3.477 \(\int \frac{\tan ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=283 \[ -\frac{2 a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{a^2 \tan ^4(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{\left (11 a^2 b^2+6 a^4+b^4\right ) \tan ^2(c+d x)}{2 b^3 d \left (a^2+b^2\right )^2}-\frac{a \left (11 a^2 b^2+6 a^4+3 b^4\right ) \tan (c+d x)}{b^4 d \left (a^2+b^2\right )^2}+\frac{a^4 \left (17 a^2 b^2+6 a^4+15 b^4\right ) \log (a+b \tan (c+d x))}{b^5 d \left (a^2+b^2\right )^3}-\frac{b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}-\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3} \]

[Out]

-((a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^3) - (b*(3*a^2 - b^2)*Log[Cos[c + d*x]])/((a^2 + b^2)^3*d) + (a^4*(6*a^4 + 1
7*a^2*b^2 + 15*b^4)*Log[a + b*Tan[c + d*x]])/(b^5*(a^2 + b^2)^3*d) - (a*(6*a^4 + 11*a^2*b^2 + 3*b^4)*Tan[c + d
*x])/(b^4*(a^2 + b^2)^2*d) + ((6*a^4 + 11*a^2*b^2 + b^4)*Tan[c + d*x]^2)/(2*b^3*(a^2 + b^2)^2*d) - (a^2*Tan[c
+ d*x]^4)/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (2*a^2*(a^2 + 2*b^2)*Tan[c + d*x]^3)/(b^2*(a^2 + b^2)^2
*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.800137, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3565, 3645, 3647, 3626, 3617, 31, 3475} \[ -\frac{2 a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{a^2 \tan ^4(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{\left (11 a^2 b^2+6 a^4+b^4\right ) \tan ^2(c+d x)}{2 b^3 d \left (a^2+b^2\right )^2}-\frac{a \left (11 a^2 b^2+6 a^4+3 b^4\right ) \tan (c+d x)}{b^4 d \left (a^2+b^2\right )^2}+\frac{a^4 \left (17 a^2 b^2+6 a^4+15 b^4\right ) \log (a+b \tan (c+d x))}{b^5 d \left (a^2+b^2\right )^3}-\frac{b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}-\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + b*Tan[c + d*x])^3,x]

[Out]

-((a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^3) - (b*(3*a^2 - b^2)*Log[Cos[c + d*x]])/((a^2 + b^2)^3*d) + (a^4*(6*a^4 + 1
7*a^2*b^2 + 15*b^4)*Log[a + b*Tan[c + d*x]])/(b^5*(a^2 + b^2)^3*d) - (a*(6*a^4 + 11*a^2*b^2 + 3*b^4)*Tan[c + d
*x])/(b^4*(a^2 + b^2)^2*d) + ((6*a^4 + 11*a^2*b^2 + b^4)*Tan[c + d*x]^2)/(2*b^3*(a^2 + b^2)^2*d) - (a^2*Tan[c
+ d*x]^4)/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (2*a^2*(a^2 + 2*b^2)*Tan[c + d*x]^3)/(b^2*(a^2 + b^2)^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 3645

