Integrand size = 23, antiderivative size = 154 \[ \int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}{3 b^5 d}-\frac {8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}{5 b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{9/2}}{9 b^5 d}+\frac {2 (a+b \sin (c+d x))^{11/2}}{11 b^5 d} \]
2/3*(a^2-b^2)^2*(a+b*sin(d*x+c))^(3/2)/b^5/d-8/5*a*(a^2-b^2)*(a+b*sin(d*x+ c))^(5/2)/b^5/d+4/7*(3*a^2-b^2)*(a+b*sin(d*x+c))^(7/2)/b^5/d-8/9*a*(a+b*si n(d*x+c))^(9/2)/b^5/d+2/11*(a+b*sin(d*x+c))^(11/2)/b^5/d
Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.76 \[ \int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 (a+b \sin (c+d x))^{3/2} \left (315 b^4 \cos ^4(c+d x)+8 \left (16 a^4-66 a^2 b^2+105 b^4+\left (-24 a^3 b+99 a b^3\right ) \sin (c+d x)+15 b^2 \left (2 a^2-3 b^2\right ) \sin ^2(c+d x)-35 a b^3 \sin ^3(c+d x)\right )\right )}{3465 b^5 d} \]
(2*(a + b*Sin[c + d*x])^(3/2)*(315*b^4*Cos[c + d*x]^4 + 8*(16*a^4 - 66*a^2 *b^2 + 105*b^4 + (-24*a^3*b + 99*a*b^3)*Sin[c + d*x] + 15*b^2*(2*a^2 - 3*b ^2)*Sin[c + d*x]^2 - 35*a*b^3*Sin[c + d*x]^3)))/(3465*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.
\(\displaystyle \int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^5 \sqrt {a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int \sqrt {a+b \sin (c+d x)} \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left ((a+b \sin (c+d x))^{9/2}-4 a (a+b \sin (c+d x))^{7/2}+2 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}-4 \left (a^3-a b^2\right ) (a+b \sin (c+d x))^{3/2}+\left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}\right )d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {4}{7} \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}-\frac {8}{5} a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}+\frac {2}{3} \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}+\frac {2}{11} (a+b \sin (c+d x))^{11/2}-\frac {8}{9} a (a+b \sin (c+d x))^{9/2}}{b^5 d}\) |
((2*(a^2 - b^2)^2*(a + b*Sin[c + d*x])^(3/2))/3 - (8*a*(a^2 - b^2)*(a + b* Sin[c + d*x])^(5/2))/5 + (4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(7/2))/7 - (8*a*(a + b*Sin[c + d*x])^(9/2))/9 + (2*(a + b*Sin[c + d*x])^(11/2))/11)/( b^5*d)
3.5.73.3.1 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 = 0.67 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a +b \right )^{2}+\left (-2 a -2 b \right ) \left (-2 a +2 b \right )+\left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a +b \right )^{2} \left (-2 a +2 b \right )+\left (-2 a -2 b \right ) \left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}}{d \,b^{5}}\) | \(150\) |
default | \(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a +b \right )^{2}+\left (-2 a -2 b \right ) \left (-2 a +2 b \right )+\left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a +b \right )^{2} \left (-2 a +2 b \right )+\left (-2 a -2 b \right ) \left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}}{d \,b^{5}}\) | \(150\) |
2/d/b^5*(1/11*(a+b*sin(d*x+c))^(11/2)-4/9*a*(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*( -2*a+2*b)+(-2*a-2*b)*(a-b)^2)*(a+b*sin(d*x+c))^(5/2)+1/3*(a+b)^2*(a-b)^2*( a+b*sin(d*x+c))^(3/2))
Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.92 \[ \int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (35 \, a b^{4} \cos \left (d x + c\right )^{4} + 128 \, a^{5} - 480 \, a^{3} b^{2} + 992 \, a b^{4} - 16 \, {\left (3 \, a^{3} b^{2} - 8 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (315 \, b^{5} \cos \left (d x + c\right )^{4} - 64 \, a^{4} b + 224 \, a^{2} b^{3} + 480 \, b^{5} + 40 \, {\left (a^{2} b^{3} + 9 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{3465 \, b^{5} d} \]
2/3465*(35*a*b^4*cos(d*x + c)^4 + 128*a^5 - 480*a^3*b^2 + 992*a*b^4 - 16*( 3*a^3*b^2 - 8*a*b^4)*cos(d*x + c)^2 + (315*b^5*cos(d*x + c)^4 - 64*a^4*b + 224*a^2*b^3 + 480*b^5 + 40*(a^2*b^3 + 9*b^5)*cos(d*x + c)^2)*sin(d*x + c) )*sqrt(b*sin(d*x + c) + a)/(b^5*d)
Timed out. \[ \int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.75 \[ \int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (315 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 1540 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 990 \, {\left (3 \, a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 2772 \, {\left (a^{3} - a b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} + 1155 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}\right )}}{3465 \, b^{5} d} \]
2/3465*(315*(b*sin(d*x + c) + a)^(11/2) - 1540*(b*sin(d*x + c) + a)^(9/2)* a + 990*(3*a^2 - b^2)*(b*sin(d*x + c) + a)^(7/2) - 2772*(a^3 - a*b^2)*(b*s in(d*x + c) + a)^(5/2) + 1155*(a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x + c) + a) ^(3/2))/(b^5*d)
Time = 0.31 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (315 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 1540 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 2970 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 2772 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} - 990 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} b^{2} + 2772 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a b^{2} - 2310 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} b^{2} + 1155 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b^{4}\right )}}{3465 \, b^{5} d} \]
2/3465*(315*(b*sin(d*x + c) + a)^(11/2) - 1540*(b*sin(d*x + c) + a)^(9/2)* a + 2970*(b*sin(d*x + c) + a)^(7/2)*a^2 - 2772*(b*sin(d*x + c) + a)^(5/2)* a^3 + 1155*(b*sin(d*x + c) + a)^(3/2)*a^4 - 990*(b*sin(d*x + c) + a)^(7/2) *b^2 + 2772*(b*sin(d*x + c) + a)^(5/2)*a*b^2 - 2310*(b*sin(d*x + c) + a)^( 3/2)*a^2*b^2 + 1155*(b*sin(d*x + c) + a)^(3/2)*b^4)/(b^5*d)
Timed out. \[ \int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^5\,\sqrt {a+b\,\sin \left (c+d\,x\right )} \,d x \]