3.23 \(\int \cot ^6(c+d x) (a+b \tan (c+d x))^3 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=191 \[ \frac {a \left (2 a^2 B-6 a b C-5 b^2 B\right ) \cot ^2(c+d x)}{4 d}-\frac {a^2 (2 a C+3 b B) \cot ^3(c+d x)}{6 d}+\frac {\left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot (c+d x)}{d}+\frac {\left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right ) \log (\sin (c+d x))}{d}+x \left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right )-\frac {a B \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d} \]

[Out]

(3*B*a^2*b-B*b^3+C*a^3-3*C*a*b^2)*x+(3*B*a^2*b-B*b^3+C*a^3-3*C*a*b^2)*cot(d*x+c)/d+1/4*a*(2*B*a^2-5*B*b^2-6*C*
a*b)*cot(d*x+c)^2/d-1/6*a^2*(3*B*b+2*C*a)*cot(d*x+c)^3/d+(B*a^3-3*B*a*b^2-3*C*a^2*b+C*b^3)*ln(sin(d*x+c))/d-1/
4*a*B*cot(d*x+c)^4*(a+b*tan(d*x+c))^2/d

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Rubi [A]  time = 0.51, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {3632, 3605, 3635, 3628, 3529, 3531, 3475} \[ \frac {a \left (2 a^2 B-6 a b C-5 b^2 B\right ) \cot ^2(c+d x)}{4 d}+\frac {\left (3 a^2 b B+a^3 C-3 a b^2 C-b^3 B\right ) \cot (c+d x)}{d}+\frac {\left (-3 a^2 b C+a^3 B-3 a b^2 B+b^3 C\right ) \log (\sin (c+d x))}{d}+x \left (3 a^2 b B+a^3 C-3 a b^2 C-b^3 B\right )-\frac {a^2 (2 a C+3 b B) \cot ^3(c+d x)}{6 d}-\frac {a B \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

(3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*x + ((3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*Cot[c + d*x])/d + (a*(2*a
^2*B - 5*b^2*B - 6*a*b*C)*Cot[c + d*x]^2)/(4*d) - (a^2*(3*b*B + 2*a*C)*Cot[c + d*x]^3)/(6*d) + ((a^3*B - 3*a*b
^2*B - 3*a^2*b*C + b^3*C)*Log[Sin[c + d*x]])/d - (a*B*Cot[c + d*x]^4*(a + b*Tan[c + d*x])^2)/(4*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 3632

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] :> Dist[1/b^2, Int[(a + b*Tan[e + f*x])
^(m + 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx &=\int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (B+C \tan (c+d x)) \, dx\\ &=-\frac {a B \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {1}{4} \int \cot ^4(c+d x) (a+b \tan (c+d x)) \left (2 a (3 b B+2 a C)-4 \left (a^2 B-b^2 B-2 a b C\right ) \tan (c+d x)-2 b (a B-2 b C) \tan ^2(c+d x)\right ) \, dx\\ &=-\frac {a^2 (3 b B+2 a C) \cot ^3(c+d x)}{6 d}-\frac {a B \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {1}{4} \int \cot ^3(c+d x) \left (-2 a \left (2 a^2 B-5 b^2 B-6 a b C\right )-4 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \tan (c+d x)-2 b^2 (a B-2 b C) \tan ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (2 a^2 B-5 b^2 B-6 a b C\right ) \cot ^2(c+d x)}{4 d}-\frac {a^2 (3 b B+2 a C) \cot ^3(c+d x)}{6 d}-\frac {a B \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {1}{4} \int \cot ^2(c+d x) \left (-4 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right )+4 \left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) \tan (c+d x)\right ) \, dx\\ &=\frac {\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \cot (c+d x)}{d}+\frac {a \left (2 a^2 B-5 b^2 B-6 a b C\right ) \cot ^2(c+d x)}{4 d}-\frac {a^2 (3 b B+2 a C) \cot ^3(c+d x)}{6 d}-\frac {a B \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\frac {1}{4} \int \cot (c+d x) \left (4 \left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right )+4 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \tan (c+d x)\right ) \, dx\\ &=\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) x+\frac {\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \cot (c+d x)}{d}+\frac {a \left (2 a^2 B-5 b^2 B-6 a b C\right ) \cot ^2(c+d x)}{4 d}-\frac {a^2 (3 b B+2 a C) \cot ^3(c+d x)}{6 d}-\frac {a B \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}+\left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) \int \cot (c+d x) \, dx\\ &=\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) x+\frac {\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \cot (c+d x)}{d}+\frac {a \left (2 a^2 B-5 b^2 B-6 a b C\right ) \cot ^2(c+d x)}{4 d}-\frac {a^2 (3 b B+2 a C) \cot ^3(c+d x)}{6 d}+\frac {\left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) \log (\sin (c+d x))}{d}-\frac {a B \cot ^4(c+d x) (a+b \tan (c+d x))^2}{4 d}\\ \end {align*}

