\(\int \frac {\cot ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [304]

Optimal result

Integrand size = 34, antiderivative size = 30 \[ \int \frac {\cot ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \cot ^2(c+d x)}{2 d}-\frac {B \log (\sin (c+d x))}{d} \]

[Out]

-1/2*B*cot(d*x+c)^2/d-B*ln(sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {21, 3554, 3556} \[ \int \frac {\cot ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \cot ^2(c+d x)}{2 d}-\frac {B \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

-1/2*(B*Cot[c + d*x]^2)/d - (B*Log[Sin[c + d*x]])/d

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3556

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

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {\cot ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right )}{2 d} \]

[In]

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

[Out]

-1/2*(B*(Cot[c + d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))/d

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {B \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(26\)
default	\(\frac {B \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(26\)
parallelrisch	\(-\frac {B \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec ^{2}\left (d x +c \right )\right )+\cot ^{2}\left (d x +c \right )\right )}{2 d}\)	\(36\)
norman	\(-\frac {B}{2 d \tan \left (d x +c \right )^{2}}-\frac {B \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(46\)
risch	\(i B x +\frac {2 i B c}{d}+\frac {2 B \,{\mathrm e}^{2 i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {B \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(61\)

[In]

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

[Out]

1/d*B*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int \frac {\cot ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {{\left (B \cos \left (2 \, d x + 2 \, c\right ) - B\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, d x + 2 \, c\right ) + \frac {1}{2}\right ) - 2 \, B}{2 \, {\left (d \cos \left (2 \, d x + 2 \, c\right ) - d\right )}} \]

[In]

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

[Out]

-1/2*((B*cos(2*d*x + 2*c) - B)*log(-1/2*cos(2*d*x + 2*c) + 1/2) - 2*B)/(d*cos(2*d*x + 2*c) - d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (26) = 52\).

Time = 0.92 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.53 \[ \int \frac {\cot ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\begin {cases} \tilde {\infty } B x & \text {for}\: c = 0 \wedge d = 0 \\\frac {x \left (B a + B b \tan {\left (c \right )}\right ) \cot ^{3}{\left (c \right )}}{a + b \tan {\left (c \right )}} & \text {for}\: d = 0 \\\tilde {\infty } B x & \text {for}\: c = - d x \\\frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B}{2 d \tan ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*B*x, Eq(c, 0) & Eq(d, 0)), (x*(B*a + B*b*tan(c))*cot(c)**3/(a + b*tan(c)), Eq(d, 0)), (zoo*B*x,
Eq(c, -d*x)), (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 = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {\cot ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {B \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, B \log \left (\tan \left (d x + c\right )\right ) - \frac {B}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(B*log(tan(d*x + c)^2 + 1) - 2*B*log(tan(d*x + c)) - B/tan(d*x + c)^2)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (28) = 56\).

Time = 0.39 (sec) , antiderivative size = 124, normalized size of antiderivative = 4.13 \[ \int \frac {\cot ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {4 \, B \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 8 \, B \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac {{\left (B + \frac {4 \, B {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1} - \frac {B {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{8 \, d} \]

[In]

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

[Out]

-1/8*(4*B*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 8*B*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) +
1) + 1)) - (B + 4*B*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - B*(cos(d*x
+ c) - 1)/(cos(d*x + c) + 1))/d

Mupad [B] (verification not implemented)

Time = 7.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {\cot ^3(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B\,\left ({\mathrm {cot}\left (c+d\,x\right )}^2-\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )+2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\right )}{2\,d} \]

[In]

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

[Out]

-(B*(2*log(tan(c + d*x)) - log(tan(c + d*x)^2 + 1) + cot(c + d*x)^2))/(2*d)