\(\int \cot (e+f x) (a+b \sec ^3(e+f x)) \, dx\) [455]

Optimal result

Integrand size = 19, antiderivative size = 54 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {(a+b) \log (1-\cos (e+f x))}{2 f}+\frac {(a-b) \log (1+\cos (e+f x))}{2 f}+\frac {b \sec (e+f x)}{f} \]

[Out]

1/2*(a+b)*ln(1-cos(f*x+e))/f+1/2*(a-b)*ln(1+cos(f*x+e))/f+b*sec(f*x+e)/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 54, 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 = {4223, 1816} \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {(a+b) \log (1-\cos (e+f x))}{2 f}+\frac {(a-b) \log (\cos (e+f x)+1)}{2 f}+\frac {b \sec (e+f x)}{f} \]

[In]

Int[Cot[e + f*x]*(a + b*Sec[e + f*x]^3),x]

[Out]

((a + b)*Log[1 - Cos[e + f*x]])/(2*f) + ((a - b)*Log[1 + Cos[e + f*x]])/(2*f) + (b*Sec[e + f*x])/f

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 4223

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, Dist[-(f*ff^(m + n*p - 1))^(-1), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*
(ff*x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {b+a x^3}{x^2 \left (1-x^2\right )} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {-a-b}{2 (-1+x)}+\frac {b}{x^2}+\frac {-a+b}{2 (1+x)}\right ) \, dx,x,\cos (e+f x)\right )}{f} \\ & = \frac {(a+b) \log (1-\cos (e+f x))}{2 f}+\frac {(a-b) \log (1+\cos (e+f x))}{2 f}+\frac {b \sec (e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.28 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=-\frac {b \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{f}+\frac {a \log (\cos (e+f x))}{f}+\frac {b \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f}+\frac {a \log (\tan (e+f x))}{f}+\frac {b \sec (e+f x)}{f} \]

[In]

Integrate[Cot[e + f*x]*(a + b*Sec[e + f*x]^3),x]

[Out]

-((b*Log[Cos[(e + f*x)/2]])/f) + (a*Log[Cos[e + f*x]])/f + (b*Log[Sin[(e + f*x)/2]])/f + (a*Log[Tan[e + f*x]])
/f + (b*Sec[e + f*x])/f

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {a \ln \left (\sin \left (f x +e \right )\right )+b \left (\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{f}\)	\(42\)
default	\(\frac {a \ln \left (\sin \left (f x +e \right )\right )+b \left (\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{f}\)	\(42\)
risch	\(-i a x -\frac {2 i a e}{f}+\frac {2 b \,{\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{f}\)	\(112\)

[In]

int(cot(f*x+e)*(a+b*sec(f*x+e)^3),x,method=_RETURNVERBOSE)

[Out]

1/f*(a*ln(sin(f*x+e))+b*(1/cos(f*x+e)+ln(csc(f*x+e)-cot(f*x+e))))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.13 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {{\left (a - b\right )} \cos \left (f x + e\right ) \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + {\left (a + b\right )} \cos \left (f x + e\right ) \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 2 \, b}{2 \, f \cos \left (f x + e\right )} \]

[In]

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

[Out]

1/2*((a - b)*cos(f*x + e)*log(1/2*cos(f*x + e) + 1/2) + (a + b)*cos(f*x + e)*log(-1/2*cos(f*x + e) + 1/2) + 2*
b)/(f*cos(f*x + e))

Sympy [F]

\[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\int \left (a + b \sec ^{3}{\left (e + f x \right )}\right ) \cot {\left (e + f x \right )}\, dx \]

[In]

integrate(cot(f*x+e)*(a+b*sec(f*x+e)**3),x)

[Out]

Integral((a + b*sec(e + f*x)**3)*cot(e + f*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {{\left (a - b\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) + {\left (a + b\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) + \frac {2 \, b}{\cos \left (f x + e\right )}}{2 \, f} \]

[In]

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

[Out]

1/2*((a - b)*log(cos(f*x + e) + 1) + (a + b)*log(cos(f*x + e) - 1) + 2*b/cos(f*x + e))/f

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.61 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {{\left (a + b\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right ) - 2 \, a \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right ) + \frac {4 \, b}{\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1}}{2 \, f} \]

[In]

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

[Out]

1/2*((a + b)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1)) - 2*a*log(abs(-(cos(f*x + e) - 1)/(cos(f*x + e)
+ 1) + 1)) + 4*b/((cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1))/f

Mupad [B] (verification not implemented)

Time = 20.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.33 \[ \int \cot (e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}{f}-\frac {2\,b}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f} \]

[In]

int(cot(e + f*x)*(a + b/cos(e + f*x)^3),x)

[Out]

(a*log(tan(e/2 + (f*x)/2)))/f - (a*log(tan(e/2 + (f*x)/2)^2 + 1))/f - (2*b)/(f*(tan(e/2 + (f*x)/2)^2 - 1)) + (
b*log(tan(e/2 + (f*x)/2)))/f