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

Optimal. Leaf size=105 \[ -\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{2 \sin (c+d x)}{a d \sqrt{a \cos (c+d x)+a}}+\frac{\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

(-7*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + Sin[c + d*x]/(
2*d*(a + a*Cos[c + d*x])^(3/2)) + (2*Sin[c + d*x])/(a*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.133844, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2758, 2751, 2649, 206} \[ -\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{2 \sin (c+d x)}{a d \sqrt{a \cos (c+d x)+a}}+\frac{\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + Sin[c + d*x]/(
2*d*(a + a*Cos[c + d*x])^(3/2)) + (2*Sin[c + d*x])/(a*d*Sqrt[a + a*Cos[c + d*x]])

Rule 2758

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

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, 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{\cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=\frac{\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{-\frac{3 a}{2}+2 a \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac{\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{2 \sin (c+d x)}{a d \sqrt{a+a \cos (c+d x)}}-\frac{7 \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=\frac{\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{2 \sin (c+d x)}{a d \sqrt{a+a \cos (c+d x)}}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{2 a d}\\ &=-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{2 \sin (c+d x)}{a d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.45932, size = 164, normalized size = 1.56 \[ \frac{\cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (16 \sin \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right )+16 \cos \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+\frac{1}{\left (\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )\right )^2}-\frac{1}{\left (\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )\right )^2}+14 \log \left (\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )\right )-14 \log \left (\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )\right )\right )}{2 d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^3*(14*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] - 14*Log[Cos[(c + d*x)/4] + Sin[(c + d*x)/4]]
+ 16*Cos[(d*x)/2]*Sin[c/2] + 16*Cos[c/2]*Sin[(d*x)/2] + (Cos[(c + d*x)/4] - Sin[(c + d*x)/4])^(-2) - (Cos[(c
+ d*x)/4] + Sin[(c + d*x)/4])^(-2)))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2))

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Maple [A]  time = 1.456, size = 174, normalized size = 1.7 \begin{align*} -{\frac{1}{4\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 7\,\sqrt{2}\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-8\,\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{a}^{-{\frac{5}{2}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(7*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/
2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^2*a-8*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*
x+1/2*c)^2-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2))/a^(5/2)/sin(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)^2*a)
^(1/2)/d

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]  time = 1.64861, size = 440, normalized size = 4.19 \begin{align*} \frac{7 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, \sqrt{a \cos \left (d x + c\right ) + a}{\left (4 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{8 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(7*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x
+ c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*sqrt(a*co
s(d*x + c) + a)*(4*cos(d*x + c) + 5)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 2.05735, size = 138, normalized size = 1.31 \begin{align*} \frac{\frac{{\left (\frac{\sqrt{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a} + \frac{9 \, \sqrt{2}}{a}\right )} \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}} + \frac{7 \, \sqrt{2} \log \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 |}\right )}{a^{\frac{3}{2}}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((sqrt(2)*tan(1/2*d*x + 1/2*c)^2/a + 9*sqrt(2)/a)*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)
+ 7*sqrt(2)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/a^(3/2))/d