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

Optimal. Leaf size=78 \[ \frac {3 \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac {\tan ^3(c+d x) (4-3 \sec (c+d x))}{12 a d}+\frac {\tan (c+d x) (8-3 \sec (c+d x))}{8 a d}-\frac {x}{a} \]

[Out]

-x/a+3/8*arctanh(sin(d*x+c))/a/d+1/8*(8-3*sec(d*x+c))*tan(d*x+c)/a/d-1/12*(4-3*sec(d*x+c))*tan(d*x+c)^3/a/d

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Rubi [A]  time = 0.11, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3888, 3881, 3770} \[ \frac {3 \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac {\tan ^3(c+d x) (4-3 \sec (c+d x))}{12 a d}+\frac {\tan (c+d x) (8-3 \sec (c+d x))}{8 a d}-\frac {x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a) + (3*ArcTanh[Sin[c + d*x]])/(8*a*d) + ((8 - 3*Sec[c + d*x])*Tan[c + d*x])/(8*a*d) - ((4 - 3*Sec[c + d*x
])*Tan[c + d*x]^3)/(12*a*d)

Rule 3770

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

Rule 3881

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(e*(e*Cot[
c + d*x])^(m - 1)*(a*m + b*(m - 1)*Csc[c + d*x]))/(d*m*(m - 1)), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m - 2)
*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

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)} \, dx &=\frac {\int (-a+a \sec (c+d x)) \tan ^4(c+d x) \, dx}{a^2}\\ &=-\frac {(4-3 \sec (c+d x)) \tan ^3(c+d x)}{12 a d}-\frac {\int (-4 a+3 a \sec (c+d x)) \tan ^2(c+d x) \, dx}{4 a^2}\\ &=\frac {(8-3 \sec (c+d x)) \tan (c+d x)}{8 a d}-\frac {(4-3 \sec (c+d x)) \tan ^3(c+d x)}{12 a d}+\frac {\int (-8 a+3 a \sec (c+d x)) \, dx}{8 a^2}\\ &=-\frac {x}{a}+\frac {(8-3 \sec (c+d x)) \tan (c+d x)}{8 a d}-\frac {(4-3 \sec (c+d x)) \tan ^3(c+d x)}{12 a d}+\frac {3 \int \sec (c+d x) \, dx}{8 a}\\ &=-\frac {x}{a}+\frac {3 \tanh ^{-1}(\sin (c+d x))}{8 a d}+\frac {(8-3 \sec (c+d x)) \tan (c+d x)}{8 a d}-\frac {(4-3 \sec (c+d x)) \tan ^3(c+d x)}{12 a d}\\ \end {align*}

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Mathematica [B]  time = 6.45, size = 893, normalized size = 11.45 \[ -\frac {2 x \sec (c+d x) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{\sec (c+d x) a+a}-\frac {3 \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec (c+d x) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{4 d (\sec (c+d x) a+a)}+\frac {3 \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec (c+d x) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{4 d (\sec (c+d x) a+a)}+\frac {8 \sec (c+d x) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d (\sec (c+d x) a+a) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {8 \sec (c+d x) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d (\sec (c+d x) a+a) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\sec (c+d x) \left (11 \sin \left (\frac {c}{2}\right )-19 \cos \left (\frac {c}{2}\right )\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{24 d (\sec (c+d x) a+a) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\sec (c+d x) \left (19 \cos \left (\frac {c}{2}\right )+11 \sin \left (\frac {c}{2}\right )\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{24 d (\sec (c+d x) a+a) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}-\frac {\sec (c+d x) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d (\sec (c+d x) a+a) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}-\frac {\sec (c+d x) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d (\sec (c+d x) a+a) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {\sec (c+d x) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 d (\sec (c+d x) a+a) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^4}-\frac {\sec (c+d x) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 d (\sec (c+d x) a+a) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*x*Cos[c/2 + (d*x)/2]^2*Sec[c + d*x])/(a + a*Sec[c + d*x]) - (3*Cos[c/2 + (d*x)/2]^2*Log[Cos[c/2 + (d*x)/2]
- Sin[c/2 + (d*x)/2]]*Sec[c + d*x])/(4*d*(a + a*Sec[c + d*x])) + (3*Cos[c/2 + (d*x)/2]^2*Log[Cos[c/2 + (d*x)/
2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x])/(4*d*(a + a*Sec[c + d*x])) + (Cos[c/2 + (d*x)/2]^2*Sec[c + d*x])/(8*d*(
a + a*Sec[c + d*x])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^4) - (Cos[c/2 + (d*x)/2]^2*Sec[c + d*x]*Sin[(d*x
)/2])/(3*d*(a + a*Sec[c + d*x])*(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]^2*Sec[c + d*x]*(-19*Cos[c/2] + 11*Sin[c/2]))/(24*d*(a + a*Sec[c + d*x])*(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]^2*Sec[c + d*x]*Sin[(d*x)/2])/(3*d*(a + a*Sec[c
+ d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) - (Cos[c/2 + (d*x)/2]^2*Sec[c + d*x]
)/(8*d*(a + a*Sec[c + d*x])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^4) - (Cos[c/2 + (d*x)/2]^2*Sec[c + d*x]*
Sin[(d*x)/2])/(3*d*(a + a*Sec[c + d*x])*(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]^2*Sec[c + d*x]*(19*Cos[c/2] + 11*Sin[c/2]))/(24*d*(a + a*Sec[c + d*x])*(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]^2*Sec[c + d*x]*Sin[(d*x)/2])/(3*d*(a +
a*Sec[c + d*x])*(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 = 0.81, size = 107, normalized size = 1.37 \[ -\frac {48 \, d x \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + 9 \, \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (32 \, \cos \left (d x + c\right )^{3} - 15 \, \cos \left (d x + c\right )^{2} - 8 \, \cos \left (d x + c\right ) + 6\right )} \sin \left (d x + c\right )}{48 \, a d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(48*d*x*cos(d*x + c)^4 - 9*cos(d*x + c)^4*log(sin(d*x + c) + 1) + 9*cos(d*x + c)^4*log(-sin(d*x + c) + 1
) - 2*(32*cos(d*x + c)^3 - 15*cos(d*x + c)^2 - 8*cos(d*x + c) + 6)*sin(d*x + c))/(a*d*cos(d*x + c)^4)

