3.532 \(\int \cot ^5(c+d x) \csc (c+d x) (a+a \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=146 \[ \frac {a^4 \sin ^3(c+d x)}{3 d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin (c+d x)}{d}-\frac {a^4 \csc ^5(c+d x)}{5 d}-\frac {a^4 \csc ^4(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}+\frac {2 a^4 \csc ^2(c+d x)}{d}+\frac {10 a^4 \csc (c+d x)}{d}-\frac {4 a^4 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 146, 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} \[ \frac {a^4 \sin ^3(c+d x)}{3 d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin (c+d x)}{d}-\frac {a^4 \csc ^5(c+d x)}{5 d}-\frac {a^4 \csc ^4(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}+\frac {2 a^4 \csc ^2(c+d x)}{d}+\frac {10 a^4 \csc (c+d x)}{d}-\frac {4 a^4 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^4,x]

[Out]

(10*a^4*Csc[c + d*x])/d + (2*a^4*Csc[c + d*x]^2)/d - (4*a^4*Csc[c + d*x]^3)/(3*d) - (a^4*Csc[c + d*x]^4)/d - (
a^4*Csc[c + d*x]^5)/(5*d) - (4*a^4*Log[Sin[c + d*x]])/d + (4*a^4*Sin[c + d*x])/d + (2*a^4*Sin[c + d*x]^2)/d +
(a^4*Sin[c + d*x]^3)/(3*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))^4 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^6 (a-x)^2 (a+x)^6}{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)^6}{x^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a \operatorname {Subst}\left (\int \left (4 a^2+\frac {a^8}{x^6}+\frac {4 a^7}{x^5}+\frac {4 a^6}{x^4}-\frac {4 a^5}{x^3}-\frac {10 a^4}{x^2}-\frac {4 a^3}{x}+4 a x+x^2\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {10 a^4 \csc (c+d x)}{d}+\frac {2 a^4 \csc ^2(c+d x)}{d}-\frac {4 a^4 \csc ^3(c+d x)}{3 d}-\frac {a^4 \csc ^4(c+d x)}{d}-\frac {a^4 \csc ^5(c+d x)}{5 d}-\frac {4 a^4 \log (\sin (c+d x))}{d}+\frac {4 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin ^2(c+d x)}{d}+\frac {a^4 \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 96, normalized size = 0.66 \[ \frac {a^4 \left (5 \sin ^3(c+d x)+30 \sin ^2(c+d x)+60 \sin (c+d x)-3 \csc ^5(c+d x)-15 \csc ^4(c+d x)-20 \csc ^3(c+d x)+30 \csc ^2(c+d x)+150 \csc (c+d x)-60 \log (\sin (c+d x))\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*Csc[c + d*x]*(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*(150*Csc[c + d*x] + 30*Csc[c + d*x]^2 - 20*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 - 3*Csc[c + d*x]^5 - 60*Log
[Sin[c + d*x]] + 60*Sin[c + d*x] + 30*Sin[c + d*x]^2 + 5*Sin[c + d*x]^3))/(15*d)

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fricas [A]  time = 0.59, size = 192, normalized size = 1.32 \[ \frac {5 \, a^{4} \cos \left (d x + c\right )^{8} - 80 \, a^{4} \cos \left (d x + c\right )^{6} + 360 \, a^{4} \cos \left (d x + c\right )^{4} - 480 \, a^{4} \cos \left (d x + c\right )^{2} + 192 \, a^{4} - 60 \, {\left (a^{4} \cos \left (d x + c\right )^{4} - 2 \, a^{4} \cos \left (d x + c\right )^{2} + a^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 15 \, {\left (2 \, a^{4} \cos \left (d x + c\right )^{6} - 5 \, a^{4} \cos \left (d x + c\right )^{4} + 6 \, a^{4} \cos \left (d x + c\right )^{2} - 2 \, a^{4}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

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))^4,x, algorithm="fricas")

[Out]

1/15*(5*a^4*cos(d*x + c)^8 - 80*a^4*cos(d*x + c)^6 + 360*a^4*cos(d*x + c)^4 - 480*a^4*cos(d*x + c)^2 + 192*a^4
- 60*(a^4*cos(d*x + c)^4 - 2*a^4*cos(d*x + c)^2 + a^4)*log(1/2*sin(d*x + c))*sin(d*x + c) - 15*(2*a^4*cos(d*x
+ c)^6 - 5*a^4*cos(d*x + c)^4 + 6*a^4*cos(d*x + c)^2 - 2*a^4)*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 = 0.45, size = 134, normalized size = 0.92 \[ \frac {5 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a^{4} \sin \left (d x + c\right ) + \frac {137 \, a^{4} \sin \left (d x + c\right )^{5} + 150 \, a^{4} \sin \left (d x + c\right )^{4} + 30 \, a^{4} \sin \left (d x + c\right )^{3} - 20 \, a^{4} \sin \left (d x + c\right )^{2} - 15 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{5}}}{15 \, d} \]

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))^4,x, algorithm="giac")

[Out]

