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

Optimal. Leaf size=76 \[ \frac{2 \cos (c+d x)}{a d \sqrt{a \sin (c+d x)+a}}-\frac{2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{3/2} d} \]

[Out]

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

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Rubi [A]  time = 0.0806601, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2679, 2649, 206} \[ \frac{2 \cos (c+d x)}{a d \sqrt{a \sin (c+d x)+a}}-\frac{2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

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{\cos ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac{2 \cos (c+d x)}{a d \sqrt{a+a \sin (c+d x)}}+\frac{2 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{a}\\ &=\frac{2 \cos (c+d x)}{a d \sqrt{a+a \sin (c+d x)}}-\frac{4 \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 )}{a d}\\ &=-\frac{2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{a^{3/2} d}+\frac{2 \cos (c+d x)}{a d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.145741, size = 84, normalized size = 1.11 \[ \frac{2 \cos ^3(c+d x) \left (\sqrt{1-\sin (c+d x)}-\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right )\right )}{d (1-\sin (c+d x))^{3/2} (a (\sin (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.134, size = 94, normalized size = 1.2 \begin{align*} 2\,{\frac{ \left ( 1+\sin \left ( dx+c \right ) \right ) \sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{{a}^{2}\cos \left ( dx+c \right ) \sqrt{a+a\sin \left ( dx+c \right ) }d} \left ( \sqrt{a-a\sin \left ( dx+c \right ) }-\sqrt{a}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/a^2*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*((a-a*sin(d*x+c))^(1/2)-a^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*
x+c))^(1/2)*2^(1/2)/a^(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{\cos \left (d x + c\right )^{2}}{{\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(cos(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.26083, size = 536, normalized size = 7.05 \begin{align*} \frac{\frac{\sqrt{2}{\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac{2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a}{\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt{a}} + 3 \, \cos \left (d x + c\right ) + 2}{\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}} + 2 \, \sqrt{a \sin \left (d x + c\right ) + a}{\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [B]  time = 1.90494, size = 257, normalized size = 3.38 \begin{align*} -\frac{2 \,{\left (\frac{\frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )} - \frac{1}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}}{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} + \frac{\sqrt{2}{\left (2 \, \sqrt{a} \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) + \sqrt{-a}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{\sqrt{-a} a^{\frac{3}{2}}} - \frac{2 \, \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 )}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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