Integrand size = 28, antiderivative size = 119 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {a^3 \log (\cos (c+d x))}{d}-\frac {3 a^2 b \sec (c+d x)}{d}+\frac {a \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{3 d}+\frac {3 a b^2 \sec ^4(c+d x)}{4 d}+\frac {b^3 \sec ^5(c+d x)}{5 d} \] Output:
a^3*ln(cos(d*x+c))/d-3*a^2*b*sec(d*x+c)/d+1/2*a*(a^2-3*b^2)*sec(d*x+c)^2/d +1/3*b*(3*a^2-b^2)*sec(d*x+c)^3/d+3/4*a*b^2*sec(d*x+c)^4/d+1/5*b^3*sec(d*x +c)^5/d
Time = 0.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {60 a^3 \log (\cos (c+d x))-180 a^2 b \sec (c+d x)+30 a \left (a^2-3 b^2\right ) \sec ^2(c+d x)-20 b \left (-3 a^2+b^2\right ) \sec ^3(c+d x)+45 a b^2 \sec ^4(c+d x)+12 b^3 \sec ^5(c+d x)}{60 d} \] Input:
Integrate[Sec[c + d*x]^3*(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
(60*a^3*Log[Cos[c + d*x]] - 180*a^2*b*Sec[c + d*x] + 30*a*(a^2 - 3*b^2)*Se c[c + d*x]^2 - 20*b*(-3*a^2 + b^2)*Sec[c + d*x]^3 + 45*a*b^2*Sec[c + d*x]^ 4 + 12*b^3*Sec[c + d*x]^5)/(60*d)
Time = 0.49 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4897, 3042, 25, 3316, 27, 522, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^3 (a \sin (c+d x)+b \tan (c+d x))^3dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \tan ^3(c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^3dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^3}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^6}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle -\frac {\int (b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^6(c+d x)d(a \cos (c+d x))}{a^3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^3 \int \frac {(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^6(c+d x)}{a^6}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle -\frac {a^3 \int \left (\frac {b^3 \sec ^6(c+d x)}{a^4}+\frac {3 b^2 \sec ^5(c+d x)}{a^3}+\frac {\left (3 a^2 b-b^3\right ) \sec ^4(c+d x)}{a^4}+\frac {\left (a^2-3 b^2\right ) \sec ^3(c+d x)}{a^3}-\frac {3 b \sec ^2(c+d x)}{a^2}-\frac {\sec (c+d x)}{a}\right )d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^3 \left (-\frac {b^3 \sec ^5(c+d x)}{5 a^3}-\frac {3 b^2 \sec ^4(c+d x)}{4 a^2}-\frac {\left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 a^2}-\frac {b \left (3 a^2-b^2\right ) \sec ^3(c+d x)}{3 a^3}+\frac {3 b \sec (c+d x)}{a}-\log (a \cos (c+d x))\right )}{d}\) |
Input:
Int[Sec[c + d*x]^3*(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
-((a^3*(-Log[a*Cos[c + d*x]] + (3*b*Sec[c + d*x])/a - ((a^2 - 3*b^2)*Sec[c + d*x]^2)/(2*a^2) - (b*(3*a^2 - b^2)*Sec[c + d*x]^3)/(3*a^3) - (3*b^2*Sec [c + d*x]^4)/(4*a^2) - (b^3*Sec[c + d*x]^5)/(5*a^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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 1.86 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{3} \sec \left (d x +c \right )^{5}}{5}+\frac {3 a \,b^{2} \sec \left (d x +c \right )^{4}}{4}+a^{2} b \sec \left (d x +c \right )^{3}-\frac {b^{3} \sec \left (d x +c \right )^{3}}{3}+\frac {\sec \left (d x +c \right )^{2} a^{3}}{2}-\frac {3 a \,b^{2} \sec \left (d x +c \right )^{2}}{2}-3 a^{2} b \sec \left (d x +c \right )-a^{3} \ln \left (\sec \left (d x +c \right )\right )}{d}\) | \(110\) |
| default | \(\frac {\frac {b^{3} \sec \left (d x +c \right )^{5}}{5}+\frac {3 a \,b^{2} \sec \left (d x +c \right )^{4}}{4}+a^{2} b \sec \left (d x +c \right )^{3}-\frac {b^{3} \sec \left (d x +c \right )^{3}}{3}+\frac {\sec \left (d x +c \right )^{2} a^{3}}{2}-\frac {3 a \,b^{2} \sec \left (d x +c \right )^{2}}{2}-3 a^{2} b \sec \left (d x +c \right )-a^{3} \ln \left (\sec \left (d x +c \right )\right )}{d}\) | \(110\) |
| risch | \(-i a^{3} x -\frac {2 i a^{3} c}{d}+\frac {-6 a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}+2 \,{\mathrm e}^{8 i \left (d x +c \right )} a^{3}-6 a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-16 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-\frac {8 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}}{3}+6 \,{\mathrm e}^{6 i \left (d x +c \right )} a^{3}-6 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-20 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+\frac {16 \,{\mathrm e}^{5 i \left (d x +c \right )} b^{3}}{15}+6 \,{\mathrm e}^{4 i \left (d x +c \right )} a^{3}-6 \,{\mathrm e}^{4 i \left (d x +c \right )} a \,b^{2}-16 \,{\mathrm e}^{3 i \left (d x +c \right )} a^{2} b -\frac {8 \,{\mathrm e}^{3 i \left (d x +c \right )} b^{3}}{3}+2 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-6 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 \,{\mathrm e}^{i \left (d x +c \right )} a^{2} b}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(290\) |
Input:
int(sec(d*x+c)^3*(a*sin(d*x+c)+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(1/5*b^3*sec(d*x+c)^5+3/4*a*b^2*sec(d*x+c)^4+a^2*b*sec(d*x+c)^3-1/3*b^ 3*sec(d*x+c)^3+1/2*sec(d*x+c)^2*a^3-3/2*a*b^2*sec(d*x+c)^2-3*a^2*b*sec(d*x +c)-a^3*ln(sec(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.92 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {60 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (-\cos \left (d x + c\right )\right ) - 180 \, a^{2} b \cos \left (d x + c\right )^{4} + 45 \, a b^{2} \cos \left (d x + c\right ) + 30 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 12 \, b^{3} + 20 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{60 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate(sec(d*x+c)^3*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas" )
