\(\int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} (c-c \sin (c+d x)) \, dx\) [12]

Optimal result

Integrand size = 34, antiderivative size = 128 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} (c-c \sin (c+d x)) \, dx=-\frac {8 a^3 c \cos ^3(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a^2 c \cos ^3(c+d x)}{21 d \sqrt {a+a \sin (c+d x)}}+\frac {4 a c \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 c \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 d} \]

[Out]

-8/63*a^3*c*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-2/9*c*cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d-2/21*a^2*c*cos(d
*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)+4/21*a*c*cos(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3055, 3060, 2838, 2830, 2725} \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} (c-c \sin (c+d x)) \, dx=\frac {2 a^2 c \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a^2 c \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}+\frac {2 a c \sin ^3(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{9 d}-\frac {2 c \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{21 d}+\frac {4 a c \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{63 d} \]

[In]

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

[Out]

(-2*a^2*c*Cos[c + d*x])/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (2*a^2*c*Cos[c + d*x]*Sin[c + d*x]^3)/(63*d*Sqrt[a +
a*Sin[c + d*x]]) + (4*a*c*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(63*d) + (2*a*c*Cos[c + d*x]*Sin[c + d*x]^3*S
qrt[a + a*Sin[c + d*x]])/(9*d) - (2*c*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(21*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 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 {2 a c \cos (c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{9 d}+\frac {2}{9} \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {3 a c}{2}-\frac {1}{2} a c \sin (c+d x)\right ) \, dx \\ & = \frac {2 a^2 c \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a c \cos (c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{9 d}+\frac {1}{21} (5 a c) \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {2 a^2 c \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a c \cos (c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{9 d}-\frac {2 c \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{21 d}+\frac {1}{21} (2 c) \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {2 a^2 c \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {4 a c \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{63 d}+\frac {2 a c \cos (c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{9 d}-\frac {2 c \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{21 d}+\frac {1}{9} (a c) \int \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {2 a^2 c \cos (c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a^2 c \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {4 a c \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{63 d}+\frac {2 a c \cos (c+d x) \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{9 d}-\frac {2 c \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{21 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.75 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.54 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} (c-c \sin (c+d x)) \, dx=-\frac {2 a c \sec (c+d x) (-1+\sin (c+d x))^2 \sqrt {a (1+\sin (c+d x))} \left (8+12 \sin (c+d x)+15 \sin ^2(c+d x)+7 \sin ^3(c+d x)\right )}{63 d} \]

[In]

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

[Out]

(-2*a*c*Sec[c + d*x]*(-1 + Sin[c + d*x])^2*Sqrt[a*(1 + Sin[c + d*x])]*(8 + 12*Sin[c + d*x] + 15*Sin[c + d*x]^2
+ 7*Sin[c + d*x]^3))/(63*d)

Maple [A] (verified)

Time = 1.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.61

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{2} c \left (7 \left (\sin ^{3}\left (d x +c \right )\right )+15 \left (\sin ^{2}\left (d x +c \right )\right )+12 \sin \left (d x +c \right )+8\right )}{63 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(78\)
parts	\(\frac {2 c \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right ) \left (15 \left (\sin ^{3}\left (d x +c \right )\right )+39 \left (\sin ^{2}\left (d x +c \right )\right )+52 \sin \left (d x +c \right )+104\right )}{105 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}-\frac {2 c \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right ) \left (35 \left (\sin ^{4}\left (d x +c \right )\right )+85 \left (\sin ^{3}\left (d x +c \right )\right )+102 \left (\sin ^{2}\left (d x +c \right )\right )+136 \sin \left (d x +c \right )+272\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(162\)

[In]

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

[Out]

-2/63*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^2*c*(7*sin(d*x+c)^3+15*sin(d*x+c)^2+12*sin(d*x+c)+8)/cos(d*x+c)/(a+a*s
in(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.21 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} (c-c \sin (c+d x)) \, dx=\frac {2 \, {\left (7 \, a c \cos \left (d x + c\right )^{5} - a c \cos \left (d x + c\right )^{4} - 11 \, a c \cos \left (d x + c\right )^{3} + a c \cos \left (d x + c\right )^{2} - 4 \, a c \cos \left (d x + c\right ) - 8 \, a c - {\left (7 \, a c \cos \left (d x + c\right )^{4} + 8 \, a c \cos \left (d x + c\right )^{3} - 3 \, a c \cos \left (d x + c\right )^{2} - 4 \, a c \cos \left (d x + c\right ) - 8 \, a c\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{63 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/63*(7*a*c*cos(d*x + c)^5 - a*c*cos(d*x + c)^4 - 11*a*c*cos(d*x + c)^3 + a*c*cos(d*x + c)^2 - 4*a*c*cos(d*x +
c) - 8*a*c - (7*a*c*cos(d*x + c)^4 + 8*a*c*cos(d*x + c)^3 - 3*a*c*cos(d*x + c)^2 - 4*a*c*cos(d*x + c) - 8*a*c
)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

Sympy [F]

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

[In]

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

[Out]

-c*(Integral(-a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**2, x) + Integral(a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x
)**4, x))

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.77 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} (c-c \sin (c+d x)) \, dx=\frac {\sqrt {2} {\left (126 \, a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 7 \, a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )} \sqrt {a}}{504 \, d} \]

[In]

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

[Out]

1/504*sqrt(2)*(126*a*c*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) - 9*a*c*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x + 7/2*c) - 7*a*c*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. \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} (c-c \sin (c+d x)) \, dx=\int {\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}\,\left (c-c\,\sin \left (c+d\,x\right )\right ) \,d x \]

[In]

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

[Out]

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