3.6.24 \(\int \frac {1}{(3+5 \sec (c+d x))^2} \, dx\) [524]

Optimal. Leaf size=56 \[ \frac {29 x}{576}+\frac {35 \text {ArcTan}\left (\frac {\sin (c+d x)}{3+\cos (c+d x)}\right )}{288 d}-\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))} \]

[Out]

29/576*x+35/288*arctan(sin(d*x+c)/(3+cos(d*x+c)))/d-25/48*tan(d*x+c)/d/(3+5*sec(d*x+c))

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Rubi [A]
time = 0.06, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3870, 4004, 3916, 2736} \begin {gather*} \frac {35 \text {ArcTan}\left (\frac {\sin (c+d x)}{\cos (c+d x)+3}\right )}{288 d}-\frac {25 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}+\frac {29 x}{576} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Sec[c + d*x])^(-2),x]

[Out]

(29*x)/576 + (35*ArcTan[Sin[c + d*x]/(3 + Cos[c + d*x])])/(288*d) - (25*Tan[c + d*x])/(48*d*(3 + 5*Sec[c + d*x
]))

Rule 2736

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(2/(d*q))*ArcTan[b*(Cos[c + d*x]/(a + q + b*Sin[c + d*x]))], x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && PosQ[a]

Rule 3870

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 3916

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

Rule 4004

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}{(3+5 \sec (c+d x))^2} \, dx &=-\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))}+\frac {1}{48} \int \frac {16+15 \sec (c+d x)}{3+5 \sec (c+d x)} \, dx\\ &=\frac {x}{9}-\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))}-\frac {35}{144} \int \frac {\sec (c+d x)}{3+5 \sec (c+d x)} \, dx\\ &=\frac {x}{9}-\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))}-\frac {7}{144} \int \frac {1}{1+\frac {3}{5} \cos (c+d x)} \, dx\\ &=\frac {29 x}{576}+\frac {35 \tan ^{-1}\left (\frac {\sin (c+d x)}{3+\cos (c+d x)}\right )}{288 d}-\frac {25 \tan (c+d x)}{48 d (3+5 \sec (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 73, normalized size = 1.30 \begin {gather*} \frac {160 (c+d x)+96 (c+d x) \cos (c+d x)+35 \text {ArcTan}\left (2 \cot \left (\frac {1}{2} (c+d x)\right )\right ) (5+3 \cos (c+d x))-150 \sin (c+d x)}{288 d (5+3 \cos (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(160*(c + d*x) + 96*(c + d*x)*Cos[c + d*x] + 35*ArcTan[2*Cot[(c + d*x)/2]]*(5 + 3*Cos[c + d*x]) - 150*Sin[c +
d*x])/(288*d*(5 + 3*Cos[c + d*x]))

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Maple [A]
time = 0.07, size = 58, normalized size = 1.04

method	result	size
derivativedivides	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{48 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+4\right )}-\frac {35 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{288}}{d}\)	\(58\)
default	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{48 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+4\right )}-\frac {35 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{288}}{d}\)	\(58\)
risch	\(\frac {x}{9}-\frac {25 i \left (5 \,{\mathrm e}^{i \left (d x +c \right )}+3\right )}{72 d \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+10 \,{\mathrm e}^{i \left (d x +c \right )}+3\right )}+\frac {35 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {1}{3}\right )}{576 d}-\frac {35 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+3\right )}{576 d}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(2/9*arctan(tan(1/2*d*x+1/2*c))-25/48*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2+4)-35/288*arctan(1/2*tan(1/
2*d*x+1/2*c)))

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Maxima [A]
time = 0.46, size = 88, normalized size = 1.57 \begin {gather*} -\frac {\frac {150 \, \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 )}} - 64 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 35 \, \arctan \left (\frac {\sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}{288 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/288*(150*sin(d*x + c)/((sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4)*(cos(d*x + c) + 1)) - 64*arctan(sin(d*x +
c)/(cos(d*x + c) + 1)) + 35*arctan(1/2*sin(d*x + c)/(cos(d*x + c) + 1)))/d

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Fricas [A]
time = 2.58, size = 73, normalized size = 1.30 \begin {gather*} \frac {192 \, d x \cos \left (d x + c\right ) + 320 \, d x + 35 \, {\left (3 \, \cos \left (d x + c\right ) + 5\right )} \arctan \left (\frac {5 \, \cos \left (d x + c\right ) + 3}{4 \, \sin \left (d x + c\right )}\right ) - 300 \, \sin \left (d x + c\right )}{576 \, {\left (3 \, d \cos \left (d x + c\right ) + 5 \, d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/576*(192*d*x*cos(d*x + c) + 320*d*x + 35*(3*cos(d*x + c) + 5)*arctan(1/4*(5*cos(d*x + c) + 3)/sin(d*x + c))
- 300*sin(d*x + c))/(3*d*cos(d*x + c) + 5*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 59, normalized size = 1.05 \begin {gather*} \frac {29 \, d x + 29 \, c - \frac {300 \, \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} + 70 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 3}\right )}{576 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/576*(29*d*x + 29*c - 300*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 4) + 70*arctan(sin(d*x + c)/(cos(d*x
+ c) + 3)))/d

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Mupad [B]
time = 0.86, size = 52, normalized size = 0.93 \begin {gather*} \frac {x}{9}-\frac {\frac {35\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{288}+\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{48\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5/cos(c + d*x) + 3)^2,x)

[Out]

x/9 - ((35*atan(tan(c/2 + (d*x)/2)/2))/288 + (25*tan(c/2 + (d*x)/2))/(48*(tan(c/2 + (d*x)/2)^2 + 4)))/d

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