Integrand size = 21, antiderivative size = 58 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=-2 a b x-\frac {2 a b \cot (c+d x)}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {\left (a^2-b^2\right ) \log (\sin (c+d x))}{d} \] Output:
-2*a*b*x-2*a*b*cot(d*x+c)/d-1/2*a^2*cot(d*x+c)^2/d-(a^2-b^2)*ln(sin(d*x+c) )/d
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.59 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {-4 a b \cot (c+d x)-a^2 \cot ^2(c+d x)+(a+i b)^2 \log (i-\tan (c+d x))-2 (a-b) (a+b) \log (\tan (c+d x))+(a-i b)^2 \log (i+\tan (c+d x))}{2 d} \] Input:
Integrate[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]
Output:
(-4*a*b*Cot[c + d*x] - a^2*Cot[c + d*x]^2 + (a + I*b)^2*Log[I - Tan[c + d* x]] - 2*(a - b)*(a + b)*Log[Tan[c + d*x]] + (a - I*b)^2*Log[I + Tan[c + d* x]])/(2*d)
Time = 0.47 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4025, 3042, 4012, 25, 3042, 4014, 3042, 25, 3956}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^2}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 4025 |
\(\displaystyle \int \cot ^2(c+d x) \left (2 a b-\left (a^2-b^2\right ) \tan (c+d x)\right )dx-\frac {a^2 \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {2 a b-\left (a^2-b^2\right ) \tan (c+d x)}{\tan (c+d x)^2}dx-\frac {a^2 \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \int -\cot (c+d x) \left (a^2+2 b \tan (c+d x) a-b^2\right )dx-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot (c+d x) \left (a^2+2 b \tan (c+d x) a-b^2\right )dx-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \frac {a^2+2 b \tan (c+d x) a-b^2}{\tan (c+d x)}dx-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle -\left (a^2-b^2\right ) \int \cot (c+d x)dx-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot (c+d x)}{d}-2 a b x\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\left (a^2-b^2\right ) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot (c+d x)}{d}-2 a b x\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \left (a^2-b^2\right ) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot (c+d x)}{d}-2 a b x\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {\left (a^2-b^2\right ) \log (-\sin (c+d x))}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot (c+d x)}{d}-2 a b x\) |
Input:
Int[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]
Output:
-2*a*b*x - (2*a*b*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^2)/(2*d) - ((a^2 - b ^2)*Log[-Sin[c + d*x]])/d
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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] + Simp[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 ]
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] + Simp[(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] && N eQ[a*c + b*d, 0]
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] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[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)*Ta n[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]
Time = 1.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(\frac {b^{2} \ln \left (\sin \left (d x +c \right )\right )+2 a b \left (-\cot \left (d x +c \right )-d x -c \right )+a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(61\) |
default | \(\frac {b^{2} \ln \left (\sin \left (d x +c \right )\right )+2 a b \left (-\cot \left (d x +c \right )-d x -c \right )+a^{2} \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(61\) |
parallelrisch | \(-\frac {a^{2} \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )-b^{2} \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )+4 a b d x +\cot \left (d x +c \right )^{2} a^{2}+4 \cot \left (d x +c \right ) a b}{2 d}\) | \(86\) |
norman | \(\frac {-\frac {a^{2}}{2 d}-2 a b x \tan \left (d x +c \right )^{2}-\frac {2 a b \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(92\) |
risch | \(-2 a b x +i a^{2} x -i b^{2} x +\frac {2 i a^{2} c}{d}-\frac {2 i b^{2} c}{d}+\frac {2 a \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{d}\) | \(128\) |
Input:
int(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(b^2*ln(sin(d*x+c))+2*a*b*(-cot(d*x+c)-d*x-c)+a^2*(-1/2*cot(d*x+c)^2-l n(sin(d*x+c))))
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.48 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {{\left (a^{2} - b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + 4 \, a b \tan \left (d x + c\right ) + {\left (4 \, a b d x + a^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2}}{2 \, d \tan \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
-1/2*((a^2 - b^2)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 + 4*a*b*tan(d*x + c) + (4*a*b*d*x + a^2)*tan(d*x + c)^2 + a^2)/(d*tan(d*x + c)^2)
Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (53) = 106\).
Time = 0.54 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.22 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\begin {cases} \tilde {\infty } a^{2} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } a^{2} x & \text {for}\: c = - d x \\\frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - 2 a b x - \frac {2 a b}{d \tan {\left (c + d x \right )}} - \frac {b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} & \text {otherwise} \end {cases} \] Input:
integrate(cot(d*x+c)**3*(a+b*tan(d*x+c))**2,x)
Output:
Piecewise((zoo*a**2*x, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))**2*cot(c)** 3, Eq(d, 0)), (zoo*a**2*x, Eq(c, -d*x)), (a**2*log(tan(c + d*x)**2 + 1)/(2 *d) - a**2*log(tan(c + d*x))/d - a**2/(2*d*tan(c + d*x)**2) - 2*a*b*x - 2* a*b/(d*tan(c + d*x)) - b**2*log(tan(c + d*x)**2 + 1)/(2*d) + b**2*log(tan( c + d*x))/d, True))
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.34 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {4 \, {\left (d x + c\right )} a b - {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {4 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
-1/2*(4*(d*x + c)*a*b - (a^2 - b^2)*log(tan(d*x + c)^2 + 1) + 2*(a^2 - b^2 )*log(tan(d*x + c)) + (4*a*b*tan(d*x + c) + a^2)/tan(d*x + c)^2)/d
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.50 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (d x + c\right )} a b}{d} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} - \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {4 \, a b \tan \left (d x + c\right ) + a^{2}}{2 \, d \tan \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
-2*(d*x + c)*a*b/d + 1/2*(a^2 - b^2)*log(tan(d*x + c)^2 + 1)/d - (a^2 - b^ 2)*log(abs(tan(d*x + c)))/d - 1/2*(4*a*b*tan(d*x + c) + a^2)/(d*tan(d*x + c)^2)
Time = 1.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.67 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {a^2}{2}+2\,b\,\mathrm {tan}\left (c+d\,x\right )\,a\right )}{d} \] Input:
int(cot(c + d*x)^3*(a + b*tan(c + d*x))^2,x)
Output:
(log(tan(c + d*x) - 1i)*(a + b*1i)^2)/(2*d) - (log(tan(c + d*x) + 1i)*(a*1 i + b)^2)/(2*d) - (log(tan(c + d*x))*(a^2 - b^2))/d - (cot(c + d*x)^2*(a^2 /2 + 2*a*b*tan(c + d*x)))/d
Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.78 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {-8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} b^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{2}+\sin \left (d x +c \right )^{2} a^{2}-8 \sin \left (d x +c \right )^{2} a b d x -2 a^{2}}{4 \sin \left (d x +c \right )^{2} d} \] Input:
int(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x)
Output:
( - 8*cos(c + d*x)*sin(c + d*x)*a*b + 4*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**2 - 4*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*b**2 - 4* log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2 + 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**2 + sin(c + d*x)**2*a**2 - 8*sin(c + d*x)**2*a*b*d*x - 2*a** 2)/(4*sin(c + d*x)**2*d)