\(\int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx\) [421]

Optimal result

Integrand size = 19, antiderivative size = 60 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=a x+\frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \log (\sin (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3610, 3612, 3556} \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {a \cot ^3(c+d x)}{3 d}+\frac {a \cot (c+d x)}{d}+a x-\frac {b \cot ^2(c+d x)}{2 d}-\frac {b \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

a*x + (a*Cot[c + d*x])/d - (b*Cot[c + d*x]^2)/(2*d) - (a*Cot[c + d*x]^3)/(3*d) - (b*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 ^3(c+d x)}{3 d}+\int \cot ^3(c+d x) (b-a \tan (c+d x)) \, dx \\ & = -\frac {b \cot ^2(c+d x)}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}+\int \cot ^2(c+d x) (-a-b \tan (c+d x)) \, dx \\ & = \frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}+\int \cot (c+d x) (-b+a \tan (c+d x)) \, dx \\ & = a x+\frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}-b \int \cot (c+d x) \, dx \\ & = a x+\frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \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 = 0.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {a \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )}{3 d}-\frac {b \left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right )}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\(\frac {-2 \left (\cot ^{3}\left (d x +c \right )\right ) a -3 \left (\cot ^{2}\left (d x +c \right )\right ) b +6 a d x +6 \cot \left (d x +c \right ) a -6 b \ln \left (\tan \left (d x +c \right )\right )+3 b \ln \left (\sec ^{2}\left (d x +c \right )\right )}{6 d}\)	\(65\)
derivativedivides	\(\frac {-\frac {a}{3 \tan \left (d x +c \right )^{3}}-\frac {b}{2 \tan \left (d x +c \right )^{2}}+\frac {a}{\tan \left (d x +c \right )}-b \ln \left (\tan \left (d x +c \right )\right )+\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(71\)
default	\(\frac {-\frac {a}{3 \tan \left (d x +c \right )^{3}}-\frac {b}{2 \tan \left (d x +c \right )^{2}}+\frac {a}{\tan \left (d x +c \right )}-b \ln \left (\tan \left (d x +c \right )\right )+\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(71\)
norman	\(\frac {a x \left (\tan ^{3}\left (d x +c \right )\right )+\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a}{3 d}-\frac {b \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {b \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(84\)
risch	\(i b x +a x +\frac {2 i b c}{d}+\frac {4 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+2 b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+\frac {8 i a}{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(109\)

[In]

int(cot(d*x+c)^4*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.48 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {3 \, b \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} - 3 \, {\left (2 \, a d x - b\right )} \tan \left (d x + c\right )^{3} - 6 \, a \tan \left (d x + c\right )^{2} + 3 \, b \tan \left (d x + c\right ) + 2 \, a}{6 \, d \tan \left (d x + c\right )^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.57 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } a x & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right ) \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } a x & \text {for}\: c = - d x \\a x + \frac {a}{d \tan {\left (c + d x \right )}} - \frac {a}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {b}{2 d \tan ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {6 \, {\left (d x + c\right )} a + 3 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, b \log \left (\tan \left (d x + c\right )\right ) + \frac {6 \, a \tan \left (d x + c\right )^{2} - 3 \, b \tan \left (d x + c\right ) - 2 \, a}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (56) = 112\).

Time = 0.53 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.33 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, {\left (d x + c\right )} a + 24 \, b \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 24 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {44 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

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

[Out]

1/24*(a*tan(1/2*d*x + 1/2*c)^3 - 3*b*tan(1/2*d*x + 1/2*c)^2 + 24*(d*x + c)*a + 24*b*log(tan(1/2*d*x + 1/2*c)^2
+ 1) - 24*b*log(abs(tan(1/2*d*x + 1/2*c))) - 15*a*tan(1/2*d*x + 1/2*c) + (44*b*tan(1/2*d*x + 1/2*c)^3 + 15*a*
tan(1/2*d*x + 1/2*c)^2 - 3*b*tan(1/2*d*x + 1/2*c) - a)/tan(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 4.77 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.60 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-\frac {b}{2}+\frac {a\,1{}\mathrm {i}}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (\frac {b}{2}+\frac {a\,1{}\mathrm {i}}{2}\right )}{d}-\frac {b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (-a\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{2}+\frac {a}{3}\right )}{d} \]

[In]

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

[Out]

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