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

Optimal. Leaf size=56 \[ -\frac{25 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}+\frac{35 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+3}\right )}{288 d}+\frac{29 x}{576} \]

[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
]))

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Rubi [A]  time = 0.0796837, antiderivative size = 56, normalized size of antiderivative = 1., 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 = {3785, 3919, 3831, 2657} \[ -\frac{25 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}+\frac{35 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+3}\right )}{288 d}+\frac{29 x}{576} \]

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 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 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]

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 2657

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

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*}

Mathematica [A]  time = 0.171145, size = 73, normalized size = 1.3 \[ \frac{160 (c+d x)-150 \sin (c+d x)+96 (c+d x) \cos (c+d x)+35 (3 \cos (c+d x)+5) \tan ^{-1}\left (2 \cot \left (\frac{1}{2} (c+d x)\right )\right )}{288 d (3 \cos (c+d x)+5)} \]

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 = 0.044, size = 63, normalized size = 1.1 \begin{align*}{\frac{2}{9\,d}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{25}{48\,d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+4 \right ) ^{-1}}-{\frac{35}{288\,d}\arctan \left ({\frac{1}{2}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.24934, size = 119, normalized size = 2.12 \begin{align*} -\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{align*}

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 = 1.63778, size = 211, normalized size = 3.77 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (5 \sec{\left (c + d x \right )} + 3\right )^{2}}\, dx \end{align*}

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 = 1.12016, size = 80, normalized size = 1.43 \begin{align*} \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{align*}

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