3.256 \(\int \cot ^6(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=233 \[ \frac {a \left (5 a^2 A-15 a b B-12 A b^2\right ) \cot ^3(c+d x)}{15 d}-\frac {a^2 (5 a B+7 A b) \cot ^4(c+d x)}{20 d}+\frac {\left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \cot ^2(c+d x)}{2 d}-\frac {\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \cot (c+d x)}{d}+\frac {\left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \log (\sin (c+d x))}{d}-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d} \]

[Out]

-(A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3)*x-(A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3)*cot(d*x+c)/d+1/2*(3*A*a^2*b-A*b^3+B*a^3
-3*B*a*b^2)*cot(d*x+c)^2/d+1/15*a*(5*A*a^2-12*A*b^2-15*B*a*b)*cot(d*x+c)^3/d-1/20*a^2*(7*A*b+5*B*a)*cot(d*x+c)
^4/d+(3*A*a^2*b-A*b^3+B*a^3-3*B*a*b^2)*ln(sin(d*x+c))/d-1/5*a*A*cot(d*x+c)^5*(a+b*tan(d*x+c))^2/d

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Rubi [A]  time = 0.50, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3605, 3635, 3628, 3529, 3531, 3475} \[ \frac {a \left (5 a^2 A-15 a b B-12 A b^2\right ) \cot ^3(c+d x)}{15 d}+\frac {\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \cot ^2(c+d x)}{2 d}-\frac {\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \cot (c+d x)}{d}+\frac {\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \log (\sin (c+d x))}{d}-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )-\frac {a^2 (5 a B+7 A b) \cot ^4(c+d x)}{20 d}-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*x) - ((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Cot[c + d*x])/d + ((3*
a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Cot[c + d*x]^2)/(2*d) + (a*(5*a^2*A - 12*A*b^2 - 15*a*b*B)*Cot[c + d*x]^3
)/(15*d) - (a^2*(7*A*b + 5*a*B)*Cot[c + d*x]^4)/(20*d) + ((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Log[Sin[c +
d*x]])/d - (a*A*Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2)/(5*d)

Rule 3475

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

Rule 3529

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

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3605

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(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)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3628

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

Rule 3635

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(c^2*C - B*c*d + A*d^2)*
(c + d*Tan[e + f*x])^(n + 1))/(d^2*f*(n + 1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x
])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d +
a*C*d)*Tan[e + f*x] + b*C*(c^2 + d^2)*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[c^2 + d^2, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \cot ^6(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (a (7 A b+5 a B)-5 \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-b (3 a A-5 b B) \tan ^2(c+d x)\right ) \, dx\\ &=-\frac {a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \cot ^4(c+d x) \left (-a \left (5 a^2 A-12 A b^2-15 a b B\right )-5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-b^2 (3 a A-5 b B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac {a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \cot ^3(c+d x) \left (-5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+5 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)\right ) \, dx\\ &=\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot ^2(c+d x)}{2 d}+\frac {a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac {a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \cot ^2(c+d x) \left (5 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)\right ) \, dx\\ &=-\frac {\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \cot (c+d x)}{d}+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot ^2(c+d x)}{2 d}+\frac {a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac {a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \cot (c+d x) \left (5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-5 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)\right ) \, dx\\ &=-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x-\frac {\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \cot (c+d x)}{d}+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot ^2(c+d x)}{2 d}+\frac {a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac {a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \int \cot (c+d x) \, dx\\ &=-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x-\frac {\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \cot (c+d x)}{d}+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot ^2(c+d x)}{2 d}+\frac {a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac {a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \log (\sin (c+d x))}{d}-\frac {a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\\ \end {align*}

