Integrand size = 21, antiderivative size = 75 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (a^2+b^2\right ) (a+b \tan (c+d x))^3}{3 b^3 d}-\frac {a (a+b \tan (c+d x))^4}{2 b^3 d}+\frac {(a+b \tan (c+d x))^5}{5 b^3 d} \] Output:
1/3*(a^2+b^2)*(a+b*tan(d*x+c))^3/b^3/d-1/2*a*(a+b*tan(d*x+c))^4/b^3/d+1/5* (a+b*tan(d*x+c))^5/b^3/d
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.72 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {(a+b \tan (c+d x))^3 \left (a^2+10 b^2-3 a b \tan (c+d x)+6 b^2 \tan ^2(c+d x)\right )}{30 b^3 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]
Output:
((a + b*Tan[c + d*x])^3*(a^2 + 10*b^2 - 3*a*b*Tan[c + d*x] + 6*b^2*Tan[c + d*x]^2))/(30*b^3*d)
Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.85, 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.
\(\displaystyle \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^4 (a+b \tan (c+d x))^2dx\) |
\(\Big \downarrow \) 3987 |
\(\displaystyle \frac {\int \frac {(a+b \tan (c+d x))^2 \left (\tan ^2(c+d x) b^2+b^2\right )}{b^2}d(b \tan (c+d x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a+b \tan (c+d x))^2 \left (\tan ^2(c+d x) b^2+b^2\right )d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left ((a+b \tan (c+d x))^4-2 a (a+b \tan (c+d x))^3+\left (a^2+b^2\right ) (a+b \tan (c+d x))^2\right )d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} \left (a^2+b^2\right ) (a+b \tan (c+d x))^3+\frac {1}{5} (a+b \tan (c+d x))^5-\frac {1}{2} a (a+b \tan (c+d x))^4}{b^3 d}\) |
Input:
Int[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^2,x]
Output:
(((a^2 + b^2)*(a + b*Tan[c + d*x])^3)/3 - (a*(a + b*Tan[c + d*x])^4)/2 + ( a + b*Tan[c + d*x])^5/5)/(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 = 8.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {a b}{2 \cos \left (d x +c \right )^{4}}-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(82\) |
default | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {a b}{2 \cos \left (d x +c \right )^{4}}-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(82\) |
risch | \(\frac {4 i \left (-30 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+15 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-15 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-30 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+35 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+5 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+25 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+5 a^{2}-b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\) | \(143\) |
Input:
int(sec(d*x+c)^4*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(b^2*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)+1/ 2*a*b/cos(d*x+c)^4-a^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {15 \, a b \cos \left (d x + c\right ) + 2 \, {\left (2 \, {\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
1/30*(15*a*b*cos(d*x + c) + 2*(2*(5*a^2 - b^2)*cos(d*x + c)^4 + (5*a^2 - b ^2)*cos(d*x + c)^2 + 3*b^2)*sin(d*x + c))/(d*cos(d*x + c)^5)
\[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**4*(a+b*tan(d*x+c))**2,x)
Output:
Integral((a + b*tan(c + d*x))**2*sec(c + d*x)**4, x)
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {6 \, b^{2} \tan \left (d x + c\right )^{5} + 15 \, a b \tan \left (d x + c\right )^{4} + 30 \, a b \tan \left (d x + c\right )^{2} + 10 \, {\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{3} + 30 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
1/30*(6*b^2*tan(d*x + c)^5 + 15*a*b*tan(d*x + c)^4 + 30*a*b*tan(d*x + c)^2 + 10*(a^2 + b^2)*tan(d*x + c)^3 + 30*a^2*tan(d*x + c))/d
Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {6 \, b^{2} \tan \left (d x + c\right )^{5} + 15 \, a b \tan \left (d x + c\right )^{4} + 10 \, a^{2} \tan \left (d x + c\right )^{3} + 10 \, b^{2} \tan \left (d x + c\right )^{3} + 30 \, a b \tan \left (d x + c\right )^{2} + 30 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
1/30*(6*b^2*tan(d*x + c)^5 + 15*a*b*tan(d*x + c)^4 + 10*a^2*tan(d*x + c)^3 + 10*b^2*tan(d*x + c)^3 + 30*a*b*tan(d*x + c)^2 + 30*a^2*tan(d*x + c))/d
Time = 0.66 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {a^2}{3}+\frac {b^2}{3}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^4}{2}}{d} \] Input:
int((a + b*tan(c + d*x))^2/cos(c + d*x)^4,x)
Output:
(a^2*tan(c + d*x) + tan(c + d*x)^3*(a^2/3 + b^2/3) + (b^2*tan(c + d*x)^5)/ 5 + a*b*tan(c + d*x)^2 + (a*b*tan(c + d*x)^4)/2)/d
Time = 0.18 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.77 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\sin \left (d x +c \right ) \left (-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b +20 \sin \left (d x +c \right )^{4} a^{2}-4 \sin \left (d x +c \right )^{4} b^{2}-50 \sin \left (d x +c \right )^{2} a^{2}+10 \sin \left (d x +c \right )^{2} b^{2}+30 a^{2}\right )}{30 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^4*(a+b*tan(d*x+c))^2,x)
Output:
(sin(c + d*x)*( - 15*cos(c + d*x)*sin(c + d*x)**3*a*b + 30*cos(c + d*x)*si n(c + d*x)*a*b + 20*sin(c + d*x)**4*a**2 - 4*sin(c + d*x)**4*b**2 - 50*sin (c + d*x)**2*a**2 + 10*sin(c + d*x)**2*b**2 + 30*a**2))/(30*cos(c + d*x)*d *(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))