3.434 \(\int \cot ^6(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=120 \[ \frac {\left (a^2-b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {\left (a^2-b^2\right ) \cot (c+d x)}{d}-x \left (a^2-b^2\right )-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a b \cot ^4(c+d x)}{2 d}+\frac {a b \cot ^2(c+d x)}{d}+\frac {2 a b \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3542, 3529, 3531, 3475} \[ \frac {\left (a^2-b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {\left (a^2-b^2\right ) \cot (c+d x)}{d}-x \left (a^2-b^2\right )-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a b \cot ^4(c+d x)}{2 d}+\frac {a b \cot ^2(c+d x)}{d}+\frac {2 a b \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2 - b^2)*x) - ((a^2 - b^2)*Cot[c + d*x])/d + (a*b*Cot[c + d*x]^2)/d + ((a^2 - b^2)*Cot[c + d*x]^3)/(3*d)
- (a*b*Cot[c + d*x]^4)/(2*d) - (a^2*Cot[c + d*x]^5)/(5*d) + (2*a*b*Log[Sin[c + d*x]])/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 3542

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

Rubi steps

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

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Mathematica [C]  time = 1.03, size = 121, normalized size = 1.01 \[ -\frac {a^2 \cot ^5(c+d x) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\tan ^2(c+d x)\right )}{5 d}+\frac {a b \left (-\cot ^4(c+d x)+2 \cot ^2(c+d x)+4 \log (\tan (c+d x))+4 \log (\cos (c+d x))\right )}{2 d}-\frac {b^2 \cot ^3(c+d x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\tan ^2(c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(a^2*Cot[c + d*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[c + d*x]^2])/d - (b^2*Cot[c + d*x]^3*Hypergeome
tric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/(3*d) + (a*b*(2*Cot[c + d*x]^2 - Cot[c + d*x]^4 + 4*Log[Cos[c + d*x]]
+ 4*Log[Tan[c + d*x]]))/(2*d)

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fricas [A]  time = 0.52, size = 141, normalized size = 1.18 \[ \frac {30 \, a b \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{5} - 15 \, {\left (2 \, {\left (a^{2} - b^{2}\right )} d x - 3 \, a b\right )} \tan \left (d x + c\right )^{5} + 30 \, a b \tan \left (d x + c\right )^{3} - 30 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{4} - 15 \, a b \tan \left (d x + c\right ) + 10 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} - 6 \, a^{2}}{30 \, d \tan \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(30*a*b*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^5 - 15*(2*(a^2 - b^2)*d*x - 3*a*b)*tan(d*x
+ c)^5 + 30*a*b*tan(d*x + c)^3 - 30*(a^2 - b^2)*tan(d*x + c)^4 - 15*a*b*tan(d*x + c) + 10*(a^2 - b^2)*tan(d*x
+ c)^2 - 6*a^2)/(d*tan(d*x + c)^5)

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giac [B]  time = 3.41, size = 287, normalized size = 2.39 \[ \frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 35 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 180 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 960 \, a b \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 960 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 330 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 300 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 480 \, {\left (a^{2} - b^{2}\right )} {\left (d x + c\right )} - \frac {2192 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 330 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 300 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 180 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/480*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 15*a*b*tan(1/2*d*x + 1/2*c)^4 - 35*a^2*tan(1/2*d*x + 1/2*c)^3 + 20*b^2*t
an(1/2*d*x + 1/2*c)^3 + 180*a*b*tan(1/2*d*x + 1/2*c)^2 - 960*a*b*log(tan(1/2*d*x + 1/2*c)^2 + 1) + 960*a*b*log
(abs(tan(1/2*d*x + 1/2*c))) + 330*a^2*tan(1/2*d*x + 1/2*c) - 300*b^2*tan(1/2*d*x + 1/2*c) - 480*(a^2 - b^2)*(d
*x + c) - (2192*a*b*tan(1/2*d*x + 1/2*c)^5 + 330*a^2*tan(1/2*d*x + 1/2*c)^4 - 300*b^2*tan(1/2*d*x + 1/2*c)^4 -
180*a*b*tan(1/2*d*x + 1/2*c)^3 - 35*a^2*tan(1/2*d*x + 1/2*c)^2 + 20*b^2*tan(1/2*d*x + 1/2*c)^2 + 15*a*b*tan(1
/2*d*x + 1/2*c) + 3*a^2)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 0.33, size = 148, normalized size = 1.23 \[ -\frac {a^{2} \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a^{2} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} \cot \left (d x +c \right )}{d}-a^{2} x -\frac {a^{2} c}{d}-\frac {a b \left (\cot ^{4}\left (d x +c \right )\right )}{2 d}+\frac {a b \left (\cot ^{2}\left (d x +c \right )\right )}{d}+\frac {2 a b \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {b^{2} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {\cot \left (d x +c \right ) b^{2}}{d}+b^{2} x +\frac {c \,b^{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.82, size = 124, normalized size = 1.03 \[ -\frac {30 \, a b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 \, a b \log \left (\tan \left (d x + c\right )\right ) + 30 \, {\left (a^{2} - b^{2}\right )} {\left (d x + c\right )} - \frac {30 \, a b \tan \left (d x + c\right )^{3} - 30 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{4} - 15 \, a b \tan \left (d x + c\right ) + 10 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} - 6 \, a^{2}}{\tan \left (d x + c\right )^{5}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(30*a*b*log(tan(d*x + c)^2 + 1) - 60*a*b*log(tan(d*x + c)) + 30*(a^2 - b^2)*(d*x + c) - (30*a*b*tan(d*x
+ c)^3 - 30*(a^2 - b^2)*tan(d*x + c)^4 - 15*a*b*tan(d*x + c) + 10*(a^2 - b^2)*tan(d*x + c)^2 - 6*a^2)/tan(d*x
+ c)^5)/d

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mupad [B]  time = 3.97, size = 145, normalized size = 1.21 \[ \frac {2\,a\,b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^5\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2-b^2\right )+\frac {a^2}{5}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^2}{3}-\frac {b^2}{3}\right )+\frac {a\,b\,\mathrm {tan}\left (c+d\,x\right )}{2}-a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.86, size = 172, normalized size = 1.43 \[ \begin {cases} \tilde {\infty } a^{2} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan {\relax (c )}\right )^{2} \cot ^{6}{\relax (c )} & \text {for}\: d = 0 \\- a^{2} x - \frac {a^{2}}{d \tan {\left (c + d x \right )}} + \frac {a^{2}}{3 d \tan ^{3}{\left (c + d x \right )}} - \frac {a^{2}}{5 d \tan ^{5}{\left (c + d x \right )}} - \frac {a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {a b}{d \tan ^{2}{\left (c + d x \right )}} - \frac {a b}{2 d \tan ^{4}{\left (c + d x \right )}} + b^{2} x + \frac {b^{2}}{d \tan {\left (c + d x \right )}} - \frac {b^{2}}{3 d \tan ^{3}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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