3.454 \(\int \cot ^6(c+d x) (a+b \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=170 \[ -\frac {3 a^3 b \cot ^4(c+d x)}{5 d}+\frac {a^2 \left (5 a^2-27 b^2\right ) \cot ^3(c+d x)}{15 d}+\frac {2 a b \left (a^2-b^2\right ) \cot ^2(c+d x)}{d}+\frac {4 a b \left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac {a^2 \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}-\frac {\left (a^4-6 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-x \left (a^4-6 a^2 b^2+b^4\right ) \]

[Out]

-(a^4-6*a^2*b^2+b^4)*x-(a^4-6*a^2*b^2+b^4)*cot(d*x+c)/d+2*a*b*(a^2-b^2)*cot(d*x+c)^2/d+1/15*a^2*(5*a^2-27*b^2)
*cot(d*x+c)^3/d-3/5*a^3*b*cot(d*x+c)^4/d+4*a*b*(a^2-b^2)*ln(sin(d*x+c))/d-1/5*a^2*cot(d*x+c)^5*(a+b*tan(d*x+c)
)^2/d

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Rubi [A]  time = 0.38, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3565, 3635, 3628, 3529, 3531, 3475} \[ \frac {a^2 \left (5 a^2-27 b^2\right ) \cot ^3(c+d x)}{15 d}+\frac {2 a b \left (a^2-b^2\right ) \cot ^2(c+d x)}{d}-\frac {\left (-6 a^2 b^2+a^4+b^4\right ) \cot (c+d x)}{d}+\frac {4 a b \left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-x \left (-6 a^2 b^2+a^4+b^4\right )-\frac {3 a^3 b \cot ^4(c+d x)}{5 d}-\frac {a^2 \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + b*Tan[c + d*x])^4,x]

[Out]

-((a^4 - 6*a^2*b^2 + b^4)*x) - ((a^4 - 6*a^2*b^2 + b^4)*Cot[c + d*x])/d + (2*a*b*(a^2 - b^2)*Cot[c + d*x]^2)/d
+ (a^2*(5*a^2 - 27*b^2)*Cot[c + d*x]^3)/(15*d) - (3*a^3*b*Cot[c + d*x]^4)/(5*d) + (4*a*b*(a^2 - b^2)*Log[Sin[
c + d*x]])/d - (a^2*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 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 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))^4 \, dx &=-\frac {a^2 \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 (12 a^2 b-5 a \left (a^2-3 b^2\right ) \tan (c+d x)-b \left (3 a^2-5 b^2\right ) \tan ^2(c+d x)\right ) \, dx\\ &=-\frac {3 a^3 b \cot ^4(c+d x)}{5 d}-\frac {a^2 \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^2 \left (5 a^2-27 b^2\right )-20 a b \left (a^2-b^2\right ) \tan (c+d x)-b^2 \left (3 a^2-5 b^2\right ) \tan ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \left (5 a^2-27 b^2\right ) \cot ^3(c+d x)}{15 d}-\frac {3 a^3 b \cot ^4(c+d x)}{5 d}-\frac {a^2 \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \cot ^3(c+d x) \left (-20 a b \left (a^2-b^2\right )+5 \left (a^4-6 a^2 b^2+b^4\right ) \tan (c+d x)\right ) \, dx\\ &=\frac {2 a b \left (a^2-b^2\right ) \cot ^2(c+d x)}{d}+\frac {a^2 \left (5 a^2-27 b^2\right ) \cot ^3(c+d x)}{15 d}-\frac {3 a^3 b \cot ^4(c+d x)}{5 d}-\frac {a^2 \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^4-6 a^2 b^2+b^4\right )+20 a b \left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=-\frac {\left (a^4-6 a^2 b^2+b^4\right ) \cot (c+d x)}{d}+\frac {2 a b \left (a^2-b^2\right ) \cot ^2(c+d x)}{d}+\frac {a^2 \left (5 a^2-27 b^2\right ) \cot ^3(c+d x)}{15 d}-\frac {3 a^3 b \cot ^4(c+d x)}{5 d}-\frac {a^2 \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \cot (c+d x) \left (20 a b \left (a^2-b^2\right )-5 \left (a^4-6 a^2 b^2+b^4\right ) \tan (c+d x)\right ) \, dx\\ &=-\left (a^4-6 a^2 b^2+b^4\right ) x-\frac {\left (a^4-6 a^2 b^2+b^4\right ) \cot (c+d x)}{d}+\frac {2 a b \left (a^2-b^2\right ) \cot ^2(c+d x)}{d}+\frac {a^2 \left (5 a^2-27 b^2\right ) \cot ^3(c+d x)}{15 d}-\frac {3 a^3 b \cot ^4(c+d x)}{5 d}-\frac {a^2 \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\left (4 a b \left (a^2-b^2\right )\right ) \int \cot (c+d x) \, dx\\ &=-\left (a^4-6 a^2 b^2+b^4\right ) x-\frac {\left (a^4-6 a^2 b^2+b^4\right ) \cot (c+d x)}{d}+\frac {2 a b \left (a^2-b^2\right ) \cot ^2(c+d x)}{d}+\frac {a^2 \left (5 a^2-27 b^2\right ) \cot ^3(c+d x)}{15 d}-\frac {3 a^3 b \cot ^4(c+d x)}{5 d}+\frac {4 a b \left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac {a^2 \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\\ \end {align*}

