\(\int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [454]

Optimal result

Integrand size = 29, antiderivative size = 156 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {256 a^4 \cos ^5(c+d x)}{5005 d (a+a \sin (c+d x))^{5/2}}-\frac {64 a^3 \cos ^5(c+d x)}{1001 d (a+a \sin (c+d x))^{3/2}}-\frac {8 a^2 \cos ^5(c+d x)}{143 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \cos ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{143 d}-\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d} \]

[Out]

-256/5005*a^4*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(5/2)-64/1001*a^3*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)/d-8/143*a^2*cos(d*x+c)^5/d/(a+a*sin(d*x+c))^(1/2)-6/143*a*cos(d*x+c)^5*(a+a*s
in(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2935, 2753, 2752} \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {256 a^4 \cos ^5(c+d x)}{5005 d (a \sin (c+d x)+a)^{5/2}}-\frac {64 a^3 \cos ^5(c+d x)}{1001 d (a \sin (c+d x)+a)^{3/2}}-\frac {8 a^2 \cos ^5(c+d x)}{143 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}-\frac {6 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{143 d} \]

[In]

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

[Out]

(-256*a^4*Cos[c + d*x]^5)/(5005*d*(a + a*Sin[c + d*x])^(5/2)) - (64*a^3*Cos[c + d*x]^5)/(1001*d*(a + a*Sin[c +
d*x])^(3/2)) - (8*a^2*Cos[c + d*x]^5)/(143*d*Sqrt[a + a*Sin[c + d*x]]) - (6*a*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*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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac {3}{13} \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx \\ & = -\frac {6 a \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 d}+\frac {1}{143} (36 a) \int \cos ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {8 a^2 \cos ^5(c+d x)}{143 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \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 d}+\frac {1}{143} \left (32 a^2\right ) \int \frac {\cos ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {64 a^3 \cos ^5(c+d x)}{1001 d (a+a \sin (c+d x))^{3/2}}-\frac {8 a^2 \cos ^5(c+d x)}{143 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \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 d}+\frac {\left (128 a^3\right ) \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{1001} \\ & = -\frac {256 a^4 \cos ^5(c+d x)}{5005 d (a+a \sin (c+d x))^{5/2}}-\frac {64 a^3 \cos ^5(c+d x)}{1001 d (a+a \sin (c+d x))^{3/2}}-\frac {8 a^2 \cos ^5(c+d x)}{143 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \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 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.92 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (19559-12600 \cos (2 (c+d x))+385 \cos (4 (c+d x))+28230 \sin (c+d x)-3290 \sin (3 (c+d x)))}{20020 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

-1/20020*(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(19559 - 12600*Cos[2*(c + d*x)]
+ 385*Cos[4*(c + d*x)] + 28230*Sin[c + d*x] - 3290*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.56

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{3} \left (385 \left (\sin ^{4}\left (d x +c \right )\right )+1645 \left (\sin ^{3}\left (d x +c \right )\right )+2765 \left (\sin ^{2}\left (d x +c \right )\right )+2295 \sin \left (d x +c \right )+918\right )}{5005 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(87\)

[In]

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

[Out]

2/5005*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^3*(385*sin(d*x+c)^4+1645*sin(d*x+c)^3+2765*sin(d*x+c)^2+2295*sin(d*x+
c)+918)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.21 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (385 \, a \cos \left (d x + c\right )^{7} - 490 \, a \cos \left (d x + c\right )^{6} - 1015 \, a \cos \left (d x + c\right )^{5} + 20 \, a \cos \left (d x + c\right )^{4} - 32 \, a \cos \left (d x + c\right )^{3} + 64 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) - {\left (385 \, a \cos \left (d x + c\right )^{6} + 875 \, a \cos \left (d x + c\right )^{5} - 140 \, a \cos \left (d x + c\right )^{4} - 160 \, a \cos \left (d x + c\right )^{3} - 192 \, a \cos \left (d x + c\right )^{2} - 256 \, a \cos \left (d x + c\right ) - 512 \, a\right )} \sin \left (d x + c\right ) - 512 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{5005 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/5005*(385*a*cos(d*x + c)^7 - 490*a*cos(d*x + c)^6 - 1015*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) - (385*a*cos(d*x + c)^6 + 875*a*cos(d*x + c)^5 - 140*a*co
s(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)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.04 \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {64 \, \sqrt {2} {\left (770 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 3185 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 5005 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 3575 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1001 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )} \sqrt {a}}{5005 \, d} \]

[In]

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

[Out]

64/5005*sqrt(2)*(770*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 - 3185*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 + 5005*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 - 3575*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 +
1001*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. \[ \int \cos ^4(c+d x) \sin (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]

[In]

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

[Out]

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