3.279 \(\int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx\)

Optimal. Leaf size=157 \[ \frac {a^2 \tan ^5(c+d x)}{5 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}-a^2 x-\frac {5 a b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a b \tan ^5(c+d x) \sec (c+d x)}{3 d}-\frac {5 a b \tan ^3(c+d x) \sec (c+d x)}{12 d}+\frac {5 a b \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \]

[Out]

-a^2*x-5/8*a*b*arctanh(sin(d*x+c))/d+a^2*tan(d*x+c)/d+5/8*a*b*sec(d*x+c)*tan(d*x+c)/d-1/3*a^2*tan(d*x+c)^3/d-5
/12*a*b*sec(d*x+c)*tan(d*x+c)^3/d+1/5*a^2*tan(d*x+c)^5/d+1/3*a*b*sec(d*x+c)*tan(d*x+c)^5/d+1/7*b^2*tan(d*x+c)^
7/d

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Rubi [A]  time = 0.20, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3886, 3473, 8, 2611, 3770, 2607, 30} \[ \frac {a^2 \tan ^5(c+d x)}{5 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}-a^2 x-\frac {5 a b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a b \tan ^5(c+d x) \sec (c+d x)}{3 d}-\frac {5 a b \tan ^3(c+d x) \sec (c+d x)}{12 d}+\frac {5 a b \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x) - (5*a*b*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*Tan[c + d*x])/d + (5*a*b*Sec[c + d*x]*Tan[c + d*x])/(8*d
) - (a^2*Tan[c + d*x]^3)/(3*d) - (5*a*b*Sec[c + d*x]*Tan[c + d*x]^3)/(12*d) + (a^2*Tan[c + d*x]^5)/(5*d) + (a*
b*Sec[c + d*x]*Tan[c + d*x]^5)/(3*d) + (b^2*Tan[c + d*x]^7)/(7*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]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^2 \tan ^6(c+d x) \, dx &=\int \left (a^2 \tan ^6(c+d x)+2 a b \sec (c+d x) \tan ^6(c+d x)+b^2 \sec ^2(c+d x) \tan ^6(c+d x)\right ) \, dx\\ &=a^2 \int \tan ^6(c+d x) \, dx+(2 a b) \int \sec (c+d x) \tan ^6(c+d x) \, dx+b^2 \int \sec ^2(c+d x) \tan ^6(c+d x) \, dx\\ &=\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a b \sec (c+d x) \tan ^5(c+d x)}{3 d}-a^2 \int \tan ^4(c+d x) \, dx-\frac {1}{3} (5 a b) \int \sec (c+d x) \tan ^4(c+d x) \, dx+\frac {b^2 \operatorname {Subst}\left (\int x^6 \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {5 a b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d}+a^2 \int \tan ^2(c+d x) \, dx+\frac {1}{4} (5 a b) \int \sec (c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {a^2 \tan (c+d x)}{d}+\frac {5 a b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {5 a b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d}-a^2 \int 1 \, dx-\frac {1}{8} (5 a b) \int \sec (c+d x) \, dx\\ &=-a^2 x-\frac {5 a b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {5 a b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {5 a b \sec (c+d x) \tan ^3(c+d x)}{12 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a b \sec (c+d x) \tan ^5(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 1.35, size = 293, normalized size = 1.87 \[ \frac {-2100 \sec ^6(c+d x) \left (7 a^2 (c+d x)-\left (a^2+b^2\right ) \tan (c+d x)\right )-\left (\sec ^7(c+d x) \left (-3444 a^2 \sin (3 (c+d x))-1988 a^2 \sin (5 (c+d x))-644 a^2 \sin (7 (c+d x))+8820 a^2 (c+d x) \cos (3 (c+d x))+2940 a^2 (c+d x) \cos (5 (c+d x))+420 a^2 c \cos (7 (c+d x))+420 a^2 d x \cos (7 (c+d x))-980 a b \sin (4 (c+d x))-1155 a b \sin (6 (c+d x))+1260 b^2 \sin (3 (c+d x))-420 b^2 \sin (5 (c+d x))+60 b^2 \sin (7 (c+d x))\right )\right )+5950 a b \tan (c+d x) \sec ^5(c+d x)+16800 a b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{26880 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16800*a*b*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c + d*x
]^7*(8820*a^2*(c + d*x)*Cos[3*(c + d*x)] + 2940*a^2*(c + d*x)*Cos[5*(c + d*x)] + 420*a^2*c*Cos[7*(c + d*x)] +
420*a^2*d*x*Cos[7*(c + d*x)] - 3444*a^2*Sin[3*(c + d*x)] + 1260*b^2*Sin[3*(c + d*x)] - 980*a*b*Sin[4*(c + d*x)
] - 1988*a^2*Sin[5*(c + d*x)] - 420*b^2*Sin[5*(c + d*x)] - 1155*a*b*Sin[6*(c + d*x)] - 644*a^2*Sin[7*(c + d*x)
] + 60*b^2*Sin[7*(c + d*x)]) + 5950*a*b*Sec[c + d*x]^5*Tan[c + d*x] - 2100*Sec[c + d*x]^6*(7*a^2*(c + d*x) - (
a^2 + b^2)*Tan[c + d*x]))/(26880*d)

