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

Optimal. Leaf size=81 \[ -\frac {125 \tan (c+d x)}{4608 d (5 \sec (c+d x)+3)}-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}+\frac {3055 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+3}\right )}{27648 d}-\frac {1007 x}{55296} \]

[Out]

-1007/55296*x+3055/27648*arctan(sin(d*x+c)/(3+cos(d*x+c)))/d-25/96*tan(d*x+c)/d/(3+5*sec(d*x+c))^2-125/4608*ta
n(d*x+c)/d/(3+5*sec(d*x+c))

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Rubi [A]  time = 0.12, antiderivative size = 81, 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, 4060, 3919, 3831, 2657} \[ -\frac {125 \tan (c+d x)}{4608 d (5 \sec (c+d x)+3)}-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}+\frac {3055 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+3}\right )}{27648 d}-\frac {1007 x}{55296} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Sec[c + d*x])^(-3),x]

[Out]

(-1007*x)/55296 + (3055*ArcTan[Sin[c + d*x]/(3 + Cos[c + d*x])])/(27648*d) - (25*Tan[c + d*x])/(96*d*(3 + 5*Se
c[c + d*x])^2) - (125*Tan[c + d*x])/(4608*d*(3 + 5*Sec[c + d*x]))

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]

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]

Rule 4060

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1
)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{(3+5 \sec (c+d x))^3} \, dx &=-\frac {25 \tan (c+d x)}{96 d (3+5 \sec (c+d x))^2}+\frac {1}{96} \int \frac {32+30 \sec (c+d x)-25 \sec ^2(c+d x)}{(3+5 \sec (c+d x))^2} \, dx\\ &=-\frac {25 \tan (c+d x)}{96 d (3+5 \sec (c+d x))^2}-\frac {125 \tan (c+d x)}{4608 d (3+5 \sec (c+d x))}+\frac {\int \frac {512-165 \sec (c+d x)}{3+5 \sec (c+d x)} \, dx}{4608}\\ &=\frac {x}{27}-\frac {25 \tan (c+d x)}{96 d (3+5 \sec (c+d x))^2}-\frac {125 \tan (c+d x)}{4608 d (3+5 \sec (c+d x))}-\frac {3055 \int \frac {\sec (c+d x)}{3+5 \sec (c+d x)} \, dx}{13824}\\ &=\frac {x}{27}-\frac {25 \tan (c+d x)}{96 d (3+5 \sec (c+d x))^2}-\frac {125 \tan (c+d x)}{4608 d (3+5 \sec (c+d x))}-\frac {611 \int \frac {1}{1+\frac {3}{5} \cos (c+d x)} \, dx}{13824}\\ &=-\frac {1007 x}{55296}+\frac {3055 \tan ^{-1}\left (\frac {\sin (c+d x)}{3+\cos (c+d x)}\right )}{27648 d}-\frac {25 \tan (c+d x)}{96 d (3+5 \sec (c+d x))^2}-\frac {125 \tan (c+d x)}{4608 d (3+5 \sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 108, normalized size = 1.33 \[ \frac {-3750 \sin (c+d x)-4725 \sin (2 (c+d x))+30720 (c+d x) \cos (c+d x)+4608 c \cos (2 (c+d x))+4608 d x \cos (2 (c+d x))+3055 (3 \cos (c+d x)+5)^2 \tan ^{-1}\left (2 \cot \left (\frac {1}{2} (c+d x)\right )\right )+30208 c+30208 d x}{27648 d (3 \cos (c+d x)+5)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30208*c + 30208*d*x + 30720*(c + d*x)*Cos[c + d*x] + 3055*ArcTan[2*Cot[(c + d*x)/2]]*(5 + 3*Cos[c + d*x])^2 +
4608*c*Cos[2*(c + d*x)] + 4608*d*x*Cos[2*(c + d*x)] - 3750*Sin[c + d*x] - 4725*Sin[2*(c + d*x)])/(27648*d*(5
+ 3*Cos[c + d*x])^2)

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fricas [A]  time = 0.54, size = 116, normalized size = 1.43 \[ \frac {18432 \, d x \cos \left (d x + c\right )^{2} + 61440 \, d x \cos \left (d x + c\right ) + 51200 \, d x + 3055 \, {\left (9 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 25\right )} \arctan \left (\frac {5 \, \cos \left (d x + c\right ) + 3}{4 \, \sin \left (d x + c\right )}\right ) - 300 \, {\left (63 \, \cos \left (d x + c\right ) + 25\right )} \sin \left (d x + c\right )}{55296 \, {\left (9 \, d \cos \left (d x + c\right )^{2} + 30 \, d \cos \left (d x + c\right ) + 25 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/55296*(18432*d*x*cos(d*x + c)^2 + 61440*d*x*cos(d*x + c) + 51200*d*x + 3055*(9*cos(d*x + c)^2 + 30*cos(d*x +
c) + 25)*arctan(1/4*(5*cos(d*x + c) + 3)/sin(d*x + c)) - 300*(63*cos(d*x + c) + 25)*sin(d*x + c))/(9*d*cos(d*
x + c)^2 + 30*d*cos(d*x + c) + 25*d)

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giac [A]  time = 0.18, size = 75, normalized size = 0.93 \[ -\frac {1007 \, d x + 1007 \, c - \frac {300 \, {\left (19 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4\right )}^{2}} - 6110 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 3}\right )}{55296 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/55296*(1007*d*x + 1007*c - 300*(19*tan(1/2*d*x + 1/2*c)^3 - 44*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^
2 + 4)^2 - 6110*arctan(sin(d*x + c)/(cos(d*x + c) + 3)))/d

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maple [A]  time = 0.43, size = 94, normalized size = 1.16 \[ \frac {475 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4608 d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+4\right )^{2}}-\frac {275 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1152 d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+4\right )^{2}}-\frac {3055 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{27648 d}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{27 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

475/4608/d/(tan(1/2*d*x+1/2*c)^2+4)^2*tan(1/2*d*x+1/2*c)^3-275/1152/d/(tan(1/2*d*x+1/2*c)^2+4)^2*tan(1/2*d*x+1
/2*c)-3055/27648/d*arctan(1/2*tan(1/2*d*x+1/2*c))+2/27/d*arctan(tan(1/2*d*x+1/2*c))

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maxima [A]  time = 0.47, size = 131, normalized size = 1.62 \[ -\frac {\frac {150 \, {\left (\frac {44 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {19 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 16} - 2048 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 3055 \, \arctan \left (\frac {\sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}{27648 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27648*(150*(44*sin(d*x + c)/(cos(d*x + c) + 1) - 19*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(8*sin(d*x + c)^2/
(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 16) - 2048*arctan(sin(d*x + c)/(cos(d*x + c) + 1)
) + 3055*arctan(1/2*sin(d*x + c)/(cos(d*x + c) + 1)))/d

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mupad [B]  time = 0.91, size = 79, normalized size = 0.98 \[ \frac {x}{27}-\frac {3055\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{27648\,d}-\frac {\frac {275\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{1152}-\frac {475\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4608}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+16\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5/cos(c + d*x) + 3)^3,x)

[Out]

x/27 - (3055*atan(tan(c/2 + (d*x)/2)/2))/(27648*d) - ((275*tan(c/2 + (d*x)/2))/1152 - (475*tan(c/2 + (d*x)/2)^
3)/4608)/(d*(8*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 16))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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