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

Optimal. Leaf size=70 \[ \frac {3 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}-\frac {3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}+\frac {x}{5} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3783, 2659, 206} \[ \frac {3 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}-\frac {3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}+\frac {x}{5} \]

Antiderivative was successfully verified.

[In]

Int[(5 + 3*Sec[c + d*x])^(-1),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2659

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

Rubi steps

\begin {align*} \int \frac {1}{5+3 \sec (c+d x)} \, dx &=\frac {x}{5}-\frac {1}{5} \int \frac {1}{1+\frac {5}{3} \cos (c+d x)} \, dx\\ &=\frac {x}{5}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{\frac {8}{3}-\frac {2 x^2}{3}} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{5 d}\\ &=\frac {x}{5}+\frac {3 \log \left (2 \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 (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{20 d}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 69, normalized size = 0.99 \[ \frac {4 (c+d x)+3 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{20 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 52, normalized size = 0.74 \[ \frac {8 \, d x - 3 \, \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 3 \, \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right )}{40 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(8*d*x - 3*log(3/2*cos(d*x + c) + 2*sin(d*x + c) + 5/2) + 3*log(3/2*cos(d*x + c) - 2*sin(d*x + c) + 5/2))
/d

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giac [A]  time = 0.21, size = 43, normalized size = 0.61 \[ \frac {4 \, d x + 4 \, c - 3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) + 3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.42, size = 51, normalized size = 0.73 \[ \frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{20 d}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{20 d}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 70, normalized size = 1.00 \[ \frac {8 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) - 3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) + 3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(8*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 2) + 3*log(sin(d*x +
c)/(cos(d*x + c) + 1) - 2))/d

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mupad [B]  time = 0.88, size = 21, normalized size = 0.30 \[ \frac {x}{5}-\frac {3\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{10\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3/cos(c + d*x) + 5),x)

[Out]

x/5 - (3*atanh(tan(c/2 + (d*x)/2)/2))/(10*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{3 \sec {\left (c + d x \right )} + 5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(3*sec(c + d*x) + 5), x)

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