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

Optimal. Leaf size=116 \[ -\frac {152 a^2 \cos (c+d x)}{105 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 a d}+\frac {4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 d}-\frac {38 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{105 d} \]

[Out]

4/35*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-2/7*cos(d*x+c)*(a+a*sin(d*x+c))^(5/2)/a/d-152/105*a^2*cos(d*x+c)/d/(a
+a*sin(d*x+c))^(1/2)-38/105*a*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.14, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2759, 2751, 2647, 2646} \[ -\frac {152 a^2 \cos (c+d x)}{105 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 a d}+\frac {4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 d}-\frac {38 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-152*a^2*Cos[c + d*x])/(105*d*Sqrt[a + a*Sin[c + d*x]]) - (38*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(105*d
) + (4*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(35*d) - (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(7*a*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 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 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)]

Rubi steps

\begin {align*} \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}+\frac {2 \int \left (\frac {5 a}{2}-a \sin (c+d x)\right ) (a+a \sin (c+d x))^{3/2} \, dx}{7 a}\\ &=\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}+\frac {19}{35} \int (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {38 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}+\frac {1}{105} (76 a) \int \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {152 a^2 \cos (c+d x)}{105 d \sqrt {a+a \sin (c+d x)}}-\frac {38 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 141, normalized size = 1.22 \[ \frac {(a (\sin (c+d x)+1))^{3/2} \left (735 \sin \left (\frac {1}{2} (c+d x)\right )-175 \sin \left (\frac {3}{2} (c+d x)\right )-63 \sin \left (\frac {5}{2} (c+d x)\right )+15 \sin \left (\frac {7}{2} (c+d x)\right )-735 \cos \left (\frac {1}{2} (c+d x)\right )-175 \cos \left (\frac {3}{2} (c+d x)\right )+63 \cos \left (\frac {5}{2} (c+d x)\right )+15 \cos \left (\frac {7}{2} (c+d x)\right )\right )}{420 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(3/2)*(-735*Cos[(c + d*x)/2] - 175*Cos[(3*(c + d*x))/2] + 63*Cos[(5*(c + d*x))/2] + 15
*Cos[(7*(c + d*x))/2] + 735*Sin[(c + d*x)/2] - 175*Sin[(3*(c + d*x))/2] - 63*Sin[(5*(c + d*x))/2] + 15*Sin[(7*
(c + d*x))/2]))/(420*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

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fricas [A]  time = 0.48, size = 122, normalized size = 1.05 \[ \frac {2 \, {\left (15 \, a \cos \left (d x + c\right )^{4} + 39 \, a \cos \left (d x + c\right )^{3} - 43 \, a \cos \left (d x + c\right )^{2} - 143 \, a \cos \left (d x + c\right ) + {\left (15 \, a \cos \left (d x + c\right )^{3} - 24 \, a \cos \left (d x + c\right )^{2} - 67 \, a \cos \left (d x + c\right ) + 76 \, a\right )} \sin \left (d x + c\right ) - 76 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

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),x, algorithm="fricas")

[Out]

2/105*(15*a*cos(d*x + c)^4 + 39*a*cos(d*x + c)^3 - 43*a*cos(d*x + c)^2 - 143*a*cos(d*x + c) + (15*a*cos(d*x +
c)^3 - 24*a*cos(d*x + c)^2 - 67*a*cos(d*x + c) + 76*a)*sin(d*x + c) - 76*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*
x + c) + d*sin(d*x + c) + d)

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giac [B]  time = 0.60, size = 226, normalized size = 1.95 \[ \frac {1}{420} \, \sqrt {2} {\left (\frac {21 \, a \cos \left (\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {315 \, a \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} + \frac {15 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {105 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} - \frac {70 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} - \frac {42 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {420 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

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),x, algorithm="giac")

[Out]

1/420*sqrt(2)*(21*a*cos(1/4*pi + 5/2*d*x + 5/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 315*a*cos(1/4*pi + 1
/2*d*x + 1/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d + 15*a*cos(-1/4*pi + 7/2*d*x + 7/2*c)*sgn(cos(-1/4*pi +
1/2*d*x + 1/2*c))/d - 105*a*cos(-1/4*pi + 3/2*d*x + 3/2*c)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d - 70*a*sgn(co
s(-1/4*pi + 1/2*d*x + 1/2*c))*sin(1/4*pi + 3/2*d*x + 3/2*c)/d - 42*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-
1/4*pi + 5/2*d*x + 5/2*c)/d + 420*a*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 = 0.61, size = 75, normalized size = 0.65 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right ) \left (15 \left (\sin ^{3}\left (d x +c \right )\right )+39 \left (\sin ^{2}\left (d x +c \right )\right )+52 \sin \left (d x +c \right )+104\right )}{105 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

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),x)

[Out]

2/105*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)*(15*sin(d*x+c)^3+39*sin(d*x+c)^2+52*sin(d*x+c)+104)/cos(d*x+c)/(a+a*si
n(d*x+c))^(1/2)/d

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

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),x, algorithm="maxima")

[Out]

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

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

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),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sin ^{2}{\left (c + d x \right )}\, dx \]

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),x)

[Out]

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

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