Integrand size = 19, antiderivative size = 54 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\log (\cos (c+d x))}{a^2 d}-\frac {\log (a+b \sec (c+d x))}{a^2 d}+\frac {1}{a d (a+b \sec (c+d x))} \] Output:
-ln(cos(d*x+c))/a^2/d-ln(a+b*sec(d*x+c))/a^2/d+1/a/d/(a+b*sec(d*x+c))
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {b+b \log (b+a \cos (c+d x))+a \cos (c+d x) \log (b+a \cos (c+d x))}{a^2 d (b+a \cos (c+d x))} \] Input:
Integrate[Tan[c + d*x]/(a + b*Sec[c + d*x])^2,x]
Output:
-((b + b*Log[b + a*Cos[c + d*x]] + a*Cos[c + d*x]*Log[b + a*Cos[c + d*x]]) /(a^2*d*(b + a*Cos[c + d*x])))
Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 25, 4373, 54, 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.
\(\displaystyle \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle \frac {\int \frac {\cos (c+d x)}{b (a+b \sec (c+d x))^2}d(b \sec (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {\cos (c+d x)}{a^2 b}-\frac {1}{a^2 (a+b \sec (c+d x))}-\frac {1}{a (a+b \sec (c+d x))^2}\right )d(b \sec (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\log (b \sec (c+d x))}{a^2}-\frac {\log (a+b \sec (c+d x))}{a^2}+\frac {1}{a (a+b \sec (c+d x))}}{d}\) |
Input:
Int[Tan[c + d*x]/(a + b*Sec[c + d*x])^2,x]
Output:
(Log[b*Sec[c + d*x]]/a^2 - Log[a + b*Sec[c + d*x]]/a^2 + 1/(a*(a + b*Sec[c + d*x])))/d
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\sec \left (d x +c \right )\right )}{a^{2}}-\frac {\ln \left (a +b \sec \left (d x +c \right )\right )}{a^{2}}+\frac {1}{a \left (a +b \sec \left (d x +c \right )\right )}}{d}\) | \(49\) |
default | \(\frac {\frac {\ln \left (\sec \left (d x +c \right )\right )}{a^{2}}-\frac {\ln \left (a +b \sec \left (d x +c \right )\right )}{a^{2}}+\frac {1}{a \left (a +b \sec \left (d x +c \right )\right )}}{d}\) | \(49\) |
risch | \(\frac {i x}{a^{2}}+\frac {2 i c}{d \,a^{2}}-\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a^{2} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{2} d}\) | \(99\) |
Input:
int(tan(d*x+c)/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/a^2*ln(sec(d*x+c))-1/a^2*ln(a+b*sec(d*x+c))+1/a/(a+b*sec(d*x+c)))
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {{\left (a \cos \left (d x + c\right ) + b\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + b}{a^{3} d \cos \left (d x + c\right ) + a^{2} b d} \] Input:
integrate(tan(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
Output:
-((a*cos(d*x + c) + b)*log(a*cos(d*x + c) + b) + b)/(a^3*d*cos(d*x + c) + a^2*b*d)
\[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\begin {cases} \frac {\tilde {\infty } x \tan {\left (c \right )}}{\sec ^{2}{\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d} & \text {for}\: b = 0 \\- \frac {1}{2 b^{2} d \sec ^{2}{\left (c + d x \right )}} & \text {for}\: a = 0 \\\frac {\int \frac {\tan {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )} - 2 \cos {\left (c + d x \right )} \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} & \text {for}\: b = - a \cos {\left (c + d x \right )} \\\frac {x \tan {\left (c \right )}}{\left (a + b \sec {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\- \frac {2 a \log {\left (\frac {a}{b} + \sec {\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} + \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} + \frac {2 a}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} - \frac {2 b \log {\left (\frac {a}{b} + \sec {\left (c + d x \right )} \right )} \sec {\left (c + d x \right )}}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} + \frac {b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \sec {\left (c + d x \right )}}{2 a^{3} d + 2 a^{2} b d \sec {\left (c + d x \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(tan(d*x+c)/(a+b*sec(d*x+c))**2,x)
