3.663 \(\int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=114 \[ -\frac {a \sin ^6(c+d x)}{6 d}-\frac {a \sin ^5(c+d x)}{5 d}+\frac {3 a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^3(c+d x)}{d}-\frac {3 a \sin ^2(c+d x)}{2 d}-\frac {3 a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

[Out]

-a*csc(d*x+c)/d+a*ln(sin(d*x+c))/d-3*a*sin(d*x+c)/d-3/2*a*sin(d*x+c)^2/d+a*sin(d*x+c)^3/d+3/4*a*sin(d*x+c)^4/d
-1/5*a*sin(d*x+c)^5/d-1/6*a*sin(d*x+c)^6/d

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Rubi [A]  time = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2836, 12, 88} \[ -\frac {a \sin ^6(c+d x)}{6 d}-\frac {a \sin ^5(c+d x)}{5 d}+\frac {3 a \sin ^4(c+d x)}{4 d}+\frac {a \sin ^3(c+d x)}{d}-\frac {3 a \sin ^2(c+d x)}{2 d}-\frac {3 a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Csc[c + d*x])/d) + (a*Log[Sin[c + d*x]])/d - (3*a*Sin[c + d*x])/d - (3*a*Sin[c + d*x]^2)/(2*d) + (a*Sin[c
+ d*x]^3)/d + (3*a*Sin[c + d*x]^4)/(4*d) - (a*Sin[c + d*x]^5)/(5*d) - (a*Sin[c + d*x]^6)/(6*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

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

Rubi steps

\begin {align*} \int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^2 (a-x)^3 (a+x)^4}{x^2} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a-x)^3 (a+x)^4}{x^2} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-3 a^5+\frac {a^7}{x^2}+\frac {a^6}{x}-3 a^4 x+3 a^3 x^2+3 a^2 x^3-a x^4-x^5\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}-\frac {3 a \sin (c+d x)}{d}-\frac {3 a \sin ^2(c+d x)}{2 d}+\frac {a \sin ^3(c+d x)}{d}+\frac {3 a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^6(c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 102, normalized size = 0.89 \[ -\frac {a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^3(c+d x)}{d}-\frac {3 a \sin (c+d x)}{d}-\frac {a \csc (c+d x)}{d}+\frac {a \left (-2 \sin ^6(c+d x)+9 \sin ^4(c+d x)-18 \sin ^2(c+d x)+12 \log (\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Csc[c + d*x])/d) - (3*a*Sin[c + d*x])/d + (a*Sin[c + d*x]^3)/d - (a*Sin[c + d*x]^5)/(5*d) + (a*(12*Log[Si
n[c + d*x]] - 18*Sin[c + d*x]^2 + 9*Sin[c + d*x]^4 - 2*Sin[c + d*x]^6))/(12*d)

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fricas [A]  time = 0.70, size = 113, normalized size = 0.99 \[ \frac {48 \, a \cos \left (d x + c\right )^{6} + 96 \, a \cos \left (d x + c\right )^{4} + 384 \, a \cos \left (d x + c\right )^{2} + 240 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 5 \, {\left (8 \, a \cos \left (d x + c\right )^{6} + 12 \, a \cos \left (d x + c\right )^{4} + 24 \, a \cos \left (d x + c\right )^{2} - 19 \, a\right )} \sin \left (d x + c\right ) - 768 \, a}{240 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(48*a*cos(d*x + c)^6 + 96*a*cos(d*x + c)^4 + 384*a*cos(d*x + c)^2 + 240*a*log(1/2*sin(d*x + c))*sin(d*x
+ c) + 5*(8*a*cos(d*x + c)^6 + 12*a*cos(d*x + c)^4 + 24*a*cos(d*x + c)^2 - 19*a)*sin(d*x + c) - 768*a)/(d*sin(
d*x + c))

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giac [A]  time = 0.21, size = 101, normalized size = 0.89 \[ -\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 45 \, a \sin \left (d x + c\right )^{4} - 60 \, a \sin \left (d x + c\right )^{3} + 90 \, a \sin \left (d x + c\right )^{2} - 60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 180 \, a \sin \left (d x + c\right ) + \frac {60 \, {\left (a \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 45*a*sin(d*x + c)^4 - 60*a*sin(d*x + c)^3 + 90*a*sin(d*x +
c)^2 - 60*a*log(abs(sin(d*x + c))) + 180*a*sin(d*x + c) + 60*(a*sin(d*x + c) + a)/sin(d*x + c))/d

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maple [A]  time = 0.29, size = 150, normalized size = 1.32 \[ \frac {a \left (\cos ^{6}\left (d x +c \right )\right )}{6 d}+\frac {a \left (\cos ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a \left (\cos ^{8}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {16 a \sin \left (d x +c \right )}{5 d}-\frac {\left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a}{d}-\frac {6 a \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5 d}-\frac {8 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*csc(d*x+c)^2*(a+a*sin(d*x+c)),x)

[Out]

1/6*a*cos(d*x+c)^6/d+1/4*a*cos(d*x+c)^4/d+1/2*a*cos(d*x+c)^2/d+a*ln(sin(d*x+c))/d-1/d*a/sin(d*x+c)*cos(d*x+c)^
8-16/5*a*sin(d*x+c)/d-1/d*cos(d*x+c)^6*sin(d*x+c)*a-6/5/d*a*sin(d*x+c)*cos(d*x+c)^4-8/5/d*a*sin(d*x+c)*cos(d*x
+c)^2

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maxima [A]  time = 0.36, size = 91, normalized size = 0.80 \[ -\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 45 \, a \sin \left (d x + c\right )^{4} - 60 \, a \sin \left (d x + c\right )^{3} + 90 \, a \sin \left (d x + c\right )^{2} - 60 \, a \log \left (\sin \left (d x + c\right )\right ) + 180 \, a \sin \left (d x + c\right ) + \frac {60 \, a}{\sin \left (d x + c\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*csc(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 45*a*sin(d*x + c)^4 - 60*a*sin(d*x + c)^3 + 90*a*sin(d*x +
c)^2 - 60*a*log(sin(d*x + c)) + 180*a*sin(d*x + c) + 60*a/sin(d*x + c))/d

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mupad [B]  time = 9.33, size = 340, normalized size = 2.98 \[ \frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}-\frac {6\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}+\frac {18\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {104\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3\,d}+\frac {44\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}-\frac {32\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{d}+\frac {32\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{3\,d}-\frac {13\,a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {14\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {112\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {136\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {96\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {32\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^7*(a + a*sin(c + d*x)))/sin(c + d*x)^2,x)

[Out]

(a*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (a*log(1/cos(c/2 + (d*x)/2)^2))/d - (6*a*cos(c/2 + (d*x)/2)
^2)/d + (18*a*cos(c/2 + (d*x)/2)^4)/d - (104*a*cos(c/2 + (d*x)/2)^6)/(3*d) + (44*a*cos(c/2 + (d*x)/2)^8)/d - (
32*a*cos(c/2 + (d*x)/2)^10)/d + (32*a*cos(c/2 + (d*x)/2)^12)/(3*d) - (13*a*cos(c/2 + (d*x)/2))/(2*d*sin(c/2 +
(d*x)/2)) - (a*sin(c/2 + (d*x)/2))/(2*d*cos(c/2 + (d*x)/2)) + (14*a*cos(c/2 + (d*x)/2)^3)/(d*sin(c/2 + (d*x)/2
)) - (112*a*cos(c/2 + (d*x)/2)^5)/(5*d*sin(c/2 + (d*x)/2)) + (136*a*cos(c/2 + (d*x)/2)^7)/(5*d*sin(c/2 + (d*x)
/2)) - (96*a*cos(c/2 + (d*x)/2)^9)/(5*d*sin(c/2 + (d*x)/2)) + (32*a*cos(c/2 + (d*x)/2)^11)/(5*d*sin(c/2 + (d*x
)/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*csc(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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