Integrand size = 28, antiderivative size = 120 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {a^3 \tan (c+d x)}{d}+\frac {3 a^2 b \tan ^2(c+d x)}{2 d}+\frac {a \left (a^2+3 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {b \left (3 a^2+b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {3 a b^2 \tan ^5(c+d x)}{5 d}+\frac {b^3 \tan ^6(c+d x)}{6 d} \] Output:
a^3*tan(d*x+c)/d+3/2*a^2*b*tan(d*x+c)^2/d+1/3*a*(a^2+3*b^2)*tan(d*x+c)^3/d +1/4*b*(3*a^2+b^2)*tan(d*x+c)^4/d+3/5*a*b^2*tan(d*x+c)^5/d+1/6*b^3*tan(d*x +c)^6/d
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.45 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {(a+b \tan (c+d x))^4 \left (a^2+15 b^2-4 a b \tan (c+d x)+10 b^2 \tan ^2(c+d x)\right )}{60 b^3 d} \] Input:
Integrate[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[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 = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3567, 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.
| \(\displaystyle \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \cos (c+d x)+b \sin (c+d x))^3}{\cos (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3567 |
| \(\displaystyle -\frac {\int (b+a \cot (c+d x))^3 \left (\cot ^2(c+d x)+1\right ) \tan ^7(c+d x)d\cot (c+d x)}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle -\frac {\int \left (b^3 \tan ^7(c+d x)+3 a b^2 \tan ^6(c+d x)+\left (b^3+3 a^2 b\right ) \tan ^5(c+d x)+\left (a^3+3 b^2 a\right ) \tan ^4(c+d x)+3 a^2 b \tan ^3(c+d x)+a^3 \tan ^2(c+d x)\right )d\cot (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-a^3 \tan (c+d x)-\frac {1}{4} b \left (3 a^2+b^2\right ) \tan ^4(c+d x)-\frac {1}{3} a \left (a^2+3 b^2\right ) \tan ^3(c+d x)-\frac {3}{2} a^2 b \tan ^2(c+d x)-\frac {3}{5} a b^2 \tan ^5(c+d x)-\frac {1}{6} b^3 \tan ^6(c+d x)}{d}\) |
Input:
Int[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]
Output:
-((-(a^3*Tan[c + d*x]) - (3*a^2*b*Tan[c + d*x]^2)/2 - (a*(a^2 + 3*b^2)*Tan [c + d*x]^3)/3 - (b*(3*a^2 + b^2)*Tan[c + d*x]^4)/4 - (3*a*b^2*Tan[c + d*x ]^5)/5 - (b^3*Tan[c + d*x]^6)/6)/d)
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[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si n[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[x^m*((b + a*x)^n/(1 + x^2)^((m + n + 2)/2)), x], x, Cot[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] && !(GtQ[ n, 0] && GtQ[m, 1])
Time = 0.60 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.99
| method | result | size |
| parts | \(-\frac {a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{3} \left (\frac {\sec \left (d x +c \right )^{6}}{6}-\frac {\sec \left (d x +c \right )^{4}}{4}\right )}{d}+\frac {3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )}{d}+\frac {3 a^{2} b \sec \left (d x +c \right )^{4}}{4 d}\) | \(119\) |
| derivativedivides | \(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )}{d}\) | \(127\) |
| default | \(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )}{d}\) | \(127\) |
| risch | \(-\frac {4 \left (-15 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+45 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-45 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-50 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+30 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-90 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-10 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-60 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-45 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-30 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+18 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 i a^{3}+3 i a \,b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\) | \(228\) |
| parallelrisch | \(-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10} a^{3}-45 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a^{2} b -55 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a^{3}+60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a \,b^{2}+90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a^{2} b -30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b^{3}+90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a^{3}-36 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a \,b^{2}-90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a^{2} b -20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b^{3}-90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{3}+36 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a \,b^{2}+90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{2} b -30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{3}+55 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3}-60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \,b^{2}-45 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} b -15 a^{3}\right )}{15 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6}}\) | \(315\) |
Input:
int(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-a^3/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+b^3/d*(1/6*sec(d*x+c)^6-1/4*sec( d*x+c)^4)+3*a*b^2/d*(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*sec(d*x+c)^4/d
Time = 0.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {10 \, b^{3} + 15 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, {\left (5 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 9 \, a b^{2} \cos \left (d x + c\right ) + {\left (5 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{6}} \] Input:
integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(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)
Timed out. \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**7*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.02 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {20 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} + 12 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a b^{2} - \frac {5 \, {\left (3 \, \sin \left (d x + c\right )^{2} - 1\right )} b^{3}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + \frac {45 \, a^{2} b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
1/60*(20*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 + 12*(3*tan(d*x + c)^5 + 5* tan(d*x + c)^3)*a*b^2 - 5*(3*sin(d*x + c)^2 - 1)*b^3/(sin(d*x + c)^6 - 3*s in(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) + 45*a^2*b/(sin(d*x + c)^2 - 1)^2)/d
Time = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.93 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {10 \, b^{3} \tan \left (d x + c\right )^{6} + 36 \, a b^{2} \tan \left (d x + c\right )^{5} + 45 \, a^{2} b \tan \left (d x + c\right )^{4} + 15 \, b^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} \tan \left (d x + c\right )^{3} + 60 \, a b^{2} \tan \left (d x + c\right )^{3} + 90 \, a^{2} b \tan \left (d x + c\right )^{2} + 60 \, a^{3} \tan \left (d x + c\right )}{60 \, d} \] Input:
integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(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 = 16.45 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {{\cos \left (c+d\,x\right )}^3\,\left (\frac {a^3\,\sin \left (c+d\,x\right )}{3}-\frac {a\,b^2\,\sin \left (c+d\,x\right )}{5}\right )+{\cos \left (c+d\,x\right )}^5\,\left (\frac {2\,a^3\,\sin \left (c+d\,x\right )}{3}-\frac {2\,a\,b^2\,\sin \left (c+d\,x\right )}{5}\right )+{\cos \left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{4}-\frac {b^3}{4}\right )+\frac {b^3}{6}+\frac {3\,a\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{5}}{d\,{\cos \left (c+d\,x\right )}^6} \] Input:
int((a*cos(c + d*x) + b*sin(c + d*x))^3/cos(c + d*x)^7,x)
Output:
(cos(c + d*x)^3*((a^3*sin(c + d*x))/3 - (a*b^2*sin(c + d*x))/5) + cos(c + d*x)^5*((2*a^3*sin(c + d*x))/3 - (2*a*b^2*sin(c + d*x))/5) + cos(c + d*x)^ 2*((3*a^2*b)/4 - b^3/4) + b^3/6 + (3*a*b^2*cos(c + d*x)*sin(c + d*x))/5)/( d*cos(c + d*x)^6)
Time = 0.17 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.66 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {\sin \left (d x +c \right ) \left (-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3}+24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}+100 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}-60 \cos \left (d x +c \right ) a^{3}-45 \sin \left (d x +c \right )^{5} a^{2} b +5 \sin \left (d x +c \right )^{5} b^{3}+135 \sin \left (d x +c \right )^{3} a^{2} b -15 \sin \left (d x +c \right )^{3} b^{3}-90 \sin \left (d x +c \right ) a^{2} b \right )}{60 d \left (\sin \left (d x +c \right )^{6}-3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(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))