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

Optimal. Leaf size=239 \[ -\frac{a^2 \tan ^4(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{a \left (2 a^2+b^2\right ) \tan ^2(c+d x)}{b^3 d \left (a^2+b^2\right )}+\frac{\left (4 a^2+b^2\right ) \tan ^3(c+d x)}{3 b^2 d \left (a^2+b^2\right )}+\frac{\left (2 a^2 b^2+4 a^4-b^4\right ) \tan (c+d x)}{b^4 d \left (a^2+b^2\right )}-\frac{2 a^5 \left (2 a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{b^5 d \left (a^2+b^2\right )^2}-\frac{2 a b \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{x \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2} \]

[Out]

-(((a^2 - b^2)*x)/(a^2 + b^2)^2) - (2*a*b*Log[Cos[c + d*x]])/((a^2 + b^2)^2*d) - (2*a^5*(2*a^2 + 3*b^2)*Log[a
+ b*Tan[c + d*x]])/(b^5*(a^2 + b^2)^2*d) + ((4*a^4 + 2*a^2*b^2 - b^4)*Tan[c + d*x])/(b^4*(a^2 + b^2)*d) - (a*(
2*a^2 + b^2)*Tan[c + d*x]^2)/(b^3*(a^2 + b^2)*d) + ((4*a^2 + b^2)*Tan[c + d*x]^3)/(3*b^2*(a^2 + b^2)*d) - (a^2
*Tan[c + d*x]^4)/(b*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.742308, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3565, 3647, 3626, 3617, 31, 3475} \[ -\frac{a^2 \tan ^4(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{a \left (2 a^2+b^2\right ) \tan ^2(c+d x)}{b^3 d \left (a^2+b^2\right )}+\frac{\left (4 a^2+b^2\right ) \tan ^3(c+d x)}{3 b^2 d \left (a^2+b^2\right )}+\frac{\left (2 a^2 b^2+4 a^4-b^4\right ) \tan (c+d x)}{b^4 d \left (a^2+b^2\right )}-\frac{2 a^5 \left (2 a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{b^5 d \left (a^2+b^2\right )^2}-\frac{2 a b \log (\cos (c+d x))}{d \left (a^2+b^2\right )^2}-\frac{x \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - b^2)*x)/(a^2 + b^2)^2) - (2*a*b*Log[Cos[c + d*x]])/((a^2 + b^2)^2*d) - (2*a^5*(2*a^2 + 3*b^2)*Log[a
+ b*Tan[c + d*x]])/(b^5*(a^2 + b^2)^2*d) + ((4*a^4 + 2*a^2*b^2 - b^4)*Tan[c + d*x])/(b^4*(a^2 + b^2)*d) - (a*(
2*a^2 + b^2)*Tan[c + d*x]^2)/(b^3*(a^2 + b^2)*d) + ((4*a^2 + b^2)*Tan[c + d*x]^3)/(3*b^2*(a^2 + b^2)*d) - (a^2
*Tan[c + d*x]^4)/(b*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3475

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

Rubi steps

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

Mathematica [C]  time = 6.21616, size = 242, normalized size = 1.01 \[ \frac{\tan ^4(c+d x)}{3 b d (a+b \tan (c+d x))}+\frac{-\frac{2 a \tan ^3(c+d x)}{b d (a+b \tan (c+d x))}+\frac{-\frac{6 a^4 \left (2 a^2+b^2\right )}{b^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{6 \left (1-\frac{2 a^2}{b^2}\right ) \tan (c+d x)}{d}-\frac{12 a^5 \left (2 a^2+3 b^2\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^2}+\frac{3 i b^2 \log (-\tan (c+d x)+i)}{d (a+i b)^2}-\frac{3 i b^2 \log (\tan (c+d x)+i)}{d (a-i b)^2}}{2 b}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tan[c + d*x]^4/(3*b*d*(a + b*Tan[c + d*x])) + ((-2*a*Tan[c + d*x]^3)/(b*d*(a + b*Tan[c + d*x])) + (((3*I)*b^2*
Log[I - Tan[c + d*x]])/((a + I*b)^2*d) - ((3*I)*b^2*Log[I + Tan[c + d*x]])/((a - I*b)^2*d) - (12*a^5*(2*a^2 +
3*b^2)*Log[a + b*Tan[c + d*x]])/(b^3*(a^2 + b^2)^2*d) - (6*(1 - (2*a^2)/b^2)*Tan[c + d*x])/d - (6*a^4*(2*a^2 +
b^2))/(b^3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])))/(2*b))/(3*b)

