Integrand size = 21, antiderivative size = 49 \[ \int \frac {\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {x}{a}-\frac {\text {arctanh}(\sin (c+d x))}{2 a d}-\frac {(2-\sec (c+d x)) \tan (c+d x)}{2 a d} \] Output:
x/a-1/2*arctanh(sin(d*x+c))/a/d-1/2*(2-sec(d*x+c))*tan(d*x+c)/a/d
Leaf count is larger than twice the leaf count of optimal. \(241\) vs. \(2(49)=98\).
Time = 1.09 (sec) , antiderivative size = 241, normalized size of antiderivative = 4.92 \[ \int \frac {\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (4 x+\frac {2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {1}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 \sin (d x)}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{2 a (1+\sec (c+d x))} \] Input:
Integrate[Tan[c + d*x]^4/(a + a*Sec[c + d*x]),x]
Output:
(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(4*x + (2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d - (2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + 1/(d*(Cos[ (c + d*x)/2] - Sin[(c + d*x)/2])^2) - 1/(d*(Cos[(c + d*x)/2] + Sin[(c + d* x)/2])^2) - (4*Sin[d*x])/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(C os[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) ))/(2*a*(1 + Sec[c + d*x]))
Time = 0.34 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4376, 25, 3042, 4369, 2009}
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.
\(\displaystyle \int \frac {\tan ^4(c+d x)}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^4}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 4376 |
\(\displaystyle \frac {\int -\left ((a-a \sec (c+d x)) \tan ^2(c+d x)\right )dx}{a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int (a-a \sec (c+d x)) \tan ^2(c+d x)dx}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \cot \left (c+d x+\frac {\pi }{2}\right )^2 \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}\) |
\(\Big \downarrow \) 4369 |
\(\displaystyle -\frac {\frac {\tan (c+d x) (2 a-a \sec (c+d x))}{2 d}-\frac {1}{2} \int (2 a-a \sec (c+d x))dx}{a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{2} \left (\frac {a \text {arctanh}(\sin (c+d x))}{d}-2 a x\right )+\frac {\tan (c+d x) (2 a-a \sec (c+d x))}{2 d}}{a^2}\) |
Input:
Int[Tan[c + d*x]^4/(a + a*Sec[c + d*x]),x]
Output:
-(((-2*a*x + (a*ArcTanh[Sin[c + d*x]])/d)/2 + ((2*a - a*Sec[c + d*x])*Tan[ c + d*x])/(2*d))/a^2)
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-e)*(e*Cot[c + d*x])^(m - 1)*((a*m + b*(m - 1)*Csc [c + d*x])/(d*m*(m - 1))), x] - Simp[e^2/m Int[(e*Cot[c + d*x])^(m - 2)*( a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m , 1]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Simp[a^(2*n)/e^(2*n) Int[(e*Cot[c + d*x])^(m + 2* n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[a ^2 - b^2, 0] && ILtQ[n, 0]
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.12
method | result | size |
risch | \(\frac {x}{a}-\frac {i \left ({\mathrm e}^{3 i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}+2\right )}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}\) | \(104\) |
derivativedivides | \(\frac {-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\) | \(110\) |
default | \(\frac {-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\) | \(110\) |
Input:
int(tan(d*x+c)^4/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
x/a-I*(exp(3*I*(d*x+c))+2*exp(2*I*(d*x+c))-exp(I*(d*x+c))+2)/d/a/(exp(2*I* (d*x+c))+1)^2-1/2/a/d*ln(exp(I*(d*x+c))+I)+1/2/a/d*ln(exp(I*(d*x+c))-I)
Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.76 \[ \int \frac {\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {4 \, d x \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{4 \, a d \cos \left (d x + c\right )^{2}} \] Input:
integrate(tan(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
1/4*(4*d*x*cos(d*x + c)^2 - cos(d*x + c)^2*log(sin(d*x + c) + 1) + cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(2*cos(d*x + c) - 1)*sin(d*x + c))/(a*d *cos(d*x + c)^2)
\[ \int \frac {\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\tan ^{4}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(tan(d*x+c)**4/(a+a*sec(d*x+c)),x)
Output:
Integral(tan(c + d*x)**4/(sec(c + d*x) + 1), x)/a
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (43) = 86\).
Time = 0.12 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.33 \[ \int \frac {\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {4 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a}}{2 \, d} \] Input:
integrate(tan(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
-1/2*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(c os(d*x + c) + 1)^4) - 4*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + log(si n(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a)/d
Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (43) = 86\).
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.96 \[ \int \frac {\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {2 \, {\left (d x + c\right )}}{a} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \] Input:
integrate(tan(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
1/2*(2*(d*x + c)/a - log(abs(tan(1/2*d*x + 1/2*c) + 1))/a + log(abs(tan(1/ 2*d*x + 1/2*c) - 1))/a + 2*(3*tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c ))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a))/d
Time = 12.48 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.69 \[ \int \frac {\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {x}{a}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )} \] Input:
int(tan(c + d*x)^4/(a + a/cos(c + d*x)),x)
Output:
x/a - atanh(tan(c/2 + (d*x)/2))/(a*d) - (tan(c/2 + (d*x)/2) - 3*tan(c/2 + (d*x)/2)^3)/(d*(a - 2*a*tan(c/2 + (d*x)/2)^2 + a*tan(c/2 + (d*x)/2)^4))
Time = 0.17 (sec) , antiderivative size = 253, normalized size of antiderivative = 5.16 \[ \int \frac {\tan ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{3}-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x -3 \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \cos \left (d x +c \right ) \tan \left (d x +c \right )^{3}+6 \cos \left (d x +c \right ) \tan \left (d x +c \right )-6 \cos \left (d x +c \right ) d x +2 \sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right ) a d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(tan(d*x+c)^4/(a+a*sec(d*x+c)),x)
Output:
(3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 3*cos(c + d*x) *log(tan((c + d*x)/2) - 1) - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin( c + d*x)**2 + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1) + 2*cos(c + d*x)*si n(c + d*x)**2*tan(c + d*x)**3 - 6*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x ) + 6*cos(c + d*x)*sin(c + d*x)**2*d*x - 3*cos(c + d*x)*sin(c + d*x) - 2*c os(c + d*x)*tan(c + d*x)**3 + 6*cos(c + d*x)*tan(c + d*x) - 6*cos(c + d*x) *d*x + 2*sin(c + d*x)**3)/(6*cos(c + d*x)*a*d*(sin(c + d*x)**2 - 1))