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

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

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 1.06, size = 115, normalized size = 0.78 \[ \frac {-3 a^4 \csc ^4(c+d x)+4 a^3 b \csc ^3(c+d x)+6 a^2 \left (2 a^2-b^2\right ) \csc ^2(c+d x)+12 a b \left (b^2-2 a^2\right ) \csc (c+d x)+12 \left (a^2-b^2\right )^2 (\log (\sin (c+d x))-\log (a+b \sin (c+d x)))}{12 a^5 d} \]

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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fricas [A]  time = 0.47, size = 271, normalized size = 1.83 \[ \frac {9 \, a^{4} - 6 \, a^{2} b^{2} - 6 \, {\left (2 \, a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 12 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (5 \, a^{3} b - 3 \, a b^{3} - 3 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{5} d \cos \left (d x + c\right )^{4} - 2 \, a^{5} d \cos \left (d x + c\right )^{2} + a^{5} d\right )}} \]

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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giac [A]  time = 0.45, size = 201, normalized size = 1.36 \[ \frac {\frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{5}} - \frac {12 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b} - \frac {25 \, a^{4} \sin \left (d x + c\right )^{4} - 50 \, a^{2} b^{2} \sin \left (d x + c\right )^{4} + 25 \, b^{4} \sin \left (d x + c\right )^{4} + 24 \, a^{3} b \sin \left (d x + c\right )^{3} - 12 \, a b^{3} \sin \left (d x + c\right )^{3} - 12 \, a^{4} \sin \left (d x + c\right )^{2} + 6 \, a^{2} b^{2} \sin \left (d x + c\right )^{2} - 4 \, a^{3} b \sin \left (d x + c\right ) + 3 \, a^{4}}{a^{5} \sin \left (d x + c\right )^{4}}}{12 \, d} \]

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

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maple [A]  time = 0.20, size = 216, normalized size = 1.46 \[ -\frac {\ln \left (a +b \sin \left (d x +c \right )\right )}{d a}+\frac {2 \ln \left (a +b \sin \left (d x +c \right )\right ) b^{2}}{d \,a^{3}}-\frac {\ln \left (a +b \sin \left (d x +c \right )\right ) b^{4}}{d \,a^{5}}-\frac {1}{4 d a \sin \left (d x +c \right )^{4}}+\frac {1}{d a \sin \left (d x +c \right )^{2}}-\frac {b^{2}}{2 d \,a^{3} \sin \left (d x +c \right )^{2}}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a d}-\frac {2 \ln \left (\sin \left (d x +c \right )\right ) b^{2}}{d \,a^{3}}+\frac {\ln \left (\sin \left (d x +c \right )\right ) b^{4}}{d \,a^{5}}-\frac {2 b}{d \,a^{2} \sin \left (d x +c \right )}+\frac {b^{3}}{d \,a^{4} \sin \left (d x +c \right )}+\frac {b}{3 d \,a^{2} \sin \left (d x +c \right )^{3}} \]

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 = 0.61, size = 139, normalized size = 0.94 \[ -\frac {\frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5}} - \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{5}} - \frac {4 \, a^{2} b \sin \left (d x + c\right ) - 12 \, {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 3 \, a^{3} + 6 \, {\left (2 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{4} \sin \left (d x + c\right )^{4}}}{12 \, d} \]

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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mupad [B]  time = 6.41, size = 281, normalized size = 1.90 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3}{16\,a}-\frac {b^2}{8\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {b}{8\,a^2}+\frac {2\,b\,\left (\frac {3}{8\,a}-\frac {b^2}{4\,a^3}\right )}{a}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a\,b^2-3\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (14\,a^2\,b-8\,b^3\right )+\frac {a^3}{4}-\frac {2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}}{16\,a^4\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{a^5\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{a^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^2*(3/(16*a) - b^2/(8*a^3)))/d - tan(c/2 + (d*x)/2)^4/(64*a*d) - (tan(c/2 + (d*x)/2)*(b/(8*
a^2) + (2*b*(3/(8*a) - b^2/(4*a^3)))/a))/d - (tan(c/2 + (d*x)/2)^2*(2*a*b^2 - 3*a^3) + tan(c/2 + (d*x)/2)^3*(1
4*a^2*b - 8*b^3) + a^3/4 - (2*a^2*b*tan(c/2 + (d*x)/2))/3)/(16*a^4*d*tan(c/2 + (d*x)/2)^4) - (log(a + 2*b*tan(
c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a^4 + b^4 - 2*a^2*b^2))/(a^5*d) + (b*tan(c/2 + (d*x)/2)^3)/(24*a^2*d
) + (log(tan(c/2 + (d*x)/2))*(a^4 + b^4 - 2*a^2*b^2))/(a^5*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{5}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

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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