3.526 \(\int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=131 \[ \frac {a^3 \sin ^4(c+d x)}{4 d}+\frac {a^3 \sin ^3(c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d}-\frac {5 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^3(c+d x)}{3 d}-\frac {3 a^3 \csc ^2(c+d x)}{2 d}-\frac {a^3 \csc (c+d x)}{d}-\frac {5 a^3 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 131, 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} \[ \frac {a^3 \sin ^4(c+d x)}{4 d}+\frac {a^3 \sin ^3(c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d}-\frac {5 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^3(c+d x)}{3 d}-\frac {3 a^3 \csc ^2(c+d x)}{2 d}-\frac {a^3 \csc (c+d x)}{d}-\frac {5 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

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

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Mathematica [A]  time = 0.26, size = 86, normalized size = 0.66 \[ -\frac {a^3 \left (-3 \sin ^4(c+d x)-12 \sin ^3(c+d x)-6 \sin ^2(c+d x)+60 \sin (c+d x)+4 \csc ^3(c+d x)+18 \csc ^2(c+d x)+12 \csc (c+d x)+60 \log (\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

-1/12*(a^3*(12*Csc[c + d*x] + 18*Csc[c + d*x]^2 + 4*Csc[c + d*x]^3 + 60*Log[Sin[c + d*x]] + 60*Sin[c + d*x] -
6*Sin[c + d*x]^2 - 12*Sin[c + d*x]^3 - 3*Sin[c + d*x]^4))/d

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fricas [A]  time = 0.77, size = 159, normalized size = 1.21 \[ \frac {96 \, a^{3} \cos \left (d x + c\right )^{6} + 192 \, a^{3} \cos \left (d x + c\right )^{4} - 768 \, a^{3} \cos \left (d x + c\right )^{2} + 512 \, a^{3} - 480 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \, {\left (8 \, a^{3} \cos \left (d x + c\right )^{6} - 40 \, a^{3} \cos \left (d x + c\right )^{4} + 45 \, a^{3} \cos \left (d x + c\right )^{2} + 35 \, a^{3}\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(96*a^3*cos(d*x + c)^6 + 192*a^3*cos(d*x + c)^4 - 768*a^3*cos(d*x + c)^2 + 512*a^3 - 480*(a^3*cos(d*x + c
)^2 - a^3)*log(1/2*sin(d*x + c))*sin(d*x + c) + 3*(8*a^3*cos(d*x + c)^6 - 40*a^3*cos(d*x + c)^4 + 45*a^3*cos(d
*x + c)^2 + 35*a^3)*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [A]  time = 0.32, size = 122, normalized size = 0.93 \[ \frac {3 \, a^{3} \sin \left (d x + c\right )^{4} + 12 \, a^{3} \sin \left (d x + c\right )^{3} + 6 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 60 \, a^{3} \sin \left (d x + c\right ) + \frac {2 \, {\left (55 \, a^{3} \sin \left (d x + c\right )^{3} - 6 \, a^{3} \sin \left (d x + c\right )^{2} - 9 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )}}{\sin \left (d x + c\right )^{3}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/12*(3*a^3*sin(d*x + c)^4 + 12*a^3*sin(d*x + c)^3 + 6*a^3*sin(d*x + c)^2 - 60*a^3*log(abs(sin(d*x + c))) - 60
*a^3*sin(d*x + c) + 2*(55*a^3*sin(d*x + c)^3 - 6*a^3*sin(d*x + c)^2 - 9*a^3*sin(d*x + c) - 2*a^3)/sin(d*x + c)
^3)/d

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maple [A]  time = 0.43, size = 179, normalized size = 1.37 \[ -\frac {5 \left (\cos ^{4}\left (d x +c \right )\right ) a^{3}}{4 d}-\frac {5 a^{3} \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}-\frac {5 a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {2 a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {16 a^{3} \sin \left (d x +c \right )}{3 d}-\frac {2 a^{3} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {8 a^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}-\frac {3 a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x)

[Out]

-5/4/d*cos(d*x+c)^4*a^3-5/2/d*a^3*cos(d*x+c)^2-5*a^3*ln(sin(d*x+c))/d-2/d*a^3/sin(d*x+c)*cos(d*x+c)^6-16/3*a^3
*sin(d*x+c)/d-2/d*a^3*cos(d*x+c)^4*sin(d*x+c)-8/3/d*a^3*cos(d*x+c)^2*sin(d*x+c)-3/2/d*a^3/sin(d*x+c)^2*cos(d*x
+c)^6-1/3/d*a^3/sin(d*x+c)^3*cos(d*x+c)^6

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maxima [A]  time = 0.45, size = 108, normalized size = 0.82 \[ \frac {3 \, a^{3} \sin \left (d x + c\right )^{4} + 12 \, a^{3} \sin \left (d x + c\right )^{3} + 6 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) - 60 \, a^{3} \sin \left (d x + c\right ) - \frac {2 \, {\left (6 \, a^{3} \sin \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) + 2 \, a^{3}\right )}}{\sin \left (d x + c\right )^{3}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/12*(3*a^3*sin(d*x + c)^4 + 12*a^3*sin(d*x + c)^3 + 6*a^3*sin(d*x + c)^2 - 60*a^3*log(sin(d*x + c)) - 60*a^3*
sin(d*x + c) - 2*(6*a^3*sin(d*x + c)^2 + 9*a^3*sin(d*x + c) + 2*a^3)/sin(d*x + c)^3)/d

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mupad [B]  time = 8.98, size = 333, normalized size = 2.54 \[ \frac {5\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {5\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {5\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {85\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {589\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}-52\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {622\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+102\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {19\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^3)/sin(c + d*x)^4,x)

[Out]

(5*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a^3*tan(c/2 + (d*x)/2)^3)/(24*d) - (5*a^3*log(tan(c/2 + (d*x)/2)))/
d - (5*a^3*tan(c/2 + (d*x)/2))/(8*d) - (3*a^3*tan(c/2 + (d*x)/2)^2)/(8*d) - ((19*a^3*tan(c/2 + (d*x)/2)^2)/3 +
12*a^3*tan(c/2 + (d*x)/2)^3 + 102*a^3*tan(c/2 + (d*x)/2)^4 + 2*a^3*tan(c/2 + (d*x)/2)^5 + (622*a^3*tan(c/2 +
(d*x)/2)^6)/3 - 52*a^3*tan(c/2 + (d*x)/2)^7 + (589*a^3*tan(c/2 + (d*x)/2)^8)/3 - 13*a^3*tan(c/2 + (d*x)/2)^9 +
85*a^3*tan(c/2 + (d*x)/2)^10 + a^3/3 + 3*a^3*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d*x)/2)^3 + 32*tan(c/2 + (d
*x)/2)^5 + 48*tan(c/2 + (d*x)/2)^7 + 32*tan(c/2 + (d*x)/2)^9 + 8*tan(c/2 + (d*x)/2)^11))

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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)**5*csc(d*x+c)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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