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

3.5.54.1 Optimal result
3.5.54.2 Mathematica [A] (verified)
3.5.54.3 Rubi [A] (verified)
3.5.54.4 Maple [A] (verified)
3.5.54.5 Fricas [A] (verification not implemented)
3.5.54.6 Sympy [F(-1)]
3.5.54.7 Maxima [F]
3.5.54.8 Giac [A] (verification not implemented)
3.5.54.9 Mupad [F(-1)]

3.5.54.1 Optimal result

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

output
-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-8/1 
43*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*sin 
(d*x+c))^(1/2)/d
 
3.5.54.2 Mathematica [A] (verified)

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

input
Integrate[Cos[c + d*x]^4*Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
 
output
-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]))
 
3.5.54.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x) \cos (c+d x)^4 (a \sin (c+d x)+a)^{3/2}dx\)

\(\Big \downarrow \) 3335

\(\displaystyle \frac {3}{13} \int \cos ^4(c+d x) (\sin (c+d x) a+a)^{3/2}dx-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{13} \int \cos (c+d x)^4 (\sin (c+d x) a+a)^{3/2}dx-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \int \cos ^4(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \int \cos (c+d x)^4 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos ^4(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \int \frac {\cos (c+d x)^4}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos ^4(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\cos (c+d x)^4}{(\sin (c+d x) a+a)^{3/2}}dx-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {3}{13} \left (\frac {12}{11} a \left (\frac {8}{9} a \left (-\frac {8 a^2 \cos ^5(c+d x)}{35 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{3/2}}\right )-\frac {2 a \cos ^5(c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\right )-\frac {2 \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d}\)

input
Int[Cos[c + d*x]^4*Sin[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]
 
output
(-2*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*d) + (3*((-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*S 
in[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
 

3.5.54.3.1 Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3152
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]
 

rule 3153
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]
 

rule 3335
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]
 
3.5.54.4 Maple [A] (verified)

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

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

input
int(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE 
)
 
output
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
 
3.5.54.5 Fricas [A] (verification not implemented)

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

input
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="fri 
cas")
 
output
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*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)
 
3.5.54.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)**4*sin(d*x+c)*(a+a*sin(d*x+c))**(3/2),x)
 
output
Timed out
 
3.5.54.7 Maxima [F]

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

input
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="max 
ima")
 
output
integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^4*sin(d*x + c), x)
 
3.5.54.8 Giac [A] (verification not implemented)

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

input
integrate(cos(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="gia 
c")
 
output
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*p 
i + 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))*s 
in(-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
 
3.5.54.9 Mupad [F(-1)]

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

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