Output:
Piecewise((zoo*x*tan(c)/sec(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (log(t an(c + d*x)**2 + 1)/(2*a**2*d), Eq(b, 0)), (-1/(2*b**2*d*sec(c + d*x)**2), Eq(a, 0)), (Integral(tan(c + d*x)/(cos(c + d*x)**2*sec(c + d*x)**2 - 2*co s(c + d*x)*sec(c + d*x) + 1), x)/a**2, Eq(b, -a*cos(c + d*x))), (x*tan(c)/ (a + b*sec(c))**2, Eq(d, 0)), (-2*a*log(a/b + sec(c + d*x))/(2*a**3*d + 2* a**2*b*d*sec(c + d*x)) + a*log(tan(c + d*x)**2 + 1)/(2*a**3*d + 2*a**2*b*d *sec(c + d*x)) + 2*a/(2*a**3*d + 2*a**2*b*d*sec(c + d*x)) - 2*b*log(a/b + sec(c + d*x))*sec(c + d*x)/(2*a**3*d + 2*a**2*b*d*sec(c + d*x)) + b*log(ta n(c + d*x)**2 + 1)*sec(c + d*x)/(2*a**3*d + 2*a**2*b*d*sec(c + d*x)), True ))
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.76 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {b}{a^{3} \cos \left (d x + c\right ) + a^{2} b} + \frac {\log \left (a \cos \left (d x + c\right ) + b\right )}{a^{2}}}{d} \] Input:
integrate(tan(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-(b/(a^3*cos(d*x + c) + a^2*b) + log(a*cos(d*x + c) + b)/a^2)/d
Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{2} d} - \frac {b}{{\left (a \cos \left (d x + c\right ) + b\right )} a^{2} d} \] Input:
integrate(tan(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="giac")
Output:
-log(abs(a*cos(d*x + c) + b))/(a^2*d) - b/((a*cos(d*x + c) + b)*a^2*d)
Time = 10.69 (sec) , antiderivative size = 257, normalized size of antiderivative = 4.76 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {a}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}\right )}{a^2\,d}-\frac {b\,\left (a+a\,\cos \left (c+d\,x\right )-2\,a\,\mathrm {atanh}\left (\frac {a}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}\right )+2\,a\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {a}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}\right )+\frac {2\,\mathrm {atanh}\left (\frac {a}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}-\frac {a\,\cos \left (c+d\,x\right )}{2\,\left (\frac {a}{2}+b+\frac {a\,\cos \left (c+d\,x\right )}{2}\right )}\right )\,\left (a^3\,d-a^3\,d\,\cos \left (c+d\,x\right )\right )}{a^2\,d}\right )}{a^2\,d\,\left (a-b\right )\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \] Input:
int(tan(c + d*x)/(a + b/cos(c + d*x))^2,x)
Output:
(2*atanh(a/(2*(a/2 + b + (a*cos(c + d*x))/2)) - (a*cos(c + d*x))/(2*(a/2 + b + (a*cos(c + d*x))/2))))/(a^2*d) - (b*(a + a*cos(c + d*x) - 2*a*atanh(a /(2*(a/2 + b + (a*cos(c + d*x))/2)) - (a*cos(c + d*x))/(2*(a/2 + b + (a*co s(c + d*x))/2))) + 2*a*cos(c + d*x)*atanh(a/(2*(a/2 + b + (a*cos(c + d*x)) /2)) - (a*cos(c + d*x))/(2*(a/2 + b + (a*cos(c + d*x))/2))) + (2*atanh(a/( 2*(a/2 + b + (a*cos(c + d*x))/2)) - (a*cos(c + d*x))/(2*(a/2 + b + (a*cos( c + d*x))/2)))*(a^3*d - a^3*d*cos(c + d*x)))/(a^2*d)))/(a^2*d*(a - b)*(b + a*cos(c + d*x)))
Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.78 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \sec \left (d x +c \right ) b +\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a -2 \,\mathrm {log}\left (\sec \left (d x +c \right ) b +a \right ) \sec \left (d x +c \right ) b -2 \,\mathrm {log}\left (\sec \left (d x +c \right ) b +a \right ) a -2 \sec \left (d x +c \right ) b}{2 a^{2} d \left (\sec \left (d x +c \right ) b +a \right )} \] Input:
int(tan(d*x+c)/(a+b*sec(d*x+c))^2,x)
Output:
(log(tan(c + d*x)**2 + 1)*sec(c + d*x)*b + log(tan(c + d*x)**2 + 1)*a - 2* log(sec(c + d*x)*b + a)*sec(c + d*x)*b - 2*log(sec(c + d*x)*b + a)*a - 2*s ec(c + d*x)*b)/(2*a**2*d*(sec(c + d*x)*b + a))