Integrand size = 23, antiderivative size = 154 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\frac {2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{5/2}}{5 b^5 d}-\frac {8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}}{9 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{11/2}}{11 b^5 d}+\frac {2 (a+b \sin (c+d x))^{13/2}}{13 b^5 d} \]
2/5*(a^2-b^2)^2*(a+b*sin(d*x+c))^(5/2)/b^5/d-8/7*a*(a^2-b^2)*(a+b*sin(d*x+ c))^(7/2)/b^5/d+4/9*(3*a^2-b^2)*(a+b*sin(d*x+c))^(9/2)/b^5/d-8/11*a*(a+b*s in(d*x+c))^(11/2)/b^5/d+2/13*(a+b*sin(d*x+c))^(13/2)/b^5/d
Time = 0.46 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.85 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\frac {2 \left (\frac {1}{5} \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{5/2}-\frac {4}{7} a (a-b) (a+b) (a+b \sin (c+d x))^{7/2}+\frac {2}{9} \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}-\frac {4}{11} a (a+b \sin (c+d x))^{11/2}+\frac {1}{13} (a+b \sin (c+d x))^{13/2}\right )}{b^5 d} \]
(2*(((a^2 - b^2)^2*(a + b*Sin[c + d*x])^(5/2))/5 - (4*a*(a - b)*(a + b)*(a + b*Sin[c + d*x])^(7/2))/7 + (2*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(9/2)) /9 - (4*a*(a + b*Sin[c + d*x])^(11/2))/11 + (a + b*Sin[c + d*x])^(13/2)/13 ))/(b^5*d)
Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3147, 476, 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.
\(\displaystyle \int \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^5 (a+b \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int (a+b \sin (c+d x))^{3/2} \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left ((a+b \sin (c+d x))^{11/2}-4 a (a+b \sin (c+d x))^{9/2}+2 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}-4 \left (a^3-a b^2\right ) (a+b \sin (c+d x))^{5/2}+\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}\right )d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {4}{9} \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}-\frac {8}{7} a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}+\frac {2}{5} \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{5/2}+\frac {2}{13} (a+b \sin (c+d x))^{13/2}-\frac {8}{11} a (a+b \sin (c+d x))^{11/2}}{b^5 d}\) |
((2*(a^2 - b^2)^2*(a + b*Sin[c + d*x])^(5/2))/5 - (8*a*(a^2 - b^2)*(a + b* Sin[c + d*x])^(7/2))/7 + (4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(9/2))/9 - (8*a*(a + b*Sin[c + d*x])^(11/2))/11 + (2*(a + b*Sin[c + d*x])^(13/2))/13) /(b^5*d)
3.5.83.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.60 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (a +b \right )^{2}+\left (-2 a -2 b \right ) \left (-2 a +2 b \right )+\left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a +b \right )^{2} \left (-2 a +2 b \right )+\left (-2 a -2 b \right ) \left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}}{d \,b^{5}}\) | \(150\) |
default | \(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (a +b \right )^{2}+\left (-2 a -2 b \right ) \left (-2 a +2 b \right )+\left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a +b \right )^{2} \left (-2 a +2 b \right )+\left (-2 a -2 b \right ) \left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}}{d \,b^{5}}\) | \(150\) |
2/d/b^5*(1/13*(a+b*sin(d*x+c))^(13/2)-4/11*a*(a+b*sin(d*x+c))^(11/2)+1/9*( (a+b)^2+(-2*a-2*b)*(-2*a+2*b)+(a-b)^2)*(a+b*sin(d*x+c))^(9/2)+1/7*((a+b)^2 *(-2*a+2*b)+(-2*a-2*b)*(a-b)^2)*(a+b*sin(d*x+c))^(7/2)+1/5*(a+b)^2*(a-b)^2 *(a+b*sin(d*x+c))^(5/2))
Time = 0.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.19 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (3465 \, b^{6} \cos \left (d x + c\right )^{6} - 384 \, a^{6} + 2144 \, a^{4} b^{2} - 8256 \, a^{2} b^{4} - 2464 \, b^{6} - 35 \, {\left (3 \, a^{2} b^{4} + 11 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (18 \, a^{4} b^{2} - 81 \, a^{2} b^{4} - 77 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2205 \, a b^{5} \cos \left (d x + c\right )^{4} - 96 \, a^{5} b + 512 \, a^{3} b^{3} + 4064 \, a b^{5} + 20 \, {\left (3 \, a^{3} b^{3} + 137 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{45045 \, b^{5} d} \]
-2/45045*(3465*b^6*cos(d*x + c)^6 - 384*a^6 + 2144*a^4*b^2 - 8256*a^2*b^4 - 2464*b^6 - 35*(3*a^2*b^4 + 11*b^6)*cos(d*x + c)^4 + 8*(18*a^4*b^2 - 81*a ^2*b^4 - 77*b^6)*cos(d*x + c)^2 - 2*(2205*a*b^5*cos(d*x + c)^4 - 96*a^5*b + 512*a^3*b^3 + 4064*a*b^5 + 20*(3*a^3*b^3 + 137*a*b^5)*cos(d*x + c)^2)*si n(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^5*d)
Timed out. \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.75 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (3465 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 16380 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 10010 \, {\left (3 \, a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 25740 \, {\left (a^{3} - a b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} + 9009 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\right )}}{45045 \, b^{5} d} \]
2/45045*(3465*(b*sin(d*x + c) + a)^(13/2) - 16380*(b*sin(d*x + c) + a)^(11 /2)*a + 10010*(3*a^2 - b^2)*(b*sin(d*x + c) + a)^(9/2) - 25740*(a^3 - a*b^ 2)*(b*sin(d*x + c) + a)^(7/2) + 9009*(a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x + c) + a)^(5/2))/(b^5*d)
\[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^5\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]