Integrand size = 40, antiderivative size = 72 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\left (\left (a^2 B-b^2 B-2 a b C\right ) x\right )-\frac {a^2 B \cot (c+d x)}{d}-\frac {b^2 C \log (\cos (c+d x))}{d}+\frac {a (2 b B+a C) \log (\sin (c+d x))}{d} \]
-(B*a^2-B*b^2-2*C*a*b)*x-a^2*B*cot(d*x+c)/d-b^2*C*ln(cos(d*x+c))/d+a*(2*B* b+C*a)*ln(sin(d*x+c))/d
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.39 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {-2 a^2 B \cot (c+d x)+i (a+i b)^2 (B+i C) \log (i-\tan (c+d x))+2 a (2 b B+a C) \log (\tan (c+d x))-(a-i b)^2 (i B+C) \log (i+\tan (c+d x))}{2 d} \]
(-2*a^2*B*Cot[c + d*x] + I*(a + I*b)^2*(B + I*C)*Log[I - Tan[c + d*x]] + 2 *a*(2*b*B + a*C)*Log[Tan[c + d*x]] - (a - I*b)^2*(I*B + C)*Log[I + Tan[c + d*x]])/(2*d)
Time = 0.60 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {3042, 4115, 3042, 4087, 3042, 4107, 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.
\(\displaystyle \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan (c+d x)^2\right )}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 4115 |
\(\displaystyle \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (B+C \tan (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^2 (B+C \tan (c+d x))}{\tan (c+d x)^2}dx\) |
\(\Big \downarrow \) 4087 |
\(\displaystyle \int \cot (c+d x) \left (b^2 C \tan ^2(c+d x)-\left (B a^2-2 b C a-b^2 B\right ) \tan (c+d x)+a (2 b B+a C)\right )dx-\frac {a^2 B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {b^2 C \tan (c+d x)^2-\left (B a^2-2 b C a-b^2 B\right ) \tan (c+d x)+a (2 b B+a C)}{\tan (c+d x)}dx-\frac {a^2 B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 4107 |
\(\displaystyle a (a C+2 b B) \int \cot (c+d x)dx+b^2 C \int \tan (c+d x)dx-x \left (a^2 B-2 a b C-b^2 B\right )-\frac {a^2 B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a (a C+2 b B) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx+b^2 C \int \tan (c+d x)dx-x \left (a^2 B-2 a b C-b^2 B\right )-\frac {a^2 B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a (a C+2 b B) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx+b^2 C \int \tan (c+d x)dx-x \left (a^2 B-2 a b C-b^2 B\right )-\frac {a^2 B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -x \left (a^2 B-2 a b C-b^2 B\right )-\frac {a^2 B \cot (c+d x)}{d}+\frac {a (a C+2 b B) \log (-\sin (c+d x))}{d}-\frac {b^2 C \log (\cos (c+d x))}{d}\) |
-((a^2*B - b^2*B - 2*a*b*C)*x) - (a^2*B*Cot[c + d*x])/d - (b^2*C*Log[Cos[c + d*x]])/d + (a*(2*b*B + a*C)*Log[-Sin[c + d*x]])/d
3.1.13.3.1 Defintions of rubi rules used
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_)])^2*((A_.) + (B_.)*tan[(e_.) + (f _.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ (-(B*c - A*d))*(b*c - a*d)^2*((c + d*Tan[e + f*x])^(n + 1)/(f*d^2*(n + 1)*( c^2 + d^2))), x] + Simp[1/(d*(c^2 + d^2)) Int[(c + d*Tan[e + f*x])^(n + 1 )*Simp[B*(b*c - a*d)^2 + A*d*(a^2*c - b^2*c + 2*a*b*d) + d*(B*(a^2*c - b^2* c + 2*a*b*d) + A*(2*a*b*c - a^2*d + b^2*d))*Tan[e + f*x] + b^2*B*(c^2 + d^2 )*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b *c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]
Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2 )/tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[B*x, x] + (Simp[A Int[1/Tan[ e + f*x], x], x] + Simp[C Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A, B, C}, x] && NeQ[A, C]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.35 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.17
method | result | size |
derivativedivides | \(\frac {B \,b^{2} \left (d x +c \right )-C \,b^{2} \ln \left (\cos \left (d x +c \right )\right )+2 B a b \ln \left (\sin \left (d x +c \right )\right )+2 C a b \left (d x +c \right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+C \,a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}\) | \(84\) |
default | \(\frac {B \,b^{2} \left (d x +c \right )-C \,b^{2} \ln \left (\cos \left (d x +c \right )\right )+2 B a b \ln \left (\sin \left (d x +c \right )\right )+2 C a b \left (d x +c \right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+C \,a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}\) | \(84\) |
