Integrand size = 25, antiderivative size = 259 \[ \int \frac {\cosh ^5(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=-\frac {2 a \left (a^4+b^4\right )^2 \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{b^{10} d}+\frac {2 \left (a^4+b^4\right )^2 \sqrt {\sinh (c+d x)}}{b^9 d}-\frac {a^3 \left (a^4+2 b^4\right ) \sinh (c+d x)}{b^8 d}+\frac {2 a^2 \left (a^4+2 b^4\right ) \sinh ^{\frac {3}{2}}(c+d x)}{3 b^7 d}-\frac {a \left (a^4+2 b^4\right ) \sinh ^2(c+d x)}{2 b^6 d}+\frac {2 \left (a^4+2 b^4\right ) \sinh ^{\frac {5}{2}}(c+d x)}{5 b^5 d}-\frac {a^3 \sinh ^3(c+d x)}{3 b^4 d}+\frac {2 a^2 \sinh ^{\frac {7}{2}}(c+d x)}{7 b^3 d}-\frac {a \sinh ^4(c+d x)}{4 b^2 d}+\frac {2 \sinh ^{\frac {9}{2}}(c+d x)}{9 b d} \]
-2*a*(a^4+b^4)^2*ln(a+b*sinh(d*x+c)^(1/2))/b^10/d-a^3*(a^4+2*b^4)*sinh(d*x +c)/b^8/d+2/3*a^2*(a^4+2*b^4)*sinh(d*x+c)^(3/2)/b^7/d-1/2*a*(a^4+2*b^4)*si nh(d*x+c)^2/b^6/d+2/5*(a^4+2*b^4)*sinh(d*x+c)^(5/2)/b^5/d-1/3*a^3*sinh(d*x +c)^3/b^4/d+2/7*a^2*sinh(d*x+c)^(7/2)/b^3/d-1/4*a*sinh(d*x+c)^4/b^2/d+2/9* sinh(d*x+c)^(9/2)/b/d+2*(a^4+b^4)^2*sinh(d*x+c)^(1/2)/b^9/d
Time = 0.31 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.85 \[ \int \frac {\cosh ^5(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\frac {-2520 a \left (a^4+b^4\right )^2 \log \left (a+b \sqrt {\sinh (c+d x)}\right )+2520 b \left (a^4+b^4\right )^2 \sqrt {\sinh (c+d x)}-1260 a^3 b^2 \left (a^4+2 b^4\right ) \sinh (c+d x)+840 a^2 b^3 \left (a^4+2 b^4\right ) \sinh ^{\frac {3}{2}}(c+d x)-630 a b^4 \left (a^4+2 b^4\right ) \sinh ^2(c+d x)+504 b^5 \left (a^4+2 b^4\right ) \sinh ^{\frac {5}{2}}(c+d x)-420 a^3 b^6 \sinh ^3(c+d x)+360 a^2 b^7 \sinh ^{\frac {7}{2}}(c+d x)-315 a b^8 \sinh ^4(c+d x)+280 b^9 \sinh ^{\frac {9}{2}}(c+d x)}{1260 b^{10} d} \]
(-2520*a*(a^4 + b^4)^2*Log[a + b*Sqrt[Sinh[c + d*x]]] + 2520*b*(a^4 + b^4) ^2*Sqrt[Sinh[c + d*x]] - 1260*a^3*b^2*(a^4 + 2*b^4)*Sinh[c + d*x] + 840*a^ 2*b^3*(a^4 + 2*b^4)*Sinh[c + d*x]^(3/2) - 630*a*b^4*(a^4 + 2*b^4)*Sinh[c + d*x]^2 + 504*b^5*(a^4 + 2*b^4)*Sinh[c + d*x]^(5/2) - 420*a^3*b^6*Sinh[c + d*x]^3 + 360*a^2*b^7*Sinh[c + d*x]^(7/2) - 315*a*b^8*Sinh[c + d*x]^4 + 28 0*b^9*Sinh[c + d*x]^(9/2))/(1260*b^10*d)
Time = 0.53 (sec) , antiderivative size = 235, 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.
