Integrand size = 28, antiderivative size = 111 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {a^3 \cos (c+d x)}{d}+\frac {3 a^2 b \log (\cos (c+d x))}{d}+\frac {a \left (a^2-3 b^2\right ) \sec (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {a b^2 \sec ^3(c+d x)}{d}+\frac {b^3 \sec ^4(c+d x)}{4 d} \] Output:
a^3*cos(d*x+c)/d+3*a^2*b*ln(cos(d*x+c))/d+a*(a^2-3*b^2)*sec(d*x+c)/d+1/2*b *(3*a^2-b^2)*sec(d*x+c)^2/d+a*b^2*sec(d*x+c)^3/d+1/4*b^3*sec(d*x+c)^4/d
Time = 1.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {4 a^3 \cos (c+d x)+12 a^2 b \log (\cos (c+d x))+4 a \left (a^2-3 b^2\right ) \sec (c+d x)+\left (6 a^2 b-2 b^3\right ) \sec ^2(c+d x)+4 a b^2 \sec ^3(c+d x)+b^3 \sec ^4(c+d x)}{4 d} \] Input:
Integrate[Sec[c + d*x]^2*(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
(4*a^3*Cos[c + d*x] + 12*a^2*b*Log[Cos[c + d*x]] + 4*a*(a^2 - 3*b^2)*Sec[c + d*x] + (6*a^2*b - 2*b^3)*Sec[c + d*x]^2 + 4*a*b^2*Sec[c + d*x]^3 + b^3* Sec[c + d*x]^4)/(4*d)
Time = 0.47 (sec) , antiderivative size = 111, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^2 (a \sin (c+d x)+b \tan (c+d x))^3dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \tan ^3(c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b)^3dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^3}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle -\frac {\int (b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^5(c+d x)d(a \cos (c+d x))}{a^3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \int \frac {(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{a^5}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle -\frac {a^2 \int \left (\frac {b^3 \sec ^5(c+d x)}{a^3}+\frac {3 b^2 \sec ^4(c+d x)}{a^2}+\frac {\left (3 a^2 b-b^3\right ) \sec ^3(c+d x)}{a^3}+\frac {\left (a^2-3 b^2\right ) \sec ^2(c+d x)}{a^2}-\frac {3 b \sec (c+d x)}{a}-1\right )d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \left (-\frac {b^3 \sec ^4(c+d x)}{4 a^2}-\frac {b \left (3 a^2-b^2\right ) \sec ^2(c+d x)}{2 a^2}-\frac {\left (a^2-3 b^2\right ) \sec (c+d x)}{a}-\frac {b^2 \sec ^3(c+d x)}{a}-3 b \log (a \cos (c+d x))-a \cos (c+d x)\right )}{d}\) |
Input:
Int[Sec[c + d*x]^2*(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
Output:
-((a^2*(-(a*Cos[c + d*x]) - 3*b*Log[a*Cos[c + d*x]] - ((a^2 - 3*b^2)*Sec[c + d*x])/a - (b*(3*a^2 - b^2)*Sec[c + d*x]^2)/(2*a^2) - (b^2*Sec[c + d*x]^ 3)/a - (b^3*Sec[c + d*x]^4)/(4*a^2)))/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 = 7.82 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{3} \sec \left (d x +c \right )^{4}}{4}+a \,b^{2} \sec \left (d x +c \right )^{3}+\frac {3 a^{2} b \sec \left (d x +c \right )^{2}}{2}-\frac {\sec \left (d x +c \right )^{2} b^{3}}{2}+\sec \left (d x +c \right ) a^{3}-3 \sec \left (d x +c \right ) a \,b^{2}-3 a^{2} b \ln \left (\sec \left (d x +c \right )\right )+\frac {a^{3}}{\sec \left (d x +c \right )}}{d}\) | \(106\) |
| default | \(\frac {\frac {b^{3} \sec \left (d x +c \right )^{4}}{4}+a \,b^{2} \sec \left (d x +c \right )^{3}+\frac {3 a^{2} b \sec \left (d x +c \right )^{2}}{2}-\frac {\sec \left (d x +c \right )^{2} b^{3}}{2}+\sec \left (d x +c \right ) a^{3}-3 \sec \left (d x +c \right ) a \,b^{2}-3 a^{2} b \ln \left (\sec \left (d x +c \right )\right )+\frac {a^{3}}{\sec \left (d x +c \right )}}{d}\) | \(106\) |
| risch | \(-3 i a^{2} b x +\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {6 i b \,a^{2} c}{d}+\frac {2 a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-6 a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-2 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+6 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-10 a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-10 a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 a^{3} {\mathrm e}^{i \left (d x +c \right )}-6 a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}\) | \(282\) |
Input:
int(sec(d*x+c)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(1/4*b^3*sec(d*x+c)^4+a*b^2*sec(d*x+c)^3+3/2*a^2*b*sec(d*x+c)^2-1/2*se c(d*x+c)^2*b^3+sec(d*x+c)*a^3-3*sec(d*x+c)*a*b^2-3*a^2*b*ln(sec(d*x+c))+a^ 3/sec(d*x+c))
Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {4 \, a^{3} \cos \left (d x + c\right )^{5} + 12 \, a^{2} b \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) + 4 \, a b^{2} \cos \left (d x + c\right ) + 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + b^{3} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{4 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas" )
