\(\int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [32]

Optimal result

Integrand size = 23, antiderivative size = 158 \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {32 a \cos (c+d x)}{45 d \sqrt {a+a \sin (c+d x)}}-\frac {16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {64 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}-\frac {32 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a d} \]

[Out]

-32/105*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a/d-32/45*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-16/63*a*cos(d*x+c)*s
in(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-2/9*a*cos(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)+64/315*cos(d*x+c)*
(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2849, 2838, 2830, 2725} \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 a \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}-\frac {16 a \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt {a \sin (c+d x)+a}}-\frac {32 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{105 a d}+\frac {64 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{315 d}-\frac {32 a \cos (c+d x)}{45 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-32*a*Cos[c + d*x])/(45*d*Sqrt[a + a*Sin[c + d*x]]) - (16*a*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a + a*Sin
[c + d*x]]) - (2*a*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (64*Cos[c + d*x]*Sqrt[a + a*S
in[c + d*x]])/(315*d) - (32*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(105*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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {8}{9} \int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {16}{21} \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}-\frac {32 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a d}+\frac {32 \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{105 a} \\ & = -\frac {16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {64 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}-\frac {32 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a d}+\frac {16}{45} \int \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {32 a \cos (c+d x)}{45 d \sqrt {a+a \sin (c+d x)}}-\frac {16 a \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {64 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 d}-\frac {32 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.04 \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {a (1+\sin (c+d x))} \left (-1890 \cos \left (\frac {1}{2} (c+d x)\right )-420 \cos \left (\frac {3}{2} (c+d x)\right )+252 \cos \left (\frac {5}{2} (c+d x)\right )+45 \cos \left (\frac {7}{2} (c+d x)\right )-35 \cos \left (\frac {9}{2} (c+d x)\right )+1890 \sin \left (\frac {1}{2} (c+d x)\right )-420 \sin \left (\frac {3}{2} (c+d x)\right )-252 \sin \left (\frac {5}{2} (c+d x)\right )+45 \sin \left (\frac {7}{2} (c+d x)\right )+35 \sin \left (\frac {9}{2} (c+d x)\right )\right )}{2520 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

Integrate[Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*(-1890*Cos[(c + d*x)/2] - 420*Cos[(3*(c + d*x))/2] + 252*Cos[(5*(c + d*x))/2] + 45
*Cos[(7*(c + d*x))/2] - 35*Cos[(9*(c + d*x))/2] + 1890*Sin[(c + d*x)/2] - 420*Sin[(3*(c + d*x))/2] - 252*Sin[(
5*(c + d*x))/2] + 45*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520*d*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2]))

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.53

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right ) \left (35 \left (\sin ^{4}\left (d x +c \right )\right )+40 \left (\sin ^{3}\left (d x +c \right )\right )+48 \left (\sin ^{2}\left (d x +c \right )\right )+64 \sin \left (d x +c \right )+128\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(83\)

[In]

int(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/315*(1+sin(d*x+c))*a*(sin(d*x+c)-1)*(35*sin(d*x+c)^4+40*sin(d*x+c)^3+48*sin(d*x+c)^2+64*sin(d*x+c)+128)/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 = 132, normalized size of antiderivative = 0.84 \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{4} - 118 \, \cos \left (d x + c\right )^{3} + 26 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} + 40 \, \cos \left (d x + c\right )^{3} - 78 \, \cos \left (d x + c\right )^{2} - 104 \, \cos \left (d x + c\right ) + 107\right )} \sin \left (d x + c\right ) + 211 \, \cos \left (d x + c\right ) + 107\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-2/315*(35*cos(d*x + c)^5 - 5*cos(d*x + c)^4 - 118*cos(d*x + c)^3 + 26*cos(d*x + c)^2 - (35*cos(d*x + c)^4 + 4
0*cos(d*x + c)^3 - 78*cos(d*x + c)^2 - 104*cos(d*x + c) + 107)*sin(d*x + c) + 211*cos(d*x + c) + 107)*sqrt(a*s
in(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

Sympy [F]

\[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sin ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(sin(d*x+c)**4*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*sin(c + d*x)**4, x)

Maxima [F]

\[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \sin \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.93 \[ \int \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (1890 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 420 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 252 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 45 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 35 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )} \sqrt {a}}{2520 \, d} \]

[In]

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/2520*sqrt(2)*(1890*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + 420*sgn(cos(-1/4*pi
+ 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + 252*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi + 5/2
*d*x + 5/2*c) + 45*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x + 7/2*c) + 35*sgn(cos(-1/4*pi + 1
/2*d*x + 1/2*c))*sin(-9/4*pi + 9/2*d*x + 9/2*c))*sqrt(a)/d

Mupad [F(-1)]

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

[In]

int(sin(c + d*x)^4*(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(sin(c + d*x)^4*(a + a*sin(c + d*x))^(1/2), x)