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

Optimal. Leaf size=200 \[ \frac{\left (3 a^2-2 b^2\right ) \tan (c+d x)}{b^4 d}-\frac{a \left (4 a^2-5 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^5 d}+\frac{\left (a^2-b^2\right )^2 \sin (c+d x)}{a b^4 d (a \cos (c+d x)+b)}+\frac{2 (a-b)^{3/2} (a+b)^{3/2} \left (4 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 b^5 d}-\frac{x}{a^2}-\frac{a \tan (c+d x) \sec (c+d x)}{b^3 d}+\frac{\tan ^3(c+d x)}{3 b^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.430917, antiderivative size = 283, normalized size of antiderivative = 1.42, number of steps used = 16, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {3898, 2897, 2664, 12, 2659, 208, 3770, 3767, 8, 3768} \[ \frac{3 \left (a^2-b^2\right ) \tan (c+d x)}{b^4 d}-\frac{2 a \left (2 a^2-3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^5 d}+\frac{\left (a^2-b^2\right )^2 \sin (c+d x)}{a b^4 d (a \cos (c+d x)+b)}-\frac{2 (a-b)^{3/2} (a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 b^3 d}+\frac{4 (a-b)^{3/2} (a+b)^{3/2} \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 b^5 d}-\frac{x}{a^2}-\frac{a \tanh ^{-1}(\sin (c+d x))}{b^3 d}-\frac{a \tan (c+d x) \sec (c+d x)}{b^3 d}+\frac{\tan ^3(c+d x)}{3 b^2 d}+\frac{\tan (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^2) - (a*ArcTanh[Sin[c + d*x]])/(b^3*d) - (2*a*(2*a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/(b^5*d) - (2*(a - b
)^(3/2)*(a + b)^(3/2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*b^3*d) + (4*(a - b)^(3/2)*(a +
b)^(3/2)*(2*a^2 + b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*b^5*d) + ((a^2 - b^2)^2*Sin[
c + d*x])/(a*b^4*d*(b + a*Cos[c + d*x])) + Tan[c + d*x]/(b^2*d) + (3*(a^2 - b^2)*Tan[c + d*x])/(b^4*d) - (a*Se
c[c + d*x]*Tan[c + d*x])/(b^3*d) + 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 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

