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

Optimal. Leaf size=68 \[ \frac{3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{20 d}-\frac{3 \log \left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{20 d}+\frac{x}{5} \]

[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)

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Rubi [A]  time = 0.0390231, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3783, 2660, 616, 31} \[ \frac{3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{20 d}-\frac{3 \log \left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{20 d}+\frac{x}{5} \]

Antiderivative was successfully verified.

[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 3783

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

Rule 2660

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 616

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 31

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

Rubi steps

\begin{align*} \int \frac{1}{5+3 \csc (c+d x)} \, dx &=\frac{x}{5}-\frac{1}{5} \int \frac{1}{1+\frac{5}{3} \sin (c+d x)} \, dx\\ &=\frac{x}{5}-\frac{2 \operatorname{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 \operatorname{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 \operatorname{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]  time = 0.0468918, size = 67, normalized size = 0.99 \[ \frac{4 (c+d x)+3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )\right )-3 \log \left (3 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{20 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.045, size = 53, normalized size = 0.8 \begin{align*}{\frac{2}{5\,d}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{3}{20\,d}\ln \left ( 3\,\tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }+{\frac{3}{20\,d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +3 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.46285, size = 96, normalized size = 1.41 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 0.496222, size = 144, normalized size = 2.12 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{3 \csc{\left (c + d x \right )} + 5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(3*csc(c + d*x) + 5), x)

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Giac [A]  time = 1.37382, size = 61, normalized size = 0.9 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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