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Mathematica [C]  time = 0.79, size = 199, normalized size = 1.04 \[ \frac {-3 a^3 B \cot ^4(c+d x)+6 a \left (a^2 B-3 a b C-3 b^2 B\right ) \cot ^2(c+d x)-4 a^2 (a C+3 b B) \cot ^3(c+d x)+12 \left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right ) \cot (c+d x)+12 \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right ) \log (\tan (c+d x))-6 (a+i b)^3 (B+i C) \log (-\tan (c+d x)+i)-6 (a-i b)^3 (B-i C) \log (\tan (c+d x)+i)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

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fricas [A]  time = 0.74, size = 225, normalized size = 1.18 \[ \frac {6 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 3 \, {\left (3 \, B a^{3} - 6 \, C a^{2} b - 6 \, B a b^{2} + 4 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{4} - 3 \, B a^{3} + 12 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \tan \left (d x + c\right )^{2} - 4 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(6*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^4 + 3*(3
*B*a^3 - 6*C*a^2*b - 6*B*a*b^2 + 4*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*d*x)*tan(d*x + c)^4 - 3*B*a^3 + 12*
(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*tan(d*x + c)^3 + 6*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2)*tan(d*x + c)^2 - 4*
(C*a^3 + 3*B*a^2*b)*tan(d*x + c))/(d*tan(d*x + c)^4)

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giac [B]  time = 90.46, size = 528, normalized size = 2.76 \[ -\frac {3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 288 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 192 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} {\left (d x + c\right )} + 192 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 192 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {400 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1200 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1200 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 400 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 360 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 288 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 96 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, 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*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

-1/192*(3*B*a^3*tan(1/2*d*x + 1/2*c)^4 - 8*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 24*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 -
36*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 72*B*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 120*
C*a^3*tan(1/2*d*x + 1/2*c) + 360*B*a^2*b*tan(1/2*d*x + 1/2*c) - 288*C*a*b^2*tan(1/2*d*x + 1/2*c) - 96*B*b^3*ta
n(1/2*d*x + 1/2*c) - 192*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*(d*x + c) + 192*(B*a^3 - 3*C*a^2*b - 3*B*a*b^
2 + C*b^3)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 192*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*log(abs(tan(1/2*d*x +
1/2*c))) + (400*B*a^3*tan(1/2*d*x + 1/2*c)^4 - 1200*C*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 1200*B*a*b^2*tan(1/2*d*x
+ 1/2*c)^4 + 400*C*b^3*tan(1/2*d*x + 1/2*c)^4 - 120*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 360*B*a^2*b*tan(1/2*d*x +
1/2*c)^3 + 288*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 96*B*b^3*tan(1/2*d*x + 1/2*c)^3 - 36*B*a^3*tan(1/2*d*x + 1/2*c
)^2 + 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 72*B*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*C*a^3*tan(1/2*d*x + 1/2*c) + 2
4*B*a^2*b*tan(1/2*d*x + 1/2*c) + 3*B*a^3)/tan(1/2*d*x + 1/2*c)^4)/d

