\(\int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [45]

Optimal result

Integrand size = 23, antiderivative size = 116 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {152 a^2 \cos (c+d x)}{105 d \sqrt {a+a \sin (c+d x)}}-\frac {38 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 d}+\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{5/2}}{7 a d} \]

[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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2838, 2830, 2726, 2725} \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {152 a^2 \cos (c+d x)}{105 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{5/2}}{7 a d}+\frac {4 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 d}-\frac {38 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{105 d} \]

[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 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 2726

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] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 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)]

Rubi steps \begin{align*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 1.57 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.25 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-735-910 \cos (c+d x)+832 \sqrt {2} \sqrt {1+\cos (c+d x)}-112 \cos (2 (c+d x))+78 \cos (3 (c+d x))+15 \cos (4 (c+d x))+560 \sin (c+d x)-238 \sin (2 (c+d x))-48 \sin (3 (c+d x))+15 \sin (4 (c+d x))\right )}{840 d \left (1+\tan \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

(a*Sec[(c + d*x)/2]^2*Sqrt[a*(1 + Sin[c + d*x])]*(-735 - 910*Cos[c + d*x] + 832*Sqrt[2]*Sqrt[1 + Cos[c + d*x]]
- 112*Cos[2*(c + d*x)] + 78*Cos[3*(c + d*x)] + 15*Cos[4*(c + d*x)] + 560*Sin[c + d*x] - 238*Sin[2*(c + d*x)]
- 48*Sin[3*(c + d*x)] + 15*Sin[4*(c + d*x)]))/(840*d*(1 + Tan[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right ) \left (15 \left (\sin ^{3}\left (d x +c \right )\right )+39 \left (\sin ^{2}\left (d x +c \right )\right )+52 \sin \left (d x +c \right )+104\right )}{105 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(75\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (15 \, a \cos \left (d x + c\right )^{4} + 39 \, a \cos \left (d x + c\right )^{3} - 43 \, a \cos \left (d x + c\right )^{2} - 143 \, a \cos \left (d x + c\right ) + {\left (15 \, a \cos \left (d x + c\right )^{3} - 24 \, a \cos \left (d x + c\right )^{2} - 67 \, a \cos \left (d x + c\right ) + 76 \, a\right )} \sin \left (d x + c\right ) - 76 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[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)

Sympy [F]

\[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sin ^{2}{\left (c + d x \right )}\, dx \]

[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)

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.55 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.07 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {2} {\left (735 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 175 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 63 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 15 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )} \sqrt {a}}{420 \, d} \]

[In]

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

[Out]

1/420*sqrt(2)*(735*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + 175*a*sgn(cos(-1/4*p
i + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + 63*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi +
5/2*d*x + 5/2*c) + 15*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-7/4*pi + 7/2*d*x + 7/2*c))*sqrt(a)/d

Mupad [F(-1)]

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

[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)