### 3.6 $$\int \frac{1}{(a+a \sin (c+d x))^{3/2}} \, dx$$

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

[Out]

-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))

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Rubi [A]  time = 0.0419245, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.214, Rules used = {2650, 2649, 206} $-\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + a*Sin[c + d*x])^(-3/2),x]

[Out]

-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 2650

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

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{1}{(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{\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{\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{\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 = 0.171792, size = 108, normalized size = 1.4 $\frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+(1+i) (-1)^{3/4} (\sin (c+d x)+1) \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{2 d (a (\sin (c+d x)+1))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(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] + (1 + 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))

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Maple [A]  time = 0.082, size = 125, normalized size = 1.6 \begin{align*} -{\frac{1}{4\,d\cos \left ( dx+c \right ) } \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{2}\sin \left ( dx+c \right ) +2\,\sqrt{a-a\sin \left ( dx+c \right ) }{a}^{3/2}+\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{2} \right ) \sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{-{\frac{7}{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(1/(a+a*sin(d*x+c))^(3/2),x)

[Out]

-1/4/a^(7/2)*(2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2*sin(d*x+c)+2*(a-a*sin(d*x+c))^(1
/2)*a^(3/2)+2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2)*(-a*(sin(d*x+c)-1))^(1/2)/cos(d*x
+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.76907, size = 666, normalized size = 8.65 \begin{align*} \frac{\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(1/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/8*(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*cos
(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))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.01637, size = 396, normalized size = 5.14 \begin{align*} \frac{\frac{\sqrt{2} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} + \sqrt{a}\right )}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )} + \frac{2 \,{\left (3 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{3} +{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} \sqrt{a} -{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )} a + a^{\frac{3}{2}}\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )} \sqrt{a} - a\right )}^{2} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}}{2 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/2*(sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + sqrt(a))
/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*
x + 1/2*c)^2 + a))^3 + (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*sqrt(a) - (sqrt(a
)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a + a^(3/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sq
rt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*sq
rt(a) - a)^2*a*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d