1/15*(5*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^2 - 60*a^4*log(abs(sin(d*x + c))) + 60*a^4*sin(d*x + c) + (13
7*a^4*sin(d*x + c)^5 + 150*a^4*sin(d*x + c)^4 + 30*a^4*sin(d*x + c)^3 - 20*a^4*sin(d*x + c)^2 - 15*a^4*sin(d*x
+ c) - 3*a^4)/sin(d*x + c)^5)/d

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maple [A]  time = 0.43, size = 235, normalized size = 1.61 \[ \frac {24 a^{4} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )}+\frac {64 a^{4} \sin \left (d x +c \right )}{5 d}+\frac {24 a^{4} \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5 d}+\frac {32 a^{4} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{5 d}-\frac {2 a^{4} \left (\cos ^{6}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{2}}-\frac {2 a^{4} \left (\cos ^{4}\left (d x +c \right )\right )}{d}-\frac {4 a^{4} \left (\cos ^{2}\left (d x +c \right )\right )}{d}-\frac {4 a^{4} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {29 a^{4} \left (\cos ^{6}\left (d x +c \right )\right )}{15 d \sin \left (d x +c \right )^{3}}-\frac {a^{4} \left (\cot ^{4}\left (d x +c \right )\right )}{d}+\frac {2 a^{4} \left (\cot ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{4} \left (\cos ^{6}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}} \]

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))^4,x)

[Out]

24/5/d*a^4/sin(d*x+c)*cos(d*x+c)^6+64/5*a^4*sin(d*x+c)/d+24/5/d*a^4*sin(d*x+c)*cos(d*x+c)^4+32/5/d*a^4*cos(d*x
+c)^2*sin(d*x+c)-2/d*a^4/sin(d*x+c)^2*cos(d*x+c)^6-2/d*a^4*cos(d*x+c)^4-4/d*a^4*cos(d*x+c)^2-4*a^4*ln(sin(d*x+
c))/d-29/15/d*a^4/sin(d*x+c)^3*cos(d*x+c)^6-1/d*a^4*cot(d*x+c)^4+2/d*a^4*cot(d*x+c)^2-1/5/d*a^4/sin(d*x+c)^5*c
os(d*x+c)^6

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maxima [A]  time = 0.57, size = 120, normalized size = 0.82 \[ \frac {5 \, a^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} \sin \left (d x + c\right )^{2} - 60 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) + 60 \, a^{4} \sin \left (d x + c\right ) + \frac {150 \, a^{4} \sin \left (d x + c\right )^{4} + 30 \, a^{4} \sin \left (d x + c\right )^{3} - 20 \, a^{4} \sin \left (d x + c\right )^{2} - 15 \, a^{4} \sin \left (d x + c\right ) - 3 \, a^{4}}{\sin \left (d x + c\right )^{5}}}{15 \, d} \]

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))^4,x, algorithm="maxima")

[Out]

1/15*(5*a^4*sin(d*x + c)^3 + 30*a^4*sin(d*x + c)^2 - 60*a^4*log(sin(d*x + c)) + 60*a^4*sin(d*x + c) + (150*a^4
*sin(d*x + c)^4 + 30*a^4*sin(d*x + c)^3 - 20*a^4*sin(d*x + c)^2 - 15*a^4*sin(d*x + c) - 3*a^4)/sin(d*x + c)^5)
/d

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mupad [B]  time = 9.09, size = 357, normalized size = 2.45 \[ \frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {19\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {4\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {398\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+264\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+1017\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+278\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {3314\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+18\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {612\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {104\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^4}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+96\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {71\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}+\frac {4\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^4)/sin(c + d*x)^6,x)

[Out]

(a^4*tan(c/2 + (d*x)/2)^2)/(4*d) - (19*a^4*tan(c/2 + (d*x)/2)^3)/(96*d) - (a^4*tan(c/2 + (d*x)/2)^4)/(16*d) -
(a^4*tan(c/2 + (d*x)/2)^5)/(160*d) - (4*a^4*log(tan(c/2 + (d*x)/2)))/d + (2*a^4*tan(c/2 + (d*x)/2)^3 - (104*a^
4*tan(c/2 + (d*x)/2)^2)/15 + (612*a^4*tan(c/2 + (d*x)/2)^4)/5 + 18*a^4*tan(c/2 + (d*x)/2)^5 + (3314*a^4*tan(c/
2 + (d*x)/2)^6)/5 + 278*a^4*tan(c/2 + (d*x)/2)^7 + 1017*a^4*tan(c/2 + (d*x)/2)^8 + 264*a^4*tan(c/2 + (d*x)/2)^
9 + 398*a^4*tan(c/2 + (d*x)/2)^10 - a^4/5 - 2*a^4*tan(c/2 + (d*x)/2))/(d*(32*tan(c/2 + (d*x)/2)^5 + 96*tan(c/2
+ (d*x)/2)^7 + 96*tan(c/2 + (d*x)/2)^9 + 32*tan(c/2 + (d*x)/2)^11)) + (71*a^4*tan(c/2 + (d*x)/2))/(16*d) + (4
*a^4*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))**4,x)

[Out]

Timed out

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