| \(\displaystyle \int \frac {\cosh ^5(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (i c+i d x)^5}{a+b \sqrt {-i \sin (i c+i d x)}}dx\) |
| \(\Big \downarrow \) 3702 |
| \(\displaystyle \frac {\int \frac {\left (\sinh ^2(c+d x)+1\right )^2}{a+b \sqrt {\sinh (c+d x)}}d\sinh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2429 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {\sinh (c+d x)} \left (\sinh ^2(c+d x)+1\right )^2}{a+b \sqrt {\sinh (c+d x)}}d\sqrt {\sinh (c+d x)}}{d}\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \frac {2 \int \left (\frac {\sinh ^4(c+d x)}{b}-\frac {a \sinh ^{\frac {7}{2}}(c+d x)}{b^2}+\frac {a^2 \sinh ^3(c+d x)}{b^3}-\frac {a^3 \sinh ^{\frac {5}{2}}(c+d x)}{b^4}+\frac {\left (a^4+2 b^4\right ) \sinh ^2(c+d x)}{b^5}-\frac {a \left (a^4+2 b^4\right ) \sinh ^{\frac {3}{2}}(c+d x)}{b^6}+\frac {a^2 \left (a^4+2 b^4\right ) \sinh (c+d x)}{b^7}-\frac {a^3 \left (a^4+2 b^4\right ) \sqrt {\sinh (c+d x)}}{b^8}+\frac {\left (a^4+b^4\right )^2}{b^9}-\frac {a \left (a^4+b^4\right )^2}{b^9 \left (a+b \sqrt {\sinh (c+d x)}\right )}\right )d\sqrt {\sinh (c+d x)}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (-\frac {a \left (a^4+b^4\right )^2 \log \left (a+b \sqrt {\sinh (c+d x)}\right )}{b^{10}}+\frac {\left (a^4+b^4\right )^2 \sqrt {\sinh (c+d x)}}{b^9}-\frac {a \left (a^4+2 b^4\right ) \sinh ^2(c+d x)}{4 b^6}+\frac {\left (a^4+2 b^4\right ) \sinh ^{\frac {5}{2}}(c+d x)}{5 b^5}-\frac {a^3 \sinh ^3(c+d x)}{6 b^4}+\frac {a^2 \sinh ^{\frac {7}{2}}(c+d x)}{7 b^3}-\frac {a^3 \left (a^4+2 b^4\right ) \sinh (c+d x)}{2 b^8}+\frac {a^2 \left (a^4+2 b^4\right ) \sinh ^{\frac {3}{2}}(c+d x)}{3 b^7}-\frac {a \sinh ^4(c+d x)}{8 b^2}+\frac {\sinh ^{\frac {9}{2}}(c+d x)}{9 b}\right )}{d}\) |
(2*(-((a*(a^4 + b^4)^2*Log[a + b*Sqrt[Sinh[c + d*x]]])/b^10) + ((a^4 + b^4 )^2*Sqrt[Sinh[c + d*x]])/b^9 - (a^3*(a^4 + 2*b^4)*Sinh[c + d*x])/(2*b^8) + (a^2*(a^4 + 2*b^4)*Sinh[c + d*x]^(3/2))/(3*b^7) - (a*(a^4 + 2*b^4)*Sinh[c + d*x]^2)/(4*b^6) + ((a^4 + 2*b^4)*Sinh[c + d*x]^(5/2))/(5*b^5) - (a^3*Si nh[c + d*x]^3)/(6*b^4) + (a^2*Sinh[c + d*x]^(7/2))/(7*b^3) - (a*Sinh[c + d *x]^4)/(8*b^2) + Sinh[c + d*x]^(9/2)/(9*b)))/d
3.5.11.3.1 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 = 0.22 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.01
