Integrand size = 23, antiderivative size = 154 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d}-\frac {8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}}{9 b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{11/2}}{11 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{13/2}}{13 b^5 d}+\frac {2 (a+b \sin (c+d x))^{15/2}}{15 b^5 d} \]
2/7*(a^2-b^2)^2*(a+b*sin(d*x+c))^(7/2)/b^5/d-8/9*a*(a^2-b^2)*(a+b*sin(d*x+ c))^(9/2)/b^5/d+4/11*(3*a^2-b^2)*(a+b*sin(d*x+c))^(11/2)/b^5/d-8/13*a*(a+b *sin(d*x+c))^(13/2)/b^5/d+2/15*(a+b*sin(d*x+c))^(15/2)/b^5/d
Time = 0.38 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.73 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2 (a+b \sin (c+d x))^{7/2} \left (6435 \left (a^2-b^2\right )^2-20020 a (a-b) (a+b) (a+b \sin (c+d x))+8190 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^2-13860 a (a+b \sin (c+d x))^3+3003 (a+b \sin (c+d x))^4\right )}{45045 b^5 d} \]
(2*(a + b*Sin[c + d*x])^(7/2)*(6435*(a^2 - b^2)^2 - 20020*a*(a - b)*(a + b )*(a + b*Sin[c + d*x]) + 8190*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^2 - 13860 *a*(a + b*Sin[c + d*x])^3 + 3003*(a + b*Sin[c + d*x])^4))/(45045*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.
| \(\displaystyle \int \cos ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^5 (a+b \sin (c+d x))^{5/2}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {\int (a+b \sin (c+d x))^{5/2} \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \frac {\int \left ((a+b \sin (c+d x))^{13/2}-4 a (a+b \sin (c+d x))^{11/2}+2 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}-4 \left (a^3-a b^2\right ) (a+b \sin (c+d x))^{7/2}+\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{5/2}\right )d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {4}{11} \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{11/2}-\frac {8}{9} a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}+\frac {2}{7} \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{7/2}+\frac {2}{15} (a+b \sin (c+d x))^{15/2}-\frac {8}{13} a (a+b \sin (c+d x))^{13/2}}{b^5 d}\) |
((2*(a^2 - b^2)^2*(a + b*Sin[c + d*x])^(7/2))/7 - (8*a*(a^2 - b^2)*(a + b* Sin[c + d*x])^(9/2))/9 + (4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(11/2))/11 - (8*a*(a + b*Sin[c + d*x])^(13/2))/13 + (2*(a + b*Sin[c + d*x])^(15/2))/1 5)/(b^5*d)
3.5.94.3.1 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 = 78.86 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {15}{2}}}{15}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {2 \left (\left (a +b \right )^{2}+\left (-2 a -2 b \right ) \left (-2 a +2 b \right )+\left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (a +b \right )^{2} \left (-2 a +2 b \right )+\left (-2 a -2 b \right ) \left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}}{d \,b^{5}}\) | \(150\) |
| default | \(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {15}{2}}}{15}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {2 \left (\left (a +b \right )^{2}+\left (-2 a -2 b \right ) \left (-2 a +2 b \right )+\left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (a +b \right )^{2} \left (-2 a +2 b \right )+\left (-2 a -2 b \right ) \left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}}{d \,b^{5}}\) | \(150\) |
2/d/b^5*(1/15*(a+b*sin(d*x+c))^(15/2)-4/13*a*(a+b*sin(d*x+c))^(13/2)+1/11* ((a+b)^2+(-2*a-2*b)*(-2*a+2*b)+(a-b)^2)*(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*(a-b) ^2*(a+b*sin(d*x+c))^(7/2))
Time = 0.32 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.45 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (7161 \, a b^{6} \cos \left (d x + c\right )^{6} - 128 \, a^{7} + 992 \, a^{5} b^{2} - 6080 \, a^{3} b^{4} - 5536 \, a b^{6} - 7 \, {\left (5 \, a^{3} b^{4} + 79 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} + 16 \, {\left (3 \, a^{5} b^{2} - 20 \, a^{3} b^{4} - 67 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + {\left (3003 \, b^{7} \cos \left (d x + c\right )^{6} + 64 \, a^{6} b - 480 \, a^{4} b^{3} - 9088 \, a^{2} b^{5} - 1248 \, b^{7} - 63 \, {\left (71 \, a^{2} b^{5} + 13 \, b^{7}\right )} \cos \left (d x + c\right )^{4} - 8 \, {\left (5 \, a^{4} b^{3} + 718 \, a^{2} b^{5} + 117 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{45045 \, b^{5} d} \]
-2/45045*(7161*a*b^6*cos(d*x + c)^6 - 128*a^7 + 992*a^5*b^2 - 6080*a^3*b^4 - 5536*a*b^6 - 7*(5*a^3*b^4 + 79*a*b^6)*cos(d*x + c)^4 + 16*(3*a^5*b^2 - 20*a^3*b^4 - 67*a*b^6)*cos(d*x + c)^2 + (3003*b^7*cos(d*x + c)^6 + 64*a^6* b - 480*a^4*b^3 - 9088*a^2*b^5 - 1248*b^7 - 63*(71*a^2*b^5 + 13*b^7)*cos(d *x + c)^4 - 8*(5*a^4*b^3 + 718*a^2*b^5 + 117*b^7)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^5*d)
Leaf count of result is larger than twice the leaf count of optimal. 714 vs. \(2 (138) = 276\).
