Integrand size = 23, antiderivative size = 97 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {16 (a+a \sin (c+d x))^{11/2}}{11 a^4 d}-\frac {24 (a+a \sin (c+d x))^{13/2}}{13 a^5 d}+\frac {4 (a+a \sin (c+d x))^{15/2}}{5 a^6 d}-\frac {2 (a+a \sin (c+d x))^{17/2}}{17 a^7 d} \] Output:
16/11*(a+a*sin(d*x+c))^(11/2)/a^4/d-24/13*(a+a*sin(d*x+c))^(13/2)/a^5/d+4/ 5*(a+a*sin(d*x+c))^(15/2)/a^6/d-2/17*(a+a*sin(d*x+c))^(17/2)/a^7/d
Time = 0.41 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.63 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 (1+\sin (c+d x))^4 (a (1+\sin (c+d x)))^{3/2} \left (-1767+3641 \sin (c+d x)-2717 \sin ^2(c+d x)+715 \sin ^3(c+d x)\right )}{12155 d} \] Input:
Integrate[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^(3/2),x]
Output:
(-2*(1 + Sin[c + d*x])^4*(a*(1 + Sin[c + d*x]))^(3/2)*(-1767 + 3641*Sin[c + d*x] - 2717*Sin[c + d*x]^2 + 715*Sin[c + d*x]^3))/(12155*d)
Time = 0.27 (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.
\(\displaystyle \int \cos ^7(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^7 (a \sin (c+d x)+a)^{3/2}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^{9/2}d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left (-(\sin (c+d x) a+a)^{15/2}+6 a (\sin (c+d x) a+a)^{13/2}-12 a^2 (\sin (c+d x) a+a)^{11/2}+8 a^3 (\sin (c+d x) a+a)^{9/2}\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {16}{11} a^3 (a \sin (c+d x)+a)^{11/2}-\frac {24}{13} a^2 (a \sin (c+d x)+a)^{13/2}-\frac {2}{17} (a \sin (c+d x)+a)^{17/2}+\frac {4}{5} a (a \sin (c+d x)+a)^{15/2}}{a^7 d}\) |
Input:
Int[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^(3/2),x]
Output:
((16*a^3*(a + a*Sin[c + d*x])^(11/2))/11 - (24*a^2*(a + a*Sin[c + d*x])^(1 3/2))/13 + (4*a*(a + a*Sin[c + d*x])^(15/2))/5 - (2*(a + a*Sin[c + d*x])^( 17/2))/17)/(a^7*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 = 0.86 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.59
method | result | size |
default | \(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {11}{2}} \left (715 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-2717 \cos \left (d x +c \right )^{2}-4356 \sin \left (d x +c \right )+4484\right )}{12155 a^{4} d}\) | \(57\) |
Input:
int(cos(d*x+c)^7*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/12155/a^4*(a+a*sin(d*x+c))^(11/2)*(715*cos(d*x+c)^2*sin(d*x+c)-2717*cos( d*x+c)^2-4356*sin(d*x+c)+4484)/d
Time = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.13 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (715 \, a \cos \left (d x + c\right )^{8} - 66 \, a \cos \left (d x + c\right )^{6} - 112 \, a \cos \left (d x + c\right )^{4} - 256 \, a \cos \left (d x + c\right )^{2} - 2 \, {\left (429 \, a \cos \left (d x + c\right )^{6} + 504 \, a \cos \left (d x + c\right )^{4} + 640 \, a \cos \left (d x + c\right )^{2} + 1024 \, a\right )} \sin \left (d x + c\right ) - 2048 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{12155 \, d} \] Input:
integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")
Output:
-2/12155*(715*a*cos(d*x + c)^8 - 66*a*cos(d*x + c)^6 - 112*a*cos(d*x + c)^ 4 - 256*a*cos(d*x + c)^2 - 2*(429*a*cos(d*x + c)^6 + 504*a*cos(d*x + c)^4 + 640*a*cos(d*x + c)^2 + 1024*a)*sin(d*x + c) - 2048*a)*sqrt(a*sin(d*x + c ) + a)/d
Timed out. \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**7*(a+a*sin(d*x+c))**(3/2),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (715 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {17}{2}} - 4862 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {15}{2}} a + 11220 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a^{2} - 8840 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{3}\right )}}{12155 \, a^{7} d} \] Input:
integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")
Output:
-2/12155*(715*(a*sin(d*x + c) + a)^(17/2) - 4862*(a*sin(d*x + c) + a)^(15/ 2)*a + 11220*(a*sin(d*x + c) + a)^(13/2)*a^2 - 8840*(a*sin(d*x + c) + a)^( 11/2)*a^3)/(a^7*d)
Time = 0.15 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.36 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {512 \, \sqrt {2} {\left (715 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{17} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 2431 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2805 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 1105 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{12155 \, d} \] Input:
integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")
Output:
-512/12155*sqrt(2)*(715*a*cos(-1/4*pi + 1/2*d*x + 1/2*c)^17*sgn(cos(-1/4*p i + 1/2*d*x + 1/2*c)) - 2431*a*cos(-1/4*pi + 1/2*d*x + 1/2*c)^15*sgn(cos(- 1/4*pi + 1/2*d*x + 1/2*c)) + 2805*a*cos(-1/4*pi + 1/2*d*x + 1/2*c)^13*sgn( cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 1105*a*cos(-1/4*pi + 1/2*d*x + 1/2*c)^11 *sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))*sqrt(a)/d
Timed out. \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^7\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:
int(cos(c + d*x)^7*(a + a*sin(c + d*x))^(3/2),x)
Output:
int(cos(c + d*x)^7*(a + a*sin(c + d*x))^(3/2), x)
\[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{7} \sin \left (d x +c \right )d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{7}d x \right ) \] Input:
int(cos(d*x+c)^7*(a+a*sin(d*x+c))^(3/2),x)
Output:
sqrt(a)*a*(int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**7*sin(c + d*x),x) + in t(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**7,x))