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

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))^3} \, dx &=-\frac{a^2 \tan ^4(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{\tan ^3(c+d x) \left (4 a^2-2 a b \tan (c+d x)+2 \left (2 a^2+b^2\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac{a^2 \tan ^4(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{2 a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\tan ^2(c+d x) \left (12 a^2 \left (a^2+2 b^2\right )-4 a b^3 \tan (c+d x)+2 \left (6 a^4+11 a^2 b^2+b^4\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=\frac{\left (6 a^4+11 a^2 b^2+b^4\right ) \tan ^2(c+d x)}{2 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^4(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{2 a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{\tan (c+d x) \left (-4 a \left (6 a^4+11 a^2 b^2+b^4\right )+4 b^3 \left (a^2-b^2\right ) \tan (c+d x)-4 a \left (6 a^4+11 a^2 b^2+3 b^4\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{4 b^3 \left (a^2+b^2\right )^2}\\ &=-\frac{a \left (6 a^4+11 a^2 b^2+3 b^4\right ) \tan (c+d x)}{b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (6 a^4+11 a^2 b^2+b^4\right ) \tan ^2(c+d x)}{2 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^4(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{2 a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{4 a^2 \left (6 a^4+11 a^2 b^2+3 b^4\right )+8 a b^5 \tan (c+d x)+4 \left (6 a^2-b^2\right ) \left (a^2+b^2\right )^2 \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{4 b^4 \left (a^2+b^2\right )^2}\\ &=-\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac{a \left (6 a^4+11 a^2 b^2+3 b^4\right ) \tan (c+d x)}{b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (6 a^4+11 a^2 b^2+b^4\right ) \tan ^2(c+d x)}{2 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^4(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{2 a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (b \left (3 a^2-b^2\right )\right ) \int \tan (c+d x) \, dx}{\left (a^2+b^2\right )^3}+\frac{\left (a^4 \left (6 a^4+17 a^2 b^2+15 b^4\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 )^3}\\ &=-\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac{b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{a \left (6 a^4+11 a^2 b^2+3 b^4\right ) \tan (c+d x)}{b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (6 a^4+11 a^2 b^2+b^4\right ) \tan ^2(c+d x)}{2 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^4(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{2 a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (a^4 \left (6 a^4+17 a^2 b^2+15 b^4\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 )^3 d}\\ &=-\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac{b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac{a^4 \left (6 a^4+17 a^2 b^2+15 b^4\right ) \log (a+b \tan (c+d x))}{b^5 \left (a^2+b^2\right )^3 d}-\frac{a \left (6 a^4+11 a^2 b^2+3 b^4\right ) \tan (c+d x)}{b^4 \left (a^2+b^2\right )^2 d}+\frac{\left (6 a^4+11 a^2 b^2+b^4\right ) \tan ^2(c+d x)}{2 b^3 \left (a^2+b^2\right )^2 d}-\frac{a^2 \tan ^4(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{2 a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 3.39238, size = 243, normalized size = 0.86 \[ \frac{-\frac{a^4 \left (6 a^2+5 b^2\right )}{b^4 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{4 a^3 \left (11 a^2 b^2+6 a^4+4 b^4\right )}{b^4 \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{2 a^4 \left (17 a^2 b^2+6 a^4+15 b^4\right ) \log (a+b \tan (c+d x))}{b^4 \left (a^2+b^2\right )^3}-\frac{4 a \tan ^3(c+d x)}{b (a+b \tan (c+d x))^2}+\frac{\tan ^4(c+d x)}{(a+b \tan (c+d x))^2}+\frac{i b \log (-\tan (c+d x)+i)}{(a+i b)^3}-\frac{b \log (\tan (c+d x)+i)}{(b+i a)^3}}{2 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + b*Tan[c + d*x])^3,x]

[Out]

((I*b*Log[I - Tan[c + d*x]])/(a + I*b)^3 - (b*Log[I + Tan[c + d*x]])/(I*a + b)^3 + (2*a^4*(6*a^4 + 17*a^2*b^2
+ 15*b^4)*Log[a + b*Tan[c + d*x]])/(b^4*(a^2 + b^2)^3) - (a^4*(6*a^2 + 5*b^2))/(b^4*(a^2 + b^2)*(a + b*Tan[c +
d*x])^2) - (4*a*Tan[c + d*x]^3)/(b*(a + b*Tan[c + d*x])^2) + Tan[c + d*x]^4/(a + b*Tan[c + d*x])^2 + (4*a^3*(
6*a^4 + 11*a^2*b^2 + 4*b^4))/(b^4*(a^2 + b^2)^2*(a + b*Tan[c + d*x])))/(2*b*d)

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Maple [A]  time = 0.035, size = 328, normalized size = 1.2 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,{b}^{3}d}}-3\,{\frac{a\tan \left ( dx+c \right ) }{{b}^{4}d}}+{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) b{a}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){b}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{{a}^{6}}{2\,d{b}^{5} \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+6\,{\frac{{a}^{8}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d{b}^{5} \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+17\,{\frac{{a}^{6}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{b}^{3}d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+15\,{\frac{{a}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{bd \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+4\,{\frac{{a}^{7}}{d{b}^{5} \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+6\,{\frac{{a}^{5}}{{b}^{3}d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }} \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))^3,x)

[Out]

1/2*tan(d*x+c)^2/b^3/d-3*a*tan(d*x+c)/b^4/d+3/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*b*a^2-1/2/d/(a^2+b^2)^3*ln(1+
tan(d*x+c)^2)*b^3-1/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a^3+3/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a*b^2-1/2/d/b^5*a^
6/(a^2+b^2)/(a+b*tan(d*x+c))^2+6/d/b^5*a^8/(a^2+b^2)^3*ln(a+b*tan(d*x+c))+17/d/b^3*a^6/(a^2+b^2)^3*ln(a+b*tan(
d*x+c))+15/d/b*a^4/(a^2+b^2)^3*ln(a+b*tan(d*x+c))+4/d/b^5*a^7/(a^2+b^2)^2/(a+b*tan(d*x+c))+6/d/b^3*a^5/(a^2+b^
2)^2/(a+b*tan(d*x+c))

