3.285 \(\int (a \cot (x)+b \csc (x))^3 \, dx\)

Optimal. Leaf size=77 \[ -\frac {1}{2} a^2 b \cos (x)-\frac {1}{4} (2 a-b) (a+b)^2 \log (1-\cos (x))-\frac {1}{4} (a-b)^2 (2 a+b) \log (\cos (x)+1)-\frac {1}{2} \csc ^2(x) (a \cos (x)+b)^2 (a+b \cos (x)) \]

[Out]

-1/2*a^2*b*cos(x)-1/2*(b+a*cos(x))^2*(a+b*cos(x))*csc(x)^2-1/4*(2*a-b)*(a+b)^2*ln(1-cos(x))-1/4*(a-b)^2*(2*a+b
)*ln(1+cos(x))

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Rubi [A]  time = 0.14, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {4392, 2668, 739, 774, 633, 31} \[ -\frac {1}{2} a^2 b \cos (x)-\frac {1}{4} (2 a-b) (a+b)^2 \log (1-\cos (x))-\frac {1}{4} (a-b)^2 (2 a+b) \log (\cos (x)+1)-\frac {1}{2} \csc ^2(x) (a \cos (x)+b)^2 (a+b \cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[(a*Cot[x] + b*Csc[x])^3,x]

[Out]

-(a^2*b*Cos[x])/2 - ((b + a*Cos[x])^2*(a + b*Cos[x])*Csc[x]^2)/2 - ((2*a - b)*(a + b)^2*Log[1 - Cos[x]])/4 - (
(a - b)^2*(2*a + b)*Log[1 + Cos[x]])/4

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4392

Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b_.))^(p_)*(u_.), x_Symbol] :> Int[A
ctivateTrig[u]*Csc[c + d*x]^(n*p)*(b + a*Cos[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int (a \cot (x)+b \csc (x))^3 \, dx &=\int (b+a \cos (x))^3 \csc ^3(x) \, dx\\ &=-\left (a^3 \operatorname {Subst}\left (\int \frac {(b+x)^3}{\left (a^2-x^2\right )^2} \, dx,x,a \cos (x)\right )\right )\\ &=-\frac {1}{2} (b+a \cos (x))^2 (a+b \cos (x)) \csc ^2(x)+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {(b+x) \left (2 a^2-b^2+b x\right )}{a^2-x^2} \, dx,x,a \cos (x)\right )\\ &=-\frac {1}{2} a^2 b \cos (x)-\frac {1}{2} (b+a \cos (x))^2 (a+b \cos (x)) \csc ^2(x)-\frac {1}{2} a \operatorname {Subst}\left (\int \frac {-a^2 b-b \left (2 a^2-b^2\right )-2 a^2 x}{a^2-x^2} \, dx,x,a \cos (x)\right )\\ &=-\frac {1}{2} a^2 b \cos (x)-\frac {1}{2} (b+a \cos (x))^2 (a+b \cos (x)) \csc ^2(x)+\frac {1}{4} \left ((2 a-b) (a+b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-x} \, dx,x,a \cos (x)\right )+\frac {1}{4} \left ((a-b)^2 (2 a+b)\right ) \operatorname {Subst}\left (\int \frac {1}{-a-x} \, dx,x,a \cos (x)\right )\\ &=-\frac {1}{2} a^2 b \cos (x)-\frac {1}{2} (b+a \cos (x))^2 (a+b \cos (x)) \csc ^2(x)-\frac {1}{4} (2 a-b) (a+b)^2 \log (1-\cos (x))-\frac {1}{4} (a-b)^2 (2 a+b) \log (1+\cos (x))\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 79, normalized size = 1.03 \[ \frac {1}{8} \left (-(a+b)^3 \csc ^2\left (\frac {x}{2}\right )+(a-b)^3 \left (-\sec ^2\left (\frac {x}{2}\right )\right )-4 (2 a-b) (a+b)^2 \log \left (\sin \left (\frac {x}{2}\right )\right )-4 (2 a+b) (a-b)^2 \log \left (\cos \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cot[x] + b*Csc[x])^3,x]

[Out]

(-((a + b)^3*Csc[x/2]^2) - 4*(a - b)^2*(2*a + b)*Log[Cos[x/2]] - 4*(2*a - b)*(a + b)^2*Log[Sin[x/2]] - (a - b)
^3*Sec[x/2]^2)/8

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fricas [A]  time = 0.97, size = 128, normalized size = 1.66 \[ \frac {2 \, a^{3} + 6 \, a b^{2} + 2 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \relax (x) + {\left (2 \, a^{3} - 3 \, a^{2} b + b^{3} - {\left (2 \, a^{3} - 3 \, a^{2} b + b^{3}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3} - {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{4 \, {\left (\cos \relax (x)^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cot(x)+b*csc(x))^3,x, algorithm="fricas")

[Out]

1/4*(2*a^3 + 6*a*b^2 + 2*(3*a^2*b + b^3)*cos(x) + (2*a^3 - 3*a^2*b + b^3 - (2*a^3 - 3*a^2*b + b^3)*cos(x)^2)*l
og(1/2*cos(x) + 1/2) + (2*a^3 + 3*a^2*b - b^3 - (2*a^3 + 3*a^2*b - b^3)*cos(x)^2)*log(-1/2*cos(x) + 1/2))/(cos
(x)^2 - 1)