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Mathematica [C]  time = 1.21, size = 237, normalized size = 1.02 \[ \frac {-12 a^3 A \cot ^5(c+d x)+20 a \left (a^2 A-3 a b B-3 A b^2\right ) \cot ^3(c+d x)-15 a^2 (a B+3 A b) \cot ^4(c+d x)+30 \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \cot ^2(c+d x)-60 \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \cot (c+d x)+60 \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \log (\tan (c+d x))+30 i (a+i b)^3 (A+i B) \log (-\tan (c+d x)+i)+30 (b+i a)^3 (A-i B) \log (\tan (c+d x)+i)}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-60*(a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Cot[c + d*x] + 30*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Cot[c +
d*x]^2 + 20*a*(a^2*A - 3*A*b^2 - 3*a*b*B)*Cot[c + d*x]^3 - 15*a^2*(3*A*b + a*B)*Cot[c + d*x]^4 - 12*a^3*A*Cot
[c + d*x]^5 + (30*I)*(a + I*b)^3*(A + I*B)*Log[I - Tan[c + d*x]] + 60*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*
Log[Tan[c + d*x]] + 30*(I*a + b)^3*(A - I*B)*Log[I + Tan[c + d*x]])/(60*d)

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fricas [A]  time = 0.55, size = 266, normalized size = 1.14 \[ \frac {30 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{5} + 15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b - 6 \, B a b^{2} - 2 \, A b^{3} - 4 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{5} - 60 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \tan \left (d x + c\right )^{4} - 12 \, A a^{3} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{3} + 20 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} - 15 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{60 \, d \tan \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^5 + 15*
(3*B*a^3 + 9*A*a^2*b - 6*B*a*b^2 - 2*A*b^3 - 4*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*d*x)*tan(d*x + c)^5 - 6
0*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*tan(d*x + c)^4 - 12*A*a^3 + 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^
3)*tan(d*x + c)^3 + 20*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*tan(d*x + c)^2 - 15*(B*a^3 + 3*A*a^2*b)*tan(d*x + c))/(
d*tan(d*x + c)^5)

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giac [B]  time = 6.11, size = 670, normalized size = 2.88 \[ \frac {6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 45 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 70 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 180 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 540 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 360 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 660 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1800 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1800 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 480 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 960 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} {\left (d x + c\right )} - 960 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 960 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2192 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6576 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6576 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2192 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 660 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1800 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1800 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 480 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 180 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 540 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 70 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 45 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/960*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 15*B*a^3*tan(1/2*d*x + 1/2*c)^4 - 45*A*a^2*b*tan(1/2*d*x + 1/2*c)^4 -
70*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 120*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 18
0*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 540*A*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 360*B*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 120
*A*b^3*tan(1/2*d*x + 1/2*c)^2 + 660*A*a^3*tan(1/2*d*x + 1/2*c) - 1800*B*a^2*b*tan(1/2*d*x + 1/2*c) - 1800*A*a*
b^2*tan(1/2*d*x + 1/2*c) + 480*B*b^3*tan(1/2*d*x + 1/2*c) - 960*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*(d*x +
c) - 960*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(tan(1/2*d*x + 1/2*c)^2 + 1) + 960*(B*a^3 + 3*A*a^2*b - 3
*B*a*b^2 - A*b^3)*log(abs(tan(1/2*d*x + 1/2*c))) - (2192*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 6576*A*a^2*b*tan(1/2*d
*x + 1/2*c)^5 - 6576*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 2192*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 660*A*a^3*tan(1/2*d*
x + 1/2*c)^4 - 1800*B*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 1800*A*a*b^2*tan(1/2*d*x + 1/2*c)^4 + 480*B*b^3*tan(1/2*d
*x + 1/2*c)^4 - 180*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 540*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 360*B*a*b^2*tan(1/2*d*
x + 1/2*c)^3 + 120*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 70*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 120*B*a^2*b*tan(1/2*d*x +
1/2*c)^2 + 120*A*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 15*B*a^3*tan(1/2*d*x + 1/2*c) + 45*A*a^2*b*tan(1/2*d*x + 1/2*c
) + 6*A*a^3)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 0.44, size = 376, normalized size = 1.61 \[ -\frac {a^{2} b B \left (\cot ^{3}\left (d x +c \right )\right )}{d}-\frac {3 B a \,b^{2} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {A \,b^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} B \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {A \,a^{3} c}{d}-\frac {B \,b^{3} c}{d}+3 A x a \,b^{2}+3 B x \,a^{2} b -\frac {3 A \,a^{2} b \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}-\frac {A a \,b^{2} \left (\cot ^{3}\left (d x +c \right )\right )}{d}-\frac {A \,b^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {A \,a^{3} \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}-\frac {a^{3} B \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}-\frac {B \cot \left (d x +c \right ) b^{3}}{d}-\frac {3 B a \,b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {A \cot \left (d x +c \right ) a^{3}}{d}+\frac {a^{3} B \ln \left (\sin \left (d x +c \right )\right )}{d}-A \,a^{3} x -B x \,b^{3}+\frac {3 A \,a^{2} b \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 A \,a^{2} b \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {3 A a \,b^{2} c}{d}+\frac {3 B \,a^{2} b c}{d}+\frac {3 B \cot \left (d x +c \right ) a^{2} b}{d}+\frac {3 A \cot \left (d x +c \right ) a \,b^{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

