3.63 \(\int \frac {\cos ^5(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=153 \[ -\frac {152 \sin (c+d x)}{15 a^3 d}-\frac {76 \sin (c+d x) \cos ^2(c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {13 \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac {13 x}{2 a^3}-\frac {\sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {11 \sin (c+d x) \cos ^3(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

13/2*x/a^3-152/15*sin(d*x+c)/a^3/d+13/2*cos(d*x+c)*sin(d*x+c)/a^3/d-1/5*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x
+c))^3-11/15*cos(d*x+c)^3*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2-76/15*cos(d*x+c)^2*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c
))

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Rubi [A]  time = 0.26, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2765, 2977, 2734} \[ -\frac {152 \sin (c+d x)}{15 a^3 d}-\frac {76 \sin (c+d x) \cos ^2(c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {13 \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac {13 x}{2 a^3}-\frac {\sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {11 \sin (c+d x) \cos ^3(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + a*Cos[c + d*x])^3,x]

[Out]

(13*x)/(2*a^3) - (152*Sin[c + d*x])/(15*a^3*d) + (13*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d) - (Cos[c + d*x]^4*Si
n[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - (11*Cos[c + d*x]^3*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) -
(76*Cos[c + d*x]^2*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

Rule 2734

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

Rule 2765

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 2977

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])

Rubi steps

\begin {align*} \int \frac {\cos ^5(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {\cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {\int \frac {\cos ^3(c+d x) (4 a-7 a \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {11 \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {\int \frac {\cos ^2(c+d x) \left (33 a^2-43 a^2 \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac {\cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {11 \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {76 \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {\int \cos (c+d x) \left (152 a^3-195 a^3 \cos (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac {13 x}{2 a^3}-\frac {152 \sin (c+d x)}{15 a^3 d}+\frac {13 \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {\cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {11 \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {76 \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 173, normalized size = 1.13 \[ \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (15 (-12 \sin (c+d x)+\sin (2 (c+d x))+26 d x) \cos ^5\left (\frac {1}{2} (c+d x)\right )+46 \tan \left (\frac {c}{2}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )-3 \tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )-3 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-508 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+46 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )\right )}{15 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + a*Cos[c + d*x])^3,x]

[Out]

(2*Cos[(c + d*x)/2]*(-3*Sec[c/2]*Sin[(d*x)/2] + 46*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] - 508*Cos[(c + d*x
)/2]^4*Sec[c/2]*Sin[(d*x)/2] + 15*Cos[(c + d*x)/2]^5*(26*d*x - 12*Sin[c + d*x] + Sin[2*(c + d*x)]) - 3*Cos[(c
+ d*x)/2]*Tan[c/2] + 46*Cos[(c + d*x)/2]^3*Tan[c/2]))/(15*a^3*d*(1 + Cos[c + d*x])^3)

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fricas [A]  time = 1.02, size = 135, normalized size = 0.88 \[ \frac {195 \, d x \cos \left (d x + c\right )^{3} + 585 \, d x \cos \left (d x + c\right )^{2} + 585 \, d x \cos \left (d x + c\right ) + 195 \, d x + {\left (15 \, \cos \left (d x + c\right )^{4} - 45 \, \cos \left (d x + c\right )^{3} - 479 \, \cos \left (d x + c\right )^{2} - 717 \, \cos \left (d x + c\right ) - 304\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/30*(195*d*x*cos(d*x + c)^3 + 585*d*x*cos(d*x + c)^2 + 585*d*x*cos(d*x + c) + 195*d*x + (15*cos(d*x + c)^4 -
45*cos(d*x + c)^3 - 479*cos(d*x + c)^2 - 717*cos(d*x + c) - 304)*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)

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giac [A]  time = 0.38, size = 113, normalized size = 0.74 \[ \frac {\frac {390 \, {\left (d x + c\right )}}{a^{3}} - \frac {60 \, {\left (7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(390*(d*x + c)/a^3 - 60*(7*tan(1/2*d*x + 1/2*c)^3 + 5*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)
^2*a^3) - (3*a^12*tan(1/2*d*x + 1/2*c)^5 - 40*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*a^12*tan(1/2*d*x + 1/2*c))/a^1
5)/d

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maple [A]  time = 0.06, size = 141, normalized size = 0.92 \[ -\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d \,a^{3}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}-\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {13 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a+a*cos(d*x+c))^3,x)

[Out]

-1/20/d/a^3*tan(1/2*d*x+1/2*c)^5+2/3/d/a^3*tan(1/2*d*x+1/2*c)^3-31/4/d/a^3*tan(1/2*d*x+1/2*c)-7/d/a^3/(1+tan(1
/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-5/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+13/d/a^3*arctan(
tan(1/2*d*x+1/2*c))

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maxima [A]  time = 1.26, size = 184, normalized size = 1.20 \[ -\frac {\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) - 4
0*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 780*arctan(sin(d*x + c)/(
cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 0.47, size = 137, normalized size = 0.90 \[ -\frac {3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-46\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+508\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-390\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (c+d\,x\right )}{60\,a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(a + a*cos(c + d*x))^3,x)

[Out]

-(3*sin(c/2 + (d*x)/2) - 46*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) + 508*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)
/2) + 420*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) - 120*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2) - 390*cos(c/2
+ (d*x)/2)^5*(c + d*x))/(60*a^3*d*cos(c/2 + (d*x)/2)^5)

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sympy [A]  time = 13.65, size = 473, normalized size = 3.09 \[ \begin {cases} \frac {390 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 120 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {780 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 120 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {390 d x}{60 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 120 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {3 \tan ^{9}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 120 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} + \frac {34 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 120 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {388 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 120 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {1310 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 120 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} - \frac {765 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{60 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 120 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 60 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{5}{\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((390*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a*
*3*d) + 780*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*
d) + 390*d*x/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 3*tan(c/2 + d*x/2)
**9/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 34*tan(c/2 + d*x/2)**7/(60*
a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 388*tan(c/2 + d*x/2)**5/(60*a**3*d*
tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 1310*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/
2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 765*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)
**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d), Ne(d, 0)), (x*cos(c)**5/(a*cos(c) + a)**3, True))

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