Integrand size = 21, antiderivative size = 69 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\log (a+b \tan (c+d x))}{b^3 d}-\frac {a^2+b^2}{2 b^3 d (a+b \tan (c+d x))^2}+\frac {2 a}{b^3 d (a+b \tan (c+d x))} \] Output:
ln(a+b*tan(d*x+c))/b^3/d-1/2*(a^2+b^2)/b^3/d/(a+b*tan(d*x+c))^2+2*a/b^3/d/ (a+b*tan(d*x+c))
Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\log (a+b \tan (c+d x))-\frac {a^2+b^2}{2 (a+b \tan (c+d x))^2}+\frac {2 a}{a+b \tan (c+d x)}}{b^3 d} \] Input:
Integrate[Sec[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]
Output:
(Log[a + b*Tan[c + d*x]] - (a^2 + b^2)/(2*(a + b*Tan[c + d*x])^2) + (2*a)/ (a + b*Tan[c + d*x]))/(b^3*d)
Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83, 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.
\(\displaystyle \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^4}{(a+b \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 3987 |
\(\displaystyle \frac {\int \frac {\tan ^2(c+d x) b^2+b^2}{b^2 (a+b \tan (c+d x))^3}d(b \tan (c+d x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\tan ^2(c+d x) b^2+b^2}{(a+b \tan (c+d x))^3}d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (-\frac {2 a}{(a+b \tan (c+d x))^2}+\frac {1}{a+b \tan (c+d x)}+\frac {a^2+b^2}{(a+b \tan (c+d x))^3}\right )d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^2+b^2}{2 (a+b \tan (c+d x))^2}+\frac {2 a}{a+b \tan (c+d x)}+\log (a+b \tan (c+d x))}{b^3 d}\) |
Input:
Int[Sec[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]
Output:
(Log[a + b*Tan[c + d*x]] - (a^2 + b^2)/(2*(a + b*Tan[c + d*x])^2) + (2*a)/ (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 = 36.72 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2}+b^{2}}{2 b^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {2 a}{b^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(63\) |
default | \(\frac {-\frac {a^{2}+b^{2}}{2 b^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {2 a}{b^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(63\) |
risch | \(\frac {-2 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a^{2}-2 i a b}{b^{2} \left (i a +b \right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{3} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{3} d}\) | \(160\) |
Input:
int(sec(d*x+c)^4/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/2*(a^2+b^2)/b^3/(a+b*tan(d*x+c))^2+2/b^3*a/(a+b*tan(d*x+c))+1/b^3* ln(a+b*tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (67) = 134\).
Time = 0.11 (sec) , antiderivative size = 284, normalized size of antiderivative = 4.12 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {4 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b^{2} - b^{4} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} b^{2} + b^{4} + {\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - {\left (a^{2} b^{2} + b^{4} + {\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right )}{2 \, {\left ({\left (a^{4} b^{3} - b^{7}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} b^{5} + b^{7}\right )} d\right )}} \] Input:
integrate(sec(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/2*(4*a^2*b^2*cos(d*x + c)^2 - 3*a^2*b^2 - b^4 - 2*(a^3*b - a*b^3)*cos(d* x + c)*sin(d*x + c) + (a^2*b^2 + b^4 + (a^4 - b^4)*cos(d*x + c)^2 + 2*(a^3 *b + a*b^3)*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*b^2 + b^4 + (a^4 - b^4)*cos(d* x + c)^2 + 2*(a^3*b + a*b^3)*cos(d*x + c)*sin(d*x + c))*log(cos(d*x + c)^2 ))/((a^4*b^3 - b^7)*d*cos(d*x + c)^2 + 2*(a^3*b^4 + a*b^6)*d*cos(d*x + c)* sin(d*x + c) + (a^2*b^5 + b^7)*d)
\[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(sec(d*x+c)**4/(a+b*tan(d*x+c))**3,x)
Output:
Integral(sec(c + d*x)**4/(a + b*tan(c + d*x))**3, x)
Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {4 \, a b \tan \left (d x + c\right ) + 3 \, a^{2} - b^{2}}{b^{5} \tan \left (d x + c\right )^{2} + 2 \, a b^{4} \tan \left (d x + c\right ) + a^{2} b^{3}} + \frac {2 \, \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{3}}}{2 \, d} \] Input:
integrate(sec(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/2*((4*a*b*tan(d*x + c) + 3*a^2 - b^2)/(b^5*tan(d*x + c)^2 + 2*a*b^4*tan( d*x + c) + a^2*b^3) + 2*log(b*tan(d*x + c) + a)/b^3)/d
Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{3} d} + \frac {4 \, a \tan \left (d x + c\right ) + \frac {3 \, a^{2} - b^{2}}{b}}{2 \, {\left (b \tan \left (d x + c\right ) + a\right )}^{2} b^{2} d} \] Input:
integrate(sec(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
log(abs(b*tan(d*x + c) + a))/(b^3*d) + 1/2*(4*a*tan(d*x + c) + (3*a^2 - b^ 2)/b)/((b*tan(d*x + c) + a)^2*b^2*d)
Time = 0.72 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.16 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {3\,a^2-b^2}{2\,b^3}+\frac {2\,a\,\mathrm {tan}\left (c+d\,x\right )}{b^2}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{b^3\,d} \] Input:
int(1/(cos(c + d*x)^4*(a + b*tan(c + d*x))^3),x)
Output:
((3*a^2 - b^2)/(2*b^3) + (2*a*tan(c + d*x))/b^2)/(d*(a^2 + b^2*tan(c + d*x )^2 + 2*a*b*tan(c + d*x))) + log(a + b*tan(c + d*x))/(b^3*d)
Time = 0.19 (sec) , antiderivative size = 3133, normalized size of antiderivative = 45.41 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4/(a+b*tan(d*x+c))^3,x)
Output:
( - 4*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*tan(c + d*x)* *2*a*b**3 - 8*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*tan(c + d*x)*a**2*b**2 - 4*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)* *3*a**3*b + 4*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*tan(c + d*x)**2*a*b**3 + 8*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*tan (c + d*x)*a**2*b**2 + 4*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x )*a**3*b - 4*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*tan(c + d*x)**2*a*b**3 - 8*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)** 3*tan(c + d*x)*a**2*b**2 - 4*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a**3*b + 4*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*t an(c + d*x)**2*a*b**3 + 8*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d *x)*tan(c + d*x)*a**2*b**2 + 4*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin( c + d*x)*a**3*b + 4*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)**2*a*b**3 + 8*cos(c + d*x)*log(t an((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**3*tan(c + d *x)*a**2*b**2 + 4*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**3*b - 4*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)**2*a*b**3 - 8*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**2*b**2 - 4*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - ...