\(\int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [453]

Optimal result

Integrand size = 31, antiderivative size = 188 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {256 a^4 \cos ^5(c+d x)}{6435 d (a+a \sin (c+d x))^{5/2}}-\frac {64 a^3 \cos ^5(c+d x)}{1287 d (a+a \sin (c+d x))^{3/2}}-\frac {56 a^2 \cos ^5(c+d x)}{1287 d \sqrt {a+a \sin (c+d x)}}-\frac {14 a \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{429 d}+\frac {4 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{39 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{15 a d} \]

[Out]

-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-2/15*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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2957, 2935, 2753, 2752} \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {256 a^4 \cos ^5(c+d x)}{6435 d (a \sin (c+d x)+a)^{5/2}}-\frac {64 a^3 \cos ^5(c+d x)}{1287 d (a \sin (c+d x)+a)^{3/2}}-\frac {56 a^2 \cos ^5(c+d x)}{1287 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{15 a d}+\frac {4 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{39 d}-\frac {14 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{429 d} \]

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-256*a^4*Cos[c + d*x]^5)/(6435*d*(a + a*Sin[c + d*x])^(5/2)) - (64*a^3*Cos[c + d*x]^5)/(1287*d*(a + a*Sin[c +
d*x])^(3/2)) - (56*a^2*Cos[c + d*x]^5)/(1287*d*Sqrt[a + a*Sin[c + d*x]]) - (14*a*Cos[c + d*x]^5*Sqrt[a + a*Si
n[c + d*x]])/(429*d) + (4*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(39*d) - (2*Cos[c + d*x]^5*(a + a*Sin[c +
d*x])^(5/2))/(15*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))^{5/2}}{15 a d}+\frac {2 \int \cos ^4(c+d x) \left (\frac {5 a}{2}-5 a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{15 a} \\ & = \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}+\frac {7}{39} \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx \\ & = -\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}+\frac {1}{143} (28 a) \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\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}+\frac {\left (224 a^2\right ) \int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{1287} \\ & = -\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}+\frac {\left (128 a^3\right ) \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{1287} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.64 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (43122-36640 \cos (2 (c+d x))+3630 \cos (4 (c+d x))+66470 \sin (c+d x)-14445 \sin (3 (c+d x))+429 \sin (5 (c+d x)))}{51480 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^2*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-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]))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.52

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{3} \left (429 \left (\sin ^{5}\left (d x +c \right )\right )+1815 \left (\sin ^{4}\left (d x +c \right )\right )+3075 \left (\sin ^{3}\left (d x +c \right )\right )+2765 \left (\sin ^{2}\left (d x +c \right )\right )+1580 \sin \left (d x +c \right )+632\right )}{6435 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(97\)

[In]

int(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.12 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (429 \, a \cos \left (d x + c\right )^{8} + 957 \, a \cos \left (d x + c\right )^{7} - 633 \, a \cos \left (d x + c\right )^{6} - 1301 \, a \cos \left (d x + c\right )^{5} + 20 \, a \cos \left (d x + c\right )^{4} - 32 \, a \cos \left (d x + c\right )^{3} + 64 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) + {\left (429 \, a \cos \left (d x + c\right )^{7} - 528 \, a \cos \left (d x + c\right )^{6} - 1161 \, a \cos \left (d x + c\right )^{5} + 140 \, a \cos \left (d x + c\right )^{4} + 160 \, a \cos \left (d x + c\right )^{3} + 192 \, a \cos \left (d x + c\right )^{2} + 256 \, a \cos \left (d x + c\right ) + 512 \, a\right )} \sin \left (d x + c\right ) - 512 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6435 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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*co
s(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 + 25
6*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)

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**2*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2} \,d x } \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.02 \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {64 \, \sqrt {2} {\left (1716 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} - 7920 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 14625 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 13585 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 6435 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1287 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{6435 \, d} \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-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(co
s(-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))*
sin(-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

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]

[In]

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + a*sin(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^4*sin(c + d*x)^2*(a + a*sin(c + d*x))^(3/2), x)