3.1318 \(\int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=179 \[ \frac {b \csc ^4(c+d x)}{4 a^2 d}-\frac {b \left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^6 d}+\frac {b \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^6 d}-\frac {\left (a^2-b^2\right )^2 \csc (c+d x)}{a^5 d}-\frac {b \left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^4 d}+\frac {\left (2 a^2-b^2\right ) \csc ^3(c+d x)}{3 a^3 d}-\frac {\csc ^5(c+d x)}{5 a d} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 179, 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 = {2837, 12, 894} \[ \frac {\left (2 a^2-b^2\right ) \csc ^3(c+d x)}{3 a^3 d}-\frac {b \left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^4 d}-\frac {\left (a^2-b^2\right )^2 \csc (c+d x)}{a^5 d}-\frac {b \left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^6 d}+\frac {b \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^6 d}+\frac {b \csc ^4(c+d x)}{4 a^2 d}-\frac {\csc ^5(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - b^2)^2*Csc[c + d*x])/(a^5*d)) - (b*(2*a^2 - b^2)*Csc[c + d*x]^2)/(2*a^4*d) + ((2*a^2 - b^2)*Csc[c +
d*x]^3)/(3*a^3*d) + (b*Csc[c + d*x]^4)/(4*a^2*d) - Csc[c + d*x]^5/(5*a*d) - (b*(a^2 - b^2)^2*Log[Sin[c + d*x]]
)/(a^6*d) + (b*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/(a^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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 6.12, size = 179, normalized size = 1.00 \[ \frac {b \csc ^4(c+d x)}{4 a^2 d}-\frac {b \left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^6 d}+\frac {b \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^6 d}-\frac {\left (a^2-b^2\right )^2 \csc (c+d x)}{a^5 d}-\frac {b \left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^4 d}+\frac {\left (2 a^2-b^2\right ) \csc ^3(c+d x)}{3 a^3 d}-\frac {\csc ^5(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.00, size = 346, normalized size = 1.93 \[ -\frac {32 \, a^{5} - 100 \, a^{3} b^{2} + 60 \, a b^{4} + 60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{4} - 20 \, {\left (4 \, a^{5} - 11 \, a^{3} b^{2} + 6 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + 60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 15 \, {\left (3 \, a^{4} b - 2 \, a^{2} b^{3} - 2 \, {\left (2 \, a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{6} d \cos \left (d x + c\right )^{4} - 2 \, a^{6} d \cos \left (d x + c\right )^{2} + a^{6} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(32*a^5 - 100*a^3*b^2 + 60*a*b^4 + 60*(a^5 - 2*a^3*b^2 + a*b^4)*cos(d*x + c)^4 - 20*(4*a^5 - 11*a^3*b^2
+ 6*a*b^4)*cos(d*x + c)^2 - 60*(a^4*b - 2*a^2*b^3 + b^5 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)^4 - 2*(a^4*b
- 2*a^2*b^3 + b^5)*cos(d*x + c)^2)*log(b*sin(d*x + c) + a)*sin(d*x + c) + 60*(a^4*b - 2*a^2*b^3 + b^5 + (a^4*b
- 2*a^2*b^3 + b^5)*cos(d*x + c)^4 - 2*(a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)^2)*log(1/2*sin(d*x + c))*sin(d*x
+ c) + 15*(3*a^4*b - 2*a^2*b^3 - 2*(2*a^4*b - a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/((a^6*d*cos(d*x + c)^4 -
2*a^6*d*cos(d*x + c)^2 + a^6*d)*sin(d*x + c))

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giac [A]  time = 0.21, size = 251, normalized size = 1.40 \[ -\frac {\frac {60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{6}} - \frac {60 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b} - \frac {137 \, a^{4} b \sin \left (d x + c\right )^{5} - 274 \, a^{2} b^{3} \sin \left (d x + c\right )^{5} + 137 \, b^{5} \sin \left (d x + c\right )^{5} - 60 \, a^{5} \sin \left (d x + c\right )^{4} + 120 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} - 60 \, a b^{4} \sin \left (d x + c\right )^{4} - 60 \, a^{4} b \sin \left (d x + c\right )^{3} + 30 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} + 40 \, a^{5} \sin \left (d x + c\right )^{2} - 20 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 15 \, a^{4} b \sin \left (d x + c\right ) - 12 \, a^{5}}{a^{6} \sin \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*(a^4*b - 2*a^2*b^3 + b^5)*log(abs(sin(d*x + c)))/a^6 - 60*(a^4*b^2 - 2*a^2*b^4 + b^6)*log(abs(b*sin(
d*x + c) + a))/(a^6*b) - (137*a^4*b*sin(d*x + c)^5 - 274*a^2*b^3*sin(d*x + c)^5 + 137*b^5*sin(d*x + c)^5 - 60*
a^5*sin(d*x + c)^4 + 120*a^3*b^2*sin(d*x + c)^4 - 60*a*b^4*sin(d*x + c)^4 - 60*a^4*b*sin(d*x + c)^3 + 30*a^2*b
^3*sin(d*x + c)^3 + 40*a^5*sin(d*x + c)^2 - 20*a^3*b^2*sin(d*x + c)^2 + 15*a^4*b*sin(d*x + c) - 12*a^5)/(a^6*s
in(d*x + c)^5))/d

