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

Optimal. Leaf size=77 \[ \frac{\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.0569845, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2750, 2649, 206} \[ \frac{\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d} \]

Antiderivative was successfully verified.

[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 2750

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)]

Rule 2649

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 206

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

Rubi steps

\begin{align*} \int \frac{\sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\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 \operatorname{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 \tanh ^{-1}\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]  time = 0.194569, size = 108, normalized size = 1.4 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+(3+3 i) (-1)^{3/4} (\sin (c+d x)+1) \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{2 d (a (\sin (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[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]  time = 0.438, size = 123, normalized size = 1.6 \begin{align*} -{\frac{1}{4\,d\cos \left ( dx+c \right ) } \left ( 3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a\sin \left ( dx+c \right ) +3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a-2\,\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{a} \right ) \sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [B]  time = 1.80934, size = 668, normalized size = 8.68 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (c + d x \right )}}{\left (a \left (\sin{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2