\(\int \cot ^5(c+d x) (a+b \tan (c+d x)) \, dx\) [422]

Optimal result

Integrand size = 19, antiderivative size = 75 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \, dx=b x+\frac {b \cot (c+d x)}{d}+\frac {a \cot ^2(c+d x)}{2 d}-\frac {b \cot ^3(c+d x)}{3 d}-\frac {a \cot ^4(c+d x)}{4 d}+\frac {a \log (\sin (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3610, 3612, 3556} \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {a \cot ^4(c+d x)}{4 d}+\frac {a \cot ^2(c+d x)}{2 d}+\frac {a \log (\sin (c+d x))}{d}-\frac {b \cot ^3(c+d x)}{3 d}+\frac {b \cot (c+d x)}{d}+b x \]

[In]

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

[Out]

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

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3610

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 3612

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]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.21 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \cot ^2(c+d x)}{2 d}-\frac {a \cot ^4(c+d x)}{4 d}-\frac {b \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )}{3 d}+\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\tan (c+d x))}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01

method	result	size
parallelrisch	\(\frac {-3 \left (\cot ^{4}\left (d x +c \right )\right ) a -4 \left (\cot ^{3}\left (d x +c \right )\right ) b +6 \left (\cot ^{2}\left (d x +c \right )\right ) a +12 b d x +12 a \ln \left (\tan \left (d x +c \right )\right )-6 a \ln \left (\sec ^{2}\left (d x +c \right )\right )+12 \cot \left (d x +c \right ) b}{12 d}\)	\(76\)
derivativedivides	\(\frac {-\frac {a}{4 \tan \left (d x +c \right )^{4}}-\frac {b}{3 \tan \left (d x +c \right )^{3}}+a \ln \left (\tan \left (d x +c \right )\right )+\frac {a}{2 \tan \left (d x +c \right )^{2}}+\frac {b}{\tan \left (d x +c \right )}-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(81\)
default	\(\frac {-\frac {a}{4 \tan \left (d x +c \right )^{4}}-\frac {b}{3 \tan \left (d x +c \right )^{3}}+a \ln \left (\tan \left (d x +c \right )\right )+\frac {a}{2 \tan \left (d x +c \right )^{2}}+\frac {b}{\tan \left (d x +c \right )}-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+b \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(81\)
norman	\(\frac {b x \left (\tan ^{4}\left (d x +c \right )\right )+\frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{d}-\frac {a}{4 d}+\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b \tan \left (d x +c \right )}{3 d}}{\tan \left (d x +c \right )^{4}}+\frac {a \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(97\)
risch	\(b x -i a x -\frac {2 i a c}{d}+\frac {4 i \left (3 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+3 b \,{\mathrm e}^{6 i \left (d x +c \right )}-3 i a \,{\mathrm e}^{4 i \left (d x +c \right )}-6 b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+5 b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(133\)

[In]

int(cot(d*x+c)^5*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/12*(-3*cot(d*x+c)^4*a-4*cot(d*x+c)^3*b+6*cot(d*x+c)^2*a+12*b*d*x+12*a*ln(tan(d*x+c))-6*a*ln(sec(d*x+c)^2)+12
*cot(d*x+c)*b)/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.33 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {6 \, a \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 3 \, {\left (4 \, b d x + 3 \, a\right )} \tan \left (d x + c\right )^{4} + 12 \, b \tan \left (d x + c\right )^{3} + 6 \, a \tan \left (d x + c\right )^{2} - 4 \, b \tan \left (d x + c\right ) - 3 \, a}{12 \, d \tan \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/12*(6*a*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^4 + 3*(4*b*d*x + 3*a)*tan(d*x + c)^4 + 12*b*ta
n(d*x + c)^3 + 6*a*tan(d*x + c)^2 - 4*b*tan(d*x + c) - 3*a)/(d*tan(d*x + c)^4)

Sympy [A] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.43 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } a x & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right ) \cot ^{5}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } a x & \text {for}\: c = - d x \\- \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {a}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {a}{4 d \tan ^{4}{\left (c + d x \right )}} + b x + \frac {b}{d \tan {\left (c + d x \right )}} - \frac {b}{3 d \tan ^{3}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {12 \, {\left (d x + c\right )} b - 6 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, a \log \left (\tan \left (d x + c\right )\right ) + \frac {12 \, b \tan \left (d x + c\right )^{3} + 6 \, a \tan \left (d x + c\right )^{2} - 4 \, b \tan \left (d x + c\right ) - 3 \, a}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

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

[Out]

1/12*(12*(d*x + c)*b - 6*a*log(tan(d*x + c)^2 + 1) + 12*a*log(tan(d*x + c)) + (12*b*tan(d*x + c)^3 + 6*a*tan(d
*x + c)^2 - 4*b*tan(d*x + c) - 3*a)/tan(d*x + c)^4)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (69) = 138\).

Time = 0.59 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.25 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 192 \, {\left (d x + c\right )} b + 192 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 192 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 120 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {400 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[In]

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

[Out]

-1/192*(3*a*tan(1/2*d*x + 1/2*c)^4 - 8*b*tan(1/2*d*x + 1/2*c)^3 - 36*a*tan(1/2*d*x + 1/2*c)^2 - 192*(d*x + c)*
b + 192*a*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 192*a*log(abs(tan(1/2*d*x + 1/2*c))) + 120*b*tan(1/2*d*x + 1/2*c)
+ (400*a*tan(1/2*d*x + 1/2*c)^4 - 120*b*tan(1/2*d*x + 1/2*c)^3 - 36*a*tan(1/2*d*x + 1/2*c)^2 + 8*b*tan(1/2*d*x
+ 1/2*c) + 3*a)/tan(1/2*d*x + 1/2*c)^4)/d

Mupad [B] (verification not implemented)

Time = 5.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.43 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (\frac {a}{2}-\frac {b\,1{}\mathrm {i}}{2}\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left (-b\,{\mathrm {tan}\left (c+d\,x\right )}^3-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{3}+\frac {a}{4}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (\frac {a}{2}+\frac {b\,1{}\mathrm {i}}{2}\right )}{d} \]

[In]

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

[Out]

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