\(\int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [141]

Optimal result

Integrand size = 23, antiderivative size = 107 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {19 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \]

[Out]

1/4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-13/16*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+19/32*arctanh(1/2*sin(d*x+
c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2837, 2829, 2728, 212} \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {19 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {13 \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}+\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[In]

Int[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(19*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Sin[c + d*x]/
(4*d*(a + a*Cos[c + d*x])^(5/2)) - (13*Sin[c + d*x])/(16*a*d*(a + a*Cos[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)]

Rule 2837

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)]

Rubi steps \begin{align*} \text {integral}& = \frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {-\frac {5 a}{2}+4 a \cos (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = \frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {19 \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2} \\ & = \frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {19 \text {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 )}{16 a^2 d} \\ & = \frac {19 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {13 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\left (76 \sqrt {2} \text {arctanh}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )-2 \sqrt {1-\cos (c+d x)} (9+13 \cos (c+d x))\right ) \sin (c+d x)}{32 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{5/2}} \]

[In]

Integrate[Cos[c + d*x]^2/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

((76*Sqrt[2]*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2]]*Cos[(c + d*x)/2]^4 - 2*Sqrt[1 - Cos[c + d*x]]*(9 + 13*Cos[c + d
*x]))*Sin[c + d*x])/(32*d*Sqrt[1 - Cos[c + d*x]]*(a*(1 + Cos[c + d*x]))^(5/2))

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.63

method	result	size
default	\(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (19 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-13 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}+2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(174\)

[In]

int(cos(d*x+c)^2/(a+cos(d*x+c)*a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/32/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(19*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))*a*cos(1/2*d*x+1/2*c)^4-13*cos(1/2*d*x+1/2*c)^2*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*2
^(1/2)*a^(1/2)+2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(7/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2
*c)^2)^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (88) = 176\).

Time = 0.26 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.76 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {19 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (13 \, \cos \left (d x + c\right ) + 9\right )} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/64*(19*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*s
qrt(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*cos(d*x + c) + a)*(13*cos(d*x + c) + 9)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d
*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 1.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\frac {19 \, \sqrt {2} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {19 \, \sqrt {2} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} {\left (13 \, \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11 \, \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \]

[In]

integrate(cos(d*x+c)^2/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/64*(19*sqrt(2)*log(sin(1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos(1/2*d*x + 1/2*c))) - 19*sqrt(2)*log(-sin(1/2*d
*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos(1/2*d*x + 1/2*c))) + 2*sqrt(2)*(13*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 11*sqrt(
a)*sin(1/2*d*x + 1/2*c))/((sin(1/2*d*x + 1/2*c)^2 - 1)^2*a^3*sgn(cos(1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(cos(c + d*x)^2/(a + a*cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^2/(a + a*cos(c + d*x))^(5/2), x)