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fricas [A]  time = 0.51, size = 184, normalized size = 1.17 \[ -\frac {1680 \, a^{2} d x \cos \left (d x + c\right )^{7} + 525 \, a b \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 525 \, a b \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (1155 \, a b \cos \left (d x + c\right )^{5} + 8 \, {\left (161 \, a^{2} - 15 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 910 \, a b \cos \left (d x + c\right )^{3} - 8 \, {\left (77 \, a^{2} - 45 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 280 \, a b \cos \left (d x + c\right ) + 24 \, {\left (7 \, a^{2} - 15 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 120 \, b^{2}\right )} \sin \left (d x + c\right )}{1680 \, d \cos \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1680*(1680*a^2*d*x*cos(d*x + c)^7 + 525*a*b*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 525*a*b*cos(d*x + c)^7*l
og(-sin(d*x + c) + 1) - 2*(1155*a*b*cos(d*x + c)^5 + 8*(161*a^2 - 15*b^2)*cos(d*x + c)^6 - 910*a*b*cos(d*x + c
)^3 - 8*(77*a^2 - 45*b^2)*cos(d*x + c)^4 + 280*a*b*cos(d*x + c) + 24*(7*a^2 - 15*b^2)*cos(d*x + c)^2 + 120*b^2
)*sin(d*x + c))/(d*cos(d*x + c)^7)

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giac [A]  time = 9.20, size = 282, normalized size = 1.80 \[ -\frac {840 \, {\left (d x + c\right )} a^{2} + 525 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 525 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (840 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 525 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 6160 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 3500 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 19768 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 9905 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 28896 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7680 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 19768 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9905 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6160 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3500 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 840 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 525 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{7}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(840*(d*x + c)*a^2 + 525*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 525*a*b*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(840*a^2*tan(1/2*d*x + 1/2*c)^13 - 525*a*b*tan(1/2*d*x + 1/2*c)^13 - 6160*a^2*tan(1/2*d*x + 1/2*c)^1
1 + 3500*a*b*tan(1/2*d*x + 1/2*c)^11 + 19768*a^2*tan(1/2*d*x + 1/2*c)^9 - 9905*a*b*tan(1/2*d*x + 1/2*c)^9 - 28
896*a^2*tan(1/2*d*x + 1/2*c)^7 + 7680*b^2*tan(1/2*d*x + 1/2*c)^7 + 19768*a^2*tan(1/2*d*x + 1/2*c)^5 + 9905*a*b
*tan(1/2*d*x + 1/2*c)^5 - 6160*a^2*tan(1/2*d*x + 1/2*c)^3 - 3500*a*b*tan(1/2*d*x + 1/2*c)^3 + 840*a^2*tan(1/2*
d*x + 1/2*c) + 525*a*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d

