3.184 \(\int \frac {1}{(a+a \tan ^2(c+d x))^3} \, dx\)

Optimal. Leaf size=79 \[ \frac {\sin (c+d x) \cos ^5(c+d x)}{6 a^3 d}+\frac {5 \sin (c+d x) \cos ^3(c+d x)}{24 a^3 d}+\frac {5 \sin (c+d x) \cos (c+d x)}{16 a^3 d}+\frac {5 x}{16 a^3} \]

[Out]

5/16*x/a^3+5/16*cos(d*x+c)*sin(d*x+c)/a^3/d+5/24*cos(d*x+c)^3*sin(d*x+c)/a^3/d+1/6*cos(d*x+c)^5*sin(d*x+c)/a^3
/d

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Rubi [A]  time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3657, 12, 2635, 8} \[ \frac {\sin (c+d x) \cos ^5(c+d x)}{6 a^3 d}+\frac {5 \sin (c+d x) \cos ^3(c+d x)}{24 a^3 d}+\frac {5 \sin (c+d x) \cos (c+d x)}{16 a^3 d}+\frac {5 x}{16 a^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tan[c + d*x]^2)^(-3),x]

[Out]

(5*x)/(16*a^3) + (5*Cos[c + d*x]*Sin[c + d*x])/(16*a^3*d) + (5*Cos[c + d*x]^3*Sin[c + d*x])/(24*a^3*d) + (Cos[
c + d*x]^5*Sin[c + d*x])/(6*a^3*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^3} \, dx &=\int \frac {\cos ^6(c+d x)}{a^3} \, dx\\ &=\frac {\int \cos ^6(c+d x) \, dx}{a^3}\\ &=\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^3 d}+\frac {5 \int \cos ^4(c+d x) \, dx}{6 a^3}\\ &=\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a^3 d}+\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^3 d}+\frac {5 \int \cos ^2(c+d x) \, dx}{8 a^3}\\ &=\frac {5 \cos (c+d x) \sin (c+d x)}{16 a^3 d}+\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a^3 d}+\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^3 d}+\frac {5 \int 1 \, dx}{16 a^3}\\ &=\frac {5 x}{16 a^3}+\frac {5 \cos (c+d x) \sin (c+d x)}{16 a^3 d}+\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a^3 d}+\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 46, normalized size = 0.58 \[ \frac {45 \sin (2 (c+d x))+9 \sin (4 (c+d x))+\sin (6 (c+d x))+60 c+60 d x}{192 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tan[c + d*x]^2)^(-3),x]

[Out]

(60*c + 60*d*x + 45*Sin[2*(c + d*x)] + 9*Sin[4*(c + d*x)] + Sin[6*(c + d*x)])/(192*a^3*d)

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fricas [A]  time = 0.39, size = 120, normalized size = 1.52 \[ \frac {15 \, d x \tan \left (d x + c\right )^{6} + 45 \, d x \tan \left (d x + c\right )^{4} + 15 \, \tan \left (d x + c\right )^{5} + 45 \, d x \tan \left (d x + c\right )^{2} + 40 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 33 \, \tan \left (d x + c\right )}{48 \, {\left (a^{3} d \tan \left (d x + c\right )^{6} + 3 \, a^{3} d \tan \left (d x + c\right )^{4} + 3 \, a^{3} d \tan \left (d x + c\right )^{2} + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*d*x*tan(d*x + c)^6 + 45*d*x*tan(d*x + c)^4 + 15*tan(d*x + c)^5 + 45*d*x*tan(d*x + c)^2 + 40*tan(d*x +
c)^3 + 15*d*x + 33*tan(d*x + c))/(a^3*d*tan(d*x + c)^6 + 3*a^3*d*tan(d*x + c)^4 + 3*a^3*d*tan(d*x + c)^2 + a^
3*d)

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giac [A]  time = 1.63, size = 61, normalized size = 0.77 \[ \frac {\frac {15 \, {\left (d x + c\right )}}{a^{3}} + \frac {15 \, \tan \left (d x + c\right )^{5} + 40 \, \tan \left (d x + c\right )^{3} + 33 \, \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{3} a^{3}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*(d*x + c)/a^3 + (15*tan(d*x + c)^5 + 40*tan(d*x + c)^3 + 33*tan(d*x + c))/((tan(d*x + c)^2 + 1)^3*a^3
))/d

