\(\int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx\) [276]

Optimal result

Integrand size = 19, antiderivative size = 61 \[ \int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a^2 \log (\cos (c+d x))}{d}+\frac {(a+b)^2 \log (1-\sec (c+d x))}{2 d}+\frac {(a-b)^2 \log (1+\sec (c+d x))}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3970, 1816} \[ \int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a^2 \log (\cos (c+d x))}{d}+\frac {(a+b)^2 \log (1-\sec (c+d x))}{2 d}+\frac {(a-b)^2 \log (\sec (c+d x)+1)}{2 d} \]

[In]

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

[Out]

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

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 3970

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2 a^2 \log (\cos (c+d x))+(a+b)^2 \log (1-\sec (c+d x))+(a-b)^2 \log (1+\sec (c+d x))}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 a b \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+b^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(48\)
default	\(\frac {a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 a b \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+b^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(48\)
risch	\(-i a^{2} x -\frac {2 i a^{2} c}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) a b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) a b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{d}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(153\)

[In]

int(cot(d*x+c)*(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*ln(sin(d*x+c))+2*a*b*ln(-cot(d*x+c)+csc(d*x+c))+b^2*ln(tan(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {2 \, b^{2} \log \left (-\cos \left (d x + c\right )\right ) - {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \cot {\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {2 \, b^{2} \log \left (\cos \left (d x + c\right )\right ) - {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.66 \[ \int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {2 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) + 2 \, b^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.38 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {b^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {b^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}{d}+\frac {2\,a\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

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

[Out]

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