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

Optimal. Leaf size=198 \[ -\frac{a \left (a^2-2 b^2\right ) \tan (c+d x)}{b^4 d}+\frac{\left (-20 a^2 b^2+8 a^4+15 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 b^5 d}+\frac{\left (4 a^2-7 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 b^3 d}-\frac{a \tan ^3(c+d x)}{3 b^2 d}-\frac{2 (a-b)^{5/2} (a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b^5 d}-\frac{x}{a}+\frac{\tan ^3(c+d x) \sec (c+d x)}{4 b d} \]

[Out]

-(x/a) + ((8*a^4 - 20*a^2*b^2 + 15*b^4)*ArcTanh[Sin[c + d*x]])/(8*b^5*d) - (2*(a - b)^(5/2)*(a + b)^(5/2)*ArcT
anh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*b^5*d) - (a*(a^2 - 2*b^2)*Tan[c + d*x])/(b^4*d) + ((4*a^2
- 7*b^2)*Sec[c + d*x]*Tan[c + d*x])/(8*b^3*d) - (a*Tan[c + d*x]^3)/(3*b^2*d) + (Sec[c + d*x]*Tan[c + d*x]^3)/(
4*b*d)

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Rubi [A]  time = 0.372717, antiderivative size = 271, normalized size of antiderivative = 1.37, number of steps used = 15, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3898, 2897, 2659, 208, 3770, 3767, 8, 3768} \[ -\frac{a \left (a^2-3 b^2\right ) \tan (c+d x)}{b^4 d}+\frac{\left (a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{\left (-3 a^2 b^2+a^4+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{b^5 d}+\frac{\left (a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b^3 d}-\frac{a \tan ^3(c+d x)}{3 b^2 d}-\frac{a \tan (c+d x)}{b^2 d}-\frac{2 (a-b)^{5/2} (a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b^5 d}-\frac{x}{a}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{8 b d}+\frac{\tan (c+d x) \sec ^3(c+d x)}{4 b d}+\frac{3 \tan (c+d x) \sec (c+d x)}{8 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a) + (3*ArcTanh[Sin[c + d*x]])/(8*b*d) + ((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/(2*b^3*d) + ((a^4 - 3*a^2*b
^2 + 3*b^4)*ArcTanh[Sin[c + d*x]])/(b^5*d) - (2*(a - b)^(5/2)*(a + b)^(5/2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)
/2])/Sqrt[a + b]])/(a*b^5*d) - (a*Tan[c + d*x])/(b^2*d) - (a*(a^2 - 3*b^2)*Tan[c + d*x])/(b^4*d) + (3*Sec[c +
d*x]*Tan[c + d*x])/(8*b*d) + ((a^2 - 3*b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*b^3*d) + (Sec[c + d*x]^3*Tan[c + d*x
])/(4*b*d) - (a*Tan[c + d*x]^3)/(3*b^2*d)

Rule 3898

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(Cos[c + d*x]^
m*(b + a*Sin[c + d*x])^n)/Sin[c + d*x]^(m + n), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rule 3768

