Integrand size = 23, antiderivative size = 86 \[ \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {14 a \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d} \] Output:
-14/15*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+4/15*cos(d*x+c)*(a+a*sin(d*x+ c))^(1/2)/d-2/5*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a/d
Time = 0.48 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.36 \[ \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {a (1+\sin (c+d x))} \left (30 \cos \left (\frac {1}{2} (c+d x)\right )+5 \cos \left (\frac {3}{2} (c+d x)\right )-3 \cos \left (\frac {5}{2} (c+d x)\right )-30 \sin \left (\frac {1}{2} (c+d x)\right )+5 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{30 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[Sin[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]
Output:
-1/30*(Sqrt[a*(1 + Sin[c + d*x])]*(30*Cos[(c + d*x)/2] + 5*Cos[(3*(c + d*x ))/2] - 3*Cos[(5*(c + d*x))/2] - 30*Sin[(c + d*x)/2] + 5*Sin[(3*(c + d*x)) /2] + 3*Sin[(5*(c + d*x))/2]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
Time = 0.44 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3238, 27, 3042, 3230, 3042, 3125}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^2(c+d x) \sqrt {a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 \sqrt {a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3238 |
\(\displaystyle \frac {2 \int \frac {1}{2} (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
Input:
Int[Sin[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]
Output:
(-2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*a*d) + ((-14*a^2*Cos[c + d *x])/(3*d*Sqrt[a + a*Sin[c + d*x]]) + (4*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d))/(5*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && Eq Q[a^2 - b^2, 0]
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] + Simp[(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)]
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] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*Si n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && ! LtQ[m, -2^(-1)]
Time = 0.34 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73
method | result | size |
default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a \left (\sin \left (d x +c \right )-1\right ) \left (3 \sin \left (d x +c \right )^{2}+4 \sin \left (d x +c \right )+8\right )}{15 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(63\) |
Input:
int(sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/15*(1+sin(d*x+c))*a*(sin(d*x+c)-1)*(3*sin(d*x+c)^2+4*sin(d*x+c)+8)/cos(d *x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) - 7\right )} \sin \left (d x + c\right ) - 11 \, \cos \left (d x + c\right ) - 7\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
Output:
2/15*(3*cos(d*x + c)^3 - cos(d*x + c)^2 - (3*cos(d*x + c)^2 + 4*cos(d*x + c) - 7)*sin(d*x + c) - 11*cos(d*x + c) - 7)*sqrt(a*sin(d*x + c) + a)/(d*co s(d*x + c) + d*sin(d*x + c) + d)
\[ \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sin ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(sin(d*x+c)**2*(a+a*sin(d*x+c))**(1/2),x)
Output:
Integral(sqrt(a*(sin(c + d*x) + 1))*sin(c + d*x)**2, x)
\[ \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \sin \left (d x + c\right )^{2} \,d x } \] Input:
integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(a*sin(d*x + c) + a)*sin(d*x + c)^2, x)
Time = 0.14 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08 \[ \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (30 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \] Input:
integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
Output:
1/30*sqrt(2)*(30*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + 5*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-3/4*pi + 3/2*d*x + 3/2*c) + 3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-5/4*pi + 5/2*d*x + 5/2 *c))*sqrt(a)/d
Timed out. \[ \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int {\sin \left (c+d\,x\right )}^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \] Input:
int(sin(c + d*x)^2*(a + a*sin(c + d*x))^(1/2),x)
Output:
int(sin(c + d*x)^2*(a + a*sin(c + d*x))^(1/2), x)
\[ \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{2}d x \right ) \] Input:
int(sin(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x)
Output:
sqrt(a)*int(sqrt(sin(c + d*x) + 1)*sin(c + d*x)**2,x)