3.22 \(\int \cot ^5(c+d x) (a+b \tan (c+d x))^3 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=154 \[ \frac{a \left (3 a^2 B-9 a b C-8 b^2 B\right ) \cot (c+d x)}{3 d}-\frac{\left (3 a^2 b B+a^3 C-3 a b^2 C-b^3 B\right ) \log (\sin (c+d x))}{d}+x \left (-3 a^2 b C+a^3 B-3 a b^2 B+b^3 C\right )-\frac{a^2 (3 a C+5 b B) \cot ^2(c+d x)}{6 d}-\frac{a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d} \]

[Out]

(a^3*B - 3*a*b^2*B - 3*a^2*b*C + b^3*C)*x + (a*(3*a^2*B - 8*b^2*B - 9*a*b*C)*Cot[c + d*x])/(3*d) - (a^2*(5*b*B
+ 3*a*C)*Cot[c + d*x]^2)/(6*d) - ((3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*Log[Sin[c + d*x]])/d - (a*B*Cot[c +
d*x]^3*(a + b*Tan[c + d*x])^2)/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.426631, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3632, 3605, 3635, 3628, 3531, 3475} \[ \frac{a \left (3 a^2 B-9 a b C-8 b^2 B\right ) \cot (c+d x)}{3 d}-\frac{\left (3 a^2 b B+a^3 C-3 a b^2 C-b^3 B\right ) \log (\sin (c+d x))}{d}+x \left (-3 a^2 b C+a^3 B-3 a b^2 B+b^3 C\right )-\frac{a^2 (3 a C+5 b B) \cot ^2(c+d x)}{6 d}-\frac{a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

(a^3*B - 3*a*b^2*B - 3*a^2*b*C + b^3*C)*x + (a*(3*a^2*B - 8*b^2*B - 9*a*b*C)*Cot[c + d*x])/(3*d) - (a^2*(5*b*B
+ 3*a*C)*Cot[c + d*x]^2)/(6*d) - ((3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*Log[Sin[c + d*x]])/d - (a*B*Cot[c +
d*x]^3*(a + b*Tan[c + d*x])^2)/(3*d)

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 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 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]

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 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 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 \cot ^5(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 ^4(c+d x) (a+b \tan (c+d x))^3 (B+C \tan (c+d x)) \, dx\\ &=-\frac{a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}+\frac{1}{3} \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (a (5 b B+3 a C)-3 \left (a^2 B-b^2 B-2 a b C\right ) \tan (c+d x)-b (a B-3 b C) \tan ^2(c+d x)\right ) \, dx\\ &=-\frac{a^2 (5 b B+3 a C) \cot ^2(c+d x)}{6 d}-\frac{a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}+\frac{1}{3} \int \cot ^2(c+d x) \left (-a \left (3 a^2 B-8 b^2 B-9 a b C\right )-3 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \tan (c+d x)-b^2 (a B-3 b C) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (3 a^2 B-8 b^2 B-9 a b C\right ) \cot (c+d x)}{3 d}-\frac{a^2 (5 b B+3 a C) \cot ^2(c+d x)}{6 d}-\frac{a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}+\frac{1}{3} \int \cot (c+d x) \left (-3 \left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right )+3 \left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) \tan (c+d x)\right ) \, dx\\ &=\left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) x+\frac{a \left (3 a^2 B-8 b^2 B-9 a b C\right ) \cot (c+d x)}{3 d}-\frac{a^2 (5 b B+3 a C) \cot ^2(c+d x)}{6 d}-\frac{a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}+\left (-3 a^2 b B+b^3 B-a^3 C+3 a b^2 C\right ) \int \cot (c+d x) \, dx\\ &=\left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) x+\frac{a \left (3 a^2 B-8 b^2 B-9 a b C\right ) \cot (c+d x)}{3 d}-\frac{a^2 (5 b B+3 a C) \cot ^2(c+d x)}{6 d}-\frac{\left (3 a^2 b B-b^3 B+a^3 C-3 a b^2 C\right ) \log (\sin (c+d x))}{d}-\frac{a B \cot ^3(c+d x) (a+b \tan (c+d x))^2}{3 d}\\ \end{align*}