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giac [A]  time = 4.10, size = 123, normalized size = 1.58 \[ -\frac {\frac {24 \, {\left (d x + c\right )}}{a} - \frac {9 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} + \frac {9 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 137 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 71 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4} a}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(24*(d*x + c)/a - 9*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a + 9*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a + 2*(3
3*tan(1/2*d*x + 1/2*c)^7 - 137*tan(1/2*d*x + 1/2*c)^5 + 71*tan(1/2*d*x + 1/2*c)^3 - 15*tan(1/2*d*x + 1/2*c))/(
(tan(1/2*d*x + 1/2*c)^2 - 1)^4*a))/d

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maple [B]  time = 0.53, size = 228, normalized size = 2.92 \[ \frac {1}{4 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {5}{6 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {3}{8 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {11}{8 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 a d}-\frac {1}{4 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5}{6 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3}{8 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {11}{8 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 a d}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a/d/(tan(1/2*d*x+1/2*c)-1)^4+5/6/a/d/(tan(1/2*d*x+1/2*c)-1)^3+3/8/a/d/(tan(1/2*d*x+1/2*c)-1)^2-11/8/a/d/(t
an(1/2*d*x+1/2*c)-1)-3/8/a/d*ln(tan(1/2*d*x+1/2*c)-1)-1/4/a/d/(tan(1/2*d*x+1/2*c)+1)^4+5/6/a/d/(tan(1/2*d*x+1/
2*c)+1)^3-3/8/a/d/(tan(1/2*d*x+1/2*c)+1)^2-11/8/a/d/(tan(1/2*d*x+1/2*c)+1)+3/8/a/d*ln(tan(1/2*d*x+1/2*c)+1)-2/
a/d*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.54, size = 247, normalized size = 3.17 \[ \frac {\frac {2 \, {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {71 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {137 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {33 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a - \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {48 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(2*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 71*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 137*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5 - 33*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a - 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a*
sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8) - 48*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - 9*log
(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a)/d

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mupad [B]  time = 2.00, size = 139, normalized size = 1.78 \[ \frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a\,d}-\frac {x}{a}+\frac {-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {137\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}-\frac {71\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*atanh(tan(c/2 + (d*x)/2)))/(4*a*d) - x/a + ((5*tan(c/2 + (d*x)/2))/4 - (71*tan(c/2 + (d*x)/2)^3)/12 + (137*
tan(c/2 + (d*x)/2)^5)/12 - (11*tan(c/2 + (d*x)/2)^7)/4)/(d*(a - 4*a*tan(c/2 + (d*x)/2)^2 + 6*a*tan(c/2 + (d*x)
/2)^4 - 4*a*tan(c/2 + (d*x)/2)^6 + a*tan(c/2 + (d*x)/2)^8))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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