Integrand size = 23, antiderivative size = 97 \[ \int \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {16 (a+a \sin (c+d x))^{9/2}}{9 a^4 d}-\frac {24 (a+a \sin (c+d x))^{11/2}}{11 a^5 d}+\frac {12 (a+a \sin (c+d x))^{13/2}}{13 a^6 d}-\frac {2 (a+a \sin (c+d x))^{15/2}}{15 a^7 d} \]
16/9*(a+a*sin(d*x+c))^(9/2)/a^4/d-24/11*(a+a*sin(d*x+c))^(11/2)/a^5/d+12/1 3*(a+a*sin(d*x+c))^(13/2)/a^6/d-2/15*(a+a*sin(d*x+c))^(15/2)/a^7/d
Time = 0.33 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.63 \[ \int \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 (1+\sin (c+d x))^4 \sqrt {a (1+\sin (c+d x))} \left (-1241+2367 \sin (c+d x)-1683 \sin ^2(c+d x)+429 \sin ^3(c+d x)\right )}{6435 d} \]
(-2*(1 + Sin[c + d*x])^4*Sqrt[a*(1 + Sin[c + d*x])]*(-1241 + 2367*Sin[c + d*x] - 1683*Sin[c + d*x]^2 + 429*Sin[c + d*x]^3))/(6435*d)
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3146, 53, 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.
\(\displaystyle \int \cos ^7(c+d x) \sqrt {a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^7 \sqrt {a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^{7/2}d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left (-(\sin (c+d x) a+a)^{13/2}+6 a (\sin (c+d x) a+a)^{11/2}-12 a^2 (\sin (c+d x) a+a)^{9/2}+8 a^3 (\sin (c+d x) a+a)^{7/2}\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {16}{9} a^3 (a \sin (c+d x)+a)^{9/2}-\frac {24}{11} a^2 (a \sin (c+d x)+a)^{11/2}-\frac {2}{15} (a \sin (c+d x)+a)^{15/2}+\frac {12}{13} a (a \sin (c+d x)+a)^{13/2}}{a^7 d}\) |
((16*a^3*(a + a*Sin[c + d*x])^(9/2))/9 - (24*a^2*(a + a*Sin[c + d*x])^(11/ 2))/11 + (12*a*(a + a*Sin[c + d*x])^(13/2))/13 - (2*(a + a*Sin[c + d*x])^( 15/2))/15)/(a^7*d)
3.2.1.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 0.45 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(-\frac {2 \left (\frac {\left (a +a \sin \left (d x +c \right )\right )^{\frac {15}{2}}}{15}-\frac {6 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {12 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {8 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}\right )}{d \,a^{7}}\) | \(73\) |
default | \(-\frac {2 \left (\frac {\left (a +a \sin \left (d x +c \right )\right )^{\frac {15}{2}}}{15}-\frac {6 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {12 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {8 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}\right )}{d \,a^{7}}\) | \(73\) |
-2/d/a^7*(1/15*(a+a*sin(d*x+c))^(15/2)-6/13*a*(a+a*sin(d*x+c))^(13/2)+12/1 1*a^2*(a+a*sin(d*x+c))^(11/2)-8/9*a^3*(a+a*sin(d*x+c))^(9/2))
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (33 \, \cos \left (d x + c\right )^{6} + 56 \, \cos \left (d x + c\right )^{4} + 128 \, \cos \left (d x + c\right )^{2} + {\left (429 \, \cos \left (d x + c\right )^{6} + 504 \, \cos \left (d x + c\right )^{4} + 640 \, \cos \left (d x + c\right )^{2} + 1024\right )} \sin \left (d x + c\right ) + 1024\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6435 \, d} \]
2/6435*(33*cos(d*x + c)^6 + 56*cos(d*x + c)^4 + 128*cos(d*x + c)^2 + (429* cos(d*x + c)^6 + 504*cos(d*x + c)^4 + 640*cos(d*x + c)^2 + 1024)*sin(d*x + c) + 1024)*sqrt(a*sin(d*x + c) + a)/d
Timed out. \[ \int \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (429 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {15}{2}} - 2970 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a + 7020 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2} - 5720 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{3}\right )}}{6435 \, a^{7} d} \]
-2/6435*(429*(a*sin(d*x + c) + a)^(15/2) - 2970*(a*sin(d*x + c) + a)^(13/2 )*a + 7020*(a*sin(d*x + c) + a)^(11/2)*a^2 - 5720*(a*sin(d*x + c) + a)^(9/ 2)*a^3)/(a^7*d)
Time = 0.31 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.32 \[ \int \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {256 \, \sqrt {2} {\left (429 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 1485 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 1755 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 715 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{6435 \, d} \]
-256/6435*sqrt(2)*(429*cos(-1/4*pi + 1/2*d*x + 1/2*c)^15*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 1485*cos(-1/4*pi + 1/2*d*x + 1/2*c)^13*sgn(cos(-1/4*p i + 1/2*d*x + 1/2*c)) + 1755*cos(-1/4*pi + 1/2*d*x + 1/2*c)^11*sgn(cos(-1/ 4*pi + 1/2*d*x + 1/2*c)) - 715*cos(-1/4*pi + 1/2*d*x + 1/2*c)^9*sgn(cos(-1 /4*pi + 1/2*d*x + 1/2*c)))*sqrt(a)/d
Timed out. \[ \int \cos ^7(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^7\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \]