3.528 \(\int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=133 \[ \frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^5(c+d x)}{5 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^3(c+d x)}{3 d}+\frac {5 a^3 \csc ^2(c+d x)}{2 d}+\frac {5 a^3 \csc (c+d x)}{d}+\frac {a^3 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 133, 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} \[ \frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^5(c+d x)}{5 d}-\frac {3 a^3 \csc ^4(c+d x)}{4 d}-\frac {a^3 \csc ^3(c+d x)}{3 d}+\frac {5 a^3 \csc ^2(c+d x)}{2 d}+\frac {5 a^3 \csc (c+d x)}{d}+\frac {a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

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

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Mathematica [A]  time = 0.17, size = 86, normalized size = 0.65 \[ \frac {a^3 \left (30 \sin ^2(c+d x)+180 \sin (c+d x)-12 \csc ^5(c+d x)-45 \csc ^4(c+d x)-20 \csc ^3(c+d x)+150 \csc ^2(c+d x)+300 \csc (c+d x)+60 \log (\sin (c+d x))\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(300*Csc[c + d*x] + 150*Csc[c + d*x]^2 - 20*Csc[c + d*x]^3 - 45*Csc[c + d*x]^4 - 12*Csc[c + d*x]^5 + 60*L
og[Sin[c + d*x]] + 180*Sin[c + d*x] + 30*Sin[c + d*x]^2))/(60*d)

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fricas [A]  time = 0.69, size = 179, normalized size = 1.35 \[ -\frac {180 \, a^{3} \cos \left (d x + c\right )^{6} - 840 \, a^{3} \cos \left (d x + c\right )^{4} + 1120 \, a^{3} \cos \left (d x + c\right )^{2} - 448 \, a^{3} - 60 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 15 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{6} - 5 \, a^{3} \cos \left (d x + c\right )^{4} + 14 \, a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3}\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(180*a^3*cos(d*x + c)^6 - 840*a^3*cos(d*x + c)^4 + 1120*a^3*cos(d*x + c)^2 - 448*a^3 - 60*(a^3*cos(d*x +
c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)*log(1/2*sin(d*x + c))*sin(d*x + c) + 15*(2*a^3*cos(d*x + c)^6 - 5*a^3*cos(
d*x + c)^4 + 14*a^3*cos(d*x + c)^2 - 8*a^3)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x
+ c))

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giac [A]  time = 0.35, size = 122, normalized size = 0.92 \[ \frac {30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 180 \, a^{3} \sin \left (d x + c\right ) - \frac {137 \, a^{3} \sin \left (d x + c\right )^{5} - 300 \, a^{3} \sin \left (d x + c\right )^{4} - 150 \, a^{3} \sin \left (d x + c\right )^{3} + 20 \, a^{3} \sin \left (d x + c\right )^{2} + 45 \, a^{3} \sin \left (d x + c\right ) + 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(abs(sin(d*x + c))) + 180*a^3*sin(d*x + c) - (137*a^3*sin(d*x + c)^5 -
300*a^3*sin(d*x + c)^4 - 150*a^3*sin(d*x + c)^3 + 20*a^3*sin(d*x + c)^2 + 45*a^3*sin(d*x + c) + 12*a^3)/sin(d
*x + c)^5)/d

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maple [A]  time = 0.44, size = 234, normalized size = 1.76 \[ -\frac {a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right ) a^{3}}{2 d}-\frac {a^{3} \left (\cos ^{2}\left (d x +c \right )\right )}{d}+\frac {a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {14 a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{15 d \sin \left (d x +c \right )^{3}}+\frac {14 a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )}+\frac {112 a^{3} \sin \left (d x +c \right )}{15 d}+\frac {14 a^{3} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{5 d}+\frac {56 a^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15 d}-\frac {3 a^{3} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x)

[Out]

-1/2/d*a^3/sin(d*x+c)^2*cos(d*x+c)^6-1/2/d*cos(d*x+c)^4*a^3-1/d*a^3*cos(d*x+c)^2+a^3*ln(sin(d*x+c))/d-14/15/d*
a^3/sin(d*x+c)^3*cos(d*x+c)^6+14/5/d*a^3/sin(d*x+c)*cos(d*x+c)^6+112/15*a^3*sin(d*x+c)/d+14/5/d*a^3*cos(d*x+c)
^4*sin(d*x+c)+56/15/d*a^3*cos(d*x+c)^2*sin(d*x+c)-3/4/d*a^3*cot(d*x+c)^4+3/2/d*a^3*cot(d*x+c)^2-1/5/d*a^3/sin(
d*x+c)^5*cos(d*x+c)^6

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maxima [A]  time = 0.75, size = 107, normalized size = 0.80 \[ \frac {30 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 180 \, a^{3} \sin \left (d x + c\right ) + \frac {300 \, a^{3} \sin \left (d x + c\right )^{4} + 150 \, a^{3} \sin \left (d x + c\right )^{3} - 20 \, a^{3} \sin \left (d x + c\right )^{2} - 45 \, a^{3} \sin \left (d x + c\right ) - 12 \, a^{3}}{\sin \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(30*a^3*sin(d*x + c)^2 + 60*a^3*log(sin(d*x + c)) + 180*a^3*sin(d*x + c) + (300*a^3*sin(d*x + c)^4 + 150*
a^3*sin(d*x + c)^3 - 20*a^3*sin(d*x + c)^2 - 45*a^3*sin(d*x + c) - 12*a^3)/sin(d*x + c)^5)/d

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mupad [B]  time = 8.95, size = 311, normalized size = 2.34 \[ \frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {266\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+78\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {1013\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {53\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+\frac {1037\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {41\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {a^3}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {37\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

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)^6,x)

[Out]

(7*a^3*tan(c/2 + (d*x)/2)^2)/(16*d) - (7*a^3*tan(c/2 + (d*x)/2)^3)/(96*d) - (3*a^3*tan(c/2 + (d*x)/2)^4)/(64*d
) - (a^3*tan(c/2 + (d*x)/2)^5)/(160*d) + (a^3*log(tan(c/2 + (d*x)/2)))/d + (11*a^3*tan(c/2 + (d*x)/2)^3 - (41*
a^3*tan(c/2 + (d*x)/2)^2)/15 + (1037*a^3*tan(c/2 + (d*x)/2)^4)/15 + (53*a^3*tan(c/2 + (d*x)/2)^5)/2 + (1013*a^
3*tan(c/2 + (d*x)/2)^6)/3 + 78*a^3*tan(c/2 + (d*x)/2)^7 + 266*a^3*tan(c/2 + (d*x)/2)^8 - a^3/5 - (3*a^3*tan(c/
2 + (d*x)/2))/2)/(d*(32*tan(c/2 + (d*x)/2)^5 + 64*tan(c/2 + (d*x)/2)^7 + 32*tan(c/2 + (d*x)/2)^9)) + (37*a^3*t
an(c/2 + (d*x)/2))/(16*d) - (a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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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)**6*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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