Output:
1/60*(60*a^3*cos(d*x + c)^5*log(-cos(d*x + c)) - 180*a^2*b*cos(d*x + c)^4 + 45*a*b^2*cos(d*x + c) + 30*(a^3 - 3*a*b^2)*cos(d*x + c)^3 + 12*b^3 + 20* (3*a^2*b - b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)^5)
\[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**3*(a*sin(d*x+c)+b*tan(d*x+c))**3,x)
Output:
Integral((a*sin(c + d*x) + b*tan(c + d*x))**3*sec(c + d*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=-\frac {30 \, a^{3} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - \frac {45 \, {\left (2 \, \sin \left (d x + c\right )^{2} - 1\right )} a b^{2}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + \frac {60 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2} b}{\cos \left (d x + c\right )^{3}} + \frac {4 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} b^{3}}{\cos \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^3*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima" )
Output:
-1/60*(30*a^3*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1)) - 45*(2*s in(d*x + c)^2 - 1)*a*b^2/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) + 60*(3*c os(d*x + c)^2 - 1)*a^2*b/cos(d*x + c)^3 + 4*(5*cos(d*x + c)^2 - 3)*b^3/cos (d*x + c)^5)/d
Time = 0.40 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {60 \, a^{3} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right ) - \frac {180 \, a^{2} b \cos \left (d x + c\right )^{4} - 45 \, a b^{2} \cos \left (d x + c\right ) - 30 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 12 \, b^{3} - 20 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{\cos \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^3*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(60*a^3*log(abs(cos(d*x + c))) - (180*a^2*b*cos(d*x + c)^4 - 45*a*b^2 *cos(d*x + c) - 30*(a^3 - 3*a*b^2)*cos(d*x + c)^3 - 12*b^3 - 20*(3*a^2*b - b^3)*cos(d*x + c)^2)/cos(d*x + c)^5)/d
Time = 20.18 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.85 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=-\frac {2\,a^3\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (6\,a^3+12\,a^2\,b-12\,a\,b^2+4\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-6\,a^3-28\,a^2\,b+12\,a\,b^2+\frac {4\,b^3}{3}\right )-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,a^2\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^3+20\,a^2\,b+\frac {4\,b^3}{3}\right )-\frac {4\,b^3}{15}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a*sin(c + d*x) + b*tan(c + d*x))^3/cos(c + d*x)^3,x)
Output:
- (2*a^3*atanh(tan(c/2 + (d*x)/2)^2))/d - (tan(c/2 + (d*x)/2)^6*(12*a^2*b - 12*a*b^2 + 6*a^3 + 4*b^3) + tan(c/2 + (d*x)/2)^4*(12*a*b^2 - 28*a^2*b - 6*a^3 + (4*b^3)/3) - 2*a^3*tan(c/2 + (d*x)/2)^8 - 4*a^2*b + tan(c/2 + (d*x )/2)^2*(20*a^2*b + 2*a^3 + (4*b^3)/3) - (4*b^3)/15)/(d*(5*tan(c/2 + (d*x)/ 2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d* x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))
Time = 0.17 (sec) , antiderivative size = 508, normalized size of antiderivative = 4.27 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4} a^{3}+120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a^{3}-60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{3}+60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{3}-120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3}+60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}+60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{3}-120 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3}+60 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3}+120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2} b +45 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{3}+30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b -16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{3}+120 \cos \left (d x +c \right ) a^{2} b +8 \cos \left (d x +c \right ) b^{3}-180 \sin \left (d x +c \right )^{4} a^{2} b +300 \sin \left (d x +c \right )^{2} a^{2} b +20 \sin \left (d x +c \right )^{2} b^{3}-120 a^{2} b -8 b^{3}}{60 \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)^3*(a*sin(d*x+c)+b*tan(d*x+c))^3,x)
Output:
( - 60*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*a**3 + 12 0*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**3 - 60*cos( c + d*x)*log(tan((c + d*x)/2)**2 + 1)*a**3 + 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3 - 120*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 + 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3 + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3 - 120*co s(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 + 60*cos(c + d*x )*log(tan((c + d*x)/2) + 1)*a**3 - 30*cos(c + d*x)*sin(c + d*x)**4*a**3 + 120*cos(c + d*x)*sin(c + d*x)**4*a**2*b + 45*cos(c + d*x)*sin(c + d*x)**4* a*b**2 + 8*cos(c + d*x)*sin(c + d*x)**4*b**3 + 30*cos(c + d*x)*sin(c + d*x )**2*a**3 - 240*cos(c + d*x)*sin(c + d*x)**2*a**2*b - 16*cos(c + d*x)*sin( c + d*x)**2*b**3 + 120*cos(c + d*x)*a**2*b + 8*cos(c + d*x)*b**3 - 180*sin (c + d*x)**4*a**2*b + 300*sin(c + d*x)**2*a**2*b + 20*sin(c + d*x)**2*b**3 - 120*a**2*b - 8*b**3)/(60*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d* x)**2 + 1))