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

Rubi steps

\begin{align*} \int \frac{\tan ^6(c+d x)}{a+b \sec (c+d x)} \, dx &=\int \frac{\sin (c+d x) \tan ^5(c+d x)}{b+a \cos (c+d x)} \, dx\\ &=\int \left (-\frac{1}{a}-\frac{\left (a^2-b^2\right )^3}{a b^5 (b+a \cos (c+d x))}+\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) \sec (c+d x)}{b^5}+\frac{\left (-a^3+3 a b^2\right ) \sec ^2(c+d x)}{b^4}+\frac{\left (a^2-3 b^2\right ) \sec ^3(c+d x)}{b^3}-\frac{a \sec ^4(c+d x)}{b^2}+\frac{\sec ^5(c+d x)}{b}\right ) \, dx\\ &=-\frac{x}{a}-\frac{a \int \sec ^4(c+d x) \, dx}{b^2}+\frac{\int \sec ^5(c+d x) \, dx}{b}-\frac{\left (a \left (a^2-3 b^2\right )\right ) \int \sec ^2(c+d x) \, dx}{b^4}+\frac{\left (a^2-3 b^2\right ) \int \sec ^3(c+d x) \, dx}{b^3}-\frac{\left (a^2-b^2\right )^3 \int \frac{1}{b+a \cos (c+d x)} \, dx}{a b^5}+\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) \int \sec (c+d x) \, dx}{b^5}\\ &=-\frac{x}{a}+\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{b^5 d}+\frac{\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 d}+\frac{\sec ^3(c+d x) \tan (c+d x)}{4 b d}+\frac{3 \int \sec ^3(c+d x) \, dx}{4 b}+\frac{\left (a^2-3 b^2\right ) \int \sec (c+d x) \, dx}{2 b^3}+\frac{a \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{b^2 d}+\frac{\left (a \left (a^2-3 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{b^4 d}-\frac{\left (2 \left (a^2-b^2\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a b^5 d}\\ &=-\frac{x}{a}+\frac{\left (a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{b^5 d}-\frac{2 (a-b)^{5/2} (a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b^5 d}-\frac{a \tan (c+d x)}{b^2 d}-\frac{a \left (a^2-3 b^2\right ) \tan (c+d x)}{b^4 d}+\frac{3 \sec (c+d x) \tan (c+d x)}{8 b d}+\frac{\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 d}+\frac{\sec ^3(c+d x) \tan (c+d x)}{4 b d}-\frac{a \tan ^3(c+d x)}{3 b^2 d}+\frac{3 \int \sec (c+d x) \, dx}{8 b}\\ &=-\frac{x}{a}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{8 b d}+\frac{\left (a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{b^5 d}-\frac{2 (a-b)^{5/2} (a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b^5 d}-\frac{a \tan (c+d x)}{b^2 d}-\frac{a \left (a^2-3 b^2\right ) \tan (c+d x)}{b^4 d}+\frac{3 \sec (c+d x) \tan (c+d x)}{8 b d}+\frac{\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 d}+\frac{\sec ^3(c+d x) \tan (c+d x)}{4 b d}-\frac{a \tan ^3(c+d x)}{3 b^2 d}\\ \end{align*}

