3.292 \(\int \frac{\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=157 \[ -\frac{b^4 \log (a+b \sec (c+d x))}{a d \left (a^2-b^2\right )^2}+\frac{1}{4 d (a+b) (1-\sec (c+d x))}+\frac{1}{4 d (a-b) (\sec (c+d x)+1)}-\frac{(2 a+3 b) \log (1-\sec (c+d x))}{4 d (a+b)^2}-\frac{(2 a-3 b) \log (\sec (c+d x)+1)}{4 d (a-b)^2}-\frac{\log (\cos (c+d x))}{a d} \]

[Out]

-(Log[Cos[c + d*x]]/(a*d)) - ((2*a + 3*b)*Log[1 - Sec[c + d*x]])/(4*(a + b)^2*d) - ((2*a - 3*b)*Log[1 + Sec[c
+ d*x]])/(4*(a - b)^2*d) - (b^4*Log[a + b*Sec[c + d*x]])/(a*(a^2 - b^2)^2*d) + 1/(4*(a + b)*d*(1 - Sec[c + d*x
])) + 1/(4*(a - b)*d*(1 + Sec[c + d*x]))

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Rubi [A]  time = 0.181046, antiderivative size = 157, 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 = {3885, 894} \[ -\frac{b^4 \log (a+b \sec (c+d x))}{a d \left (a^2-b^2\right )^2}+\frac{1}{4 d (a+b) (1-\sec (c+d x))}+\frac{1}{4 d (a-b) (\sec (c+d x)+1)}-\frac{(2 a+3 b) \log (1-\sec (c+d x))}{4 d (a+b)^2}-\frac{(2 a-3 b) \log (\sec (c+d x)+1)}{4 d (a-b)^2}-\frac{\log (\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/(a + b*Sec[c + d*x]),x]

[Out]

-(Log[Cos[c + d*x]]/(a*d)) - ((2*a + 3*b)*Log[1 - Sec[c + d*x]])/(4*(a + b)^2*d) - ((2*a - 3*b)*Log[1 + Sec[c
+ d*x]])/(4*(a - b)^2*d) - (b^4*Log[a + b*Sec[c + d*x]])/(a*(a^2 - b^2)^2*d) + 1/(4*(a + b)*d*(1 - Sec[c + d*x
])) + 1/(4*(a - b)*d*(1 + Sec[c + d*x]))

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A]  time = 1.0088, size = 141, normalized size = 0.9 \[ -\frac{8 b^4 \log (a \cos (c+d x)+b)+a (a-b)^2 (a+b) \csc ^2\left (\frac{1}{2} (c+d x)\right )+a (a-b) (a+b)^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )+4 a (a-b)^2 (2 a+3 b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 a (2 a-3 b) (a+b)^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 a d (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3/(a + b*Sec[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.075, size = 167, normalized size = 1.1 \begin{align*} -{\frac{{b}^{4}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}a}}-{\frac{1}{d \left ( 4\,a-4\,b \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) a}{2\,d \left ( a-b \right ) ^{2}}}+{\frac{3\,\ln \left ( \cos \left ( dx+c \right ) +1 \right ) b}{4\,d \left ( a-b \right ) ^{2}}}+{\frac{1}{d \left ( 4\,a+4\,b \right ) \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) a}{2\,d \left ( a+b \right ) ^{2}}}-{\frac{3\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{4\,d \left ( a+b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+b*sec(d*x+c)),x)

[Out]

-1/d*b^4/(a+b)^2/(a-b)^2/a*ln(b+a*cos(d*x+c))-1/d/(4*a-4*b)/(cos(d*x+c)+1)-1/2/d/(a-b)^2*ln(cos(d*x+c)+1)*a+3/
4/d/(a-b)^2*ln(cos(d*x+c)+1)*b+1/d/(4*a+4*b)/(-1+cos(d*x+c))-1/2/d/(a+b)^2*ln(-1+cos(d*x+c))*a-3/4/d/(a+b)^2*l
n(-1+cos(d*x+c))*b

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Maxima [A]  time = 1.0286, size = 194, normalized size = 1.24 \begin{align*} -\frac{\frac{4 \, b^{4} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac{{\left (2 \, a - 3 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{{\left (2 \, a + 3 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{2 \,{\left (b \cos \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.07828, size = 587, normalized size = 3.74 \begin{align*} \frac{2 \, a^{4} - 2 \, a^{2} b^{2} - 2 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) - 4 \,{\left (b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) +{\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3} -{\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3} -{\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(2*a^4 - 2*a^2*b^2 - 2*(a^3*b - a*b^3)*cos(d*x + c) - 4*(b^4*cos(d*x + c)^2 - b^4)*log(a*cos(d*x + c) + b)
+ (2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3 - (2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*
x + c) + 1/2) + (2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3 - (2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*cos(d*x + c)^2)*l
og(-1/2*cos(d*x + c) + 1/2))/((a^5 - 2*a^3*b^2 + a*b^4)*d*cos(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{3}{\left (c + d x \right )}}{a + b \sec{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3/(a+b*sec(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**3/(a + b*sec(c + d*x)), x)

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Giac [A]  time = 1.45078, size = 346, normalized size = 2.2 \begin{align*} -\frac{\frac{8 \, b^{4} \log \left ({\left | a + b + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac{2 \,{\left (2 \, a + 3 \, b\right )} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{{\left (a + b + \frac{4 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{6 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (\cos \left (d x + c\right ) - 1\right )}} - \frac{8 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a} - \frac{\cos \left (d x + c\right ) - 1}{{\left (a - b\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(8*b^4*log(abs(a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))
)/(a^5 - 2*a^3*b^2 + a*b^4) + 2*(2*a + 3*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^2 + 2*a*b + b
^2) - (a + b + 4*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 6*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x
+ c) + 1)/((a^2 + 2*a*b + b^2)*(cos(d*x + c) - 1)) - 8*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a
- (cos(d*x + c) - 1)/((a - b)*(cos(d*x + c) + 1)))/d