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

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

Mathematica [B]  time = 6.22672, size = 865, normalized size = 4.32 \[ \frac{(b+a \cos (c+d x))^2 \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x)}{6 b^2 d (a+b \sec (c+d x))^2 \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))^2 \left (9 a^2 \sin \left (\frac{1}{2} (c+d x)\right )-7 b^2 \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^2(c+d x)}{3 b^4 d (a+b \sec (c+d x))^2 \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))^2 \left (9 a^2 \sin \left (\frac{1}{2} (c+d x)\right )-7 b^2 \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^2(c+d x)}{3 b^4 d (a+b \sec (c+d x))^2 \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)) \left (\sin (c+d x) a^4-2 b^2 \sin (c+d x) a^2+b^4 \sin (c+d x)\right ) \sec ^2(c+d x)}{a b^4 d (a+b \sec (c+d x))^2}-\frac{(c+d x) (b+a \cos (c+d x))^2 \sec ^2(c+d x)}{a^2 d (a+b \sec (c+d x))^2}-\frac{2 \left (b^2-a^2\right )^2 \left (4 a^2+b^2\right ) \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))^2 \sec ^2(c+d x)}{a^2 b^5 \sqrt{a^2-b^2} d (a+b \sec (c+d x))^2}+\frac{\left (4 a^3-5 a b^2\right ) (b+a \cos (c+d x))^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^2(c+d x)}{b^5 d (a+b \sec (c+d x))^2}+\frac{\left (5 a b^2-4 a^3\right ) (b+a \cos (c+d x))^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^2(c+d x)}{b^5 d (a+b \sec (c+d x))^2}+\frac{(b-6 a) (b+a \cos (c+d x))^2 \sec ^2(c+d x)}{12 b^3 d (a+b \sec (c+d x))^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(6 a-b) (b+a \cos (c+d x))^2 \sec ^2(c+d x)}{12 b^3 d (a+b \sec (c+d x))^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(b+a \cos (c+d x))^2 \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x)}{6 b^2 d (a+b \sec (c+d x))^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((c + d*x)*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^2)/(a^2*d*(a + b*Sec[c + d*x])^2)) - (2*(-a^2 + b^2)^2*(4*a^2
+ b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^2)/(a^2*b^5*S
qrt[a^2 - b^2]*d*(a + b*Sec[c + d*x])^2) + ((4*a^3 - 5*a*b^2)*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]]*Sec[c + d*x]^2)/(b^5*d*(a + b*Sec[c + d*x])^2) + ((-4*a^3 + 5*a*b^2)*(b + a*Cos[c + d*x])^2*Lo
g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sec[c + d*x]^2)/(b^5*d*(a + b*Sec[c + d*x])^2) + ((-6*a + b)*(b + a*Cos
[c + d*x])^2*Sec[c + d*x]^2)/(12*b^3*d*(a + b*Sec[c + d*x])^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + ((b +
a*Cos[c + d*x])^2*Sec[c + d*x]^2*Sin[(c + d*x)/2])/(6*b^2*d*(a + b*Sec[c + d*x])^2*(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])^3) + ((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*Sin[(c + d*x)/2])/(6*b^2*d*(a + b*Sec[c + d*x])^2*(Cos
[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + ((6*a - b)*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^2)/(12*b^3*d*(a + b*Sec[
c + d*x])^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + ((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*(9*a^2*Sin[(c +
d*x)/2] - 7*b^2*Sin[(c + d*x)/2]))/(3*b^4*d*(a + b*Sec[c + d*x])^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + ((
b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*(9*a^2*Sin[(c + d*x)/2] - 7*b^2*Sin[(c + d*x)/2]))/(3*b^4*d*(a + b*Sec[c
+ d*x])^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + ((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(a^4*Sin[c + d*x] - 2*
a^2*b^2*Sin[c + d*x] + b^4*Sin[c + d*x]))/(a*b^4*d*(a + b*Sec[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.092, size = 723, normalized size = 3.6 \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))^2,x)

[Out]

-2/d/a^2*arctan(tan(1/2*d*x+1/2*c))-2/d/b^4*a^3*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^
2*b-a-b)+4/d/b^2*a*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)-2/d/a*tan(1/2*d*x+1/
2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)+8/d/b^5*a^4/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2
*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-14/d/b^3*a^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b
))^(1/2))+4/d/b/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+2/d*b/a^2/((a+b)*(a-
b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/3/d/b^2/(tan(1/2*d*x+1/2*c)+1)^3+1/2/d/b^2/(
tan(1/2*d*x+1/2*c)+1)^2+1/d/b^3/(tan(1/2*d*x+1/2*c)+1)^2*a-3/d/b^4/(tan(1/2*d*x+1/2*c)+1)*a^2-1/d/b^3/(tan(1/2
*d*x+1/2*c)+1)*a+2/d/b^2/(tan(1/2*d*x+1/2*c)+1)-4/d*a^3/b^5*ln(tan(1/2*d*x+1/2*c)+1)+5/d*a/b^3*ln(tan(1/2*d*x+
1/2*c)+1)-1/3/d/b^2/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)^2-1/d/b^3/(tan(1/2*d*x+1/2*c)-1)
^2*a-3/d/b^4/(tan(1/2*d*x+1/2*c)-1)*a^2-1/d/b^3/(tan(1/2*d*x+1/2*c)-1)*a+2/d/b^2/(tan(1/2*d*x+1/2*c)-1)+4/d*a^
3/b^5*ln(tan(1/2*d*x+1/2*c)-1)-5/d*a/b^3*ln(tan(1/2*d*x+1/2*c)-1)

