3.5.77 \(\int \frac {\tan ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [477]

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

[Out]

-a*(a^2-3*b^2)*x/(a^2+b^2)^3-b*(3*a^2-b^2)*ln(cos(d*x+c))/(a^2+b^2)^3/d+a^4*(6*a^4+17*a^2*b^2+15*b^4)*ln(a+b*t
an(d*x+c))/b^5/(a^2+b^2)^3/d-a*(6*a^4+11*a^2*b^2+3*b^4)*tan(d*x+c)/b^4/(a^2+b^2)^2/d+1/2*(6*a^4+11*a^2*b^2+b^4
)*tan(d*x+c)^2/b^3/(a^2+b^2)^2/d-1/2*a^2*tan(d*x+c)^4/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^2-2*a^2*(a^2+2*b^2)*tan(d
*x+c)^3/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))

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Rubi [A]
time = 0.56, antiderivative size = 283, normalized size of antiderivative = 1.00, 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 = {3646, 3726, 3728, 3707, 3698, 31, 3556} \begin {gather*} -\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 {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 {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}-\frac {a \left (6 a^4+11 a^2 b^2+3 b^4\right ) \tan (c+d x)}{b^4 d \left (a^2+b^2\right )^2}+\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 d \left (a^2+b^2\right )^3}+\frac {\left (6 a^4+11 a^2 b^2+b^4\right ) \tan ^2(c+d x)}{2 b^3 d \left (a^2+b^2\right )^2} \end {gather*}

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 31

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

Rule 3556

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

Rule 3646

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 3698

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 3707

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 3726

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*Ta
n[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 3728

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])))

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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 3.81, size = 243, normalized size = 0.86 \begin {gather*} \frac {\frac {i b \log (i-\tan (c+d x))}{(a+i b)^3}-\frac {b \log (i+\tan (c+d x))}{(i a+b)^3}+\frac {2 a^4 \left (6 a^4+17 a^2 b^2+15 b^4\right ) \log (a+b \tan (c+d x))}{b^4 \left (a^2+b^2\right )^3}-\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 \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 {4 a^3 \left (6 a^4+11 a^2 b^2+4 b^4\right )}{b^4 \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}}{2 b d} \end {gather*}

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 = 0.21, size = 203, normalized size = 0.72 Too large to display

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 308, normalized size = 1.09 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 628 vs. \(2 (279) = 558\).
time = 1.04, size = 628, normalized size = 2.22 \begin {gather*} \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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: AttributeError} \end {gather*}

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]

Exception raised: AttributeError >> 'NoneType' object has no attribute 'primitive'

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Giac [A]
time = 2.51, size = 345, normalized size = 1.22 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 4.35, size = 284, normalized size = 1.00 \begin {gather*} \frac {\frac {2\,\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a^7+3\,a^5\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {7\,a^8+11\,a^6\,b^2}{2\,b\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2\,b^4+2\,a\,b^5\,\mathrm {tan}\left (c+d\,x\right )+b^6\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {b}{{\left (a^2+b^2\right )}^2}-\frac {1}{b^3}+\frac {6\,a^2}{b^5}-\frac {4\,a^2\,b}{{\left (a^2+b^2\right )}^3}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,b^3\,d}-\frac {3\,a\,\mathrm {tan}\left (c+d\,x\right )}{b^4\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6/(a + b*tan(c + d*x))^3,x)

[Out]

((2*tan(c + d*x)*(2*a^7 + 3*a^5*b^2))/(a^4 + b^4 + 2*a^2*b^2) + (7*a^8 + 11*a^6*b^2)/(2*b*(a^4 + b^4 + 2*a^2*b
^2)))/(d*(a^2*b^4 + b^6*tan(c + d*x)^2 + 2*a*b^5*tan(c + d*x))) - log(tan(c + d*x) + 1i)/(2*d*(a*b^2*3i - 3*a^
2*b - a^3*1i + b^3)) - (log(tan(c + d*x) - 1i)*1i)/(2*d*(3*a*b^2 - a^2*b*3i - a^3 + b^3*1i)) + (log(a + b*tan(
c + d*x))*(b/(a^2 + b^2)^2 - 1/b^3 + (6*a^2)/b^5 - (4*a^2*b)/(a^2 + b^2)^3))/d + tan(c + d*x)^2/(2*b^3*d) - (3
*a*tan(c + d*x))/(b^4*d)

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