parallelrisch | \(\frac {\left (-2 B a b -C \,a^{2}+C \,b^{2}\right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (4 B a b +2 C \,a^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-2 B \,a^{2} \cot \left (d x +c \right )-2 d x \left (B \,a^{2}-B \,b^{2}-2 C a b \right )}{2 d}\) | \(87\) |
norman | \(\frac {\left (-B \,a^{2}+B \,b^{2}+2 C a b \right ) x \tan \left (d x +c \right )^{2}-\frac {B \,a^{2} \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}+\frac {a \left (2 B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (2 B a b +C \,a^{2}-C \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(107\) |
risch | \(-B \,a^{2} x +B \,b^{2} x +2 C a b x -\frac {2 i C \,a^{2} c}{d}+i C \,b^{2} x -i C \,a^{2} x -\frac {2 i B \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-2 i B a b x -\frac {4 i B a b c}{d}+\frac {2 i C \,b^{2} c}{d}+\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) C \,b^{2}}{d}\) | \(160\) |
1/d*(B*b^2*(d*x+c)-C*b^2*ln(cos(d*x+c))+2*B*a*b*ln(sin(d*x+c))+2*C*a*b*(d* x+c)+B*a^2*(-cot(d*x+c)-d*x-c)+C*a^2*ln(sin(d*x+c)))
Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.56 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {C b^{2} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + 2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} d x \tan \left (d x + c\right ) + 2 \, B a^{2} - {\left (C a^{2} + 2 \, B a b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )} \]
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")
-1/2*(C*b^2*log(1/(tan(d*x + c)^2 + 1))*tan(d*x + c) + 2*(B*a^2 - 2*C*a*b - B*b^2)*d*x*tan(d*x + c) + 2*B*a^2 - (C*a^2 + 2*B*a*b)*log(tan(d*x + c)^2 /(tan(d*x + c)^2 + 1))*tan(d*x + c))/(d*tan(d*x + c))
Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (66) = 132\).
Time = 1.01 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.19 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\- B a^{2} x - \frac {B a^{2}}{d \tan {\left (c + d x \right )}} - \frac {B a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 B a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + B b^{2} x - \frac {C a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 2 C a b x + \frac {C b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {otherwise} \end {cases} \]
Piecewise((nan, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))**2*(B*tan(c) + C*t an(c)**2)*cot(c)**3, Eq(d, 0)), (nan, Eq(c, -d*x)), (-B*a**2*x - B*a**2/(d *tan(c + d*x)) - B*a*b*log(tan(c + d*x)**2 + 1)/d + 2*B*a*b*log(tan(c + d* x))/d + B*b**2*x - C*a**2*log(tan(c + d*x)**2 + 1)/(2*d) + C*a**2*log(tan( c + d*x))/d + 2*C*a*b*x + C*b**2*log(tan(c + d*x)**2 + 1)/(2*d), True))
Time = 0.38 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.29 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )} + {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (C a^{2} + 2 \, B a b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {2 \, B a^{2}}{\tan \left (d x + c\right )}}{2 \, d} \]
integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")
-1/2*(2*(B*a^2 - 2*C*a*b - B*b^2)*(d*x + c) + (C*a^2 + 2*B*a*b - C*b^2)*lo g(tan(d*x + c)^2 + 1) - 2*(C*a^2 + 2*B*a*b)*log(tan(d*x + c)) + 2*B*a^2/ta n(d*x + c))/d
Time = 1.67 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.64 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )} + {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (C a^{2} + 2 \, B a b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + \frac {2 \, {\left (C a^{2} \tan \left (d x + c\right ) + 2 \, B a b \tan \left (d x + c\right ) + B a^{2}\right )}}{\tan \left (d x + c\right )}}{2 \, d} \]
-1/2*(2*(B*a^2 - 2*C*a*b - B*b^2)*(d*x + c) + (C*a^2 + 2*B*a*b - C*b^2)*lo g(tan(d*x + c)^2 + 1) - 2*(C*a^2 + 2*B*a*b)*log(abs(tan(d*x + c))) + 2*(C* a^2*tan(d*x + c) + 2*B*a*b*tan(d*x + c) + B*a^2)/tan(d*x + c))/d
Time = 8.56 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.39 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (C\,a^2+2\,B\,b\,a\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (C+B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {B\,a^2\,\mathrm {cot}\left (c+d\,x\right )}{d} \]