\(\int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\) [42]

Optimal result

Integrand size = 26, antiderivative size = 60 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {b \sec ^6(c+d x)}{6 d}+\frac {a \tan (c+d x)}{d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \]

[Out]

1/6*b*sec(d*x+c)^6/d+a*tan(d*x+c)/d+2/3*a*tan(d*x+c)^3/d+1/5*a*tan(d*x+c)^5/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3169, 3852, 2686, 30} \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a \tan ^5(c+d x)}{5 d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {b \sec ^6(c+d x)}{6 d} \]

[In]

Int[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^6)/(6*d) + (a*Tan[c + d*x])/d + (2*a*Tan[c + d*x]^3)/(3*d) + (a*Tan[c + d*x]^5)/(5*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3169

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a \sec ^6(c+d x)+b \sec ^6(c+d x) \tan (c+d x)\right ) \, dx \\ & = a \int \sec ^6(c+d x) \, dx+b \int \sec ^6(c+d x) \tan (c+d x) \, dx \\ & = -\frac {a \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d}+\frac {b \text {Subst}\left (\int x^5 \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {b \sec ^6(c+d x)}{6 d}+\frac {a \tan (c+d x)}{d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {b \sec ^6(c+d x)}{6 d}+\frac {a \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d} \]

[In]

Integrate[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^6)/(6*d) + (a*(Tan[c + d*x] + (2*Tan[c + d*x]^3)/3 + Tan[c + d*x]^5/5))/d

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {-a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+\frac {b}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(48\)
default	\(\frac {-a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+\frac {b}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(48\)
parts	\(-\frac {a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {b \sec \left (d x +c \right )^{6}}{6 d}\)	\(50\)
risch	\(\frac {\frac {32 i a \,{\mathrm e}^{6 i \left (d x +c \right )}}{3}+\frac {32 b \,{\mathrm e}^{6 i \left (d x +c \right )}}{3}+16 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+\frac {32 i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{5}+\frac {16 i a}{15}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(75\)
parallelrisch	\(-\frac {2 \left (a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} b -\frac {7 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3}+\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a}{5}-\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b}{3}-\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a}{5}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{3}-b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6}}\)	\(143\)
norman	\(\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {86 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {86 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d}+\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {20 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {20 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\)	\(235\)

[In]

int(sec(d*x+c)^7*(cos(d*x+c)*a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-a*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+1/6*b/cos(d*x+c)^6)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {2 \, {\left (8 \, a \cos \left (d x + c\right )^{5} + 4 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 5 \, b}{30 \, d \cos \left (d x + c\right )^{6}} \]

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(2*(8*a*cos(d*x + c)^5 + 4*a*cos(d*x + c)^3 + 3*a*cos(d*x + c))*sin(d*x + c) + 5*b)/(d*cos(d*x + c)^6)

Sympy [F(-1)]

Timed out. \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**7*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {2 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a - \frac {5 \, b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{30 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/30*(2*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a - 5*b/(sin(d*x + c)^2 - 1)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {5 \, b \tan \left (d x + c\right )^{6} + 6 \, a \tan \left (d x + c\right )^{5} + 15 \, b \tan \left (d x + c\right )^{4} + 20 \, a \tan \left (d x + c\right )^{3} + 15 \, b \tan \left (d x + c\right )^{2} + 30 \, a \tan \left (d x + c\right )}{30 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/30*(5*b*tan(d*x + c)^6 + 6*a*tan(d*x + c)^5 + 15*b*tan(d*x + c)^4 + 20*a*tan(d*x + c)^3 + 15*b*tan(d*x + c)^
2 + 30*a*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 21.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {\frac {8\,a\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^5}{15}+\frac {4\,a\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3}{15}+\frac {a\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )}{5}+\frac {b}{6}}{d\,{\cos \left (c+d\,x\right )}^6} \]

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))/cos(c + d*x)^7,x)

[Out]

(b/6 + (a*cos(c + d*x)*sin(c + d*x))/5 + (4*a*cos(c + d*x)^3*sin(c + d*x))/15 + (8*a*cos(c + d*x)^5*sin(c + d*
x))/15)/(d*cos(c + d*x)^6)