\(\int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [71]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 21, antiderivative size = 77 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cos (c+d x)}{2 d (a+a \sin (c+d x))^{3/2}} \]

[Out]

1/2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-3/4*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(
3/2)/d*2^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2829, 2728, 212} \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d} \]

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2829

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.40 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )+(3+3 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) (1+\sin (c+d x))\right )}{2 d (a (1+\sin (c+d x)))^{3/2}} \]

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2] + (3 + 3*I)*(-1)^(3/4)*ArcTanh[(1/
2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(1 + Sin[c + d*x])))/(2*d*(a*(1 + Sin[c + d*x]))^(3/2))

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.60

method result size
default \(-\frac {\left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sin \left (d x +c \right )+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -2 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {a}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{4 a^{\frac {5}{2}} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) \(123\)

[In]

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

[Out]

-1/4*(3*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a*sin(d*x+c)+3*2^(1/2)*arctanh(1/2*(a-a*si
n(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a-2*(a-a*sin(d*x+c))^(1/2)*a^(1/2))*(-a*(sin(d*x+c)-1))^(1/2)/a^(5/2)/cos(d*x
+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (62) = 124\).

Time = 0.31 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.29 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - 2 \, a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + 2 \, a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (62) = 124\).

Time = 0.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.78 \[ \int \frac {\sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\frac {3 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {3 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{8 \, d} \]

[In]

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

[Out]

1/8*(3*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 3*sqrt(
2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 2*sqrt(2)*sin(-1/4
*pi + 1/2*d*x + 1/2*c)/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)*a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

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

[In]

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

[Out]

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