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Mathematica [C]  time = 0.42, size = 154, normalized size = 0.91 \[ -\frac {\frac {1}{5} a^4 \cot ^5(c+d x)+a^3 b \cot ^4(c+d x)-\frac {1}{3} a^2 \left (a^2-6 b^2\right ) \cot ^3(c+d x)+\left (a^4-6 a^2 b^2+b^4\right ) \cot (c+d x)-2 a b (a-b) (a+b) \cot ^2(c+d x)+\frac {1}{2} i (a-i b)^4 \log (-\cot (c+d x)+i)-\frac {1}{2} i (a+i b)^4 \log (\cot (c+d x)+i)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + b*Tan[c + d*x])^4,x]

[Out]

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

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fricas [A]  time = 0.48, size = 186, normalized size = 1.09 \[ \frac {30 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{5} + 15 \, {\left (3 \, a^{3} b - 2 \, a b^{3} - {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} d x\right )} \tan \left (d x + c\right )^{5} - 15 \, a^{3} b \tan \left (d x + c\right ) - 15 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{4} - 3 \, a^{4} + 30 \, {\left (a^{3} b - a b^{3}\right )} \tan \left (d x + c\right )^{3} + 5 \, {\left (a^{4} - 6 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{15 \, d \tan \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/15*(30*(a^3*b - a*b^3)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^5 + 15*(3*a^3*b - 2*a*b^3 - (a^
4 - 6*a^2*b^2 + b^4)*d*x)*tan(d*x + c)^5 - 15*a^3*b*tan(d*x + c) - 15*(a^4 - 6*a^2*b^2 + b^4)*tan(d*x + c)^4 -
3*a^4 + 30*(a^3*b - a*b^3)*tan(d*x + c)^3 + 5*(a^4 - 6*a^2*b^2)*tan(d*x + c)^2)/(d*tan(d*x + c)^5)

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giac [B]  time = 15.11, size = 416, normalized size = 2.45 \[ \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 35 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 240 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 330 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1800 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 480 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )} - 1920 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 1920 \, {\left (a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {4384 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4384 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 330 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1800 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 360 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 240 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 30 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{4}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/480*(3*a^4*tan(1/2*d*x + 1/2*c)^5 - 30*a^3*b*tan(1/2*d*x + 1/2*c)^4 - 35*a^4*tan(1/2*d*x + 1/2*c)^3 + 120*a^
2*b^2*tan(1/2*d*x + 1/2*c)^3 + 360*a^3*b*tan(1/2*d*x + 1/2*c)^2 - 240*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 330*a^4*t
an(1/2*d*x + 1/2*c) - 1800*a^2*b^2*tan(1/2*d*x + 1/2*c) + 240*b^4*tan(1/2*d*x + 1/2*c) - 480*(a^4 - 6*a^2*b^2
+ b^4)*(d*x + c) - 1920*(a^3*b - a*b^3)*log(tan(1/2*d*x + 1/2*c)^2 + 1) + 1920*(a^3*b - a*b^3)*log(abs(tan(1/2
*d*x + 1/2*c))) - (4384*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 4384*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 330*a^4*tan(1/2*d*x
+ 1/2*c)^4 - 1800*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 + 240*b^4*tan(1/2*d*x + 1/2*c)^4 - 360*a^3*b*tan(1/2*d*x + 1
/2*c)^3 + 240*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 35*a^4*tan(1/2*d*x + 1/2*c)^2 + 120*a^2*b^2*tan(1/2*d*x + 1/2*c)^
2 + 30*a^3*b*tan(1/2*d*x + 1/2*c) + 3*a^4)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 0.38, size = 232, normalized size = 1.36 \[ -\frac {a^{4} \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a^{4} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{4} \cot \left (d x +c \right )}{d}-a^{4} x -\frac {a^{4} c}{d}-\frac {a^{3} b \left (\cot ^{4}\left (d x +c \right )\right )}{d}+\frac {2 a^{3} b \left (\cot ^{2}\left (d x +c \right )\right )}{d}+\frac {4 a^{3} b \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {2 a^{2} b^{2} \left (\cot ^{3}\left (d x +c \right )\right )}{d}+6 a^{2} b^{2} x +\frac {6 \cot \left (d x +c \right ) a^{2} b^{2}}{d}+\frac {6 a^{2} b^{2} c}{d}-\frac {2 a \,b^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{d}-\frac {4 a \,b^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-b^{4} x -\frac {\cot \left (d x +c \right ) b^{4}}{d}-\frac {b^{4} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+b*tan(d*x+c))^4,x)

