\(\int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [330]

Optimal result

Integrand size = 31, antiderivative size = 156 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {64 a^3 \cos ^3(c+d x)}{385 d (a+a \sin (c+d x))^{3/2}}-\frac {48 a^2 \cos ^3(c+d x)}{385 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{77 d}+\frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d} \]

[Out]

-64/385*a^3*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)+4/33*cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d-2/11*cos(d*x+c)^3
*(a+a*sin(d*x+c))^(5/2)/a/d-48/385*a^2*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-6/77*a*cos(d*x+c)^3*(a+a*sin(d*x+
c))^(1/2)/d

Rubi [A] (verified)

Time = 0.32 (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} \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {64 a^3 \cos ^3(c+d x)}{385 d (a \sin (c+d x)+a)^{3/2}}-\frac {48 a^2 \cos ^3(c+d x)}{385 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}+\frac {4 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{33 d}-\frac {6 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{77 d} \]

[In]

Int[Cos[c + d*x]^2*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-64*a^3*Cos[c + d*x]^3)/(385*d*(a + a*Sin[c + d*x])^(3/2)) - (48*a^2*Cos[c + d*x]^3)/(385*d*Sqrt[a + a*Sin[c
+ d*x]]) - (6*a*Cos[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(77*d) + (4*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2)
)/(33*d) - (2*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2))/(11*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 ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac {2 \int \cos ^2(c+d x) \left (\frac {5 a}{2}-3 a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{11 a} \\ & = \frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac {3}{11} \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx \\ & = -\frac {6 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{77 d}+\frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac {1}{77} (24 a) \int \cos ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {48 a^2 \cos ^3(c+d x)}{385 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{77 d}+\frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d}+\frac {1}{385} \left (96 a^2\right ) \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {64 a^3 \cos ^3(c+d x)}{385 d (a+a \sin (c+d x))^{3/2}}-\frac {48 a^2 \cos ^3(c+d x)}{385 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{77 d}+\frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.61 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \sqrt {a (1+\sin (c+d x))} (4159-2280 \cos (2 (c+d x))+105 \cos (4 (c+d x))+5076 \sin (c+d x)-700 \sin (3 (c+d x)))}{4620 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[Cos[c + d*x]^2*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-1/4620*(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(4159 - 2280*Cos[2*(c + d*x)] +
105*Cos[4*(c + d*x)] + 5076*Sin[c + d*x] - 700*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.56

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{2} \left (105 \left (\sin ^{4}\left (d x +c \right )\right )+350 \left (\sin ^{3}\left (d x +c \right )\right )+465 \left (\sin ^{2}\left (d x +c \right )\right )+372 \sin \left (d x +c \right )+248\right )}{1155 d \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}}\)	\(87\)

[In]

int(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/1155/d*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^2*(105*sin(d*x+c)^4+350*sin(d*x+c)^3+465*sin(d*x+c)^2+372*sin(d*x+
c)+248)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.06 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (105 \, a \cos \left (d x + c\right )^{6} + 245 \, a \cos \left (d x + c\right )^{5} - 185 \, a \cos \left (d x + c\right )^{4} - 397 \, a \cos \left (d x + c\right )^{3} + 24 \, a \cos \left (d x + c\right )^{2} - 96 \, a \cos \left (d x + c\right ) + {\left (105 \, a \cos \left (d x + c\right )^{5} - 140 \, a \cos \left (d x + c\right )^{4} - 325 \, a \cos \left (d x + c\right )^{3} + 72 \, a \cos \left (d x + c\right )^{2} + 96 \, a \cos \left (d x + c\right ) + 192 \, a\right )} \sin \left (d x + c\right ) - 192 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{1155 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/1155*(105*a*cos(d*x + c)^6 + 245*a*cos(d*x + c)^5 - 185*a*cos(d*x + c)^4 - 397*a*cos(d*x + c)^3 + 24*a*cos(d
*x + c)^2 - 96*a*cos(d*x + c) + (105*a*cos(d*x + c)^5 - 140*a*cos(d*x + c)^4 - 325*a*cos(d*x + c)^3 + 72*a*cos
(d*x + c)^2 + 96*a*cos(d*x + c) + 192*a)*sin(d*x + c) - 192*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*si
n(d*x + c) + d)

Sympy [F]

\[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**2*sin(d*x+c)**2*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral((a*(sin(c + d*x) + 1))**(3/2)*sin(c + d*x)**2*cos(c + d*x)**2, x)

Maxima [F]

\[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^2*sin(d*x + c)^2, x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.04 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {16 \, \sqrt {2} {\left (420 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1540 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2145 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1386 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )} \sqrt {a}}{1155 \, d} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

16/1155*sqrt(2)*(420*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 - 1540*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 + 2145*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 - 1386*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 + 3
85*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3)*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]

[In]

int(cos(c + d*x)^2*sin(c + d*x)^2*(a + a*sin(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^2*sin(c + d*x)^2*(a + a*sin(c + d*x))^(3/2), x)