Integrand size = 19, antiderivative size = 60 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=a x+\frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \log (\sin (c+d x))}{d} \] Output:
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
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {b \csc ^2(c+d x)}{2 d}-\frac {a \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )}{3 d}-\frac {b \log (\sin (c+d x))}{d} \] Input:
Integrate[Cot[c + d*x]^4*(a + b*Tan[c + d*x]),x]
Output:
-1/2*(b*Csc[c + d*x]^2)/d - (a*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, - 1/2, -Tan[c + d*x]^2])/(3*d) - (b*Log[Sin[c + d*x]])/d
Time = 0.54 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 4012, 3042, 4012, 25, 3042, 4012, 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.
\(\displaystyle \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \tan (c+d x)}{\tan (c+d x)^4}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \int \cot ^3(c+d x) (b-a \tan (c+d x))dx-\frac {a \cot ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {b-a \tan (c+d x)}{\tan (c+d x)^3}dx-\frac {a \cot ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \int -\cot ^2(c+d x) (a+b \tan (c+d x))dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot ^2(c+d x) (a+b \tan (c+d x))dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \frac {a+b \tan (c+d x)}{\tan (c+d x)^2}dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -\int \cot (c+d x) (b-a \tan (c+d x))dx-\frac {a \cot ^3(c+d x)}{3 d}+\frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \frac {b-a \tan (c+d x)}{\tan (c+d x)}dx-\frac {a \cot ^3(c+d x)}{3 d}+\frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle -b \int \cot (c+d x)dx-\frac {a \cot ^3(c+d x)}{3 d}+\frac {a \cot (c+d x)}{d}+a x-\frac {b \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -b \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a \cot ^3(c+d x)}{3 d}+\frac {a \cot (c+d x)}{d}+a x-\frac {b \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {a \cot ^3(c+d x)}{3 d}+\frac {a \cot (c+d x)}{d}+a x-\frac {b \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {a \cot ^3(c+d x)}{3 d}+\frac {a \cot (c+d x)}{d}+a x-\frac {b \cot ^2(c+d x)}{2 d}-\frac {b \log (-\sin (c+d x))}{d}\) |
Input:
Int[Cot[c + d*x]^4*(a + b*Tan[c + d*x]),x]
Output:
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
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]
Time = 0.86 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.12
method | result | size |
parallelrisch | \(\frac {-3 b \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )-2 \cot \left (d x +c \right )^{3} a -3 \cot \left (d x +c \right )^{2} b +6 a x d +6 \cot \left (d x +c \right ) a}{6 d}\) | \(67\) |
derivativedivides | \(\frac {\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )-\frac {a}{3 \tan \left (d x +c \right )^{3}}-\frac {b}{2 \tan \left (d x +c \right )^{2}}+\frac {a}{\tan \left (d x +c \right )}-b \ln \left (\tan \left (d x +c \right )\right )}{d}\) | \(71\) |
default | \(\frac {\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )-\frac {a}{3 \tan \left (d x +c \right )^{3}}-\frac {b}{2 \tan \left (d x +c \right )^{2}}+\frac {a}{\tan \left (d x +c \right )}-b \ln \left (\tan \left (d x +c \right )\right )}{d}\) | \(71\) |
norman | \(\frac {a x \tan \left (d x +c \right )^{3}+\frac {a \tan \left (d x +c \right )^{2}}{d}-\frac {a}{3 d}-\frac {b \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {b \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(84\) |
risch | \(i b x +a x +\frac {2 i b c}{d}+\frac {4 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+2 b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+\frac {8 i a}{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(109\) |
Input:
int(cot(d*x+c)^4*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/6*(-3*b*(2*ln(tan(d*x+c))-ln(sec(d*x+c)^2))-2*cot(d*x+c)^3*a-3*cot(d*x+c )^2*b+6*a*x*d+6*cot(d*x+c)*a)/d
Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.48 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {3 \, b \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} - 3 \, {\left (2 \, a d x - b\right )} \tan \left (d x + c\right )^{3} - 6 \, a \tan \left (d x + c\right )^{2} + 3 \, b \tan \left (d x + c\right ) + 2 \, a}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:
integrate(cot(d*x+c)^4*(a+b*tan(d*x+c)),x, algorithm="fricas")
Output:
-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)
Time = 0.63 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.57 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } a x & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right ) \cot ^{4}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } a x & \text {for}\: c = - d x \\a x + \frac {a}{d \tan {\left (c + d x \right )}} - \frac {a}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {b}{2 d \tan ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(cot(d*x+c)**4*(a+b*tan(d*x+c)),x)
Output:
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))
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {6 \, {\left (d x + c\right )} a + 3 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, b \log \left (\tan \left (d x + c\right )\right ) + \frac {6 \, a \tan \left (d x + c\right )^{2} - 3 \, b \tan \left (d x + c\right ) - 2 \, a}{\tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(cot(d*x+c)^4*(a+b*tan(d*x+c)),x, algorithm="maxima")
Output:
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
Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.32 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {{\left (d x + c\right )} a}{d} + \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} - \frac {b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} + \frac {6 \, a \tan \left (d x + c\right )^{2} - 3 \, b \tan \left (d x + c\right ) - 2 \, a}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:
integrate(cot(d*x+c)^4*(a+b*tan(d*x+c)),x, algorithm="giac")
Output:
(d*x + c)*a/d + 1/2*b*log(tan(d*x + c)^2 + 1)/d - b*log(abs(tan(d*x + c))) /d + 1/6*(6*a*tan(d*x + c)^2 - 3*b*tan(d*x + c) - 2*a)/(d*tan(d*x + c)^3)
Time = 1.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.60 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-\frac {b}{2}+\frac {a\,1{}\mathrm {i}}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (\frac {b}{2}+\frac {a\,1{}\mathrm {i}}{2}\right )}{d}-\frac {b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (-a\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{2}+\frac {a}{3}\right )}{d} \] Input:
int(cot(c + d*x)^4*(a + b*tan(c + d*x)),x)
Output:
(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
Time = 0.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.98 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -4 \cos \left (d x +c \right ) a +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} b +12 \sin \left (d x +c \right )^{3} a d x +3 \sin \left (d x +c \right )^{3} b -6 \sin \left (d x +c \right ) b}{12 \sin \left (d x +c \right )^{3} d} \] Input:
int(cot(d*x+c)^4*(a+b*tan(d*x+c)),x)
Output:
(16*cos(c + d*x)*sin(c + d*x)**2*a - 4*cos(c + d*x)*a + 12*log(tan((c + d* x)/2)**2 + 1)*sin(c + d*x)**3*b - 12*log(tan((c + d*x)/2))*sin(c + d*x)**3 *b + 12*sin(c + d*x)**3*a*d*x + 3*sin(c + d*x)**3*b - 6*sin(c + d*x)*b)/(1 2*sin(c + d*x)**3*d)