\(\int (a+a \sin (c+d x))^{3/2} \, dx\) [47]

Optimal result

Integrand size = 14, antiderivative size = 59 \[ \int (a+a \sin (c+d x))^{3/2} \, dx=-\frac {8 a^2 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2726, 2725} \[ \int (a+a \sin (c+d x))^{3/2} \, dx=-\frac {8 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}-\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d} \]

[In]

Int[(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-8*a^2*Cos[c + d*x])/(3*d*Sqrt[a + a*Sin[c + d*x]]) - (2*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {1}{3} (4 a) \int \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {8 a^2 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.51 \[ \int (a+a \sin (c+d x))^{3/2} \, dx=-\frac {(a (1+\sin (c+d x)))^{3/2} \left (9 \cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )-9 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

[In]

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

[Out]

-1/3*((a*(1 + Sin[c + d*x]))^(3/2)*(9*Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2] - 9*Sin[(c + d*x)/2] + Sin[(3*(c
+ d*x))/2]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right ) \left (\sin \left (d x +c \right )+5\right )}{3 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(53\)

[In]

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

[Out]

2/3*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)*(sin(d*x+c)+5)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29 \[ \int (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (a \cos \left (d x + c\right )^{2} + 5 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right ) - 4 \, a\right )} \sin \left (d x + c\right ) + 4 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.14 \[ \int (a+a \sin (c+d x))^{3/2} \, dx=\frac {\sqrt {2} {\left (9 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )} \sqrt {a}}{3 \, d} \]

[In]

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

[Out]

1/3*sqrt(2)*(9*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + a*sgn(cos(-1/4*pi + 1/2*
d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c))*sqrt(a)/d

Mupad [F(-1)]

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

[In]

int((a + a*sin(c + d*x))^(3/2),x)

[Out]

int((a + a*sin(c + d*x))^(3/2), x)