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maple [A]  time = 0.49, size = 274, normalized size = 1.53 \[ \frac {b \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{2} d}-\frac {2 b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{4}}+\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{6}}-\frac {1}{5 d a \sin \left (d x +c \right )^{5}}+\frac {2}{3 d a \sin \left (d x +c \right )^{3}}-\frac {b^{2}}{3 d \,a^{3} \sin \left (d x +c \right )^{3}}-\frac {1}{d a \sin \left (d x +c \right )}+\frac {2 b^{2}}{d \,a^{3} \sin \left (d x +c \right )}-\frac {b^{4}}{d \,a^{5} \sin \left (d x +c \right )}-\frac {b}{d \,a^{2} \sin \left (d x +c \right )^{2}}+\frac {b^{3}}{2 d \,a^{4} \sin \left (d x +c \right )^{2}}+\frac {b}{4 d \,a^{2} \sin \left (d x +c \right )^{4}}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2} d}+\frac {2 b^{3} \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{4}}-\frac {b^{5} \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 170, normalized size = 0.95 \[ \frac {\frac {60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6}} - \frac {60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{6}} + \frac {15 \, a^{3} b \sin \left (d x + c\right ) - 60 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} - 12 \, a^{4} - 30 \, {\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{3} + 20 \, {\left (2 \, a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{5} \sin \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*(a^4*b - 2*a^2*b^3 + b^5)*log(b*sin(d*x + c) + a)/a^6 - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(sin(d*x + c)
)/a^6 + (15*a^3*b*sin(d*x + c) - 60*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 - 12*a^4 - 30*(2*a^3*b - a*b^3)*sin
(d*x + c)^3 + 20*(2*a^4 - a^2*b^2)*sin(d*x + c)^2)/(a^5*sin(d*x + c)^5))/d

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mupad [B]  time = 11.92, size = 381, normalized size = 2.13 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {b^2}{8\,a^3}-\frac {5}{16\,a}+\frac {2\,b\,\left (\frac {b}{16\,a^2}+\frac {2\,b\,\left (\frac {5}{32\,a}-\frac {b^2}{8\,a^3}\right )}{a}\right )}{a}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {b}{32\,a^2}+\frac {b\,\left (\frac {5}{32\,a}-\frac {b^2}{8\,a^3}\right )}{a}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5}{96\,a}-\frac {b^2}{24\,a^3}\right )}{d}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^4\,b-2\,a^2\,b^3+b^5\right )}{a^6\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4\,b-2\,a^2\,b^3+b^5\right )}{a^6\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {5\,a^4}{3}-\frac {4\,a^2\,b^2}{3}\right )-\frac {a^4}{5}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (10\,a^4-28\,a^2\,b^2+16\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a\,b^3-6\,a^3\,b\right )+\frac {a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{32\,a^5\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(b^2/(8*a^3) - 5/(16*a) + (2*b*(b/(16*a^2) + (2*b*(5/(32*a) - b^2/(8*a^3)))/a))/a))/d - (t
an(c/2 + (d*x)/2)^2*(b/(32*a^2) + (b*(5/(32*a) - b^2/(8*a^3)))/a))/d - tan(c/2 + (d*x)/2)^5/(160*a*d) + (tan(c
/2 + (d*x)/2)^3*(5/(96*a) - b^2/(24*a^3)))/d + (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a^4*
b + b^5 - 2*a^2*b^3))/(a^6*d) + (b*tan(c/2 + (d*x)/2)^4)/(64*a^2*d) - (log(tan(c/2 + (d*x)/2))*(a^4*b + b^5 -
2*a^2*b^3))/(a^6*d) + (tan(c/2 + (d*x)/2)^2*((5*a^4)/3 - (4*a^2*b^2)/3) - a^4/5 - tan(c/2 + (d*x)/2)^4*(10*a^4
+ 16*b^4 - 28*a^2*b^2) + tan(c/2 + (d*x)/2)^3*(4*a*b^3 - 6*a^3*b) + (a^3*b*tan(c/2 + (d*x)/2))/2)/(32*a^5*d*t
an(c/2 + (d*x)/2)^5)

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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+b*sin(d*x+c)),x)

[Out]

Timed out

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