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Maxima [A]  time = 1.51564, size = 416, normalized size = 1.47 \begin{align*} -\frac{\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{2 \,{\left (6 \, a^{8} + 17 \, a^{6} b^{2} + 15 \, a^{4} b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} + b^{11}} - \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{7 \, a^{8} + 11 \, a^{6} b^{2} + 4 \,{\left (2 \, a^{7} b + 3 \, a^{5} b^{3}\right )} \tan \left (d x + c\right )}{a^{6} b^{5} + 2 \, a^{4} b^{7} + a^{2} b^{9} +{\left (a^{4} b^{7} + 2 \, a^{2} b^{9} + b^{11}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b^{6} + 2 \, a^{3} b^{8} + a b^{10}\right )} \tan \left (d x + c\right )} - \frac{b \tan \left (d x + c\right )^{2} - 6 \, a \tan \left (d x + c\right )}{b^{4}}}{2 \, 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))^3,x, algorithm="maxima")

[Out]

-1/2*(2*(a^3 - 3*a*b^2)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(6*a^8 + 17*a^6*b^2 + 15*a^4*b^4)*lo
g(b*tan(d*x + c) + a)/(a^6*b^5 + 3*a^4*b^7 + 3*a^2*b^9 + b^11) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1)/(a^6
+ 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (7*a^8 + 11*a^6*b^2 + 4*(2*a^7*b + 3*a^5*b^3)*tan(d*x + c))/(a^6*b^5 + 2*a^4*
b^7 + a^2*b^9 + (a^4*b^7 + 2*a^2*b^9 + b^11)*tan(d*x + c)^2 + 2*(a^5*b^6 + 2*a^3*b^8 + a*b^10)*tan(d*x + c)) -
(b*tan(d*x + c)^2 - 6*a*tan(d*x + c))/b^4)/d

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Fricas [B]  time = 2.68055, size = 1361, normalized size = 4.81 \begin{align*} \frac{6 \, a^{8} b^{2} + 14 \, a^{6} b^{4} + 3 \, a^{4} b^{6} + a^{2} b^{8} +{\left (a^{6} b^{4} + 3 \, a^{4} b^{6} + 3 \, a^{2} b^{8} + b^{10}\right )} \tan \left (d x + c\right )^{4} - 4 \,{\left (a^{7} b^{3} + 3 \, a^{5} b^{5} + 3 \, a^{3} b^{7} + a b^{9}\right )} \tan \left (d x + c\right )^{3} - 2 \,{\left (a^{5} b^{5} - 3 \, a^{3} b^{7}\right )} d x -{\left (18 \, a^{8} b^{2} + 45 \, a^{6} b^{4} + 30 \, a^{4} b^{6} + 8 \, a^{2} b^{8} - b^{10} + 2 \,{\left (a^{3} b^{7} - 3 \, a b^{9}\right )} d x\right )} \tan \left (d x + c\right )^{2} +{\left (6 \, a^{10} + 17 \, a^{8} b^{2} + 15 \, a^{6} b^{4} +{\left (6 \, a^{8} b^{2} + 17 \, a^{6} b^{4} + 15 \, a^{4} b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (6 \, a^{9} b + 17 \, a^{7} b^{3} + 15 \, a^{5} b^{5}\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 ) -{\left (6 \, a^{10} + 17 \, a^{8} b^{2} + 15 \, a^{6} b^{4} + 3 \, a^{4} b^{6} - a^{2} b^{8} +{\left (6 \, a^{8} b^{2} + 17 \, a^{6} b^{4} + 15 \, a^{4} b^{6} + 3 \, a^{2} b^{8} - b^{10}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (6 \, a^{9} b + 17 \, a^{7} b^{3} + 15 \, a^{5} b^{5} + 3 \, a^{3} b^{7} - a b^{9}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \,{\left (6 \, a^{9} b + 11 \, a^{7} b^{3} - a b^{9} + 2 \,{\left (a^{4} b^{6} - 3 \, a^{2} b^{8}\right )} d x\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b^{7} + 3 \, a^{4} b^{9} + 3 \, a^{2} b^{11} + b^{13}\right )} d \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b^{6} + 3 \, a^{5} b^{8} + 3 \, a^{3} b^{10} + a b^{12}\right )} d \tan \left (d x + c\right ) +{\left (a^{8} b^{5} + 3 \, a^{6} b^{7} + 3 \, a^{4} b^{9} + a^{2} b^{11}\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))^3,x, algorithm="fricas")