________________________________________________________________________________________

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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.39632, size = 1893, normalized size = 9.46 \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))^2,x, algorithm="fricas")

[Out]

[-1/6*(6*a*b^5*d*x*cos(d*x + c)^4 + 6*b^6*d*x*cos(d*x + c)^3 + 3*((4*a^5 - 3*a^3*b^2 - a*b^4)*cos(d*x + c)^4 +
(4*a^4*b - 3*a^2*b^3 - b^5)*cos(d*x + c)^3)*sqrt(a^2 - b^2)*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*((4*a^6 - 5*a^4*b^2)*cos(d*x + c)^4 + (4*a^5*b - 5*a^3*b^3)*cos(d*x + c)^3)*log(sin(d*x + c)
+ 1) - 3*((4*a^6 - 5*a^4*b^2)*cos(d*x + c)^4 + (4*a^5*b - 5*a^3*b^3)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) +
2*(2*a^3*b^3*cos(d*x + c) - a^2*b^4 - (12*a^5*b - 13*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 - (6*a^4*b^2 - 7*a^2*b
^4)*cos(d*x + c)^2)*sin(d*x + c))/(a^3*b^5*d*cos(d*x + c)^4 + a^2*b^6*d*cos(d*x + c)^3), -1/6*(6*a*b^5*d*x*cos
(d*x + c)^4 + 6*b^6*d*x*cos(d*x + c)^3 - 6*((4*a^5 - 3*a^3*b^2 - a*b^4)*cos(d*x + c)^4 + (4*a^4*b - 3*a^2*b^3
- b^5)*cos(d*x + c)^3)*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
))) + 3*((4*a^6 - 5*a^4*b^2)*cos(d*x + c)^4 + (4*a^5*b - 5*a^3*b^3)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*
((4*a^6 - 5*a^4*b^2)*cos(d*x + c)^4 + (4*a^5*b - 5*a^3*b^3)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) + 2*(2*a^3*
b^3*cos(d*x + c) - a^2*b^4 - (12*a^5*b - 13*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 - (6*a^4*b^2 - 7*a^2*b^4)*cos(d*
x + c)^2)*sin(d*x + c))/(a^3*b^5*d*cos(d*x + c)^4 + a^2*b^6*d*cos(d*x + c)^3)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{6}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, 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))**2,x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 5.04026, size = 555, normalized size = 2.78 \begin{align*} -\frac{\frac{3 \,{\left (d x + c\right )}}{a^{2}} + \frac{3 \,{\left (4 \, a^{3} - 5 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{5}} - \frac{3 \,{\left (4 \, a^{3} - 5 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{5}} + \frac{6 \,{\left (a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )} a b^{4}} - \frac{6 \,{\left (4 \, a^{6} - 7 \, a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{\sqrt{-a^{2} + b^{2}} a^{2} b^{5}} + \frac{2 \,{\left (9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 16 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, b^{2} \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 )}^{3} b^{4}}}{3 \, d} \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))^2,x, algorithm="giac")

[Out]

-1/3*(3*(d*x + c)/a^2 + 3*(4*a^3 - 5*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^5 - 3*(4*a^3 - 5*a*b^2)*log(a
bs(tan(1/2*d*x + 1/2*c) - 1))/b^5 + 6*(a^4*tan(1/2*d*x + 1/2*c) - 2*a^2*b^2*tan(1/2*d*x + 1/2*c) + b^4*tan(1/2
*d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)*a*b^4) - 6*(4*a^6 - 7*a^4*b^2 +
2*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^2*b^5) + 2*(9*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*a*b*tan(1
/2*d*x + 1/2*c)^5 - 6*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^2*tan(1/2*d*x + 1/2*c)^3 + 16*b^2*tan(1/2*d*x + 1/2*c)
^3 + 9*a^2*tan(1/2*d*x + 1/2*c) - 3*a*b*tan(1/2*d*x + 1/2*c) - 6*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2
*c)^2 - 1)^3*b^4))/d