\(\int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [443]

Optimal result

Integrand size = 31, antiderivative size = 156 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {1472 a^3 \cos ^5(c+d x)}{45045 d (a+a \sin (c+d x))^{5/2}}-\frac {368 a^2 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac {46 a \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}+\frac {20 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d} \]

[Out]

-1472/45045*a^3*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-368/9009*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(3/2)-2/13*
cos(d*x+c)^5*(a+a*sin(d*x+c))^(3/2)/a/d-46/1287*a*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)+20/143*cos(d*x+c)^5*(a
+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2957, 2935, 2753, 2752} \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {1472 a^3 \cos ^5(c+d x)}{45045 d (a \sin (c+d x)+a)^{5/2}}-\frac {368 a^2 \cos ^5(c+d x)}{9009 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 a d}+\frac {20 \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{143 d}-\frac {46 a \cos ^5(c+d x)}{1287 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-1472*a^3*Cos[c + d*x]^5)/(45045*d*(a + a*Sin[c + d*x])^(5/2)) - (368*a^2*Cos[c + d*x]^5)/(9009*d*(a + a*Sin[
c + d*x])^(3/2)) - (46*a*Cos[c + d*x]^5)/(1287*d*Sqrt[a + a*Sin[c + d*x]]) + (20*Cos[c + d*x]^5*Sqrt[a + a*Sin
[c + d*x]])/(143*d) - (2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*a*d)

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

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] + Dist[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]

Rule 2935

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
] + Dist[(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] /; F
reeQ[{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]

Rule 2957

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] + Dist[
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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d}+\frac {2 \int \cos ^4(c+d x) \left (\frac {3 a}{2}-5 a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{13 a} \\ & = \frac {20 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d}+\frac {23}{143} \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {46 a \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}+\frac {20 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d}+\frac {(184 a) \int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{1287} \\ & = -\frac {368 a^2 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac {46 a \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}+\frac {20 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d}+\frac {\left (736 a^2\right ) \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{9009} \\ & = -\frac {1472 a^3 \cos ^5(c+d x)}{45045 d (a+a \sin (c+d x))^{5/2}}-\frac {368 a^2 \cos ^5(c+d x)}{9009 d (a+a \sin (c+d x))^{3/2}}-\frac {46 a \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}+\frac {20 \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.70 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (81183-62440 \cos (2 (c+d x))+3465 \cos (4 (c+d x))+119780 \sin (c+d x)-21420 \sin (3 (c+d x)))}{180180 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

-1/180180*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(81183 - 62440*Cos[2*(c + d*x)]
+ 3465*Cos[4*(c + d*x)] + 119780*Sin[c + d*x] - 21420*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2]))

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.54

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{3} \left (3465 \left (\sin ^{4}\left (d x +c \right )\right )+10710 \left (\sin ^{3}\left (d x +c \right )\right )+12145 \left (\sin ^{2}\left (d x +c \right )\right )+6940 \sin \left (d x +c \right )+2776\right )}{45045 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(85\)

[In]

int(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/45045*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^3*(3465*sin(d*x+c)^4+10710*sin(d*x+c)^3+12145*sin(d*x+c)^2+6940*sin(d*
x+c)+2776)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.10 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (3465 \, \cos \left (d x + c\right )^{7} - 315 \, \cos \left (d x + c\right )^{6} - 4585 \, \cos \left (d x + c\right )^{5} + 115 \, \cos \left (d x + c\right )^{4} - 184 \, \cos \left (d x + c\right )^{3} + 368 \, \cos \left (d x + c\right )^{2} - {\left (3465 \, \cos \left (d x + c\right )^{6} + 3780 \, \cos \left (d x + c\right )^{5} - 805 \, \cos \left (d x + c\right )^{4} - 920 \, \cos \left (d x + c\right )^{3} - 1104 \, \cos \left (d x + c\right )^{2} - 1472 \, \cos \left (d x + c\right ) - 2944\right )} \sin \left (d x + c\right ) - 1472 \, \cos \left (d x + c\right ) - 2944\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{45045 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/45045*(3465*cos(d*x + c)^7 - 315*cos(d*x + c)^6 - 4585*cos(d*x + c)^5 + 115*cos(d*x + c)^4 - 184*cos(d*x + c
)^3 + 368*cos(d*x + c)^2 - (3465*cos(d*x + c)^6 + 3780*cos(d*x + c)^5 - 805*cos(d*x + c)^4 - 920*cos(d*x + c)^
3 - 1104*cos(d*x + c)^2 - 1472*cos(d*x + c) - 2944)*sin(d*x + c) - 1472*cos(d*x + c) - 2944)*sqrt(a*sin(d*x +
c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

Sympy [F]

\[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**2*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*sin(c + d*x)**2*cos(c + d*x)**4, x)

Maxima [F]

\[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)^4*sin(d*x + c)^2, x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {32 \, \sqrt {2} {\left (13860 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 49140 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 65065 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 38610 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9009 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{45045 \, d} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

32/45045*sqrt(2)*(13860*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^13 - 49140*sgn(cos(
-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 + 65065*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(
-1/4*pi + 1/2*d*x + 1/2*c)^9 - 38610*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 + 90
09*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

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \]

[In]

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + a*sin(c + d*x))^(1/2), x)