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {2 \sinh \left (d x +c \right )^{\frac {9}{2}} b^{8}}{9}-\frac {a \sinh \left (d x +c \right )^{4} b^{7}}{4}+\frac {2 a^{2} \sinh \left (d x +c \right )^{\frac {7}{2}} b^{6}}{7}-\frac {a^{3} \sinh \left (d x +c \right )^{3} b^{5}}{3}+\frac {2 a^{4} b^{4} \sinh \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {4 b^{8} \sinh \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {a^{5} b^{3} \sinh \left (d x +c \right )^{2}}{2}-a \,b^{7} \sinh \left (d x +c \right )^{2}+\frac {2 a^{6} b^{2} \sinh \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {4 a^{2} b^{6} \sinh \left (d x +c \right )^{\frac {3}{2}}}{3}-a^{7} b \sinh \left (d x +c \right )-2 a^{3} b^{5} \sinh \left (d x +c \right )+2 a^{8} \sqrt {\sinh \left (d x +c \right )}+4 a^{4} b^{4} \sqrt {\sinh \left (d x +c \right )}+2 b^{8} \sqrt {\sinh \left (d x +c \right )}}{b^{9}}-\frac {2 a \left (a^{8}+2 a^{4} b^{4}+b^{8}\right ) \ln \left (a +b \sqrt {\sinh \left (d x +c \right )}\right )}{b^{10}}}{d}\) | \(262\) |
| default | \(\frac {\frac {\frac {2 \sinh \left (d x +c \right )^{\frac {9}{2}} b^{8}}{9}-\frac {a \sinh \left (d x +c \right )^{4} b^{7}}{4}+\frac {2 a^{2} \sinh \left (d x +c \right )^{\frac {7}{2}} b^{6}}{7}-\frac {a^{3} \sinh \left (d x +c \right )^{3} b^{5}}{3}+\frac {2 a^{4} b^{4} \sinh \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {4 b^{8} \sinh \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {a^{5} b^{3} \sinh \left (d x +c \right )^{2}}{2}-a \,b^{7} \sinh \left (d x +c \right )^{2}+\frac {2 a^{6} b^{2} \sinh \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {4 a^{2} b^{6} \sinh \left (d x +c \right )^{\frac {3}{2}}}{3}-a^{7} b \sinh \left (d x +c \right )-2 a^{3} b^{5} \sinh \left (d x +c \right )+2 a^{8} \sqrt {\sinh \left (d x +c \right )}+4 a^{4} b^{4} \sqrt {\sinh \left (d x +c \right )}+2 b^{8} \sqrt {\sinh \left (d x +c \right )}}{b^{9}}-\frac {2 a \left (a^{8}+2 a^{4} b^{4}+b^{8}\right ) \ln \left (a +b \sqrt {\sinh \left (d x +c \right )}\right )}{b^{10}}}{d}\) | \(262\) |
1/d*(2/b^9*(1/9*sinh(d*x+c)^(9/2)*b^8-1/8*a*sinh(d*x+c)^4*b^7+1/7*a^2*sinh (d*x+c)^(7/2)*b^6-1/6*a^3*sinh(d*x+c)^3*b^5+1/5*a^4*b^4*sinh(d*x+c)^(5/2)+ 2/5*b^8*sinh(d*x+c)^(5/2)-1/4*a^5*b^3*sinh(d*x+c)^2-1/2*a*b^7*sinh(d*x+c)^ 2+1/3*a^6*b^2*sinh(d*x+c)^(3/2)+2/3*a^2*b^6*sinh(d*x+c)^(3/2)-1/2*a^7*b*si nh(d*x+c)-a^3*b^5*sinh(d*x+c)+a^8*sinh(d*x+c)^(1/2)+2*a^4*b^4*sinh(d*x+c)^ (1/2)+b^8*sinh(d*x+c)^(1/2))-2*a*(a^8+2*a^4*b^4+b^8)/b^10*ln(a+b*sinh(d*x+ c)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 2595 vs. \(2 (233) = 466\).