[Out]

1/2*(6*a^8*b^2 + 14*a^6*b^4 + 3*a^4*b^6 + a^2*b^8 + (a^6*b^4 + 3*a^4*b^6 + 3*a^2*b^8 + b^10)*tan(d*x + c)^4 -
4*(a^7*b^3 + 3*a^5*b^5 + 3*a^3*b^7 + a*b^9)*tan(d*x + c)^3 - 2*(a^5*b^5 - 3*a^3*b^7)*d*x - (18*a^8*b^2 + 45*a^
6*b^4 + 30*a^4*b^6 + 8*a^2*b^8 - b^10 + 2*(a^3*b^7 - 3*a*b^9)*d*x)*tan(d*x + c)^2 + (6*a^10 + 17*a^8*b^2 + 15*
a^6*b^4 + (6*a^8*b^2 + 17*a^6*b^4 + 15*a^4*b^6)*tan(d*x + c)^2 + 2*(6*a^9*b + 17*a^7*b^3 + 15*a^5*b^5)*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)) - (6*a^10 + 17*a^8*b^2 + 15*a
^6*b^4 + 3*a^4*b^6 - a^2*b^8 + (6*a^8*b^2 + 17*a^6*b^4 + 15*a^4*b^6 + 3*a^2*b^8 - b^10)*tan(d*x + c)^2 + 2*(6*
a^9*b + 17*a^7*b^3 + 15*a^5*b^5 + 3*a^3*b^7 - a*b^9)*tan(d*x + c))*log(1/(tan(d*x + c)^2 + 1)) - 2*(6*a^9*b +
11*a^7*b^3 - a*b^9 + 2*(a^4*b^6 - 3*a^2*b^8)*d*x)*tan(d*x + c))/((a^6*b^7 + 3*a^4*b^9 + 3*a^2*b^11 + b^13)*d*t
an(d*x + c)^2 + 2*(a^7*b^6 + 3*a^5*b^8 + 3*a^3*b^10 + a*b^12)*d*tan(d*x + c) + (a^8*b^5 + 3*a^6*b^7 + 3*a^4*b^
9 + a^2*b^11)*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 5.13054, size = 466, normalized size = 1.65 \begin{align*} -\frac{\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{2 \,{\left (6 \, a^{8} + 17 \, a^{6} b^{2} + 15 \, a^{4} b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} + b^{11}} + \frac{18 \, a^{8} b^{2} \tan \left (d x + c\right )^{2} + 51 \, a^{6} b^{4} \tan \left (d x + c\right )^{2} + 45 \, a^{4} b^{6} \tan \left (d x + c\right )^{2} + 28 \, a^{9} b \tan \left (d x + c\right ) + 82 \, a^{7} b^{3} \tan \left (d x + c\right ) + 78 \, a^{5} b^{5} \tan \left (d x + c\right ) + 11 \, a^{10} + 33 \, a^{8} b^{2} + 34 \, a^{6} b^{4}}{{\left (a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} + b^{11}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}} - \frac{b^{3} \tan \left (d x + c\right )^{2} - 6 \, a b^{2} \tan \left (d x + c\right )}{b^{6}}}{2 \, 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))^3,x, algorithm="giac")

[Out]

-1/2*(2*(a^3 - 3*a*b^2)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1
)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(6*a^8 + 17*a^6*b^2 + 15*a^4*b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*
b^5 + 3*a^4*b^7 + 3*a^2*b^9 + b^11) + (18*a^8*b^2*tan(d*x + c)^2 + 51*a^6*b^4*tan(d*x + c)^2 + 45*a^4*b^6*tan(
d*x + c)^2 + 28*a^9*b*tan(d*x + c) + 82*a^7*b^3*tan(d*x + c) + 78*a^5*b^5*tan(d*x + c) + 11*a^10 + 33*a^8*b^2
+ 34*a^6*b^4)/((a^6*b^5 + 3*a^4*b^7 + 3*a^2*b^9 + b^11)*(b*tan(d*x + c) + a)^2) - (b^3*tan(d*x + c)^2 - 6*a*b^
2*tan(d*x + c))/b^6)/d