Mathematica [B]  time = 6.16505, size = 907, normalized size = 4.58 \[ -\frac{2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right ) (b+a \cos (c+d x)) \sec (c+d x) \left (b^2-a^2\right )^3}{a b^5 \sqrt{a^2-b^2} d (a+b \sec (c+d x))}-\frac{a (b+a \cos (c+d x)) \sec (c+d x) \sin \left (\frac{1}{2} (c+d x)\right )}{6 b^2 d (a+b \sec (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{(b+a \cos (c+d x)) \sec (c+d x) \left (7 a b^2 \sin \left (\frac{1}{2} (c+d x)\right )-3 a^3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 b^4 d (a+b \sec (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{(b+a \cos (c+d x)) \sec (c+d x) \left (7 a b^2 \sin \left (\frac{1}{2} (c+d x)\right )-3 a^3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{3 b^4 d (a+b \sec (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{(c+d x) (b+a \cos (c+d x)) \sec (c+d x)}{a d (a+b \sec (c+d x))}+\frac{\left (-8 a^4+20 b^2 a^2-15 b^4\right ) (b+a \cos (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec (c+d x)}{8 b^5 d (a+b \sec (c+d x))}+\frac{\left (8 a^4-20 b^2 a^2+15 b^4\right ) (b+a \cos (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec (c+d x)}{8 b^5 d (a+b \sec (c+d x))}+\frac{\left (12 a^2-4 b a-27 b^2\right ) (b+a \cos (c+d x)) \sec (c+d x)}{48 b^3 d (a+b \sec (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{\left (-12 a^2+4 b a+27 b^2\right ) (b+a \cos (c+d x)) \sec (c+d x)}{48 b^3 d (a+b \sec (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a (b+a \cos (c+d x)) \sec (c+d x) \sin \left (\frac{1}{2} (c+d x)\right )}{6 b^2 d (a+b \sec (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{(b+a \cos (c+d x)) \sec (c+d x)}{16 b d (a+b \sec (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{(b+a \cos (c+d x)) \sec (c+d x)}{16 b d (a+b \sec (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((c + d*x)*(b + a*Cos[c + d*x])*Sec[c + d*x])/(a*d*(a + b*Sec[c + d*x]))) - (2*(-a^2 + b^2)^3*ArcTanh[((-a +
b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])*Sec[c + d*x])/(a*b^5*Sqrt[a^2 - b^2]*d*(a + b*Sec[
c + d*x])) + ((-8*a^4 + 20*a^2*b^2 - 15*b^4)*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sec
[c + d*x])/(8*b^5*d*(a + b*Sec[c + d*x])) + ((8*a^4 - 20*a^2*b^2 + 15*b^4)*(b + a*Cos[c + d*x])*Log[Cos[(c + d
*x)/2] + Sin[(c + d*x)/2]]*Sec[c + d*x])/(8*b^5*d*(a + b*Sec[c + d*x])) + ((b + a*Cos[c + d*x])*Sec[c + d*x])/
(16*b*d*(a + b*Sec[c + d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + ((12*a^2 - 4*a*b - 27*b^2)*(b + a*Cos[
c + d*x])*Sec[c + d*x])/(48*b^3*d*(a + b*Sec[c + d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (a*(b + a*Co
s[c + d*x])*Sec[c + d*x]*Sin[(c + d*x)/2])/(6*b^2*d*(a + b*Sec[c + d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
^3) - ((b + a*Cos[c + d*x])*Sec[c + d*x])/(16*b*d*(a + b*Sec[c + d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4
) - (a*(b + a*Cos[c + d*x])*Sec[c + d*x]*Sin[(c + d*x)/2])/(6*b^2*d*(a + b*Sec[c + d*x])*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^3) + ((-12*a^2 + 4*a*b + 27*b^2)*(b + a*Cos[c + d*x])*Sec[c + d*x])/(48*b^3*d*(a + b*Sec[c +
d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + ((b + a*Cos[c + d*x])*Sec[c + d*x]*(-3*a^3*Sin[(c + d*x)/2] +
7*a*b^2*Sin[(c + d*x)/2]))/(3*b^4*d*(a + b*Sec[c + d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + ((b + a*Cos
[c + d*x])*Sec[c + d*x]*(-3*a^3*Sin[(c + d*x)/2] + 7*a*b^2*Sin[(c + d*x)/2]))/(3*b^4*d*(a + b*Sec[c + d*x])*(C
os[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [B]  time = 0.082, size = 785, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/b/(tan(1/2*d*x+1/2*c)-1)^3-5/8/d/b/(tan(1/2*d*x+1/2*c)-1)^2-15/8/d/b*ln(tan(1/2*d*x+1/2*c)-1)-7/8/d/b/(t
