Integrand size = 31, antiderivative size = 188 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {256 a^4 \cos ^5(c+d x)}{6435 d (a+a \sin (c+d x))^{5/2}}-\frac {64 a^3 \cos ^5(c+d x)}{1287 d (a+a \sin (c+d x))^{3/2}}-\frac {56 a^2 \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}-\frac {14 a \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{429 d}+\frac {4 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 a d} \]
-256/6435*a^4*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-64/1287*a^3*cos(d*x+c) ^5/d/(a+a*sin(d*x+c))^(3/2)+4/39*cos(d*x+c)^5*(a+a*sin(d*x+c))^(3/2)/d-2/1 5*cos(d*x+c)^5*(a+a*sin(d*x+c))^(5/2)/a/d-56/1287*a^2*cos(d*x+c)^5/d/(a+a* sin(d*x+c))^(1/2)-14/429*a*cos(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d
Time = 6.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.64 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (43122-36640 \cos (2 (c+d x))+3630 \cos (4 (c+d x))+66470 \sin (c+d x)-14445 \sin (3 (c+d x))+429 \sin (5 (c+d x)))}{51480 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]
-1/51480*(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d* x])]*(43122 - 36640*Cos[2*(c + d*x)] + 3630*Cos[4*(c + d*x)] + 66470*Sin[c + d*x] - 14445*Sin[3*(c + d*x)] + 429*Sin[5*(c + d*x)]))/(d*(Cos[(c + d*x )/2] + Sin[(c + d*x)/2]))
Time = 1.07 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3357, 27, 3042, 3335, 3042, 3153, 3042, 3153, 3042, 3153, 3042, 3152}
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.
| \(\displaystyle \int \sin ^2(c+d x) \cos ^4(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^2 \cos (c+d x)^4 (a \sin (c+d x)+a)^{3/2}dx\) |
| \(\Big \downarrow \) 3357 |
| \(\displaystyle \frac {2 \int \frac {5}{2} \cos ^4(c+d x) (a-2 a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{15 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos ^4(c+d x) (a-2 a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \cos (c+d x)^4 (a-2 a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3335 |
| \(\displaystyle \frac {\frac {7}{13} a \int \cos ^4(c+d x) (\sin (c+d x) a+a)^{3/2}dx+\frac {4 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7}{13} a \int \cos (c+d x)^4 (\sin (c+d x) a+a)^{3/2}dx+\frac {4 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {\frac {7}{13} a \left (\frac {12}{11} a \int \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )+\frac {4 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7}{13} a \left (\frac {12}{11} a \int \cos (c+d x)^4 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )+\frac {4 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {\frac {7}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )+\frac {4 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )+\frac {4 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle \frac {\frac {7}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )+\frac {4 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )+\frac {4 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle \frac {\frac {7}{13} a \left (\frac {12}{11} a \left (\frac {8}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )+\frac {4 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}}{3 a}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}\) |
(-2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(5/2))/(15*a*d) + ((4*a*Cos[c + d* x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*d) + (7*a*((-2*a*Cos[c + d*x]^5*Sqrt[ a + a*Sin[c + d*x]])/(11*d) + (12*a*((-2*a*Cos[c + d*x]^5)/(9*d*Sqrt[a + a *Sin[c + d*x]]) + (8*a*((-8*a^2*Cos[c + d*x]^5)/(35*d*(a + a*Sin[c + d*x]) ^(5/2)) - (2*a*Cos[c + d*x]^5)/(7*d*(a + a*Sin[c + d*x])^(3/2))))/9))/11)) /13)/(3*a)
3.5.53.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^( p + 1))*((a + b*Sin[e + f*x])^(m + 1)/(b*f*g*(m + p + 2))), x] + Simp[1/(b* (m + p + 2)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(b*(m + 1) - a *(p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^ 2 - b^2, 0] && NeQ[m + p + 2, 0]
Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.52
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{3} \left (429 \left (\sin ^{5}\left (d x +c \right )\right )+1815 \left (\sin ^{4}\left (d x +c \right )\right )+3075 \left (\sin ^{3}\left (d x +c \right )\right )+2765 \left (\sin ^{2}\left (d x +c \right )\right )+1580 \sin \left (d x +c \right )+632\right )}{6435 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(97\) |
2/6435*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^3*(429*sin(d*x+c)^5+1815*sin(d*x+ c)^4+3075*sin(d*x+c)^3+2765*sin(d*x+c)^2+1580*sin(d*x+c)+632)/cos(d*x+c)/( a+a*sin(d*x+c))^(1/2)/d
Time = 0.29 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.12 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (429 \, a \cos \left (d x + c\right )^{8} + 957 \, a \cos \left (d x + c\right )^{7} - 633 \, a \cos \left (d x + c\right )^{6} - 1301 \, a \cos \left (d x + c\right )^{5} + 20 \, a \cos \left (d x + c\right )^{4} - 32 \, a \cos \left (d x + c\right )^{3} + 64 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) + {\left (429 \, a \cos \left (d x + c\right )^{7} - 528 \, a \cos \left (d x + c\right )^{6} - 1161 \, a \cos \left (d x + c\right )^{5} + 140 \, a \cos \left (d x + c\right )^{4} + 160 \, a \cos \left (d x + c\right )^{3} + 192 \, a \cos \left (d x + c\right )^{2} + 256 \, a \cos \left (d x + c\right ) + 512 \, a\right )} \sin \left (d x + c\right ) - 512 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6435 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]
2/6435*(429*a*cos(d*x + c)^8 + 957*a*cos(d*x + c)^7 - 633*a*cos(d*x + c)^6 - 1301*a*cos(d*x + c)^5 + 20*a*cos(d*x + c)^4 - 32*a*cos(d*x + c)^3 + 64* a*cos(d*x + c)^2 - 256*a*cos(d*x + c) + (429*a*cos(d*x + c)^7 - 528*a*cos( d*x + c)^6 - 1161*a*cos(d*x + c)^5 + 140*a*cos(d*x + c)^4 + 160*a*cos(d*x + c)^3 + 192*a*cos(d*x + c)^2 + 256*a*cos(d*x + c) + 512*a)*sin(d*x + c) - 512*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
Timed out. \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]
\[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2} \,d x } \]
Time = 0.39 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.02 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {64 \, \sqrt {2} {\left (1716 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} - 7920 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 14625 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 13585 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 6435 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1287 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{6435 \, d} \]
-64/6435*sqrt(2)*(1716*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^15 - 7920*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4 *pi + 1/2*d*x + 1/2*c)^13 + 14625*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*si n(-1/4*pi + 1/2*d*x + 1/2*c)^11 - 13585*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2* c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 + 6435*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 1287*a*sgn(cos(-1/4*pi + 1/2*d* x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)*sqrt(a)/d
Timed out. \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]