3.80 \(\int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=72 \[ \frac {\tan ^3(c+d x)}{3 a^2 d}+\frac {\tan (c+d x)}{a^2 d}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {\tan (c+d x) \sec (c+d x)}{a^2 d}-\frac {x}{a^2} \]

[Out]

-x/a^2+arctanh(sin(d*x+c))/a^2/d+tan(d*x+c)/a^2/d-sec(d*x+c)*tan(d*x+c)/a^2/d+1/3*tan(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.15, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3888, 3886, 3473, 8, 2611, 3770, 2607, 30} \[ \frac {\tan ^3(c+d x)}{3 a^2 d}+\frac {\tan (c+d x)}{a^2 d}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {\tan (c+d x) \sec (c+d x)}{a^2 d}-\frac {x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^2,x]

[Out]

-(x/a^2) + ArcTanh[Sin[c + d*x]]/(a^2*d) + Tan[c + d*x]/(a^2*d) - (Sec[c + d*x]*Tan[c + d*x])/(a^2*d) + Tan[c
+ d*x]^3/(3*a^2*d)

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\frac {\int (-a+a \sec (c+d x))^2 \tan ^2(c+d x) \, dx}{a^4}\\ &=\frac {\int \left (a^2 \tan ^2(c+d x)-2 a^2 \sec (c+d x) \tan ^2(c+d x)+a^2 \sec ^2(c+d x) \tan ^2(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \tan ^2(c+d x) \, dx}{a^2}+\frac {\int \sec ^2(c+d x) \tan ^2(c+d x) \, dx}{a^2}-\frac {2 \int \sec (c+d x) \tan ^2(c+d x) \, dx}{a^2}\\ &=\frac {\tan (c+d x)}{a^2 d}-\frac {\sec (c+d x) \tan (c+d x)}{a^2 d}-\frac {\int 1 \, dx}{a^2}+\frac {\int \sec (c+d x) \, dx}{a^2}+\frac {\operatorname {Subst}\left (\int x^2 \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=-\frac {x}{a^2}+\frac {\tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {\tan (c+d x)}{a^2 d}-\frac {\sec (c+d x) \tan (c+d x)}{a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 6.31, size = 767, normalized size = 10.65 \[ -\frac {4 x \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x)}{(a \sec (c+d x)+a)^2}+\frac {8 \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x)}{3 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) (a \sec (c+d x)+a)^2 \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {8 \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x)}{3 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) (a \sec (c+d x)+a)^2 \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\left (7 \sin \left (\frac {c}{2}\right )-5 \cos \left (\frac {c}{2}\right )\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x)}{3 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) (a \sec (c+d x)+a)^2 \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\left (7 \sin \left (\frac {c}{2}\right )+5 \cos \left (\frac {c}{2}\right )\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x)}{3 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) (a \sec (c+d x)+a)^2 \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {2 \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x)}{3 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) (a \sec (c+d x)+a)^2 \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {2 \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x)}{3 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) (a \sec (c+d x)+a)^2 \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}-\frac {4 \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a \sec (c+d x)+a)^2}+\frac {4 \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^2,x]

[Out]

(-4*x*Cos[c/2 + (d*x)/2]^4*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^2 - (4*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x
)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^2)/(d*(a + a*Sec[c + d*x])^2) + (4*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 +
(d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^2)/(d*(a + a*Sec[c + d*x])^2) + (2*Cos[c/2 + (d*x)/2]^4*Sec[c + d*
x]^2*Sin[(d*x)/2])/(3*d*(a + a*Sec[c + d*x])^2*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])
^3) + (Cos[c/2 + (d*x)/2]^4*Sec[c + d*x]^2*(-5*Cos[c/2] + 7*Sin[c/2]))/(3*d*(a + a*Sec[c + d*x])^2*(Cos[c/2] -
Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (8*Cos[c/2 + (d*x)/2]^4*Sec[c + d*x]^2*Sin[(d*x)/2])
/(3*d*(a + a*Sec[c + d*x])^2*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (2*Cos[c/2 + (
d*x)/2]^4*Sec[c + d*x]^2*Sin[(d*x)/2])/(3*d*(a + a*Sec[c + d*x])^2*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] +
Sin[c/2 + (d*x)/2])^3) + (Cos[c/2 + (d*x)/2]^4*Sec[c + d*x]^2*(5*Cos[c/2] + 7*Sin[c/2]))/(3*d*(a + a*Sec[c +
d*x])^2*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (8*Cos[c/2 + (d*x)/2]^4*Sec[c + d
*x]^2*Sin[(d*x)/2])/(3*d*(a + a*Sec[c + d*x])^2*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]
))

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fricas [A]  time = 0.76, size = 97, normalized size = 1.35 \[ -\frac {6 \, d x \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{6 \, a^{2} d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(6*d*x*cos(d*x + c)^3 - 3*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3*cos(d*x + c)^3*log(-sin(d*x + c) + 1)
- 2*(2*cos(d*x + c)^2 - 3*cos(d*x + c) + 1)*sin(d*x + c))/(a^2*d*cos(d*x + c)^3)

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giac [A]  time = 4.13, size = 99, normalized size = 1.38 \[ -\frac {\frac {3 \, {\left (d x + c\right )}}{a^{2}} - \frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {4 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/3*(3*(d*x + c)/a^2 - 3*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 + 3*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 +
4*(3*tan(1/2*d*x + 1/2*c)^5 - tan(1/2*d*x + 1/2*c)^3)/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^2))/d

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maple [B]  time = 0.60, size = 185, normalized size = 2.57 \[ -\frac {1}{3 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}-\frac {1}{3 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+a*sec(d*x+c))^2,x)

[Out]

-1/3/a^2/d/(tan(1/2*d*x+1/2*c)-1)^3-3/2/a^2/d/(tan(1/2*d*x+1/2*c)-1)^2-2/a^2/d/(tan(1/2*d*x+1/2*c)-1)-1/a^2/d*
ln(tan(1/2*d*x+1/2*c)-1)-1/3/a^2/d/(tan(1/2*d*x+1/2*c)+1)^3+3/2/a^2/d/(tan(1/2*d*x+1/2*c)+1)^2-2/a^2/d/(tan(1/
2*d*x+1/2*c)+1)+1/a^2/d*ln(tan(1/2*d*x+1/2*c)+1)-2/a^2/d*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.82, size = 196, normalized size = 2.72 \[ -\frac {\frac {4 \, {\left (\frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {6 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3*(4*(sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6
) + 6*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 3*log(sin
(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2)/d

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mupad [B]  time = 1.39, size = 111, normalized size = 1.54 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {x}{a^2}+\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6/(a + a/cos(c + d*x))^2,x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2)))/(a^2*d) - x/a^2 + ((4*tan(c/2 + (d*x)/2)^3)/3 - 4*tan(c/2 + (d*x)/2)^5)/(d*(3*a^
2*tan(c/2 + (d*x)/2)^2 - 3*a^2*tan(c/2 + (d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 - a^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tan ^{6}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**6/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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