\(\int \frac {1}{5+3 \csc (c+d x)} \, dx\) [53]

Optimal result

Integrand size = 12, antiderivative size = 68 \[ \int \frac {1}{5+3 \csc (c+d x)} \, dx=\frac {x}{5}+\frac {3 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}-\frac {3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{20 d} \]

[Out]

1/5*x+3/20*ln(3*cos(1/2*d*x+1/2*c)+sin(1/2*d*x+1/2*c))/d-3/20*ln(cos(1/2*d*x+1/2*c)+3*sin(1/2*d*x+1/2*c))/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3868, 2739, 630, 31} \[ \int \frac {1}{5+3 \csc (c+d x)} \, dx=\frac {3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+3 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}-\frac {3 \log \left (3 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}+\frac {x}{5} \]

[In]

Int[(5 + 3*Csc[c + d*x])^(-1),x]

[Out]

x/5 + (3*Log[3*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(20*d) - (3*Log[Cos[(c + d*x)/2] + 3*Sin[(c + d*x)/2]])/(
20*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3868

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a/b)*Sin[c
+ d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {x}{5}-\frac {1}{5} \int \frac {1}{1+\frac {5}{3} \sin (c+d x)} \, dx \\ & = \frac {x}{5}-\frac {2 \text {Subst}\left (\int \frac {1}{1+\frac {10 x}{3}+x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{5 d} \\ & = \frac {x}{5}-\frac {3 \text {Subst}\left (\int \frac {1}{\frac {1}{3}+x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}+\frac {3 \text {Subst}\left (\int \frac {1}{3+x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{20 d} \\ & = \frac {x}{5}+\frac {3 \log \left (3+\tan \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}-\frac {3 \log \left (1+3 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{20 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {1}{5+3 \csc (c+d x)} \, dx=\frac {4 (c+d x)+3 \log \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{20 d} \]

[In]

Integrate[(5 + 3*Csc[c + d*x])^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.60

method	result	size
norman	\(\frac {x}{5}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{20 d}-\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{20 d}\)	\(41\)
risch	\(\frac {x}{5}+\frac {3 \ln \left (\frac {4}{5}+\frac {3 i}{5}+{\mathrm e}^{i \left (d x +c \right )}\right )}{20 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {4}{5}+\frac {3 i}{5}\right )}{20 d}\)	\(43\)
parallelrisch	\(\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )-3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\ln \left (27\right )+4 d x}{20 d}\)	\(43\)
derivativedivides	\(\frac {-\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{20}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{20}}{d}\)	\(48\)
default	\(\frac {-\frac {3 \ln \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{20}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{20}}{d}\)	\(48\)

[In]

int(1/(5+3*csc(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/5*x+3/20/d*ln(tan(1/2*d*x+1/2*c)+3)-3/20/d*ln(3*tan(1/2*d*x+1/2*c)+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {1}{5+3 \csc (c+d x)} \, dx=\frac {8 \, d x + 3 \, \log \left (4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right ) - 3 \, \log \left (-4 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right ) + 5\right )}{40 \, d} \]

[In]

integrate(1/(5+3*csc(d*x+c)),x, algorithm="fricas")

[Out]

1/40*(8*d*x + 3*log(4*cos(d*x + c) + 3*sin(d*x + c) + 5) - 3*log(-4*cos(d*x + c) + 3*sin(d*x + c) + 5))/d

Sympy [F]

\[ \int \frac {1}{5+3 \csc (c+d x)} \, dx=\int \frac {1}{3 \csc {\left (c + d x \right )} + 5}\, dx \]

[In]

integrate(1/(5+3*csc(d*x+c)),x)

[Out]

Integral(1/(3*csc(c + d*x) + 5), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5+3 \csc (c+d x)} \, dx=\frac {8 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) - 3 \, \log \left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + 3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 3\right )}{20 \, d} \]

[In]

integrate(1/(5+3*csc(d*x+c)),x, algorithm="maxima")

[Out]

1/20*(8*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) - 3*log(3*sin(d*x + c)/(cos(d*x + c) + 1) + 1) + 3*log(sin(d*x
+ c)/(cos(d*x + c) + 1) + 3))/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.66 \[ \int \frac {1}{5+3 \csc (c+d x)} \, dx=\frac {4 \, d x + 4 \, c - 3 \, \log \left ({\left | 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \right |}\right )}{20 \, d} \]

[In]

integrate(1/(5+3*csc(d*x+c)),x, algorithm="giac")

[Out]

1/20*(4*d*x + 4*c - 3*log(abs(3*tan(1/2*d*x + 1/2*c) + 1)) + 3*log(abs(tan(1/2*d*x + 1/2*c) + 3)))/d

Mupad [B] (verification not implemented)

Time = 18.57 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.40 \[ \int \frac {1}{5+3 \csc (c+d x)} \, dx=\frac {x}{5}-\frac {3\,\mathrm {atanh}\left (\frac {1}{2\,\left (\frac {200\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{27}+\frac {20}{9}\right )}+\frac {41}{40}\right )}{10\,d} \]

[In]

int(1/(3/sin(c + d*x) + 5),x)

[Out]

x/5 - (3*atanh(1/(2*((200*tan(c/2 + (d*x)/2))/27 + 20/9)) + 41/40))/(10*d)