Mathematica [C]  time = 1.27275, size = 164, normalized size = 1.06 \[ \frac{6 a \left (a^2 B-3 a b C-3 b^2 B\right ) \cot (c+d x)-6 \left (3 a^2 b B+a^3 C-3 a b^2 C-b^3 B\right ) \log (\tan (c+d x))-3 a^2 (a C+3 b B) \cot ^2(c+d x)-2 a^3 B \cot ^3(c+d x)+3 (a-i b)^3 (C+i B) \log (\tan (c+d x)+i)+3 (a+i b)^3 (C-i B) \log (-\tan (c+d x)+i)}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.087, size = 233, normalized size = 1.5 \begin{align*}{\frac{B{b}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+C{b}^{3}x+{\frac{C{b}^{3}c}{d}}-3\,Ba{b}^{2}x-3\,{\frac{B\cot \left ( dx+c \right ) a{b}^{2}}{d}}-3\,{\frac{Ba{b}^{2}c}{d}}+3\,{\frac{Ca{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{3\,B{a}^{2}b \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-3\,{\frac{B{a}^{2}b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-3\,Cx{a}^{2}b-3\,{\frac{C\cot \left ( dx+c \right ){a}^{2}b}{d}}-3\,{\frac{C{a}^{2}bc}{d}}-{\frac{B{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{B\cot \left ( dx+c \right ){a}^{3}}{d}}+B{a}^{3}x+{\frac{B{a}^{3}c}{d}}-{\frac{C{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{C{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.69671, size = 243, normalized size = 1.58 \begin{align*} \frac{6 \,{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )}{\left (d x + c\right )} + 3 \,{\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \,{\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac{2 \, B a^{3} - 6 \,{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \,{\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.09676, size = 419, normalized size = 2.72 \begin{align*} -\frac{3 \,{\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B 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 )^{3} + 2 \, B a^{3} + 3 \,{\left (C a^{3} + 3 \, B a^{2} b - 2 \,{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} d x\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} + 3 \,{\left (C a^{3} + 3 \, B a^{2} b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 142.243, size = 330, normalized size = 2.14 \begin{align*} \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{\left (c \right )}\right )^{3} \left (B \tan{\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{5}{\left (c \right )} & \text{for}\: d = 0 \\B a^{3} x + \frac{B a^{3}}{d \tan{\left (c + d x \right )}} - \frac{B a^{3}}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac{3 B a^{2} b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac{3 B a^{2} b \log{\left (\tan{\left (c + d x \right )} \right )}}{d} - \frac{3 B a^{2} b}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 B a b^{2} x - \frac{3 B a b^{2}}{d \tan{\left (c + d x \right )}} - \frac{B b^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{B b^{3} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + \frac{C a^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac{C a^{3} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} - \frac{C a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 C a^{2} b x - \frac{3 C a^{2} b}{d \tan{\left (c + d x \right )}} - \frac{3 C a b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{3 C a b^{2} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + C b^{3} x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(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)**5, Eq(d, 0)), (B*a**3*x + B*a**3/(d*tan(c + d*x)) - B*a**3/(3*d*tan(c + d*x)**3) + 3*B*a**2*b*lo
g(tan(c + d*x)**2 + 1)/(2*d) - 3*B*a**2*b*log(tan(c + d*x))/d - 3*B*a**2*b/(2*d*tan(c + d*x)**2) - 3*B*a*b**2*
x - 3*B*a*b**2/(d*tan(c + d*x)) - B*b**3*log(tan(c + d*x)**2 + 1)/(2*d) + B*b**3*log(tan(c + d*x))/d + C*a**3*
log(tan(c + d*x)**2 + 1)/(2*d) - C*a**3*log(tan(c + d*x))/d - C*a**3/(2*d*tan(c + d*x)**2) - 3*C*a**2*b*x - 3*
C*a**2*b/(d*tan(c + d*x)) - 3*C*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) + 3*C*a*b**2*log(tan(c + d*x))/d + C*b**
3*x, True))

________________________________________________________________________________________

Giac [B]  time = 2.65894, size = 527, normalized size = 3.42 \begin{align*} \frac{B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, B a^{2} b \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 ) + 36 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 36 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \,{\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )}{\left (d x + c\right )} + 24 \,{\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 24 \,{\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{44 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 132 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 132 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 44 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 36 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 36 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(B*a^3*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^3*tan(1/2*d*x + 1/2*c)^2 - 9*B*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 15*B*
a^3*tan(1/2*d*x + 1/2*c) + 36*C*a^2*b*tan(1/2*d*x + 1/2*c) + 36*B*a*b^2*tan(1/2*d*x + 1/2*c) + 24*(B*a^3 - 3*C
*a^2*b - 3*B*a*b^2 + C*b^3)*(d*x + c) + 24*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*log(tan(1/2*d*x + 1/2*c)^2
+ 1) - 24*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*log(abs(tan(1/2*d*x + 1/2*c))) + (44*C*a^3*tan(1/2*d*x + 1/2
*c)^3 + 132*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 132*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 44*B*b^3*tan(1/2*d*x + 1/2*c
)^3 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^2
- 3*C*a^3*tan(1/2*d*x + 1/2*c) - 9*B*a^2*b*tan(1/2*d*x + 1/2*c) - B*a^3)/tan(1/2*d*x + 1/2*c)^3)/d