Integrand size = 21, antiderivative size = 61 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {2 a \log (a+b \tan (c+d x))}{b^3 d}+\frac {\tan (c+d x)}{b^2 d}-\frac {a^2+b^2}{b^3 d (a+b \tan (c+d x))} \] Output:
-2*a*ln(a+b*tan(d*x+c))/b^3/d+tan(d*x+c)/b^2/d-(a^2+b^2)/b^3/d/(a+b*tan(d* x+c))
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {-2 a \log (a+b \tan (c+d x))+b \tan (c+d x)-\frac {a^2+b^2}{a+b \tan (c+d x)}}{b^3 d} \] Input:
Integrate[Sec[c + d*x]^4/(a + b*Tan[c + d*x])^2,x]
Output:
(-2*a*Log[a + b*Tan[c + d*x]] + b*Tan[c + d*x] - (a^2 + b^2)/(a + b*Tan[c + d*x]))/(b^3*d)
Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3987, 27, 476, 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.
\(\displaystyle \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^4}{(a+b \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 3987 |
\(\displaystyle \frac {\int \frac {\tan ^2(c+d x) b^2+b^2}{b^2 (a+b \tan (c+d x))^2}d(b \tan (c+d x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\tan ^2(c+d x) b^2+b^2}{(a+b \tan (c+d x))^2}d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (-\frac {2 a}{a+b \tan (c+d x)}+\frac {a^2+b^2}{(a+b \tan (c+d x))^2}+1\right )d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))+b \tan (c+d x)}{b^3 d}\) |
Input:
Int[Sec[c + d*x]^4/(a + b*Tan[c + d*x])^2,x]
Output:
(-2*a*Log[a + b*Tan[c + d*x]] + b*Tan[c + d*x] - (a^2 + b^2)/(a + b*Tan[c + d*x]))/(b^3*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(b*f) Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 15.57 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (d x +c \right )}{b^{2}}-\frac {a^{2}+b^{2}}{b^{3} \left (a +b \tan \left (d x +c \right )\right )}-\frac {2 a \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(57\) |
default | \(\frac {\frac {\tan \left (d x +c \right )}{b^{2}}-\frac {a^{2}+b^{2}}{b^{3} \left (a +b \tan \left (d x +c \right )\right )}-\frac {2 a \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(57\) |
risch | \(-\frac {4 i \left (-i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b -i a \right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right ) b^{2} d}+\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{3} d}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{3} d}\) | \(136\) |
Input:
int(sec(d*x+c)^4/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/b^2*tan(d*x+c)-1/b^3*(a^2+b^2)/(a+b*tan(d*x+c))-2/b^3*a*ln(a+b*tan( d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (61) = 122\).
Time = 0.10 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.92 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {2 \, b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - b^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} + a b \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + a b \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right )}{a b^{3} d \cos \left (d x + c\right )^{2} + b^{4} d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \] Input:
integrate(sec(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
-(2*b^2*cos(d*x + c)^2 - 2*a*b*cos(d*x + c)*sin(d*x + c) - b^2 + (a^2*cos( d*x + c)^2 + a*b*cos(d*x + c)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - (a^2*cos(d*x + c)^2 + a*b*cos( d*x + c)*sin(d*x + c))*log(cos(d*x + c)^2))/(a*b^3*d*cos(d*x + c)^2 + b^4* d*cos(d*x + c)*sin(d*x + c))
\[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(sec(d*x+c)**4/(a+b*tan(d*x+c))**2,x)
Output:
Integral(sec(c + d*x)**4/(a + b*tan(c + d*x))**2, x)
Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {a^{2} + b^{2}}{b^{4} \tan \left (d x + c\right ) + a b^{3}} + \frac {2 \, a \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{3}} - \frac {\tan \left (d x + c\right )}{b^{2}}}{d} \] Input:
integrate(sec(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
-((a^2 + b^2)/(b^4*tan(d*x + c) + a*b^3) + 2*a*log(b*tan(d*x + c) + a)/b^3 - tan(d*x + c)/b^2)/d
Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {2 \, a \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{3} d} + \frac {\tan \left (d x + c\right )}{b^{2} d} - \frac {a^{2} + b^{2}}{{\left (b \tan \left (d x + c\right ) + a\right )} b^{3} d} \] Input:
integrate(sec(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
-2*a*log(abs(b*tan(d*x + c) + a))/(b^3*d) + tan(d*x + c)/(b^2*d) - (a^2 + b^2)/((b*tan(d*x + c) + a)*b^3*d)
Time = 0.73 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{b^2\,d}-\frac {a^2+b^2}{b\,d\,\left (\mathrm {tan}\left (c+d\,x\right )\,b^3+a\,b^2\right )}-\frac {2\,a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{b^3\,d} \] Input:
int(1/(cos(c + d*x)^4*(a + b*tan(c + d*x))^2),x)
Output:
tan(c + d*x)/(b^2*d) - (a^2 + b^2)/(b*d*(a*b^2 + b^3*tan(c + d*x))) - (2*a *log(a + b*tan(c + d*x)))/(b^3*d)
Time = 0.22 (sec) , antiderivative size = 1477, normalized size of antiderivative = 24.21 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4/(a+b*tan(d*x+c))^2,x)
Output:
(2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*tan(c + d*x)*a*b **2 + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**2*b - 2* cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*tan(c + d*x)*a*b**2 - 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**2*b + 2*cos(c + d *x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*tan(c + d*x)*a*b**2 + 2*cos( c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a**2*b - 2*cos(c + d*x) *log(tan((c + d*x)/2) + 1)*sin(c + d*x)*tan(c + d*x)*a*b**2 - 2*cos(c + d* x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**2*b - 2*cos(c + d*x)*log(tan( (c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**3*tan(c + d*x) *a*b**2 - 2*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**3*a**2*b + 2*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2 *tan((c + d*x)/2)*b - a)*sin(c + d*x)*tan(c + d*x)*a*b**2 + 2*cos(c + d*x) *log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)*a**2*b - cos(c + d*x)*sec(c + d*x)**4*sin(c + d*x)**3*b**3 + cos(c + d*x)*sec(c + d*x)**4*sin(c + d*x)*b**3 - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4* tan(c + d*x)*a**2*b - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3 + 4 *log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*tan(c + d*x)*a**2*b + 4*log(tan ((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 - 2*log(tan((c + d*x)/2) - 1)*tan( c + d*x)*a**2*b - 2*log(tan((c + d*x)/2) - 1)*a**3 - 2*log(tan((c + d*x)/2 ) + 1)*sin(c + d*x)**4*tan(c + d*x)*a**2*b - 2*log(tan((c + d*x)/2) + 1...