\(\int \sin (c+d x) (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)) \, dx\) [938]

Optimal result

Integrand size = 29, antiderivative size = 61 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\frac {c x}{2}-\frac {a \cos (c+d x)}{d}+\frac {2 b E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d}-\frac {c \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

1/2*c*x-a*cos(d*x+c)/d-2*b*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1
/4*Pi+1/2*d*x),2^(1/2))/d-1/2*c*cos(d*x+c)*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4480, 4486, 2719, 2718, 2715, 8} \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=-\frac {a \cos (c+d x)}{d}+\frac {2 b E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d}-\frac {c \sin (c+d x) \cos (c+d x)}{2 d}+\frac {c x}{2} \]

[In]

Int[Sin[c + d*x]*(a + b/Sqrt[Sin[c + d*x]] + c*Sin[c + d*x]),x]

[Out]

(c*x)/2 - (a*Cos[c + d*x])/d + (2*b*EllipticE[(c - Pi/2 + d*x)/2, 2])/d - (c*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 4480

Int[(u_)*((a_) + (b_.)*(F_)[(d_.) + (e_.)*(x_)]^(p_.) + (c_.)*(F_)[(d_.) + (e_.)*(x_)]^(q_.))^(n_.), x_Symbol]
:> Int[ActivateTrig[u*F[d + e*x]^(n*p)*(b + a/F[d + e*x]^p + c*F[d + e*x]^(q - p))^n], x] /; FreeQ[{a, b, c,
d, e, p, q}, x] && InertTrigQ[F] && IntegerQ[n] && NegQ[p]

Rule 4486

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps \begin{align*} \text {integral}& = \int \sqrt {\sin (c+d x)} \left (b+a \sqrt {\sin (c+d x)}+c \sin ^{\frac {3}{2}}(c+d x)\right ) \, dx \\ & = \int \left (b \sqrt {\sin (c+d x)}+a \sin (c+d x)+c \sin ^2(c+d x)\right ) \, dx \\ & = a \int \sin (c+d x) \, dx+b \int \sqrt {\sin (c+d x)} \, dx+c \int \sin ^2(c+d x) \, dx \\ & = -\frac {a \cos (c+d x)}{d}+\frac {2 b E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d}-\frac {c \cos (c+d x) \sin (c+d x)}{2 d}+\frac {1}{2} c \int 1 \, dx \\ & = \frac {c x}{2}-\frac {a \cos (c+d x)}{d}+\frac {2 b E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d}-\frac {c \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\frac {-4 a \cos (c+d x)-8 b E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+c (2 c+2 d x-\sin (2 (c+d x)))}{4 d} \]

[In]

Integrate[Sin[c + d*x]*(a + b/Sqrt[Sin[c + d*x]] + c*Sin[c + d*x]),x]

[Out]

(-4*a*Cos[c + d*x] - 8*b*EllipticE[(-2*c + Pi - 2*d*x)/4, 2] + c*(2*c + 2*d*x - Sin[2*(c + d*x)]))/(4*d)

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.16

method	result	size
default	\(-\frac {a \cos \left (d x +c \right )}{d}+\frac {c \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {b \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{\cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}\)	\(132\)
parts	\(-\frac {a \cos \left (d x +c \right )}{d}+\frac {c \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {b \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{\cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}\)	\(132\)

[In]

int(sin(d*x+c)*(a+c*sin(d*x+c)+b/sin(d*x+c)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

-a*cos(d*x+c)/d+c/d*(-1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)-b*(sin(d*x+c)+1)^(1/2)*(-2*sin(d*x+c)+2)^(1/2)*
(-sin(d*x+c))^(1/2)*(2*EllipticE((sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-EllipticF((sin(d*x+c)+1)^(1/2),1/2*2^(1/2))
)/cos(d*x+c)/sin(d*x+c)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.77 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=-\frac {c \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 i \, \sqrt {2} \sqrt {-i} b {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 2 i \, \sqrt {2} \sqrt {i} b {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - c \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) + 2 \, a \cos \left (d x + c\right )}{2 \, d} \]

[In]

integrate(sin(d*x+c)*(a+c*sin(d*x+c)+b/sin(d*x+c)^(1/2)),x, algorithm="fricas")

[Out]

-1/2*(c*cos(d*x + c)*sin(d*x + c) - 2*I*sqrt(2)*sqrt(-I)*b*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos
(d*x + c) + I*sin(d*x + c))) + 2*I*sqrt(2)*sqrt(I)*b*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x +
c) - I*sin(d*x + c))) - c*arctan(sin(d*x + c)/cos(d*x + c)) + 2*a*cos(d*x + c))/d

Sympy [F]

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

[In]

integrate(sin(d*x+c)*(a+c*sin(d*x+c)+b/sin(d*x+c)**(1/2)),x)

[Out]

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

Maxima [F]

\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\int { {\left (c \sin \left (d x + c\right ) + a + \frac {b}{\sqrt {\sin \left (d x + c\right )}}\right )} \sin \left (d x + c\right ) \,d x } \]

[In]

integrate(sin(d*x+c)*(a+c*sin(d*x+c)+b/sin(d*x+c)^(1/2)),x, algorithm="maxima")

[Out]

1/4*(2*c*d*x - 4*a*cos(d*x + c) + 2*d*integrate(-(((b*cos(3/2*d*x + 3/2*c) - b*cos(1/2*d*x + 1/2*c) - b*sin(3/
2*d*x + 3/2*c) - b*sin(1/2*d*x + 1/2*c))*cos(1/2*arctan2(sin(d*x + c), -cos(d*x + c) + 1)) - (b*cos(3/2*d*x +
3/2*c) - b*cos(1/2*d*x + 1/2*c) + b*sin(3/2*d*x + 3/2*c) + b*sin(1/2*d*x + 1/2*c))*sin(1/2*arctan2(sin(d*x + c
), -cos(d*x + c) + 1)))*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c) + 1)) + ((b*cos(3/2*d*x + 3/2*c) - b*cos(1/
2*d*x + 1/2*c) + b*sin(3/2*d*x + 3/2*c) + b*sin(1/2*d*x + 1/2*c))*cos(1/2*arctan2(sin(d*x + c), -cos(d*x + c)
+ 1)) + (b*cos(3/2*d*x + 3/2*c) - b*cos(1/2*d*x + 1/2*c) - b*sin(3/2*d*x + 3/2*c) - b*sin(1/2*d*x + 1/2*c))*si
n(1/2*arctan2(sin(d*x + c), -cos(d*x + c) + 1)))*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c) + 1)))/((cos(d*x +
c)^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1)^(1/4)*(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*cos(d*x + c) + 1)^(1/4
)), x) - c*sin(2*d*x + 2*c))/d

Giac [F]

\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\int { {\left (c \sin \left (d x + c\right ) + a + \frac {b}{\sqrt {\sin \left (d x + c\right )}}\right )} \sin \left (d x + c\right ) \,d x } \]

[In]

integrate(sin(d*x+c)*(a+c*sin(d*x+c)+b/sin(d*x+c)^(1/2)),x, algorithm="giac")

[Out]

integrate((c*sin(d*x + c) + a + b/sqrt(sin(d*x + c)))*sin(d*x + c), x)

Mupad [B] (verification not implemented)

Time = 27.98 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.84 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\frac {c\,x}{2}-\frac {c\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {a\,\cos \left (c+d\,x\right )}{d}+\frac {2\,b\,\mathrm {E}\left (\frac {c}{2}-\frac {\pi }{4}+\frac {d\,x}{2}\middle |2\right )}{d} \]

[In]

int(sin(c + d*x)*(a + c*sin(c + d*x) + b/sin(c + d*x)^(1/2)),x)

[Out]

(c*x)/2 - (c*sin(2*c + 2*d*x))/(4*d) - (a*cos(c + d*x))/d + (2*b*ellipticE(c/2 - pi/4 + (d*x)/2, 2))/d