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maple [A]  time = 0.41, size = 219, normalized size = 1.39 \[ \frac {a^{2} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}-\frac {a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} \tan \left (d x +c \right )}{d}-a^{2} x -\frac {a^{2} c}{d}+\frac {a b \left (\sin ^{7}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{6}}-\frac {a b \left (\sin ^{7}\left (d x +c \right )\right )}{12 d \cos \left (d x +c \right )^{4}}+\frac {a b \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {a b \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}+\frac {5 a b \left (\sin ^{3}\left (d x +c \right )\right )}{24 d}+\frac {5 a b \sin \left (d x +c \right )}{8 d}-\frac {5 a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*tan(d*x+c)^5/d-1/3*a^2*tan(d*x+c)^3/d+a^2*tan(d*x+c)/d-a^2*x-1/d*a^2*c+1/3/d*a*b*sin(d*x+c)^7/cos(d*x+
c)^6-1/12/d*a*b*sin(d*x+c)^7/cos(d*x+c)^4+1/8/d*a*b*sin(d*x+c)^7/cos(d*x+c)^2+1/8*a*b*sin(d*x+c)^5/d+5/24*a*b*
sin(d*x+c)^3/d+5/8*a*b*sin(d*x+c)/d-5/8/d*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/7/d*b^2*sin(d*x+c)^7/cos(d*x+c)^7

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maxima [A]  time = 0.69, size = 150, normalized size = 0.96 \[ \frac {240 \, b^{2} \tan \left (d x + c\right )^{7} + 112 \, {\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a^{2} - 35 \, a b {\left (\frac {2 \, {\left (33 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{1680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1680*(240*b^2*tan(d*x + c)^7 + 112*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*a
^2 - 35*a*b*(2*(33*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 15*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 +
3*sin(d*x + c)^2 - 1) + 15*log(sin(d*x + c) + 1) - 15*log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 2.64, size = 403, normalized size = 2.57 \[ \frac {\left (\frac {5\,a\,b}{4}-2\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (\frac {44\,a^2}{3}-\frac {25\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {283\,a\,b}{12}-\frac {706\,a^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {344\,a^2}{5}-\frac {128\,b^2}{7}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-\frac {706\,a^2}{15}-\frac {283\,b\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {44\,a^2}{3}+\frac {25\,b\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,a^2-\frac {5\,b\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {64\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^6+25\,a^4\,b^2}+\frac {25\,a^4\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^6+25\,a^4\,b^2}\right )}{d}-\frac {5\,a\,b\,\mathrm {atanh}\left (\frac {40\,a^5\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{40\,a^5\,b+\frac {125\,a^3\,b^3}{8}}+\frac {125\,a^3\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,\left (40\,a^5\,b+\frac {125\,a^3\,b^3}{8}\right )}\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^7*((344*a^2)/5 - (128*b^2)/7) + tan(c/2 + (d*x)/2)^13*((5*a*b)/4 - 2*a^2) + tan(c/2 + (d*x
)/2)^3*((25*a*b)/3 + (44*a^2)/3) - tan(c/2 + (d*x)/2)^11*((25*a*b)/3 - (44*a^2)/3) - tan(c/2 + (d*x)/2)^5*((28
3*a*b)/12 + (706*a^2)/15) + tan(c/2 + (d*x)/2)^9*((283*a*b)/12 - (706*a^2)/15) - tan(c/2 + (d*x)/2)*((5*a*b)/4
+ 2*a^2))/(d*(7*tan(c/2 + (d*x)/2)^2 - 21*tan(c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 - 35*tan(c/2 + (d*x)
/2)^8 + 21*tan(c/2 + (d*x)/2)^10 - 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 - 1)) - (2*a^2*atan((64*a^6
*tan(c/2 + (d*x)/2))/(64*a^6 + 25*a^4*b^2) + (25*a^4*b^2*tan(c/2 + (d*x)/2))/(64*a^6 + 25*a^4*b^2)))/d - (5*a*
b*atanh((40*a^5*b*tan(c/2 + (d*x)/2))/(40*a^5*b + (125*a^3*b^3)/8) + (125*a^3*b^3*tan(c/2 + (d*x)/2))/(8*(40*a
^5*b + (125*a^3*b^3)/8))))/(4*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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