3.304 \(\int \frac {\cot (c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=138 \[ \frac {b^2}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {b^2 \left (3 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 d \left (a^2-b^2\right )^2}+\frac {\log (\cos (c+d x))}{a^2 d}+\frac {\log (1-\sec (c+d x))}{2 d (a+b)^2}+\frac {\log (\sec (c+d x)+1)}{2 d (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3885, 894} \[ \frac {b^2}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {b^2 \left (3 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 d \left (a^2-b^2\right )^2}+\frac {\log (\cos (c+d x))}{a^2 d}+\frac {\log (1-\sec (c+d x))}{2 d (a+b)^2}+\frac {\log (\sec (c+d x)+1)}{2 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]/(a + b*Sec[c + d*x])^2,x]

[Out]

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

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Mathematica [A]  time = 0.36, size = 189, normalized size = 1.37 \[ \frac {a \cos (c+d x) \left (\left (b^4-3 a^2 b^2\right ) \log (a \cos (c+d x)+b)+a^2 (a-b)^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a^2 (a+b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+b \left ((a-b) \left (a^2 (a-b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-b^2 (a+b)\right )+\left (b^4-3 a^2 b^2\right ) \log (a \cos (c+d x)+b)+a^2 (a+b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{a^2 d (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]/(a + b*Sec[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.60, size = 234, normalized size = 1.70 \[ -\frac {2 \, a^{2} b^{3} - 2 \, b^{5} + 2 \, {\left (3 \, a^{2} b^{3} - b^{5} + {\left (3 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(2*a^2*b^3 - 2*b^5 + 2*(3*a^2*b^3 - b^5 + (3*a^3*b^2 - a*b^4)*cos(d*x + c))*log(a*cos(d*x + c) + b) - (a^
4*b + 2*a^3*b^2 + a^2*b^3 + (a^5 + 2*a^4*b + a^3*b^2)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - (a^4*b - 2*a
^3*b^2 + a^2*b^3 + (a^5 - 2*a^4*b + a^3*b^2)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^7 - 2*a^5*b^2 + a
^3*b^4)*d*cos(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)

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giac [B]  time = 0.28, size = 303, normalized size = 2.20 \[ -\frac {\frac {2 \, {\left (3 \, a^{2} b^{2} - b^{4}\right )} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}} - \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (3 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} + \frac {3 \, a^{2} b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b^{4} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{{\left (a^{5} + a^{4} b - a^{3} b^{2} - a^{2} b^{3}\right )} {\left (a + b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}} + \frac {2 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.61, size = 141, normalized size = 1.02 \[ -\frac {b^{3}}{d \,a^{2} \left (a +b \right ) \left (a -b \right ) \left (b +a \cos \left (d x +c \right )\right )}-\frac {3 b^{2} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {b^{4} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{2} \left (a -b \right )^{2} a^{2}}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{2 d \left (a +b \right )^{2}}+\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 \left (a -b \right )^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)/(a+b*sec(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.32, size = 142, normalized size = 1.03 \[ -\frac {\frac {2 \, b^{3}}{a^{4} b - a^{2} b^{3} + {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )} + \frac {2 \, {\left (3 \, a^{2} b^{2} - b^{4}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}} - \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.98, size = 160, normalized size = 1.16 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a^2+2\,a\,b+b^2\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}-\frac {b^2\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (3\,a^2-b^2\right )}{a^2\,d\,{\left (a^2-b^2\right )}^2}-\frac {2\,b^3}{a\,d\,\left (a+b\right )\,{\left (a-b\right )}^2\,\left (\left (b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)/(a + b/cos(c + d*x))^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(cot(c + d*x)/(a + b*sec(c + d*x))**2, x)

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