3.1.88 \(\int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^5} \, dx\) [88]

Optimal. Leaf size=143 \[ -\frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac {5 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac {\sin (c+d x)}{21 a^2 d (a+a \cos (c+d x))^3}+\frac {2 \sin (c+d x)}{63 a d \left (a^2+a^2 \cos (c+d x)\right )^2}+\frac {2 \sin (c+d x)}{63 d \left (a^5+a^5 \cos (c+d x)\right )} \]

[Out]

-1/9*sin(d*x+c)/d/(a+a*cos(d*x+c))^5+5/63*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^4+1/21*sin(d*x+c)/a^2/d/(a+a*cos(d*x
+c))^3+2/63*sin(d*x+c)/a/d/(a^2+a^2*cos(d*x+c))^2+2/63*sin(d*x+c)/d/(a^5+a^5*cos(d*x+c))

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Rubi [A]
time = 0.07, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2829, 2729, 2727} \begin {gather*} \frac {2 \sin (c+d x)}{63 d \left (a^5 \cos (c+d x)+a^5\right )}+\frac {2 \sin (c+d x)}{63 a d \left (a^2 \cos (c+d x)+a^2\right )^2}+\frac {\sin (c+d x)}{21 a^2 d (a \cos (c+d x)+a)^3}+\frac {5 \sin (c+d x)}{63 a d (a \cos (c+d x)+a)^4}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/9*Sin[c + d*x]/(d*(a + a*Cos[c + d*x])^5) + (5*Sin[c + d*x])/(63*a*d*(a + a*Cos[c + d*x])^4) + Sin[c + d*x]
/(21*a^2*d*(a + a*Cos[c + d*x])^3) + (2*Sin[c + d*x])/(63*a*d*(a^2 + a^2*Cos[c + d*x])^2) + (2*Sin[c + d*x])/(
63*d*(a^5 + a^5*Cos[c + d*x]))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*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*} \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^5} \, dx &=-\frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac {5 \int \frac {1}{(a+a \cos (c+d x))^4} \, dx}{9 a}\\ &=-\frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac {5 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac {5 \int \frac {1}{(a+a \cos (c+d x))^3} \, dx}{21 a^2}\\ &=-\frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac {5 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac {\sin (c+d x)}{21 a^2 d (a+a \cos (c+d x))^3}+\frac {2 \int \frac {1}{(a+a \cos (c+d x))^2} \, dx}{21 a^3}\\ &=-\frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac {5 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac {\sin (c+d x)}{21 a^2 d (a+a \cos (c+d x))^3}+\frac {2 \sin (c+d x)}{63 a^3 d (a+a \cos (c+d x))^2}+\frac {2 \int \frac {1}{a+a \cos (c+d x)} \, dx}{63 a^4}\\ &=-\frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac {5 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac {\sin (c+d x)}{21 a^2 d (a+a \cos (c+d x))^3}+\frac {2 \sin (c+d x)}{63 a^3 d (a+a \cos (c+d x))^2}+\frac {2 \sin (c+d x)}{63 d \left (a^5+a^5 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 97, normalized size = 0.68 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^9\left (\frac {1}{2} (c+d x)\right ) \left (63 \sin \left (\frac {d x}{2}\right )-63 \sin \left (c+\frac {d x}{2}\right )+84 \sin \left (c+\frac {3 d x}{2}\right )+36 \sin \left (2 c+\frac {5 d x}{2}\right )+9 \sin \left (3 c+\frac {7 d x}{2}\right )+\sin \left (4 c+\frac {9 d x}{2}\right )\right )}{8064 a^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^9*(63*Sin[(d*x)/2] - 63*Sin[c + (d*x)/2] + 84*Sin[c + (3*d*x)/2] + 36*Sin[2*c + (5*
d*x)/2] + 9*Sin[3*c + (7*d*x)/2] + Sin[4*c + (9*d*x)/2]))/(8064*a^5*d)

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Maple [A]
time = 0.13, size = 58, normalized size = 0.41

method result size
derivativedivides \(\frac {-\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\) \(58\)
default \(\frac {-\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\) \(58\)
risch \(\frac {4 i \left (63 \,{\mathrm e}^{5 i \left (d x +c \right )}+63 \,{\mathrm e}^{4 i \left (d x +c \right )}+84 \,{\mathrm e}^{3 i \left (d x +c \right )}+36 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{63 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\) \(80\)
norman \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}+\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 a d}+\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 a d}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 a d}-\frac {25 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1008 a d}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{144 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}\) \(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d/a^5*(-1/9*tan(1/2*d*x+1/2*c)^9-2/7*tan(1/2*d*x+1/2*c)^7+2/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

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Maxima [A]
time = 0.30, size = 87, normalized size = 0.61 \begin {gather*} \frac {\frac {63 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {42 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {18 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{1008 \, a^{5} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1008*(63*sin(d*x + c)/(cos(d*x + c) + 1) + 42*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 18*sin(d*x + c)^7/(cos(d
*x + c) + 1)^7 - 7*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^5*d)

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Fricas [A]
time = 0.39, size = 123, normalized size = 0.86 \begin {gather*} \frac {{\left (2 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 25 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{63 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/63*(2*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 25*cos(d*x + c) + 5)*sin(d*x + c)/(a^5*d*cos(
d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c)
 + a^5*d)

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Sympy [A]
time = 3.12, size = 85, normalized size = 0.59 \begin {gather*} \begin {cases} - \frac {\tan ^{9}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{144 a^{5} d} - \frac {\tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{56 a^{5} d} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{24 a^{5} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{16 a^{5} d} & \text {for}\: d \neq 0 \\\frac {x \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{5}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-tan(c/2 + d*x/2)**9/(144*a**5*d) - tan(c/2 + d*x/2)**7/(56*a**5*d) + tan(c/2 + d*x/2)**3/(24*a**5*
d) + tan(c/2 + d*x/2)/(16*a**5*d), Ne(d, 0)), (x*cos(c)/(a*cos(c) + a)**5, True))

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Giac [A]
time = 0.48, size = 59, normalized size = 0.41 \begin {gather*} -\frac {7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 42 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 63 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{1008 \, a^{5} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1008*(7*tan(1/2*d*x + 1/2*c)^9 + 18*tan(1/2*d*x + 1/2*c)^7 - 42*tan(1/2*d*x + 1/2*c)^3 - 63*tan(1/2*d*x + 1
/2*c))/(a^5*d)

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Mupad [B]
time = 0.40, size = 58, normalized size = 0.41 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+63\right )}{1008\,a^5\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(42*tan(c/2 + (d*x)/2)^2 - 18*tan(c/2 + (d*x)/2)^6 - 7*tan(c/2 + (d*x)/2)^8 + 63))/(1008*a
^5*d)

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