\(\int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [139]

Optimal result

Integrand size = 23, antiderivative size = 183 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {163 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {197 \sin (c+d x)}{24 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {95 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d} \]

[Out]

-1/4*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-17/16*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)
+163/32*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)-197/24*sin(d*x+c)/a^2
/d/(a+a*cos(d*x+c))^(1/2)+95/48*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a^3/d

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2844, 3056, 3047, 3102, 2830, 2728, 212} \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {163 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {95 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{48 a^3 d}-\frac {197 \sin (c+d x)}{24 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac {17 \sin (c+d x) \cos ^2(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[In]

Int[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(163*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (Cos[c + d*x
]^3*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (17*Cos[c + d*x]^2*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d
*x])^(3/2)) - (197*Sin[c + d*x])/(24*a^2*d*Sqrt[a + a*Cos[c + d*x]]) + (95*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*
x])/(48*a^3*d)

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 2830

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*S
in[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 2844

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

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3056

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3102

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {\int \frac {\cos ^2(c+d x) \left (3 a-\frac {11}{2} a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {\int \frac {\cos (c+d x) \left (17 a^2-\frac {95}{4} a^2 \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {\int \frac {17 a^2 \cos (c+d x)-\frac {95}{4} a^2 \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {95 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}-\frac {\int \frac {-\frac {95 a^3}{8}+\frac {197}{4} a^3 \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{12 a^5} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {197 \sin (c+d x)}{24 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {95 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}+\frac {163 \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2} \\ & = -\frac {\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {197 \sin (c+d x)}{24 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {95 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}-\frac {163 \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 {163 \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 {\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {197 \sin (c+d x)}{24 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {95 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\left (-978 \sqrt {2} \text {arctanh}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+\sqrt {1-\cos (c+d x)} (379+479 \cos (c+d x)+80 \cos (2 (c+d x))-8 \cos (3 (c+d x)))\right ) \sin (c+d x)}{48 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{5/2}} \]

[In]

Integrate[Cos[c + d*x]^4/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

-1/48*((-978*Sqrt[2]*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2]]*Cos[(c + d*x)/2]^4 + Sqrt[1 - Cos[c + d*x]]*(379 + 479*
Cos[c + d*x] + 80*Cos[2*(c + d*x)] - 8*Cos[3*(c + d*x)]))*Sin[c + d*x])/(d*Sqrt[1 - Cos[c + d*x]]*(a*(1 + Cos[
c + d*x]))^(5/2))

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.32

method	result	size
default	\(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (128 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+489 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-512 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-87 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}+6 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(242\)

[In]

int(cos(d*x+c)^4/(a+cos(d*x+c)*a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/96/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(128*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*c
os(1/2*d*x+1/2*c)^6+489*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)^4-512*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)^4-87*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)+6*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(7/2)/si
n(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {489 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (32 \, \cos \left (d x + c\right )^{3} - 160 \, \cos \left (d x + c\right )^{2} - 503 \, \cos \left (d x + c\right ) - 299\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/192*(489*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
*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*(32*cos(d*x + c)^3 - 160*cos(d*x + c)^2 - 503*cos(d*x + c) - 299)*sqrt(a*cos(d*x + c) + a)*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. \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 3.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left (29 \, \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {128 \, \sqrt {2} {\left (a^{\frac {13}{2}} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{\frac {13}{2}} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{9} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{96 \, d} \]

[In]

integrate(cos(d*x+c)^4/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/96*(3*sqrt(2)*(29*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 27*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))) - 128*sqrt(2)*(a^(13/2)*sin(1/2*d*x + 1/2*c)^3 + 3*a^(13/2)*sin(1/2*d*x
+ 1/2*c))/(a^9*sgn(cos(1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(cos(c + d*x)^4/(a + a*cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^4/(a + a*cos(c + d*x))^(5/2), x)