Time = 0.86 (sec) , antiderivative size = 2595, normalized size of antiderivative = 10.02 \[ \int \frac {\cosh ^5(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\text {Too large to display} \]
-1/20160*(315*a*b^8*cosh(d*x + c)^8 + 315*a*b^8*sinh(d*x + c)^8 + 840*a^3* b^6*cosh(d*x + c)^7 - 840*a^3*b^6*cosh(d*x + c) + 315*a*b^8 + 840*(3*a*b^8 *cosh(d*x + c) + a^3*b^6)*sinh(d*x + c)^7 + 1260*(2*a^5*b^4 + 3*a*b^8)*cos h(d*x + c)^6 + 420*(21*a*b^8*cosh(d*x + c)^2 + 14*a^3*b^6*cosh(d*x + c) + 6*a^5*b^4 + 9*a*b^8)*sinh(d*x + c)^6 + 2520*(4*a^7*b^2 + 7*a^3*b^6)*cosh(d *x + c)^5 + 2520*(7*a*b^8*cosh(d*x + c)^3 + 7*a^3*b^6*cosh(d*x + c)^2 + 4* a^7*b^2 + 7*a^3*b^6 + 3*(2*a^5*b^4 + 3*a*b^8)*cosh(d*x + c))*sinh(d*x + c) ^5 - 20160*((a^9 + 2*a^5*b^4 + a*b^8)*d*x + (a^9 + 2*a^5*b^4 + a*b^8)*c)*c osh(d*x + c)^4 + 210*(105*a*b^8*cosh(d*x + c)^4 + 140*a^3*b^6*cosh(d*x + c )^3 - 96*(a^9 + 2*a^5*b^4 + a*b^8)*d*x + 90*(2*a^5*b^4 + 3*a*b^8)*cosh(d*x + c)^2 - 96*(a^9 + 2*a^5*b^4 + a*b^8)*c + 60*(4*a^7*b^2 + 7*a^3*b^6)*cosh (d*x + c))*sinh(d*x + c)^4 - 2520*(4*a^7*b^2 + 7*a^3*b^6)*cosh(d*x + c)^3 + 840*(21*a*b^8*cosh(d*x + c)^5 + 35*a^3*b^6*cosh(d*x + c)^4 - 12*a^7*b^2 - 21*a^3*b^6 + 30*(2*a^5*b^4 + 3*a*b^8)*cosh(d*x + c)^3 + 30*(4*a^7*b^2 + 7*a^3*b^6)*cosh(d*x + c)^2 - 96*((a^9 + 2*a^5*b^4 + a*b^8)*d*x + (a^9 + 2* a^5*b^4 + a*b^8)*c)*cosh(d*x + c))*sinh(d*x + c)^3 + 1260*(2*a^5*b^4 + 3*a *b^8)*cosh(d*x + c)^2 + 1260*(7*a*b^8*cosh(d*x + c)^6 + 14*a^3*b^6*cosh(d* x + c)^5 + 2*a^5*b^4 + 3*a*b^8 + 15*(2*a^5*b^4 + 3*a*b^8)*cosh(d*x + c)^4 + 20*(4*a^7*b^2 + 7*a^3*b^6)*cosh(d*x + c)^3 - 96*((a^9 + 2*a^5*b^4 + a*b^ 8)*d*x + (a^9 + 2*a^5*b^4 + a*b^8)*c)*cosh(d*x + c)^2 - 6*(4*a^7*b^2 + ...
Timed out. \[ \int \frac {\cosh ^5(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cosh ^5(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\int { \frac {\cosh \left (d x + c\right )^{5}}{b \sqrt {\sinh \left (d x + c\right )} + a} \,d x } \]
\[ \int \frac {\cosh ^5(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\int { \frac {\cosh \left (d x + c\right )^{5}}{b \sqrt {\sinh \left (d x + c\right )} + a} \,d x } \]
Timed out. \[ \int \frac {\cosh ^5(c+d x)}{a+b \sqrt {\sinh (c+d x)}} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^5}{a+b\,\sqrt {\mathrm {sinh}\left (c+d\,x\right )}} \,d x \]