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]

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

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Rubi [A]  time = 0.18, antiderivative size = 157, 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 = {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 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 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]

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*}

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Mathematica [A]  time = 1.05, size = 141, normalized size = 0.90 \[ -\frac {8 b^4 \log (a \cos (c+d x)+b)+a (a-b)^2 (a+b) \csc ^2\left (\frac {1}{2} (c+d x)\right )+a (a-b) (a+b)^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+4 a (a-b)^2 (2 a+3 b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a (2 a-3 b) (a+b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 a d (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(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*Cos[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)/(
a*(a - b)^2*(a + b)^2*d)

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fricas [A]  time = 0.60, size = 263, normalized size = 1.68 \[ \frac {2 \, a^{4} - 2 \, a^{2} b^{2} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) - 4 \, {\left (b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + {\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3} - {\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3} - {\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]

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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giac [B]  time = 0.86, size = 403, normalized size = 2.57 \[ -\frac {\frac {2 \, {\left (2 \, a + 3 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {4 \, {\left (a^{3} - 2 \, a b^{2}\right )} \log \left ({\left | a + b - \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a + b + \frac {4 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {6 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + 2 \, b^{4}\right )} \log \left (\frac {{\left | 2 \, b + \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left | a \right |} \right |}}{{\left | 2 \, b + \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 2 \, {\left | a \right |} \right |}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left | a \right |}} - \frac {\cos \left (d x + c\right ) - 1}{{\left (a - b\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \]

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

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maple [A]  time = 0.88, size = 167, normalized size = 1.06 \[ -\frac {b^{4} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{2} \left (a -b \right )^{2} a}+\frac {1}{d \left (4 a +4 b \right ) \left (-1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (-1+\cos \left (d x +c \right )\right ) a}{2 d \left (a +b \right )^{2}}-\frac {3 \ln \left (-1+\cos \left (d x +c \right )\right ) b}{4 d \left (a +b \right )^{2}}-\frac {1}{d \left (4 a -4 b \right ) \left (1+\cos \left (d x +c \right )\right )}+\frac {3 \ln \left (1+\cos \left (d x +c \right )\right ) b}{4 d \left (a -b \right )^{2}}-\frac {a \ln \left (1+\cos \left (d x +c \right )\right )}{2 \left (a -b \right )^{2} d} \]

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

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maxima [A]  time = 0.34, size = 144, normalized size = 0.92 \[ -\frac {\frac {4 \, b^{4} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac {{\left (2 \, a - 3 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (2 \, a + 3 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (b \cos \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]

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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mupad [B]  time = 1.85, size = 174, normalized size = 1.11 \[ \frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2\,d\,\left (4\,a-4\,b\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a+3\,b\right )}{d\,\left (2\,a^2+4\,a\,b+2\,b^2\right )}-\frac {a-b}{2\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a+b\right )\,\left (4\,a-4\,b\right )}-\frac {b^4\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a\,d\,{\left (a^2-b^2\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

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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