Output:
1/4*(4*a^3*cos(d*x + c)^5 + 12*a^2*b*cos(d*x + c)^4*log(-cos(d*x + c)) + 4 *a*b^2*cos(d*x + c) + 4*(a^3 - 3*a*b^2)*cos(d*x + c)^3 + b^3 + 2*(3*a^2*b - b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)^4)
\[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**2*(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)**2, x)
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {b^{3} \tan \left (d x + c\right )^{4} - 6 \, a^{2} b {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} + 4 \, a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - \frac {4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a b^{2}}{\cos \left (d x + c\right )^{3}}}{4 \, d} \] Input:
integrate(sec(d*x+c)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima" )
Output:
1/4*(b^3*tan(d*x + c)^4 - 6*a^2*b*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1)) + 4*a^3*(1/cos(d*x + c) + cos(d*x + c)) - 4*(3*cos(d*x + c)^2 - 1)*a*b^2/cos(d*x + c)^3)/d
Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {4 \, a^{3} \cos \left (d x + c\right ) + 12 \, a^{2} b \log \left ({\left | \cos \left (d x + c\right ) \right |}\right ) + \frac {4 \, a b^{2} \cos \left (d x + c\right ) + 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + b^{3} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{\cos \left (d x + c\right )^{4}}}{4 \, d} \] Input:
integrate(sec(d*x+c)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/4*(4*a^3*cos(d*x + c) + 12*a^2*b*log(abs(cos(d*x + c))) + (4*a*b^2*cos(d *x + c) + 4*(a^3 - 3*a*b^2)*cos(d*x + c)^3 + b^3 + 2*(3*a^2*b - b^3)*cos(d *x + c)^2)/cos(d*x + c)^4)/d
Time = 20.34 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.01 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-12\,a^3+6\,a^2\,b+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a^3-6\,a^2\,b+4\,a\,b^2+4\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (4\,a^3+6\,a^2\,b+12\,a\,b^2-4\,b^3\right )-4\,a\,b^2+4\,a^3+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {6\,a^2\,b\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \] Input:
int((a*sin(c + d*x) + b*tan(c + d*x))^3/cos(c + d*x)^2,x)
Output:
(tan(c/2 + (d*x)/2)^2*(12*a*b^2 + 6*a^2*b - 12*a^3) + tan(c/2 + (d*x)/2)^4 *(4*a*b^2 - 6*a^2*b + 12*a^3 + 4*b^3) - tan(c/2 + (d*x)/2)^6*(12*a*b^2 + 6 *a^2*b + 4*a^3 - 4*b^3) - 4*a*b^2 + 4*a^3 + 6*a^2*b*tan(c/2 + (d*x)/2)^8)/ (d*(2*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2) ^6 - 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (6*a^2*b*atanh (tan(c/2 + (d*x)/2)^2))/d
Time = 0.16 (sec) , antiderivative size = 430, normalized size of antiderivative = 3.87 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3}-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}+12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}+8 \cos \left (d x +c \right ) a^{3}-8 \cos \left (d x +c \right ) a \,b^{2}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4} a^{2} b +24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a^{2} b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{2} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{2} b -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{2} b -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b -8 \sin \left (d x +c \right )^{4} a^{3}-6 \sin \left (d x +c \right )^{4} a^{2} b +8 \sin \left (d x +c \right )^{4} a \,b^{2}+\sin \left (d x +c \right )^{4} b^{3}+16 \sin \left (d x +c \right )^{2} a^{3}+6 \sin \left (d x +c \right )^{2} a^{2} b -16 \sin \left (d x +c \right )^{2} a \,b^{2}-8 a^{3}+8 a \,b^{2}}{4 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x)
Output:
(4*cos(c + d*x)*sin(c + d*x)**4*a**3 - 12*cos(c + d*x)*sin(c + d*x)**2*a** 3 + 12*cos(c + d*x)*sin(c + d*x)**2*a*b**2 + 8*cos(c + d*x)*a**3 - 8*cos(c + d*x)*a*b**2 - 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*a**2*b + 24*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**2*b - 12*log(tan((c + d *x)/2)**2 + 1)*a**2*b + 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2* b - 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b + 12*log(tan((c + d*x)/2) - 1)*a**2*b + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b - 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b + 12*log(tan((c + d* x)/2) + 1)*a**2*b - 8*sin(c + d*x)**4*a**3 - 6*sin(c + d*x)**4*a**2*b + 8* sin(c + d*x)**4*a*b**2 + sin(c + d*x)**4*b**3 + 16*sin(c + d*x)**2*a**3 + 6*sin(c + d*x)**2*a**2*b - 16*sin(c + d*x)**2*a*b**2 - 8*a**3 + 8*a*b**2)/ (4*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))