3.465 \(\int \frac {\cot ^3(c+d x)}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=107 \[ \frac {b x}{a^2+b^2}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\left (a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {b^4 \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 d \left (a^2+b^2\right )}-\frac {\cot ^2(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3569, 3649, 3652, 3530, 3475} \[ -\frac {\left (a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {b^4 \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 d \left (a^2+b^2\right )}+\frac {b x}{a^2+b^2}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\cot ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x)/(a^2 + b^2) + (b*Cot[c + d*x])/(a^2*d) - Cot[c + d*x]^2/(2*a*d) - ((a^2 - b^2)*Log[Sin[c + d*x]])/(a^3*d
) - (b^4*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/(a^3*(a^2 + b^2)*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 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3652

Int[((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Simp[((a*(A*c - c*C) - b*(A*d - C*d))*x)/((a^2 + b^2)*(c^2 + d^2)), x] + (Dist[
(A*b^2 + a^2*C)/((b*c - a*d)*(a^2 + b^2)), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] - Dist[(c^2*C
+ A*d^2)/((b*c - a*d)*(c^2 + d^2)), Int[(d - c*Tan[e + f*x])/(c + d*Tan[e + f*x]), x], x]) /; FreeQ[{a, b, c,
d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rubi steps

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

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Mathematica [C]  time = 0.68, size = 106, normalized size = 0.99 \[ -\frac {-\frac {2 b \cot (c+d x)}{a^2}+\frac {2 b^4 \log (a \cot (c+d x)+b)}{a^3 \left (a^2+b^2\right )}-\frac {\log (-\cot (c+d x)+i)}{a-i b}-\frac {\log (\cot (c+d x)+i)}{a+i b}+\frac {\cot ^2(c+d x)}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 180, normalized size = 1.68 \[ -\frac {b^{4} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + a^{4} + a^{2} b^{2} + {\left (a^{4} - b^{4}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} - {\left (2 \, a^{3} b d x - a^{4} - a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{2 \, {\left (a^{5} + a^{3} b^{2}\right )} d \tan \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.32, size = 155, normalized size = 1.45 \[ -\frac {\frac {2 \, b^{5} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{5} b + a^{3} b^{3}} - \frac {2 \, {\left (d x + c\right )} b}{a^{2} + b^{2}} - \frac {a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {2 \, {\left (a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {3 \, a^{2} \tan \left (d x + c\right )^{2} - 3 \, b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) - a^{2}}{a^{3} \tan \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.52, size = 144, normalized size = 1.35 \[ -\frac {b^{4} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right ) a^{3}}-\frac {1}{2 d a \tan \left (d x +c \right )^{2}}-\frac {\ln \left (\tan \left (d x +c \right )\right )}{d a}+\frac {\ln \left (\tan \left (d x +c \right )\right ) b^{2}}{d \,a^{3}}+\frac {b}{d \,a^{2} \tan \left (d x +c \right )}+\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \left (a^{2}+b^{2}\right )}+\frac {b \arctan \left (\tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.65, size = 122, normalized size = 1.14 \[ -\frac {\frac {2 \, b^{4} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{5} + a^{3} b^{2}} - \frac {2 \, {\left (d x + c\right )} b}{a^{2} + b^{2}} - \frac {a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {2 \, {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{3}} - \frac {2 \, b \tan \left (d x + c\right ) - a}{a^{2} \tan \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.20, size = 137, normalized size = 1.28 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {1}{2\,a}-\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{a^2}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{a^3\,d}-\frac {b^4\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,\left (a^5+a^3\,b^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.16, size = 1352, normalized size = 12.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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