3.443 \(\int \cot ^5(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=130 \[ \frac {a \left (a^2-3 b^2\right ) \cot ^2(c+d x)}{2 d}+\frac {b \left (3 a^2-b^2\right ) \cot (c+d x)}{d}+\frac {a \left (a^2-3 b^2\right ) \log (\sin (c+d x))}{d}+b x \left (3 a^2-b^2\right )-\frac {3 a^2 b \cot ^3(c+d x)}{4 d}-\frac {a^2 \cot ^4(c+d x) (a+b \tan (c+d x))}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3565, 3628, 3529, 3531, 3475} \[ \frac {a \left (a^2-3 b^2\right ) \cot ^2(c+d x)}{2 d}+\frac {b \left (3 a^2-b^2\right ) \cot (c+d x)}{d}+\frac {a \left (a^2-3 b^2\right ) \log (\sin (c+d x))}{d}+b x \left (3 a^2-b^2\right )-\frac {3 a^2 b \cot ^3(c+d x)}{4 d}-\frac {a^2 \cot ^4(c+d x) (a+b \tan (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [C]  time = 1.60, size = 118, normalized size = 0.91 \[ -\frac {a^3 \cot ^4(c+d x)-2 a \left (a^2-3 b^2\right ) \cot ^2(c+d x)+4 b \left (b^2-3 a^2\right ) \cot (c+d x)+4 a^2 b \cot ^3(c+d x)+2 (a-i b)^3 \log (-\cot (c+d x)+i)+2 (a+i b)^3 \log (\cot (c+d x)+i)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 152, normalized size = 1.17 \[ \frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + {\left (3 \, a^{3} - 6 \, a b^{2} + 4 \, {\left (3 \, a^{2} b - b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{4} - 4 \, a^{2} b \tan \left (d x + c\right ) + 4 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{3} - a^{3} + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{4 \, d \tan \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 7.08, size = 301, normalized size = 2.32 \[ -\frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 360 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 192 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )} + 192 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 192 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {400 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1200 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 360 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 96 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(3*a^3*tan(1/2*d*x + 1/2*c)^4 - 24*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*a^3*tan(1/2*d*x + 1/2*c)^2 + 72*a*
b^2*tan(1/2*d*x + 1/2*c)^2 + 360*a^2*b*tan(1/2*d*x + 1/2*c) - 96*b^3*tan(1/2*d*x + 1/2*c) - 192*(3*a^2*b - b^3
)*(d*x + c) + 192*(a^3 - 3*a*b^2)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 192*(a^3 - 3*a*b^2)*log(abs(tan(1/2*d*x +
1/2*c))) + (400*a^3*tan(1/2*d*x + 1/2*c)^4 - 1200*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 360*a^2*b*tan(1/2*d*x + 1/2*c
)^3 + 96*b^3*tan(1/2*d*x + 1/2*c)^3 - 36*a^3*tan(1/2*d*x + 1/2*c)^2 + 72*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 24*a^2
*b*tan(1/2*d*x + 1/2*c) + 3*a^3)/tan(1/2*d*x + 1/2*c)^4)/d

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maple [A]  time = 0.36, size = 159, normalized size = 1.22 \[ -\frac {a^{3} \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a^{2} b \left (\cot ^{3}\left (d x +c \right )\right )}{d}+3 a^{2} b x +\frac {3 a^{2} b \cot \left (d x +c \right )}{d}+\frac {3 a^{2} b c}{d}-\frac {3 b^{2} a \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 b^{2} a \ln \left (\sin \left (d x +c \right )\right )}{d}-b^{3} x -\frac {\cot \left (d x +c \right ) b^{3}}{d}-\frac {c \,b^{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.58, size = 135, normalized size = 1.04 \[ \frac {4 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )} - 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) - \frac {4 \, a^{2} b \tan \left (d x + c\right ) - 4 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{3} + a^{3} - 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{4}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.93, size = 145, normalized size = 1.12 \[ -\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (3\,a\,b^2-a^3\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3}{2\,d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {3\,a\,b^2}{2}-\frac {a^3}{2}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,a^2\,b-b^3\right )+\frac {a^3}{4}+a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.69, size = 207, normalized size = 1.59 \[ \begin {cases} \tilde {\infty } a^{3} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan {\relax (c )}\right )^{3} \cot ^{5}{\relax (c )} & \text {for}\: d = 0 \\- \frac {a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {a^{3}}{4 d \tan ^{4}{\left (c + d x \right )}} + 3 a^{2} b x + \frac {3 a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {a^{2} b}{d \tan ^{3}{\left (c + d x \right )}} + \frac {3 a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {3 a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {3 a b^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - b^{3} x - \frac {b^{3}}{d \tan {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+b*tan(d*x+c))**3,x)

[Out]

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

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