosh(d*x + c)^2 + 6*(3*b^5*cosh(d*x + c)^2 + 5*a^2 *b^3*cosh(d*x + c) + 10*a^4*b + 9*b^5)*sinh(d*x + c)^2 + 2*(6*b^5*cosh(d*x + c)^3 + 15*a^2*b^3*cosh(d*x + c)^2 - 5*a^2*b^3 + 6*(10*a^4*b + 9*b^5)...
Timed out. \[ \int \frac {\cosh ^3(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cosh ^3(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\int { \frac {\cosh \left (d x + c\right )^{3}}{b \sqrt {\sinh \left (d x + c\right )} + a} \,d x } \]
\[ \int \frac {\cosh ^3(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\int { \frac {\cosh \left (d x + c\right )^{3}}{b \sqrt {\sinh \left (d x + c\right )} + a} \,d x } \]
Timed out. \[ \int \frac {\cosh ^3(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^3}{a+b\,\sqrt {\mathrm {sinh}\left (c+d\,x\right )}} \,d x \]