-1/d*a^2*b*B*cot(d*x+c)^3-3/4/d*A*a^2*b*cot(d*x+c)^4-1/d*A*a*b^2*cot(d*x+c)^3+3/2/d*A*a^2*b*cot(d*x+c)^2+1/3/d
*A*a^3*cot(d*x+c)^3+1/2/d*a^3*B*cot(d*x+c)^2-1/d*A*b^3*ln(sin(d*x+c))-1/d*A*a^3*c-1/d*B*b^3*c+3*A*x*a*b^2+3*B*
x*a^2*b-1/5/d*A*a^3*cot(d*x+c)^5-1/4/d*a^3*B*cot(d*x+c)^4-1/d*B*cot(d*x+c)*b^3-1/2/d*A*b^3*cot(d*x+c)^2-3/d*B*
a*b^2*ln(sin(d*x+c))-1/d*A*cot(d*x+c)*a^3+1/d*a^3*B*ln(sin(d*x+c))-A*a^3*x-B*x*b^3+3/d*A*a^2*b*ln(sin(d*x+c))+
3/d*A*a*b^2*c+3/d*B*a^2*b*c+3/d*B*cot(d*x+c)*a^2*b+3/d*A*cot(d*x+c)*a*b^2-3/2/d*B*a*b^2*cot(d*x+c)^2

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maxima [A]  time = 0.78, size = 250, normalized size = 1.07 \[ -\frac {60 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} {\left (d x + c\right )} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {60 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \tan \left (d x + c\right )^{4} + 12 \, A a^{3} - 30 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{3} - 20 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 15 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(60*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*(d*x + c) + 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(t
an(d*x + c)^2 + 1) - 60*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(tan(d*x + c)) + (60*(A*a^3 - 3*B*a^2*b - 3
*A*a*b^2 + B*b^3)*tan(d*x + c)^4 + 12*A*a^3 - 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*tan(d*x + c)^3 - 20*(
A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*tan(d*x + c)^2 + 15*(B*a^3 + 3*A*a^2*b)*tan(d*x + c))/tan(d*x + c)^5)/d

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mupad [B]  time = 6.57, size = 238, normalized size = 1.02 \[ -\frac {{\mathrm {cot}\left (c+d\,x\right )}^5\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^3}{4}+\frac {3\,A\,b\,a^2}{4}\right )+\frac {A\,a^3}{5}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {A\,a^3}{3}+B\,a^2\,b+A\,a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (A\,a^3-3\,B\,a^2\,b-3\,A\,a\,b^2+B\,b^3\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-\frac {B\,a^3}{2}-\frac {3\,A\,a^2\,b}{2}+\frac {3\,B\,a\,b^2}{2}+\frac {A\,b^3}{2}\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B\,a^3-3\,A\,a^2\,b+3\,B\,a\,b^2+A\,b^3\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^6*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3,x)