an(1/2*d*x+1/2*c)-1)-1/4/d/b/(tan(1/2*d*x+1/2*c)+1)^4+1/2/d/b/(tan(1/2*d*x+1/2*c)+1)^3+5/8/d/b/(tan(1/2*d*x+1/
2*c)+1)^2+15/8/d/b*ln(tan(1/2*d*x+1/2*c)+1)-7/8/d/b/(tan(1/2*d*x+1/2*c)+1)+1/4/d/b/(tan(1/2*d*x+1/2*c)-1)^4-2/
d/a*arctan(tan(1/2*d*x+1/2*c))+2/d*b/a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2
))+1/d/b^4/(tan(1/2*d*x+1/2*c)-1)*a^3+1/2/d/b^3/(tan(1/2*d*x+1/2*c)-1)*a^2-2/d/b^2/(tan(1/2*d*x+1/2*c)-1)*a+1/
3/d/b^2/(tan(1/2*d*x+1/2*c)+1)^3*a-1/2/d/b^3/(tan(1/2*d*x+1/2*c)+1)^2*a^2-1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2*a
+1/d/b^5*ln(tan(1/2*d*x+1/2*c)+1)*a^4-5/2/d/b^3*ln(tan(1/2*d*x+1/2*c)+1)*a^2+1/d/b^4/(tan(1/2*d*x+1/2*c)+1)*a^
3+1/2/d/b^3/(tan(1/2*d*x+1/2*c)+1)*a^2-2/d/b^2/(tan(1/2*d*x+1/2*c)+1)*a+1/3/d/b^2/(tan(1/2*d*x+1/2*c)-1)^3*a+1
/2/d/b^3/(tan(1/2*d*x+1/2*c)-1)^2*a^2+1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)^2*a-1/d/b^5*ln(tan(1/2*d*x+1/2*c)-1)*a^
4+5/2/d/b^3*ln(tan(1/2*d*x+1/2*c)-1)*a^2-2/d/b^5*a^5/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+
b)*(a-b))^(1/2))+6/d/b^3*a^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-6/d/b*a
/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.8257, size = 1423, normalized size = 7.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(48*b^5*d*x*cos(d*x + c)^4 - 24*(a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)*cos(d*x + c)^4*log((2*a*b*cos(d
*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a
^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 3*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^4*log(sin(d*x
+ c) + 1) + 3*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*a^2*b^3*cos(d*x + c
) - 6*a*b^4 + 8*(3*a^4*b - 7*a^2*b^3)*cos(d*x + c)^3 - 3*(4*a^3*b^2 - 9*a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(
a*b^5*d*cos(d*x + c)^4), -1/48*(48*b^5*d*x*cos(d*x + c)^4 + 48*(a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)*arctan
(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^4 - 3*(8*a^5 - 20*a^3*b^2 + 1
5*a*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + 3*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^4*log(-sin(d*x
+ c) + 1) + 2*(8*a^2*b^3*cos(d*x + c) - 6*a*b^4 + 8*(3*a^4*b - 7*a^2*b^3)*cos(d*x + c)^3 - 3*(4*a^3*b^2 - 9*a*
b^4)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^5*d*cos(d*x + c)^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{6}{\left (c + d x \right )}}{a + b \sec{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 4.70169, size = 639, normalized size = 3.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(24*(d*x + c)/a - 3*(8*a^4 - 20*a^2*b^2 + 15*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^5 + 3*(8*a^4 - 20
*a^2*b^2 + 15*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^5 + 48*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(pi*floor(1
/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 +
b^2)))/(sqrt(-a^2 + b^2)*a*b^5) - 2*(24*a^3*tan(1/2*d*x + 1/2*c)^7 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 48*a*b
^2*tan(1/2*d*x + 1/2*c)^7 - 21*b^3*tan(1/2*d*x + 1/2*c)^7 - 72*a^3*tan(1/2*d*x + 1/2*c)^5 - 12*a^2*b*tan(1/2*d
*x + 1/2*c)^5 + 176*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 45*b^3*tan(1/2*d*x + 1/2*c)^5 + 72*a^3*tan(1/2*d*x + 1/2*c)
^3 - 12*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 176*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 45*b^3*tan(1/2*d*x + 1/2*c)^3 - 24*a
^3*tan(1/2*d*x + 1/2*c) + 12*a^2*b*tan(1/2*d*x + 1/2*c) + 48*a*b^2*tan(1/2*d*x + 1/2*c) - 21*b^3*tan(1/2*d*x +
1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^4*b^4))/d