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maple [A]  time = 0.34, size = 95, normalized size = 1.20 \[ \frac {\tan \left (d x +c \right )}{6 d \,a^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}+\frac {5 \tan \left (d x +c \right )}{24 d \,a^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {5 \tan \left (d x +c \right )}{16 d \,a^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*tan(d*x+c)^2)^3,x)

[Out]

1/6/d/a^3*tan(d*x+c)/(1+tan(d*x+c)^2)^3+5/24/d/a^3*tan(d*x+c)/(1+tan(d*x+c)^2)^2+5/16/d/a^3*tan(d*x+c)/(1+tan(
d*x+c)^2)+5/16/d/a^3*arctan(tan(d*x+c))

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maxima [A]  time = 0.94, size = 90, normalized size = 1.14 \[ \frac {\frac {15 \, \tan \left (d x + c\right )^{5} + 40 \, \tan \left (d x + c\right )^{3} + 33 \, \tan \left (d x + c\right )}{a^{3} \tan \left (d x + c\right )^{6} + 3 \, a^{3} \tan \left (d x + c\right )^{4} + 3 \, a^{3} \tan \left (d x + c\right )^{2} + a^{3}} + \frac {15 \, {\left (d x + c\right )}}{a^{3}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*((15*tan(d*x + c)^5 + 40*tan(d*x + c)^3 + 33*tan(d*x + c))/(a^3*tan(d*x + c)^6 + 3*a^3*tan(d*x + c)^4 + 3
*a^3*tan(d*x + c)^2 + a^3) + 15*(d*x + c)/a^3)/d

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mupad [B]  time = 11.93, size = 51, normalized size = 0.65 \[ \frac {5\,x}{16\,a^3}+\frac {{\cos \left (c+d\,x\right )}^6\,\left (\frac {5\,{\mathrm {tan}\left (c+d\,x\right )}^5}{16}+\frac {5\,{\mathrm {tan}\left (c+d\,x\right )}^3}{6}+\frac {11\,\mathrm {tan}\left (c+d\,x\right )}{16}\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tan(c + d*x)^2)^3,x)

[Out]

(5*x)/(16*a^3) + (cos(c + d*x)^6*((11*tan(c + d*x))/16 + (5*tan(c + d*x)^3)/6 + (5*tan(c + d*x)^5)/16))/(a^3*d
)

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sympy [A]  time = 1.40, size = 454, normalized size = 5.75 \[ \begin {cases} \frac {15 d x \tan ^{6}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {45 d x \tan ^{4}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {45 d x \tan ^{2}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {15 d x}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {15 \tan ^{5}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {40 \tan ^{3}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {33 \tan {\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x}{\left (a \tan ^{2}{\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tan(d*x+c)**2)**3,x)

[Out]

Piecewise((15*d*x*tan(c + d*x)**6/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*tan(c + d*x)**4 + 144*a**3*d*tan(c +
d*x)**2 + 48*a**3*d) + 45*d*x*tan(c + d*x)**4/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*tan(c + d*x)**4 + 144*a
**3*d*tan(c + d*x)**2 + 48*a**3*d) + 45*d*x*tan(c + d*x)**2/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*tan(c + d*
x)**4 + 144*a**3*d*tan(c + d*x)**2 + 48*a**3*d) + 15*d*x/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*tan(c + d*x)*
*4 + 144*a**3*d*tan(c + d*x)**2 + 48*a**3*d) + 15*tan(c + d*x)**5/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*tan(
c + d*x)**4 + 144*a**3*d*tan(c + d*x)**2 + 48*a**3*d) + 40*tan(c + d*x)**3/(48*a**3*d*tan(c + d*x)**6 + 144*a*
*3*d*tan(c + d*x)**4 + 144*a**3*d*tan(c + d*x)**2 + 48*a**3*d) + 33*tan(c + d*x)/(48*a**3*d*tan(c + d*x)**6 +
144*a**3*d*tan(c + d*x)**4 + 144*a**3*d*tan(c + d*x)**2 + 48*a**3*d), Ne(d, 0)), (x/(a*tan(c)**2 + a)**3, True
))

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