[Out]

(log(tan(c + d*x) - 1i)*(A + B*1i)*(a + b*1i)^3*1i)/(2*d) - (log(tan(c + d*x))*(A*b^3 - B*a^3 - 3*A*a^2*b + 3*
B*a*b^2))/d - (cot(c + d*x)^5*(tan(c + d*x)*((B*a^3)/4 + (3*A*a^2*b)/4) + (A*a^3)/5 + tan(c + d*x)^2*(A*a*b^2
- (A*a^3)/3 + B*a^2*b) + tan(c + d*x)^4*(A*a^3 + B*b^3 - 3*A*a*b^2 - 3*B*a^2*b) + tan(c + d*x)^3*((A*b^3)/2 -
(B*a^3)/2 - (3*A*a^2*b)/2 + (3*B*a*b^2)/2)))/d - (log(tan(c + d*x) + 1i)*(A - B*1i)*(a - b*1i)^3*1i)/(2*d)

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sympy [A]  time = 9.15, size = 471, normalized size = 2.02 \[ \begin {cases} \tilde {\infty } A a^{3} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right )^{3} \cot ^{6}{\relax (c )} & \text {for}\: d = 0 \\- A a^{3} x - \frac {A a^{3}}{d \tan {\left (c + d x \right )}} + \frac {A a^{3}}{3 d \tan ^{3}{\left (c + d x \right )}} - \frac {A a^{3}}{5 d \tan ^{5}{\left (c + d x \right )}} - \frac {3 A a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 A a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {3 A a^{2} b}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {3 A a^{2} b}{4 d \tan ^{4}{\left (c + d x \right )}} + 3 A a b^{2} x + \frac {3 A a b^{2}}{d \tan {\left (c + d x \right )}} - \frac {A a b^{2}}{d \tan ^{3}{\left (c + d x \right )}} + \frac {A b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {A b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A b^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {B a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {B a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {B a^{3}}{4 d \tan ^{4}{\left (c + d x \right )}} + 3 B a^{2} b x + \frac {3 B a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {B a^{2} b}{d \tan ^{3}{\left (c + d x \right )}} + \frac {3 B a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 B a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 B a b^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - B b^{3} x - \frac {B b^{3}}{d \tan {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(a+b*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

Piecewise((zoo*A*a**3*x, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(A + B*tan(c))*(a + b*tan(c)
)**3*cot(c)**6, Eq(d, 0)), (-A*a**3*x - A*a**3/(d*tan(c + d*x)) + A*a**3/(3*d*tan(c + d*x)**3) - A*a**3/(5*d*t
an(c + d*x)**5) - 3*A*a**2*b*log(tan(c + d*x)**2 + 1)/(2*d) + 3*A*a**2*b*log(tan(c + d*x))/d + 3*A*a**2*b/(2*d
*tan(c + d*x)**2) - 3*A*a**2*b/(4*d*tan(c + d*x)**4) + 3*A*a*b**2*x + 3*A*a*b**2/(d*tan(c + d*x)) - A*a*b**2/(
d*tan(c + d*x)**3) + A*b**3*log(tan(c + d*x)**2 + 1)/(2*d) - A*b**3*log(tan(c + d*x))/d - A*b**3/(2*d*tan(c +
d*x)**2) - B*a**3*log(tan(c + d*x)**2 + 1)/(2*d) + B*a**3*log(tan(c + d*x))/d + B*a**3/(2*d*tan(c + d*x)**2) -
B*a**3/(4*d*tan(c + d*x)**4) + 3*B*a**2*b*x + 3*B*a**2*b/(d*tan(c + d*x)) - B*a**2*b/(d*tan(c + d*x)**3) + 3*
B*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) - 3*B*a*b**2*log(tan(c + d*x))/d - 3*B*a*b**2/(2*d*tan(c + d*x)**2) -
B*b**3*x - B*b**3/(d*tan(c + d*x)), True))

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