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

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

[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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Rubi [A]  time = 0.204636, antiderivative size = 179, normalized size of antiderivative = 1., 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 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]

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]))

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*}

Mathematica [A]  time = 6.12829, size = 179, normalized size = 1. \[ \frac{\left (2 a^2-b^2\right ) \csc ^3(c+d x)}{3 a^3 d}-\frac{b \left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^4 d}-\frac{\left (a^2-b^2\right )^2 \csc (c+d x)}{a^5 d}-\frac{b \left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^6 d}+\frac{b \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^6 d}+\frac{b \csc ^4(c+d x)}{4 a^2 d}-\frac{\csc ^5(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

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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Maple [A]  time = 0.098, size = 274, normalized size = 1.5 \begin{align*}{\frac{b\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{a}^{2}}}-2\,{\frac{{b}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{a}^{4}}}+{\frac{{b}^{5}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{a}^{6}}}-{\frac{1}{5\,da \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2}{3\,da \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{2}}{3\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{da\sin \left ( dx+c \right ) }}+2\,{\frac{{b}^{2}}{d{a}^{3}\sin \left ( dx+c \right ) }}-{\frac{{b}^{4}}{d{a}^{5}\sin \left ( dx+c \right ) }}-{\frac{b}{d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3}}{2\,d{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{b}{4\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{2}}}+2\,{\frac{{b}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{4}}}-{\frac{{b}^{5}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{6}}} \end{align*}

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]

1/d/a^2*b*ln(a+b*sin(d*x+c))-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 = 1.00004, size = 230, normalized size = 1.28 \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 1.52689, size = 807, normalized size = 4.51 \begin{align*} -\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 )} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.20982, size = 339, normalized size = 1.89 \begin{align*} -\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} \end{align*}

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