Integrand size = 38, antiderivative size = 43 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-((a B-b C) x)-\frac {a B \cot (c+d x)}{d}+\frac {(b B+a C) \log (\sin (c+d x))}{d} \]
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.05 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=b C x-\frac {a B \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )}{d}+\frac {b B \log (\cos (c+d x))}{d}+\frac {a C \log (\cos (c+d x))}{d}+\frac {b B \log (\tan (c+d x))}{d}+\frac {a C \log (\tan (c+d x))}{d} \]
b*C*x - (a*B*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] )/d + (b*B*Log[Cos[c + d*x]])/d + (a*C*Log[Cos[c + d*x]])/d + (b*B*Log[Tan [c + d*x]])/d + (a*C*Log[Tan[c + d*x]])/d
Time = 0.49 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {3042, 4115, 3042, 4074, 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.
\(\displaystyle \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan (c+d x)^2\right )}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 4115 |
\(\displaystyle \int \cot ^2(c+d x) (a+b \tan (c+d x)) (B+C \tan (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x)) (B+C \tan (c+d x))}{\tan (c+d x)^2}dx\) |
\(\Big \downarrow \) 4074 |
\(\displaystyle \int \cot (c+d x) (b B+a C-(a B-b C) \tan (c+d x))dx-\frac {a B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {b B+a C-(a B-b C) \tan (c+d x)}{\tan (c+d x)}dx-\frac {a B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle (a C+b B) \int \cot (c+d x)dx-(x (a B-b C))-\frac {a B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a C+b B) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-(x (a B-b C))-\frac {a B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -(a C+b B) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-(x (a B-b C))-\frac {a B \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {(a C+b B) \log (-\sin (c+d x))}{d}-(x (a B-b C))-\frac {a B \cot (c+d x)}{d}\) |
3.1.5.3.1 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[((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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b *c - a*d)*(A*b - a*B)*((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*A*c + b*B*c + A*b*d - a*B*d - (A*b*c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[m , -1] && NeQ[a^2 + b^2, 0]
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 = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.23
method | result | size |
derivativedivides | \(\frac {B b \ln \left (\sin \left (d x +c \right )\right )+C b \left (d x +c \right )+B a \left (-\cot \left (d x +c \right )-d x -c \right )+C a \ln \left (\sin \left (d x +c \right )\right )}{d}\) | \(53\) |
default | \(\frac {B b \ln \left (\sin \left (d x +c \right )\right )+C b \left (d x +c \right )+B a \left (-\cot \left (d x +c \right )-d x -c \right )+C a \ln \left (\sin \left (d x +c \right )\right )}{d}\) | \(53\) |
parallelrisch | \(\frac {\left (-B b -C a \right ) \ln \left (\sec \left (d x +c \right )^{2}\right )+\left (2 B b +2 C a \right ) \ln \left (\tan \left (d x +c \right )\right )-2 B \cot \left (d x +c \right ) a -2 d x \left (B a -C b \right )}{2 d}\) | \(65\) |
norman | \(\frac {\left (-B a +C b \right ) x \tan \left (d x +c \right )^{2}-\frac {B a \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}+\frac {\left (B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (B b +C a \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(84\) |
risch | \(-i B b x -i C a x -B a x +C b x -\frac {2 i B b c}{d}-\frac {2 i C a c}{d}-\frac {2 i B a}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C a}{d}\) | \(98\) |
Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.70 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (B a - C b\right )} d x \tan \left (d x + c\right ) - {\left (C a + B b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + 2 \, B a}{2 \, d \tan \left (d x + c\right )} \]
-1/2*(2*(B*a - C*b)*d*x*tan(d*x + c) - (C*a + B*b)*log(tan(d*x + c)^2/(tan (d*x + c)^2 + 1))*tan(d*x + c) + 2*B*a)/(d*tan(d*x + c))
Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (36) = 72\).
Time = 0.72 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.70 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right ) \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\- B a x - \frac {B a}{d \tan {\left (c + d x \right )}} - \frac {B b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {C a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + C b x & \text {otherwise} \end {cases} \]
Piecewise((nan, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))*(B*tan(c) + C*tan( c)**2)*cot(c)**3, Eq(d, 0)), (nan, Eq(c, -d*x)), (-B*a*x - B*a/(d*tan(c + d*x)) - B*b*log(tan(c + d*x)**2 + 1)/(2*d) + B*b*log(tan(c + d*x))/d - C*a *log(tan(c + d*x)**2 + 1)/(2*d) + C*a*log(tan(c + d*x))/d + C*b*x, True))
Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.58 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (B a - C b\right )} {\left (d x + c\right )} + {\left (C a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (C a + B b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {2 \, B a}{\tan \left (d x + c\right )}}{2 \, d} \]
-1/2*(2*(B*a - C*b)*(d*x + c) + (C*a + B*b)*log(tan(d*x + c)^2 + 1) - 2*(C *a + B*b)*log(tan(d*x + c)) + 2*B*a/tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (43) = 86\).
Time = 1.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.77 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, {\left (B a - C b\right )} {\left (d x + c\right )} - 2 \, {\left (C a + B b\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 2 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]
1/2*(B*a*tan(1/2*d*x + 1/2*c) - 2*(B*a - C*b)*(d*x + c) - 2*(C*a + B*b)*lo g(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) )) - (2*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c) + B*a)/tan(1 /2*d*x + 1/2*c))/d
Time = 8.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.02 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,b+C\,a\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )}{2\,d}-\frac {B\,a\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d} \]