[Out]

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

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maxima [A]  time = 0.83, size = 170, normalized size = 1.00 \[ -\frac {15 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )} + 30 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 \, {\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {15 \, a^{3} b \tan \left (d x + c\right ) + 15 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{4} + 3 \, a^{4} - 30 \, {\left (a^{3} b - a b^{3}\right )} \tan \left (d x + c\right )^{3} - 5 \, {\left (a^{4} - 6 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/15*(15*(a^4 - 6*a^2*b^2 + b^4)*(d*x + c) + 30*(a^3*b - a*b^3)*log(tan(d*x + c)^2 + 1) - 60*(a^3*b - a*b^3)*
log(tan(d*x + c)) + (15*a^3*b*tan(d*x + c) + 15*(a^4 - 6*a^2*b^2 + b^4)*tan(d*x + c)^4 + 3*a^4 - 30*(a^3*b - a
*b^3)*tan(d*x + c)^3 - 5*(a^4 - 6*a^2*b^2)*tan(d*x + c)^2)/tan(d*x + c)^5)/d

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mupad [B]  time = 4.06, size = 174, normalized size = 1.02 \[ \frac {4\,a\,b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^5\,\left ({\mathrm {tan}\left (c+d\,x\right )}^3\,\left (2\,a\,b^3-2\,a^3\,b\right )-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^4}{3}-2\,a^2\,b^2\right )+\frac {a^4}{5}+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^4-6\,a^2\,b^2+b^4\right )+a^3\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^4\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^4\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^6*(a + b*tan(c + d*x))^4,x)

[Out]

(log(tan(c + d*x) - 1i)*(a*1i - b)^4*1i)/(2*d) - (cot(c + d*x)^5*(tan(c + d*x)^3*(2*a*b^3 - 2*a^3*b) - tan(c +
d*x)^2*(a^4/3 - 2*a^2*b^2) + a^4/5 + tan(c + d*x)^4*(a^4 + b^4 - 6*a^2*b^2) + a^3*b*tan(c + d*x)))/d - (log(t
an(c + d*x) + 1i)*(a - b*1i)^4*1i)/(2*d) + (4*a*b*log(tan(c + d*x))*(a^2 - b^2))/d

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sympy [A]  time = 9.33, size = 265, normalized size = 1.56 \[ \begin {cases} \tilde {\infty } a^{4} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan {\relax (c )}\right )^{4} \cot ^{6}{\relax (c )} & \text {for}\: d = 0 \\- a^{4} x - \frac {a^{4}}{d \tan {\left (c + d x \right )}} + \frac {a^{4}}{3 d \tan ^{3}{\left (c + d x \right )}} - \frac {a^{4}}{5 d \tan ^{5}{\left (c + d x \right )}} - \frac {2 a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {4 a^{3} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {2 a^{3} b}{d \tan ^{2}{\left (c + d x \right )}} - \frac {a^{3} b}{d \tan ^{4}{\left (c + d x \right )}} + 6 a^{2} b^{2} x + \frac {6 a^{2} b^{2}}{d \tan {\left (c + d x \right )}} - \frac {2 a^{2} b^{2}}{d \tan ^{3}{\left (c + d x \right )}} + \frac {2 a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - \frac {4 a b^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {2 a b^{3}}{d \tan ^{2}{\left (c + d x \right )}} - b^{4} x - \frac {b^{4}}{d \tan {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(a+b*tan(d*x+c))**4,x)

[Out]

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

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