Integrand size = 21, antiderivative size = 75 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\left (a^2+b^2\right ) (a+b \tan (c+d x))^4}{4 b^3 d}-\frac {2 a (a+b \tan (c+d x))^5}{5 b^3 d}+\frac {(a+b \tan (c+d x))^6}{6 b^3 d} \] Output:
1/4*(a^2+b^2)*(a+b*tan(d*x+c))^4/b^3/d-2/5*a*(a+b*tan(d*x+c))^5/b^3/d+1/6* (a+b*tan(d*x+c))^6/b^3/d
Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.72 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {(a+b \tan (c+d x))^4 \left (a^2+15 b^2-4 a b \tan (c+d x)+10 b^2 \tan ^2(c+d x)\right )}{60 b^3 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^3,x]
Output:
((a + b*Tan[c + d*x])^4*(a^2 + 15*b^2 - 4*a*b*Tan[c + d*x] + 10*b^2*Tan[c + d*x]^2))/(60*b^3*d)
Time = 0.28 (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.
\(\displaystyle \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^4 (a+b \tan (c+d x))^3dx\) |
\(\Big \downarrow \) 3987 |
\(\displaystyle \frac {\int \frac {(a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )}{b^2}d(b \tan (c+d x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left ((a+b \tan (c+d x))^5-2 a (a+b \tan (c+d x))^4+\left (a^2+b^2\right ) (a+b \tan (c+d x))^3\right )d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} \left (a^2+b^2\right ) (a+b \tan (c+d x))^4+\frac {1}{6} (a+b \tan (c+d x))^6-\frac {2}{5} a (a+b \tan (c+d x))^5}{b^3 d}\) |
Input:
Int[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^3,x]
Output:
(((a^2 + b^2)*(a + b*Tan[c + d*x])^4)/4 - (2*a*(a + b*Tan[c + d*x])^5)/5 + (a + b*Tan[c + d*x])^6/6)/(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 = 15.34 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.69
method | result | size |
derivativedivides | \(\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(127\) |
default | \(\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(127\) |
risch | \(-\frac {4 \left (-15 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+45 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-45 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-50 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+30 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-90 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-10 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-60 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-45 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-30 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+18 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 i a^{3}+3 i a \,b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\) | \(228\) |
Input:
int(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(b^3*(1/6*sin(d*x+c)^4/cos(d*x+c)^6+1/12*sin(d*x+c)^4/cos(d*x+c)^4)+3* a*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)+3/4*a ^2*b/cos(d*x+c)^4-a^3*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.40 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {10 \, b^{3} + 15 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, {\left (5 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 9 \, a b^{2} \cos \left (d x + c\right ) + {\left (5 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{6}} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/60*(10*b^3 + 15*(3*a^2*b - b^3)*cos(d*x + c)^2 + 4*(2*(5*a^3 - 3*a*b^2)* cos(d*x + c)^5 + 9*a*b^2*cos(d*x + c) + (5*a^3 - 3*a*b^2)*cos(d*x + c)^3)* sin(d*x + c))/(d*cos(d*x + c)^6)
\[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**4*(a+b*tan(d*x+c))**3,x)
Output:
Integral((a + b*tan(c + d*x))**3*sec(c + d*x)**4, x)
Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.31 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {10 \, b^{3} \tan \left (d x + c\right )^{6} + 36 \, a b^{2} \tan \left (d x + c\right )^{5} + 90 \, a^{2} b \tan \left (d x + c\right )^{2} + 15 \, {\left (3 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{4} + 60 \, a^{3} \tan \left (d x + c\right ) + 20 \, {\left (a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{3}}{60 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/60*(10*b^3*tan(d*x + c)^6 + 36*a*b^2*tan(d*x + c)^5 + 90*a^2*b*tan(d*x + c)^2 + 15*(3*a^2*b + b^3)*tan(d*x + c)^4 + 60*a^3*tan(d*x + c) + 20*(a^3 + 3*a*b^2)*tan(d*x + c)^3)/d
Time = 0.23 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.49 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {10 \, b^{3} \tan \left (d x + c\right )^{6} + 36 \, a b^{2} \tan \left (d x + c\right )^{5} + 45 \, a^{2} b \tan \left (d x + c\right )^{4} + 15 \, b^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} \tan \left (d x + c\right )^{3} + 60 \, a b^{2} \tan \left (d x + c\right )^{3} + 90 \, a^{2} b \tan \left (d x + c\right )^{2} + 60 \, a^{3} \tan \left (d x + c\right )}{60 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(10*b^3*tan(d*x + c)^6 + 36*a*b^2*tan(d*x + c)^5 + 45*a^2*b*tan(d*x + c)^4 + 15*b^3*tan(d*x + c)^4 + 20*a^3*tan(d*x + c)^3 + 60*a*b^2*tan(d*x + c)^3 + 90*a^2*b*tan(d*x + c)^2 + 60*a^3*tan(d*x + c))/d
Time = 0.65 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.29 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {a^3}{3}+a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {3\,a^2\,b}{4}+\frac {b^3}{4}\right )+a^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^6}{6}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}}{d} \] Input:
int((a + b*tan(c + d*x))^3/cos(c + d*x)^4,x)
Output:
(tan(c + d*x)^3*(a*b^2 + a^3/3) + tan(c + d*x)^4*((3*a^2*b)/4 + b^3/4) + a ^3*tan(c + d*x) + (b^3*tan(c + d*x)^6)/6 + (3*a^2*b*tan(c + d*x)^2)/2 + (3 *a*b^2*tan(c + d*x)^5)/5)/d
Time = 0.18 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.65 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\sin \left (d x +c \right ) \left (-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3}+24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}+100 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-60 \cos \left (d x +c \right ) a^{3}-45 \sin \left (d x +c \right )^{5} a^{2} b +5 \sin \left (d x +c \right )^{5} b^{3}+135 \sin \left (d x +c \right )^{3} a^{2} b -15 \sin \left (d x +c \right )^{3} b^{3}-90 \sin \left (d x +c \right ) a^{2} b \right )}{60 d \left (\sin \left (d x +c \right )^{6}-3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x)
Output:
(sin(c + d*x)*( - 40*cos(c + d*x)*sin(c + d*x)**4*a**3 + 24*cos(c + d*x)*s in(c + d*x)**4*a*b**2 + 100*cos(c + d*x)*sin(c + d*x)**2*a**3 - 60*cos(c + d*x)*sin(c + d*x)**2*a*b**2 - 60*cos(c + d*x)*a**3 - 45*sin(c + d*x)**5*a **2*b + 5*sin(c + d*x)**5*b**3 + 135*sin(c + d*x)**3*a**2*b - 15*sin(c + d *x)**3*b**3 - 90*sin(c + d*x)*a**2*b))/(60*d*(sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3*sin(c + d*x)**2 - 1))