\(\int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [322]

Optimal result

Integrand size = 31, antiderivative size = 193 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {76 a \cos (c+d x)}{495 d \sqrt {a+a \sin (c+d x)}}-\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {152 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3465 d}+\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {76 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{1155 a d} \]

[Out]

-76/1155*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a/d-76/495*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-38/693*a*cos(d*x+c
)*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)+2/99*a*cos(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)+152/3465*cos(d
*x+c)*(a+a*sin(d*x+c))^(1/2)/d+2/11*cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2958, 3055, 3060, 2849, 2838, 2830, 2725} \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}+\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{99 d \sqrt {a \sin (c+d x)+a}}-\frac {38 a \sin ^3(c+d x) \cos (c+d x)}{693 d \sqrt {a \sin (c+d x)+a}}-\frac {76 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{1155 a d}+\frac {152 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3465 d}-\frac {76 a \cos (c+d x)}{495 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Cos[c + d*x]^2*Sin[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-76*a*Cos[c + d*x])/(495*d*Sqrt[a + a*Sin[c + d*x]]) - (38*a*Cos[c + d*x]*Sin[c + d*x]^3)/(693*d*Sqrt[a + a*S
in[c + d*x]]) + (2*a*Cos[c + d*x]*Sin[c + d*x]^4)/(99*d*Sqrt[a + a*Sin[c + d*x]]) + (152*Cos[c + d*x]*Sqrt[a +
a*Sin[c + d*x]])/(3465*d) + (2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*d) - (76*Cos[c + d*x
]*(a + a*Sin[c + d*x])^(3/2))/(1155*a*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 2958

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

Rule 3055

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*Sin[e + f*x], x], 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] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sin ^3(c+d x) (a-a \sin (c+d x)) (a+a \sin (c+d x))^{3/2} \, dx}{a^2} \\ & = \frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}+\frac {2 \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {3 a^2}{2}-\frac {1}{2} a^2 \sin (c+d x)\right ) \, dx}{11 a^2} \\ & = \frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 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 d}+\frac {19}{99} \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 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 d}+\frac {38}{231} \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 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 d}-\frac {76 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{1155 a d}+\frac {76 \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{1155 a} \\ & = -\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {152 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3465 d}+\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {76 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{1155 a d}+\frac {38}{495} \int \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {76 a \cos (c+d x)}{495 d \sqrt {a+a \sin (c+d x)}}-\frac {38 a \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {152 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3465 d}+\frac {2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {76 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{1155 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.56 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \sqrt {a (1+\sin (c+d x))} (5657-3540 \cos (2 (c+d x))+315 \cos (4 (c+d x))+7638 \sin (c+d x)-1330 \sin (3 (c+d x)))}{13860 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[Cos[c + d*x]^2*Sin[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

-1/13860*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(5657 - 3540*Cos[2*(c + d*x)] + 3
15*Cos[4*(c + d*x)] + 7638*Sin[c + d*x] - 1330*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.44

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right )^{2} \left (315 \left (\sin ^{4}\left (d x +c \right )\right )+665 \left (\sin ^{3}\left (d x +c \right )\right )+570 \left (\sin ^{2}\left (d x +c \right )\right )+456 \sin \left (d x +c \right )+304\right )}{3465 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(85\)

[In]

int(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3465*(1+sin(d*x+c))*a*(sin(d*x+c)-1)^2*(315*sin(d*x+c)^4+665*sin(d*x+c)^3+570*sin(d*x+c)^2+456*sin(d*x+c)+3
04)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (315 \, \cos \left (d x + c\right )^{6} + 350 \, \cos \left (d x + c\right )^{5} - 500 \, \cos \left (d x + c\right )^{4} - 586 \, \cos \left (d x + c\right )^{3} + 17 \, \cos \left (d x + c\right )^{2} + {\left (315 \, \cos \left (d x + c\right )^{5} - 35 \, \cos \left (d x + c\right )^{4} - 535 \, \cos \left (d x + c\right )^{3} + 51 \, \cos \left (d x + c\right )^{2} + 68 \, \cos \left (d x + c\right ) + 136\right )} \sin \left (d x + c\right ) - 68 \, \cos \left (d x + c\right ) - 136\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3465 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/3465*(315*cos(d*x + c)^6 + 350*cos(d*x + c)^5 - 500*cos(d*x + c)^4 - 586*cos(d*x + c)^3 + 17*cos(d*x + c)^2
+ (315*cos(d*x + c)^5 - 35*cos(d*x + c)^4 - 535*cos(d*x + c)^3 + 51*cos(d*x + c)^2 + 68*cos(d*x + c) + 136)*si
n(d*x + c) - 68*cos(d*x + c) - 136)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.81 \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {8 \, \sqrt {2} {\left (2520 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 7700 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 8910 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4851 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1155 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )} \sqrt {a}}{3465 \, d} \]

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^3*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

8/3465*sqrt(2)*(2520*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 - 7700*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 + 8910*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*p
i + 1/2*d*x + 1/2*c)^7 - 4851*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 1155*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. \[ \int \cos ^2(c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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