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maple [A]  time = 0.55, size = 302, normalized size = 1.58 \[ -\frac {a^{3} B \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{3} B \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{3} B \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {C \,a^{3} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {C \cot \left (d x +c \right ) a^{3}}{d}+a^{3} C x +\frac {C \,a^{3} c}{d}-\frac {a^{2} b B \left (\cot ^{3}\left (d x +c \right )\right )}{d}+3 B x \,a^{2} b +\frac {3 B \cot \left (d x +c \right ) a^{2} b}{d}+\frac {3 B \,a^{2} b c}{d}-\frac {3 C \,a^{2} b \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 C \,a^{2} b \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {3 B a \,b^{2} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 B a \,b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-3 a \,b^{2} C x -\frac {3 C \cot \left (d x +c \right ) a \,b^{2}}{d}-\frac {3 C a \,b^{2} c}{d}-B x \,b^{3}-\frac {B \cot \left (d x +c \right ) b^{3}}{d}-\frac {B \,b^{3} c}{d}+\frac {b^{3} C \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.74, size = 215, normalized size = 1.13 \[ \frac {12 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} {\left (d x + c\right )} - 6 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {3 \, B a^{3} - 12 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )^{3} - 6 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \tan \left (d x + c\right )^{2} + 4 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

1/12*(12*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*(d*x + c) - 6*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*log(tan
(d*x + c)^2 + 1) + 12*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*log(tan(d*x + c)) - (3*B*a^3 - 12*(C*a^3 + 3*B*a
^2*b - 3*C*a*b^2 - B*b^3)*tan(d*x + c)^3 - 6*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2)*tan(d*x + c)^2 + 4*(C*a^3 + 3*B*a
^2*b)*tan(d*x + c))/tan(d*x + c)^4)/d

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mupad [B]  time = 8.94, size = 204, normalized size = 1.07 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^3-3\,C\,a^2\,b-3\,B\,a\,b^2+C\,b^3\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {C\,a^3}{3}+B\,b\,a^2\right )+\frac {B\,a^3}{4}+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {B\,a^3}{2}+\frac {3\,C\,a^2\,b}{2}+\frac {3\,B\,a\,b^2}{2}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-C\,a^3-3\,B\,a^2\,b+3\,C\,a\,b^2+B\,b^3\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^6*(B*tan(c + d*x) + C*tan(c + d*x)^2)*(a + b*tan(c + d*x))^3,x)

[Out]

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

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sympy [A]  time = 11.01, size = 398, normalized size = 2.08 \[ \begin {cases} \text {NaN} & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan {\relax (c )}\right )^{3} \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right ) \cot ^{6}{\relax (c )} & \text {for}\: d = 0 \\- \frac {B a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {B a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {B a^{3}}{4 d \tan ^{4}{\left (c + d x \right )}} + 3 B a^{2} b x + \frac {3 B a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {B a^{2} b}{d \tan ^{3}{\left (c + d x \right )}} + \frac {3 B a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 B a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 B a b^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - B b^{3} x - \frac {B b^{3}}{d \tan {\left (c + d x \right )}} + C a^{3} x + \frac {C a^{3}}{d \tan {\left (c + d x \right )}} - \frac {C a^{3}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {3 C a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 C a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 C a^{2} b}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 C a b^{2} x - \frac {3 C a b^{2}}{d \tan {\left (c + d x \right )}} - \frac {C b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(a+b*tan(d*x+c))**3*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

Piecewise((nan, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(a + b*tan(c))**3*(B*tan(c) + C*tan(c
)**2)*cot(c)**6, Eq(d, 0)), (-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*ta
n(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)) + C*a**3*x + C*a**3/(d*tan(c + d*x)) - C*a**3/(3*d*tan(c
+ d*x)**3) + 3*C*a**2*b*log(tan(c + d*x)**2 + 1)/(2*d) - 3*C*a**2*b*log(tan(c + d*x))/d - 3*C*a**2*b/(2*d*tan
(c + d*x)**2) - 3*C*a*b**2*x - 3*C*a*b**2/(d*tan(c + d*x)) - C*b**3*log(tan(c + d*x)**2 + 1)/(2*d) + C*b**3*lo
g(tan(c + d*x))/d, True))

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