Time = 129.11 (sec) , antiderivative size = 714, normalized size of antiderivative = 4.64 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\begin {cases} a^{\frac {5}{2}} x \cos ^{5}{\left (c \right )} & \text {for}\: b = 0 \wedge d = 0 \\a^{\frac {5}{2}} \cdot \left (\frac {8 \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {\sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d}\right ) & \text {for}\: b = 0 \\x \left (a + b \sin {\left (c \right )}\right )^{\frac {5}{2}} \cos ^{5}{\left (c \right )} & \text {for}\: d = 0 \\\frac {256 a^{7} \sqrt {a + b \sin {\left (c + d x \right )}}}{45045 b^{5} d} - \frac {128 a^{6} \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}}{45045 b^{4} d} - \frac {1984 a^{5} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )}}{45045 b^{3} d} - \frac {32 a^{5} \sqrt {a + b \sin {\left (c + d x \right )}} \cos ^{2}{\left (c + d x \right )}}{693 b^{3} d} + \frac {64 a^{4} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{3}{\left (c + d x \right )}}{3003 b^{2} d} + \frac {16 a^{4} \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{693 b^{2} d} + \frac {2432 a^{3} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{4}{\left (c + d x \right )}}{9009 b d} + \frac {128 a^{3} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{231 b d} + \frac {2 a^{3} \sqrt {a + b \sin {\left (c + d x \right )}} \cos ^{4}{\left (c + d x \right )}}{7 b d} + \frac {18176 a^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{5}{\left (c + d x \right )}}{45045 d} + \frac {736 a^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{693 d} + \frac {6 a^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{7 d} + \frac {11072 a b \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{6}{\left (c + d x \right )}}{45045 d} + \frac {544 a b \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{693 d} + \frac {6 a b \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{7 d} + \frac {64 b^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{7}{\left (c + d x \right )}}{1155 d} + \frac {16 b^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{77 d} + \frac {2 b^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{7 d} & \text {otherwise} \end {cases} \]
Piecewise((a**(5/2)*x*cos(c)**5, Eq(b, 0) & Eq(d, 0)), (a**(5/2)*(8*sin(c + d*x)**5/(15*d) + 4*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + sin(c + d*x)* cos(c + d*x)**4/d), Eq(b, 0)), (x*(a + b*sin(c))**(5/2)*cos(c)**5, Eq(d, 0 )), (256*a**7*sqrt(a + b*sin(c + d*x))/(45045*b**5*d) - 128*a**6*sqrt(a + b*sin(c + d*x))*sin(c + d*x)/(45045*b**4*d) - 1984*a**5*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**2/(45045*b**3*d) - 32*a**5*sqrt(a + b*sin(c + d*x))*c os(c + d*x)**2/(693*b**3*d) + 64*a**4*sqrt(a + b*sin(c + d*x))*sin(c + d*x )**3/(3003*b**2*d) + 16*a**4*sqrt(a + b*sin(c + d*x))*sin(c + d*x)*cos(c + d*x)**2/(693*b**2*d) + 2432*a**3*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**4 /(9009*b*d) + 128*a**3*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**2*cos(c + d* x)**2/(231*b*d) + 2*a**3*sqrt(a + b*sin(c + d*x))*cos(c + d*x)**4/(7*b*d) + 18176*a**2*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**5/(45045*d) + 736*a**2 *sqrt(a + b*sin(c + d*x))*sin(c + d*x)**3*cos(c + d*x)**2/(693*d) + 6*a**2 *sqrt(a + b*sin(c + d*x))*sin(c + d*x)*cos(c + d*x)**4/(7*d) + 11072*a*b*s qrt(a + b*sin(c + d*x))*sin(c + d*x)**6/(45045*d) + 544*a*b*sqrt(a + b*sin (c + d*x))*sin(c + d*x)**4*cos(c + d*x)**2/(693*d) + 6*a*b*sqrt(a + b*sin( c + d*x))*sin(c + d*x)**2*cos(c + d*x)**4/(7*d) + 64*b**2*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**7/(1155*d) + 16*b**2*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**5*cos(c + d*x)**2/(77*d) + 2*b**2*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**3*cos(c + d*x)**4/(7*d), True))
Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.75 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2 \, {\left (3003 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {15}{2}} - 13860 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a + 8190 \, {\left (3 \, a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 20020 \, {\left (a^{3} - a b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} + 6435 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}\right )}}{45045 \, b^{5} d} \]
2/45045*(3003*(b*sin(d*x + c) + a)^(15/2) - 13860*(b*sin(d*x + c) + a)^(13 /2)*a + 8190*(3*a^2 - b^2)*(b*sin(d*x + c) + a)^(11/2) - 20020*(a^3 - a*b^ 2)*(b*sin(d*x + c) + a)^(9/2) + 6435*(a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x + c) + a)^(7/2))/(b^5*d)
\[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int {\cos \left (c+d\,x\right )}^5\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]