3.528 \(\int \frac {1}{(5+3 \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=95 \[ \frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}+\frac {123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{1600 d}-\frac {123 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{1600 d}+\frac {x}{25} \]

[Out]

1/25*x+123/1600*ln(2*cos(1/2*d*x+1/2*c)-sin(1/2*d*x+1/2*c))/d-123/1600*ln(2*cos(1/2*d*x+1/2*c)+sin(1/2*d*x+1/2
*c))/d+9/80*tan(d*x+c)/d/(5+3*sec(d*x+c))

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Rubi [A]  time = 0.09, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3785, 3919, 3831, 2659, 206} \[ \frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}+\frac {123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{1600 d}-\frac {123 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{1600 d}+\frac {x}{25} \]

Antiderivative was successfully verified.

[In]

Int[(5 + 3*Sec[c + d*x])^(-2),x]

[Out]

x/25 + (123*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(1600*d) - (123*Log[2*Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]])/(1600*d) + (9*Tan[c + d*x])/(80*d*(5 + 3*Sec[c + d*x]))

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 3785

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rubi steps

\begin {align*} \int \frac {1}{(5+3 \sec (c+d x))^2} \, dx &=\frac {9 \tan (c+d x)}{80 d (5+3 \sec (c+d x))}-\frac {1}{80} \int \frac {-16+15 \sec (c+d x)}{5+3 \sec (c+d x)} \, dx\\ &=\frac {x}{25}+\frac {9 \tan (c+d x)}{80 d (5+3 \sec (c+d x))}-\frac {123}{400} \int \frac {\sec (c+d x)}{5+3 \sec (c+d x)} \, dx\\ &=\frac {x}{25}+\frac {9 \tan (c+d x)}{80 d (5+3 \sec (c+d x))}-\frac {41}{400} \int \frac {1}{1+\frac {5}{3} \cos (c+d x)} \, dx\\ &=\frac {x}{25}+\frac {9 \tan (c+d x)}{80 d (5+3 \sec (c+d x))}-\frac {41 \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 )}{200 d}\\ &=\frac {x}{25}+\frac {123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{1600 d}-\frac {123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{1600 d}+\frac {9 \tan (c+d x)}{80 d (5+3 \sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 162, normalized size = 1.71 \[ \frac {5 \cos (c+d x) \left (64 (c+d x)+123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-123 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 \left (60 \sin (c+d x)+123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-123 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {1}{2} (c+d x)\right )\right )+64 c+64 d x\right )}{1600 d (5 \cos (c+d x)+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + 3*Sec[c + d*x])^(-2),x]

[Out]

(5*Cos[c + d*x]*(64*(c + d*x) + 123*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 123*Log[2*Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]]) + 3*(64*c + 64*d*x + 123*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 123*Log[2*Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]] + 60*Sin[c + d*x]))/(1600*d*(3 + 5*Cos[c + d*x]))

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fricas [A]  time = 0.46, size = 102, normalized size = 1.07 \[ \frac {640 \, d x \cos \left (d x + c\right ) + 384 \, d x - 123 \, {\left (5 \, \cos \left (d x + c\right ) + 3\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 123 \, {\left (5 \, \cos \left (d x + c\right ) + 3\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 360 \, \sin \left (d x + c\right )}{3200 \, {\left (5 \, d \cos \left (d x + c\right ) + 3 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3200*(640*d*x*cos(d*x + c) + 384*d*x - 123*(5*cos(d*x + c) + 3)*log(3/2*cos(d*x + c) + 2*sin(d*x + c) + 5/2)
+ 123*(5*cos(d*x + c) + 3)*log(3/2*cos(d*x + c) - 2*sin(d*x + c) + 5/2) + 360*sin(d*x + c))/(5*d*cos(d*x + c)
+ 3*d)

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giac [A]  time = 0.19, size = 69, normalized size = 0.73 \[ \frac {64 \, d x + 64 \, c - \frac {180 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4} - 123 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) + 123 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{1600 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/1600*(64*d*x + 64*c - 180*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 4) - 123*log(abs(tan(1/2*d*x + 1/2*
c) + 2)) + 123*log(abs(tan(1/2*d*x + 1/2*c) - 2)))/d

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maple [A]  time = 0.48, size = 87, normalized size = 0.92 \[ -\frac {9}{160 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}+\frac {123 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{1600 d}-\frac {9}{160 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}-\frac {123 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{1600 d}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{25 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5+3*sec(d*x+c))^2,x)

[Out]

-9/160/d/(tan(1/2*d*x+1/2*c)-2)+123/1600/d*ln(tan(1/2*d*x+1/2*c)-2)-9/160/d/(tan(1/2*d*x+1/2*c)+2)-123/1600/d*
ln(tan(1/2*d*x+1/2*c)+2)+2/25/d*arctan(tan(1/2*d*x+1/2*c))

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maxima [A]  time = 0.47, size = 111, normalized size = 1.17 \[ -\frac {\frac {180 \, \sin \left (d x + c\right )}{{\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 4\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - 128 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 123 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) - 123 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{1600 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/1600*(180*sin(d*x + c)/((sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4)*(cos(d*x + c) + 1)) - 128*arctan(sin(d*x
+ c)/(cos(d*x + c) + 1)) + 123*log(sin(d*x + c)/(cos(d*x + c) + 1) + 2) - 123*log(sin(d*x + c)/(cos(d*x + c) +
1) - 2))/d

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mupad [B]  time = 0.88, size = 52, normalized size = 0.55 \[ \frac {x}{25}-\frac {\frac {123\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{800}+\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{80\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3/cos(c + d*x) + 5)^2,x)

[Out]

x/25 - ((123*atanh(tan(c/2 + (d*x)/2)/2))/800 + (9*tan(c/2 + (d*x)/2))/(80*(tan(c/2 + (d*x)/2)^2 - 4)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (3 \sec {\left (c + d x \right )} + 5\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5+3*sec(d*x+c))**2,x)

[Out]

Integral((3*sec(c + d*x) + 5)**(-2), x)

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