\(\int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [142]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {5 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \]

[Out]

-1/4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)+5/16*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+5/32*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 = 0.10 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2829, 2729, 2728, 212} \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {5 \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

(5*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)) + (5*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 2729

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 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)]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {5 \int \frac {1}{(a+a \cos (c+d x))^{3/2}} \, dx}{8 a} \\ & = -\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {5 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {5 \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 {5 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {5 \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 {5 \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 {5 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.61 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {40 \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )+2 \sin (c+d x)+5 \sin (2 (c+d x))}{32 d (a (1+\cos (c+d x)))^{5/2}} \]

[In]

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

[Out]

(40*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 + 2*Sin[c + d*x] + 5*Sin[2*(c + d*x)])/(32*d*(a*(1 + Cos[c +
d*x]))^(5/2))

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.63

method	result	size
default	\(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (5 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}-2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(174\)

[In]

int(cos(d*x+c)/(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)*(5*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+5*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 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {5 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (5 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/64*(5*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*sq
rt(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)*(5*cos(d*x + c) + 1)*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]

\[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {\cos {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral(cos(c + d*x)/(a*(cos(c + d*x) + 1))**(5/2), x)

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.64 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.64 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\sqrt {2} {\left (5 \, \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{32 \, {\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} d \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

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

[Out]

-1/32*sqrt(2)*(5*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 3*sqrt(a)*sin(1/2*d*x + 1/2*c))/((sin(1/2*d*x + 1/2*c)^2 - 1
)^2*a^3*d*sgn(cos(1/2*d*x + 1/2*c)))

Mupad [F(-1)]

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

[In]

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

[Out]

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