Integrand size = 26, antiderivative size = 115 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {3 a^2 b \cos (c+d x)}{d}+\frac {a^3 \cos ^2(c+d x)}{2 d}-\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}+\frac {b \left (3 a^2-b^2\right ) \sec (c+d x)}{d}+\frac {3 a b^2 \sec ^2(c+d x)}{2 d}+\frac {b^3 \sec ^3(c+d x)}{3 d} \] Output:
3*a^2*b*cos(d*x+c)/d+1/2*a^3*cos(d*x+c)^2/d-a*(a^2-3*b^2)*ln(cos(d*x+c))/d +b*(3*a^2-b^2)*sec(d*x+c)/d+3/2*a*b^2*sec(d*x+c)^2/d+1/3*b^3*sec(d*x+c)^3/ d
Time = 0.90 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.87 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {36 a^2 b \cos (c+d x)+3 a^3 \cos (2 (c+d x))+2 \left (-6 a \left (a^2-3 b^2\right ) \log (\cos (c+d x))-6 b \left (-3 a^2+b^2\right ) \sec (c+d x)+9 a b^2 \sec ^2(c+d x)+2 b^3 \sec ^3(c+d x)\right )}{12 d} \] Input:
Integrate[Sec[c + d*x]*(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
(36*a^2*b*Cos[c + d*x] + 3*a^3*Cos[2*(c + d*x)] + 2*(-6*a*(a^2 - 3*b^2)*Lo g[Cos[c + d*x]] - 6*b*(-3*a^2 + b^2)*Sec[c + d*x] + 9*a*b^2*Sec[c + d*x]^2 + 2*b^3*Sec[c + d*x]^3))/(12*d)
Time = 0.48 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.93, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, 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.
| \(\displaystyle \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \tan ^3(c+d x) \sec (c+d x) (a \cos (c+d x)+b)^3dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^3}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^4}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle -\frac {\int (b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^4(c+d x)d(a \cos (c+d x))}{a^3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a \int \frac {(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a^4}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle -\frac {a \int \left (\frac {b^3 \sec ^4(c+d x)}{a^2}+\frac {3 b^2 \sec ^3(c+d x)}{a}+\frac {\left (3 a^2 b-b^3\right ) \sec ^2(c+d x)}{a^2}+\frac {\left (a^2-3 b^2\right ) \sec (c+d x)}{a}-3 b-a \cos (c+d x)\right )d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a \left (-\frac {b \left (3 a^2-b^2\right ) \sec (c+d x)}{a}+\left (a^2-3 b^2\right ) \log (a \cos (c+d x))-\frac {1}{2} a^2 \cos ^2(c+d x)-\frac {b^3 \sec ^3(c+d x)}{3 a}-3 a b \cos (c+d x)-\frac {3}{2} b^2 \sec ^2(c+d x)\right )}{d}\) |
Input:
Int[Sec[c + d*x]*(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
-((a*(-3*a*b*Cos[c + d*x] - (a^2*Cos[c + d*x]^2)/2 + (a^2 - 3*b^2)*Log[a*C os[c + d*x]] - (b*(3*a^2 - b^2)*Sec[c + d*x])/a - (3*b^2*Sec[c + d*x]^2)/2 - (b^3*Sec[c + d*x]^3)/(3*a)))/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 = 8.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{3} \sec \left (d x +c \right )^{3}}{3}+\frac {3 a \,b^{2} \sec \left (d x +c \right )^{2}}{2}+3 a^{2} b \sec \left (d x +c \right )-\sec \left (d x +c \right ) b^{3}+a \left (a^{2}-3 b^{2}\right ) \ln \left (\sec \left (d x +c \right )\right )+\frac {a^{3}}{2 \sec \left (d x +c \right )^{2}}+\frac {3 a^{2} b}{\sec \left (d x +c \right )}}{d}\) | \(101\) |
| default | \(\frac {\frac {b^{3} \sec \left (d x +c \right )^{3}}{3}+\frac {3 a \,b^{2} \sec \left (d x +c \right )^{2}}{2}+3 a^{2} b \sec \left (d x +c \right )-\sec \left (d x +c \right ) b^{3}+a \left (a^{2}-3 b^{2}\right ) \ln \left (\sec \left (d x +c \right )\right )+\frac {a^{3}}{2 \sec \left (d x +c \right )^{2}}+\frac {3 a^{2} b}{\sec \left (d x +c \right )}}{d}\) | \(101\) |
| risch | \(i a^{3} x -3 i x a \,b^{2}+\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} a^{2} b}{2 d}+\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2} b}{2 d}+\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{3} c}{d}-\frac {6 i a \,b^{2} c}{d}+\frac {2 b \left (9 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+9 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+18 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-2 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+9 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a^{2} {\mathrm e}^{i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {3 a \,b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(279\) |
Input:
int(sec(d*x+c)*(a*sin(d*x+c)+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*b^3*sec(d*x+c)^3+3/2*a*b^2*sec(d*x+c)^2+3*a^2*b*sec(d*x+c)-sec(d* x+c)*b^3+a*(a^2-3*b^2)*ln(sec(d*x+c))+1/2*a^3/sec(d*x+c)^2+3*a^2*b/sec(d*x +c))
Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.06 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {6 \, a^{3} \cos \left (d x + c\right )^{5} + 36 \, a^{2} b \cos \left (d x + c\right )^{4} - 3 \, a^{3} \cos \left (d x + c\right )^{3} - 12 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) + 18 \, a b^{2} \cos \left (d x + c\right ) + 4 \, b^{3} + 12 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate(sec(d*x+c)*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/12*(6*a^3*cos(d*x + c)^5 + 36*a^2*b*cos(d*x + c)^4 - 3*a^3*cos(d*x + c)^ 3 - 12*(a^3 - 3*a*b^2)*cos(d*x + c)^3*log(-cos(d*x + c)) + 18*a*b^2*cos(d* x + c) + 4*b^3 + 12*(3*a^2*b - b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)^3)
\[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3} \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(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), x)
Time = 0.04 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=-\frac {3 \, {\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} a^{3} + 9 \, a b^{2} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 18 \, a^{2} b {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{3}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(sec(d*x+c)*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/6*(3*(sin(d*x + c)^2 + log(sin(d*x + c)^2 - 1))*a^3 + 9*a*b^2*(1/(sin(d *x + c)^2 - 1) - log(sin(d*x + c)^2 - 1)) - 18*a^2*b*(1/cos(d*x + c) + cos (d*x + c)) + 2*(3*cos(d*x + c)^2 - 1)*b^3/cos(d*x + c)^3)/d
Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.87 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {3 \, a^{3} \cos \left (d x + c\right )^{2} + 18 \, a^{2} b \cos \left (d x + c\right ) - 6 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right ) + \frac {9 \, a b^{2} \cos \left (d x + c\right ) + 2 \, b^{3} + 6 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(sec(d*x+c)*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/6*(3*a^3*cos(d*x + c)^2 + 18*a^2*b*cos(d*x + c) - 6*(a^3 - 3*a*b^2)*log( abs(cos(d*x + c))) + (9*a*b^2*cos(d*x + c) + 2*b^3 + 6*(3*a^2*b - b^3)*cos (d*x + c)^2)/cos(d*x + c)^3)/d
Time = 20.54 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.90 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {2\,a^3\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )-6\,a\,b^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-2\,a^3-12\,a^2\,b+6\,a\,b^2+\frac {4\,b^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (6\,a^3-12\,a^2\,b+6\,a\,b^2-4\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (6\,a^3-12\,a^2\,b+6\,a\,b^2+\frac {20\,b^3}{3}\right )+12\,a^2\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (6\,a\,b^2-2\,a^3\right )-\frac {4\,b^3}{3}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^3\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \] Input:
int((a*sin(c + d*x) + b*tan(c + d*x))^3/cos(c + d*x),x)
Output:
(2*a^3*atanh(tan(c/2 + (d*x)/2)^2) - 6*a*b^2*atanh(tan(c/2 + (d*x)/2)^2))/ d - (tan(c/2 + (d*x)/2)^2*(6*a*b^2 - 12*a^2*b - 2*a^3 + (4*b^3)/3) - tan(c /2 + (d*x)/2)^6*(6*a*b^2 - 12*a^2*b + 6*a^3 - 4*b^3) + tan(c/2 + (d*x)/2)^ 4*(6*a*b^2 - 12*a^2*b + 6*a^3 + (20*b^3)/3) + 12*a^2*b - tan(c/2 + (d*x)/2 )^8*(6*a*b^2 - 2*a^3) - (4*b^3)/3)/(d*(tan(c/2 + (d*x)/2)^2 - 1)^3*(tan(c/ 2 + (d*x)/2)^2 + 1)^2)
Time = 0.16 (sec) , antiderivative size = 536, normalized size of antiderivative = 4.66 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)*(a*sin(d*x+c)+b*tan(d*x+c))^3,x)
Output:
(6*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**3 - 18*cos (c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a*b**2 - 6*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*a**3 + 18*cos(c + d*x)*log(tan((c + d*x) /2)**2 + 1)*a*b**2 - 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **2*a**3 + 18*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b** 2 + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3 - 18*cos(c + d*x)*log(ta n((c + d*x)/2) - 1)*a*b**2 - 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin( c + d*x)**2*a**3 + 18*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)* *2*a*b**2 + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**3 - 18*cos(c + d*x )*log(tan((c + d*x)/2) + 1)*a*b**2 - 3*cos(c + d*x)*sin(c + d*x)**4*a**3 + 3*cos(c + d*x)*sin(c + d*x)**2*a**3 - 36*cos(c + d*x)*sin(c + d*x)**2*a** 2*b - 9*cos(c + d*x)*sin(c + d*x)**2*a*b**2 + 4*cos(c + d*x)*sin(c + d*x)* *2*b**3 + 36*cos(c + d*x)*a**2*b - 4*cos(c + d*x)*b**3 - 18*sin(c + d*x)** 4*a**2*b + 54*sin(c + d*x)**2*a**2*b - 6*sin(c + d*x)**2*b**3 - 36*a**2*b + 4*b**3)/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))