\(\int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [472]

Optimal result

Integrand size = 31, antiderivative size = 205 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {4 \cos (c+d x)}{165 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^3(c+d x)}{231 a d \sqrt {a+a \sin (c+d x)}}+\frac {14 \cos (c+d x) \sin ^4(c+d x)}{33 a d \sqrt {a+a \sin (c+d x)}}+\frac {8 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{1155 a^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a^2 d}-\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{385 a^3 d} \]

[Out]

-4/385*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a^3/d-4/165*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-2/231*cos(d*x+c)*si
n(d*x+c)^3/a/d/(a+a*sin(d*x+c))^(1/2)+14/33*cos(d*x+c)*sin(d*x+c)^4/a/d/(a+a*sin(d*x+c))^(1/2)+8/1155*cos(d*x+
c)*(a+a*sin(d*x+c))^(1/2)/a^2/d-2/11*cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2)/a^2/d

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2959, 2849, 2838, 2830, 2725, 3125, 3060} \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{385 a^3 d}-\frac {2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 a^2 d}+\frac {8 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{1155 a^2 d}+\frac {14 \sin ^4(c+d x) \cos (c+d x)}{33 a d \sqrt {a \sin (c+d x)+a}}-\frac {2 \sin ^3(c+d x) \cos (c+d x)}{231 a d \sqrt {a \sin (c+d x)+a}}-\frac {4 \cos (c+d x)}{165 a d \sqrt {a \sin (c+d x)+a}} \]

[In]

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

[Out]

(-4*Cos[c + d*x])/(165*a*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^3)/(231*a*d*Sqrt[a + a*Sin
[c + d*x]]) + (14*Cos[c + d*x]*Sin[c + d*x]^4)/(33*a*d*Sqrt[a + a*Sin[c + d*x]]) + (8*Cos[c + d*x]*Sqrt[a + a*
Sin[c + d*x]])/(1155*a^2*d) - (2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*a^2*d) - (4*Cos[c +
d*x]*(a + a*Sin[c + d*x])^(3/2))/(385*a^3*d)

Rule 2725

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x
]])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2830

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

Rule 2838

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-Cos[e + f*x])*(
(a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) -
a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 2849

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[2*n*((b*c + a*d)
/(b*(2*n + 1))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

Rule 2959

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[-2/(a*b*d), Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 2), x], x] + Dist[1/a^2
, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 2)*(1 + Sin[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}
, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 3125

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*
sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(
n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Si
mp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^
(-1)] && NeQ[m + n + 2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \left (1+\sin ^2(c+d x)\right ) \, dx}{a^2}-\frac {2 \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a^2} \\ & = \frac {4 \cos (c+d x) \sin ^4(c+d x)}{9 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a^2 d}+\frac {2 \int \sin ^3(c+d x) \left (\frac {19 a}{2}+\frac {1}{2} a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{11 a^3}-\frac {16 \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{9 a^2} \\ & = \frac {32 \cos (c+d x) \sin ^3(c+d x)}{63 a d \sqrt {a+a \sin (c+d x)}}+\frac {14 \cos (c+d x) \sin ^4(c+d x)}{33 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a^2 d}-\frac {32 \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{21 a^2}+\frac {179 \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{99 a^2} \\ & = -\frac {2 \cos (c+d x) \sin ^3(c+d x)}{231 a d \sqrt {a+a \sin (c+d x)}}+\frac {14 \cos (c+d x) \sin ^4(c+d x)}{33 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a^2 d}+\frac {64 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^3 d}-\frac {64 \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{105 a^3}+\frac {358 \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{231 a^2} \\ & = -\frac {2 \cos (c+d x) \sin ^3(c+d x)}{231 a d \sqrt {a+a \sin (c+d x)}}+\frac {14 \cos (c+d x) \sin ^4(c+d x)}{33 a d \sqrt {a+a \sin (c+d x)}}-\frac {128 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 a^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a^2 d}-\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{385 a^3 d}+\frac {716 \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{1155 a^3}-\frac {32 \int \sqrt {a+a \sin (c+d x)} \, dx}{45 a^2} \\ & = \frac {64 \cos (c+d x)}{45 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^3(c+d x)}{231 a d \sqrt {a+a \sin (c+d x)}}+\frac {14 \cos (c+d x) \sin ^4(c+d x)}{33 a d \sqrt {a+a \sin (c+d x)}}+\frac {8 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{1155 a^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a^2 d}-\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{385 a^3 d}+\frac {358 \int \sqrt {a+a \sin (c+d x)} \, dx}{495 a^2} \\ & = -\frac {4 \cos (c+d x)}{165 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^3(c+d x)}{231 a d \sqrt {a+a \sin (c+d x)}}+\frac {14 \cos (c+d x) \sin ^4(c+d x)}{33 a d \sqrt {a+a \sin (c+d x)}}+\frac {8 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{1155 a^2 d}-\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 a^2 d}-\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{385 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.42 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.50 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \sqrt {a (1+\sin (c+d x))} (-204+140 \cos (2 (c+d x))-475 \sin (c+d x)+105 \sin (3 (c+d x)))}{2310 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*Sqrt[a*(1 + Sin[c + d*x])]*(-204 + 140*Cos[2*(c + d*x)] - 475*Sin[c +
d*x] + 105*Sin[3*(c + d*x)]))/(2310*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.38

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{3} \left (105 \left (\sin ^{3}\left (d x +c \right )\right )+70 \left (\sin ^{2}\left (d x +c \right )\right )+40 \sin \left (d x +c \right )+16\right )}{1155 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(77\)

[In]

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

[Out]

2/1155/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^3*(105*sin(d*x+c)^3+70*sin(d*x+c)^2+40*sin(d*x+c)+16)/cos(d*x+c)/(a+a*s
in(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (105 \, \cos \left (d x + c\right )^{6} - 140 \, \cos \left (d x + c\right )^{5} - 460 \, \cos \left (d x + c\right )^{4} + 274 \, \cos \left (d x + c\right )^{3} + 607 \, \cos \left (d x + c\right )^{2} + {\left (105 \, \cos \left (d x + c\right )^{5} + 245 \, \cos \left (d x + c\right )^{4} - 215 \, \cos \left (d x + c\right )^{3} - 489 \, \cos \left (d x + c\right )^{2} + 118 \, \cos \left (d x + c\right ) + 236\right )} \sin \left (d x + c\right ) - 118 \, \cos \left (d x + c\right ) - 236\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{1155 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

-2/1155*(105*cos(d*x + c)^6 - 140*cos(d*x + c)^5 - 460*cos(d*x + c)^4 + 274*cos(d*x + c)^3 + 607*cos(d*x + c)^
2 + (105*cos(d*x + c)^5 + 245*cos(d*x + c)^4 - 215*cos(d*x + c)^3 - 489*cos(d*x + c)^2 + 118*cos(d*x + c) + 23
6)*sin(d*x + c) - 118*cos(d*x + c) - 236)*sqrt(a*sin(d*x + c) + a)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) +
a^2*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.50 \[ \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {8 \, \sqrt {2} {\left (840 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1540 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 990 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 231 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}\right )}}{1155 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

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

[Out]

-8/1155*sqrt(2)*(840*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 - 1540*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9
+ 990*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 231*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)/(a^2*d*sgn(cos
(-1/4*pi + 1/2*d*x + 1/2*c)))

Mupad [F(-1)]

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

[In]

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

[Out]

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