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Maple [A]  time = 0.029, size = 233, normalized size = 1. \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,{b}^{2}d}}-{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{{b}^{3}d}}+3\,{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d{b}^{4}}}-{\frac{\tan \left ( dx+c \right ) }{{b}^{2}d}}+{\frac{ab\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{a}^{6}}{d{b}^{5} \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }}-4\,{\frac{{a}^{7}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d{b}^{5} \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-6\,{\frac{{a}^{5}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{b}^{3}d \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.65445, size = 278, normalized size = 1.16 \begin{align*} -\frac{\frac{3 \, a^{6}}{a^{3} b^{5} + a b^{7} +{\left (a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )} - \frac{3 \, a b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{3 \,{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{6 \,{\left (2 \, a^{7} + 3 \, a^{5} b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} b^{5} + 2 \, a^{2} b^{7} + b^{9}} - \frac{b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right )^{2} + 3 \,{\left (3 \, a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{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*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3*(3*a^6/(a^3*b^5 + a*b^7 + (a^2*b^6 + b^8)*tan(d*x + c)) - 3*a*b*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a^2*b^2
+ b^4) + 3*(a^2 - b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + 6*(2*a^7 + 3*a^5*b^2)*log(b*tan(d*x + c) + a)/(a^4*
b^5 + 2*a^2*b^7 + b^9) - (b^2*tan(d*x + c)^3 - 3*a*b*tan(d*x + c)^2 + 3*(3*a^2 - b^2)*tan(d*x + c))/b^4)/d

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Fricas [A]  time = 2.60257, size = 829, normalized size = 3.47 \begin{align*} -\frac{6 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 3 \, a^{2} b^{6} -{\left (a^{4} b^{4} + 2 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{4} + 2 \,{\left (a^{5} b^{3} + 2 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )^{3} + 3 \,{\left (a^{3} b^{5} - a b^{7}\right )} d x - 3 \,{\left (2 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - b^{8}\right )} \tan \left (d x + c\right )^{2} + 3 \,{\left (2 \, a^{8} + 3 \, a^{6} b^{2} +{\left (2 \, a^{7} b + 3 \, a^{5} b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 3 \,{\left (2 \, a^{8} + 3 \, a^{6} b^{2} - a^{2} b^{6} +{\left (2 \, a^{7} b + 3 \, a^{5} b^{3} - a b^{7}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 3 \,{\left (4 \, a^{7} b + 4 \, a^{5} b^{3} - a^{3} b^{5} - 2 \, a b^{7} -{\left (a^{2} b^{6} - b^{8}\right )} d x\right )} \tan \left (d x + c\right )}{3 \,{\left ({\left (a^{4} b^{6} + 2 \, a^{2} b^{8} + b^{10}\right )} d \tan \left (d x + c\right ) +{\left (a^{5} b^{5} + 2 \, a^{3} b^{7} + a b^{9}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*a^6*b^2 + 6*a^4*b^4 + 3*a^2*b^6 - (a^4*b^4 + 2*a^2*b^6 + b^8)*tan(d*x + c)^4 + 2*(a^5*b^3 + 2*a^3*b^5
+ a*b^7)*tan(d*x + c)^3 + 3*(a^3*b^5 - a*b^7)*d*x - 3*(2*a^6*b^2 + 3*a^4*b^4 - b^8)*tan(d*x + c)^2 + 3*(2*a^8
+ 3*a^6*b^2 + (2*a^7*b + 3*a^5*b^3)*tan(d*x + c))*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x
+ c)^2 + 1)) - 3*(2*a^8 + 3*a^6*b^2 - a^2*b^6 + (2*a^7*b + 3*a^5*b^3 - a*b^7)*tan(d*x + c))*log(1/(tan(d*x +
c)^2 + 1)) - 3*(4*a^7*b + 4*a^5*b^3 - a^3*b^5 - 2*a*b^7 - (a^2*b^6 - b^8)*d*x)*tan(d*x + c))/((a^4*b^6 + 2*a^2
*b^8 + b^10)*d*tan(d*x + c) + (a^5*b^5 + 2*a^3*b^7 + a*b^9)*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 5.01189, size = 339, normalized size = 1.42 \begin{align*} \frac{\frac{3 \, a b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{3 \,{\left (a^{2} - b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{6 \,{\left (2 \, a^{7} + 3 \, a^{5} b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b^{5} + 2 \, a^{2} b^{7} + b^{9}} + \frac{3 \,{\left (4 \, a^{7} b \tan \left (d x + c\right ) + 6 \, a^{5} b^{3} \tan \left (d x + c\right ) + 3 \, a^{8} + 5 \, a^{6} b^{2}\right )}}{{\left (a^{4} b^{5} + 2 \, a^{2} b^{7} + b^{9}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}} + \frac{b^{4} \tan \left (d x + c\right )^{3} - 3 \, a b^{3} \tan \left (d x + c\right )^{2} + 9 \, a^{2} b^{2} \tan \left (d x + c\right ) - 3 \, b^{4} \tan \left (d x + c\right )}{b^{6}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a*b*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) - 3*(a^2 - b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) -
6*(2*a^7 + 3*a^5*b^2)*log(abs(b*tan(d*x + c) + a))/(a^4*b^5 + 2*a^2*b^7 + b^9) + 3*(4*a^7*b*tan(d*x + c) + 6*
a^5*b^3*tan(d*x + c) + 3*a^8 + 5*a^6*b^2)/((a^4*b^5 + 2*a^2*b^7 + b^9)*(b*tan(d*x + c) + a)) + (b^4*tan(d*x +
c)^3 - 3*a*b^3*tan(d*x + c)^2 + 9*a^2*b^2*tan(d*x + c) - 3*b^4*tan(d*x + c))/b^6)/d