Integrand size = 23, antiderivative size = 73 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {8 (a+a \sin (c+d x))^{11/2}}{11 a^3 d}-\frac {8 (a+a \sin (c+d x))^{13/2}}{13 a^4 d}+\frac {2 (a+a \sin (c+d x))^{15/2}}{15 a^5 d} \] Output:
8/11*(a+a*sin(d*x+c))^(11/2)/a^3/d-8/13*(a+a*sin(d*x+c))^(13/2)/a^4/d+2/15 *(a+a*sin(d*x+c))^(15/2)/a^5/d
Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.70 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {2 (1+\sin (c+d x))^3 (a (1+\sin (c+d x)))^{5/2} \left (263-374 \sin (c+d x)+143 \sin ^2(c+d x)\right )}{2145 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(5/2),x]
Output:
(2*(1 + Sin[c + d*x])^3*(a*(1 + Sin[c + d*x]))^(5/2)*(263 - 374*Sin[c + d* x] + 143*Sin[c + d*x]^2))/(2145*d)
Time = 0.27 (sec) , antiderivative size = 66, 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.
\(\displaystyle \int \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^5 (a \sin (c+d x)+a)^{5/2}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^{9/2}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left ((\sin (c+d x) a+a)^{13/2}-4 a (\sin (c+d x) a+a)^{11/2}+4 a^2 (\sin (c+d x) a+a)^{9/2}\right )d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {8}{11} a^2 (a \sin (c+d x)+a)^{11/2}+\frac {2}{15} (a \sin (c+d x)+a)^{15/2}-\frac {8}{13} a (a \sin (c+d x)+a)^{13/2}}{a^5 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(5/2),x]
Output:
((8*a^2*(a + a*Sin[c + d*x])^(11/2))/11 - (8*a*(a + a*Sin[c + d*x])^(13/2) )/13 + (2*(a + a*Sin[c + d*x])^(15/2))/15)/(a^5*d)
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 = 69.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.56
method | result | size |
default | \(-\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {11}{2}} \left (143 \cos \left (d x +c \right )^{2}+374 \sin \left (d x +c \right )-406\right )}{2145 a^{3} d}\) | \(41\) |
Input:
int(cos(d*x+c)^5*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/2145/a^3*(a+a*sin(d*x+c))^(11/2)*(143*cos(d*x+c)^2+374*sin(d*x+c)-406)/ d
Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.56 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (341 \, a^{2} \cos \left (d x + c\right )^{6} - 28 \, a^{2} \cos \left (d x + c\right )^{4} - 64 \, a^{2} \cos \left (d x + c\right )^{2} - 512 \, a^{2} + {\left (143 \, a^{2} \cos \left (d x + c\right )^{6} - 252 \, a^{2} \cos \left (d x + c\right )^{4} - 320 \, a^{2} \cos \left (d x + c\right )^{2} - 512 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2145 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-2/2145*(341*a^2*cos(d*x + c)^6 - 28*a^2*cos(d*x + c)^4 - 64*a^2*cos(d*x + c)^2 - 512*a^2 + (143*a^2*cos(d*x + c)^6 - 252*a^2*cos(d*x + c)^4 - 320*a ^2*cos(d*x + c)^2 - 512*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/d
Timed out. \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*(a+a*sin(d*x+c))**(5/2),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {2 \, {\left (143 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {15}{2}} - 660 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a + 780 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2}\right )}}{2145 \, a^{5} d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
2/2145*(143*(a*sin(d*x + c) + a)^(15/2) - 660*(a*sin(d*x + c) + a)^(13/2)* a + 780*(a*sin(d*x + c) + a)^(11/2)*a^2)/(a^5*d)
Time = 0.14 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.48 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {256 \, \sqrt {2} {\left (143 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 330 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 195 \, a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{2145 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
256/2145*sqrt(2)*(143*a^2*cos(-1/4*pi + 1/2*d*x + 1/2*c)^15*sgn(cos(-1/4*p i + 1/2*d*x + 1/2*c)) - 330*a^2*cos(-1/4*pi + 1/2*d*x + 1/2*c)^13*sgn(cos( -1/4*pi + 1/2*d*x + 1/2*c)) + 195*a^2*cos(-1/4*pi + 1/2*d*x + 1/2*c)^11*sg n(cos(-1/4*pi + 1/2*d*x + 1/2*c)))*sqrt(a)/d
Timed out. \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int {\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:
int(cos(c + d*x)^5*(a + a*sin(c + d*x))^(5/2),x)
Output:
int(cos(c + d*x)^5*(a + a*sin(c + d*x))^(5/2), x)
\[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )d x \right )+\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{5}d x \right ) \] Input:
int(cos(d*x+c)^5*(a+a*sin(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**5*sin(c + d*x)**2,x ) + 2*int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**5*sin(c + d*x),x) + int(sqr t(sin(c + d*x) + 1)*cos(c + d*x)**5,x))