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giac [A]  time = 0.15, size = 86, normalized size = 1.12 \[ -\frac {1}{4} \, {\left (2 \, a^{3} - 3 \, a^{2} b + b^{3}\right )} \log \left (\cos \relax (x) + 1\right ) - \frac {1}{4} \, {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} \log \left (-\cos \relax (x) + 1\right ) + \frac {a^{3} + 3 \, a b^{2} + {\left (3 \, a^{2} b + b^{3}\right )} \cos \relax (x)}{2 \, {\left (\cos \relax (x) + 1\right )} {\left (\cos \relax (x) - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cot(x)+b*csc(x))^3,x, algorithm="giac")

[Out]

-1/4*(2*a^3 - 3*a^2*b + b^3)*log(cos(x) + 1) - 1/4*(2*a^3 + 3*a^2*b - b^3)*log(-cos(x) + 1) + 1/2*(a^3 + 3*a*b
^2 + (3*a^2*b + b^3)*cos(x))/((cos(x) + 1)*(cos(x) - 1))

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maple [A]  time = 0.07, size = 87, normalized size = 1.13 \[ -\frac {a^{3} \left (\cot ^{2}\relax (x )\right )}{2}-a^{3} \ln \left (\sin \relax (x )\right )-\frac {3 a^{2} b \left (\cos ^{3}\relax (x )\right )}{2 \sin \relax (x )^{2}}-\frac {3 a^{2} b \cos \relax (x )}{2}-\frac {3 a^{2} b \ln \left (\csc \relax (x )-\cot \relax (x )\right )}{2}-\frac {3 a \,b^{2}}{2 \sin \relax (x )^{2}}-\frac {b^{3} \csc \relax (x ) \cot \relax (x )}{2}+\frac {b^{3} \ln \left (\csc \relax (x )-\cot \relax (x )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cot(x)+b*csc(x))^3,x)

[Out]

-1/2*a^3*cot(x)^2-a^3*ln(sin(x))-3/2*a^2*b/sin(x)^2*cos(x)^3-3/2*a^2*b*cos(x)-3/2*a^2*b*ln(csc(x)-cot(x))-3/2*
a*b^2/sin(x)^2-1/2*b^3*csc(x)*cot(x)+1/2*b^3*ln(csc(x)-cot(x))

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maxima [A]  time = 0.31, size = 87, normalized size = 1.13 \[ -\frac {3}{2} \, a b^{2} \cot \relax (x)^{2} + \frac {3}{4} \, a^{2} b {\left (\frac {2 \, \cos \relax (x)}{\cos \relax (x)^{2} - 1} + \log \left (\cos \relax (x) + 1\right ) - \log \left (\cos \relax (x) - 1\right )\right )} + \frac {1}{4} \, b^{3} {\left (\frac {2 \, \cos \relax (x)}{\cos \relax (x)^{2} - 1} - \log \left (\cos \relax (x) + 1\right ) + \log \left (\cos \relax (x) - 1\right )\right )} - \frac {1}{2} \, a^{3} {\left (\frac {1}{\sin \relax (x)^{2}} + \log \left (\sin \relax (x)^{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cot(x)+b*csc(x))^3,x, algorithm="maxima")

[Out]

-3/2*a*b^2*cot(x)^2 + 3/4*a^2*b*(2*cos(x)/(cos(x)^2 - 1) + log(cos(x) + 1) - log(cos(x) - 1)) + 1/4*b^3*(2*cos
(x)/(cos(x)^2 - 1) - log(cos(x) + 1) + log(cos(x) - 1)) - 1/2*a^3*(1/sin(x)^2 + log(sin(x)^2))

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mupad [B]  time = 2.45, size = 82, normalized size = 1.06 \[ a^3\,\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )-\frac {\frac {a^3}{8}+\frac {3\,a^2\,b}{8}+\frac {3\,a\,b^2}{8}+\frac {b^3}{8}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2}-\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (a^3+\frac {3\,a^2\,b}{2}-\frac {b^3}{2}\right )-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,{\left (a-b\right )}^3}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/sin(x) + a*cot(x))^3,x)

[Out]

a^3*log(tan(x/2)^2 + 1) - ((3*a*b^2)/8 + (3*a^2*b)/8 + a^3/8 + b^3/8)/tan(x/2)^2 - log(tan(x/2))*((3*a^2*b)/2
+ a^3 - b^3/2) - (tan(x/2)^2*(a - b)^3)/8

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sympy [A]  time = 14.11, size = 124, normalized size = 1.61 \[ \frac {a^{3} \log {\left (- \csc ^{2}{\relax (x )} \right )}}{2} - \frac {a^{3} \csc ^{2}{\relax (x )}}{2} - \frac {3 a^{2} b \log {\left (\cos {\relax (x )} - 1 \right )}}{4} + \frac {3 a^{2} b \log {\left (\cos {\relax (x )} + 1 \right )}}{4} + \frac {3 a^{2} b \cos {\relax (x )}}{2 \cos ^{2}{\relax (x )} - 2} - \frac {3 a b^{2} \csc ^{2}{\relax (x )}}{2} + \frac {b^{3} \log {\left (\cos {\relax (x )} - 1 \right )}}{4} - \frac {b^{3} \log {\left (\cos {\relax (x )} + 1 \right )}}{4} + \frac {b^{3} \cos {\relax (x )}}{2 \cos ^{2}{\relax (x )} - 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cot(x)+b*csc(x))**3,x)

[Out]

a**3*log(-csc(x)**2)/2 - a**3*csc(x)**2/2 - 3*a**2*b*log(cos(x) - 1)/4 + 3*a**2*b*log(cos(x) + 1)/4 + 3*a**2*b
*cos(x)/(2*cos(x)**2 - 2) - 3*a*b**2*csc(x)**2/2 + b**3*log(cos(x) - 1)/4 - b**3*log(cos(x) + 1)/4 + b**3*cos(
x)/(2*cos(x)**2 - 2)

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