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

Optimal. Leaf size=128 \[ -\frac {8 a^3 c \cos ^3(c+d x)}{63 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 a^2 c \cos ^3(c+d x)}{21 d \sqrt {a \sin (c+d x)+a}}-\frac {2 c \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 d}+\frac {4 a c \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{21 d} \]

[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

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Rubi [A]  time = 0.35, antiderivative size = 165, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {2976, 2981, 2759, 2751, 2646} \[ \frac {2 a^2 c \sin ^3(c+d x) \cos (c+d x)}{63 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a^2 c \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}+\frac {2 a c \sin ^3(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{9 d}-\frac {2 c \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{21 d}+\frac {4 a c \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{63 d} \]

Antiderivative was successfully verified.

[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 2646

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 2751

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*Sin
[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 2759

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 2976

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)))*S
in[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 2981

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)*Sqr
t[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*} \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 \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*}

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Mathematica [A]  time = 0.81, size = 101, normalized size = 0.79 \[ \frac {a c \sqrt {a (\sin (c+d x)+1)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (-69 \sin (c+d x)+7 \sin (3 (c+d x))+30 \cos (2 (c+d x))-62)}{126 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(-62 + 30*Cos[2*(c + d*x)] - 69*Sin[c
+ d*x] + 7*Sin[3*(c + d*x)]))/(126*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.43, size = 155, normalized size = 1.21 \[ \frac {2 \, {\left (7 \, a c \cos \left (d x + c\right )^{5} - a c \cos \left (d x + c\right )^{4} - 11 \, a c \cos \left (d x + c\right )^{3} + a c \cos \left (d x + c\right )^{2} - 4 \, a c \cos \left (d x + c\right ) - 8 \, a c - {\left (7 \, a c \cos \left (d x + c\right )^{4} + 8 \, a c \cos \left (d x + c\right )^{3} - 3 \, a c \cos \left (d x + c\right )^{2} - 4 \, a c \cos \left (d x + c\right ) - 8 \, a c\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{63 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.28, size = 105, normalized size = 0.82 \[ -\frac {1}{504} \, \sqrt {2} {\left (\frac {9 \, a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {7 \, a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} - \frac {126 \, a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)*(9*a*c*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 7/2*d*x + 7/2*c)/d + 7*a*c*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 9/2*d*x + 9/2*c)/d - 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)/d)*sqrt(a)

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maple [A]  time = 1.07, size = 78, normalized size = 0.61 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{2} c \left (7 \left (\sin ^{3}\left (d x +c \right )\right )+15 \left (\sin ^{2}\left (d x +c \right )\right )+12 \sin \left (d x +c \right )+8\right )}{63 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (c \sin \left (d x + c\right ) - c\right )} \sin \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}\,\left (c-c\,\sin \left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - c \left (\int \left (- a \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )}\right )\, dx + \int a \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{4}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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