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

Optimal. Leaf size=148 \[ \frac{\left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^3 d}-\frac{b \left (2 a^2-b^2\right ) \csc (c+d x)}{a^4 d}+\frac{\left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^5 d}-\frac{\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^5 d}+\frac{b \csc ^3(c+d x)}{3 a^2 d}-\frac{\csc ^4(c+d x)}{4 a d} \]

[Out]

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

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Rubi [A]  time = 0.138352, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2721, 894} \[ \frac{\left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^3 d}-\frac{b \left (2 a^2-b^2\right ) \csc (c+d x)}{a^4 d}+\frac{\left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^5 d}-\frac{\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^5 d}+\frac{b \csc ^3(c+d x)}{3 a^2 d}-\frac{\csc ^4(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

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

Mathematica [A]  time = 3.84689, size = 115, normalized size = 0.78 \[ \frac{6 a^2 \left (2 a^2-b^2\right ) \csc ^2(c+d x)+12 a b \left (b^2-2 a^2\right ) \csc (c+d x)+12 \left (a^2-b^2\right )^2 (\log (\sin (c+d x))-\log (a+b \sin (c+d x)))+4 a^3 b \csc ^3(c+d x)-3 a^4 \csc ^4(c+d x)}{12 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a*b*(-2*a^2 + b^2)*Csc[c + d*x] + 6*a^2*(2*a^2 - b^2)*Csc[c + d*x]^2 + 4*a^3*b*Csc[c + d*x]^3 - 3*a^4*Csc[
c + d*x]^4 + 12*(a^2 - b^2)^2*(Log[Sin[c + d*x]] - Log[a + b*Sin[c + d*x]]))/(12*a^5*d)

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Maple [A]  time = 0.063, size = 216, normalized size = 1.5 \begin{align*} -{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{da}}+2\,{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{2}}{d{a}^{3}}}-{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{4}}{d{a}^{5}}}-{\frac{1}{4\,da \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{1}{da \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2}}{2\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{da}}-2\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ){b}^{2}}{d{a}^{3}}}+{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ){b}^{4}}{d{a}^{5}}}-2\,{\frac{b}{d{a}^{2}\sin \left ( dx+c \right ) }}+{\frac{{b}^{3}}{d{a}^{4}\sin \left ( dx+c \right ) }}+{\frac{b}{3\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.46289, size = 188, normalized size = 1.27 \begin{align*} -\frac{\frac{12 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5}} - \frac{12 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{5}} - \frac{4 \, a^{2} b \sin \left (d x + c\right ) - 12 \,{\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 3 \, a^{3} + 6 \,{\left (2 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{4} \sin \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.663, size = 629, normalized size = 4.25 \begin{align*} \frac{9 \, a^{4} - 6 \, a^{2} b^{2} - 6 \,{\left (2 \, a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - 12 \,{\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 12 \,{\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 4 \,{\left (5 \, a^{3} b - 3 \, a b^{3} - 3 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \,{\left (a^{5} d \cos \left (d x + c\right )^{4} - 2 \, a^{5} d \cos \left (d x + c\right )^{2} + a^{5} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(9*a^4 - 6*a^2*b^2 - 6*(2*a^4 - a^2*b^2)*cos(d*x + c)^2 - 12*((a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^4 + a^
4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*log(b*sin(d*x + c) + a) + 12*((a^4 - 2*a^2*b^2
+ b^4)*cos(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*log(-1/2*sin(d*x +
c)) - 4*(5*a^3*b - 3*a*b^3 - 3*(2*a^3*b - a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^5*d*cos(d*x + c)^4 - 2*a^5*d
*cos(d*x + c)^2 + a^5*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{5}{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5/(a+b*sin(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**5/(a + b*sin(c + d*x)), x)

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Giac [A]  time = 1.53765, size = 271, normalized size = 1.83 \begin{align*} \frac{\frac{12 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{5}} - \frac{12 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b} - \frac{25 \, a^{4} \sin \left (d x + c\right )^{4} - 50 \, a^{2} b^{2} \sin \left (d x + c\right )^{4} + 25 \, b^{4} \sin \left (d x + c\right )^{4} + 24 \, a^{3} b \sin \left (d x + c\right )^{3} - 12 \, a b^{3} \sin \left (d x + c\right )^{3} - 12 \, a^{4} \sin \left (d x + c\right )^{2} + 6 \, a^{2} b^{2} \sin \left (d x + c\right )^{2} - 4 \, a^{3} b \sin \left (d x + c\right ) + 3 \, a^{4}}{a^{5} \sin \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(a^4 - 2*a^2*b^2 + b^4)*log(abs(sin(d*x + c)))/a^5 - 12*(a^4*b - 2*a^2*b^3 + b^5)*log(abs(b*sin(d*x +
c) + a))/(a^5*b) - (25*a^4*sin(d*x + c)^4 - 50*a^2*b^2*sin(d*x + c)^4 + 25*b^4*sin(d*x + c)^4 + 24*a^3*b*sin(
d*x + c)^3 - 12*a*b^3*sin(d*x + c)^3 - 12*a^4*sin(d*x + c)^2 + 6*a^2*b^2*sin(d*x + c)^2 - 4*a^3*b*sin(d*x + c)
+ 3*a^4)/(a^5*sin(d*x + c)^4))/d