\(\int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\) [40]

Optimal result

Integrand size = 26, antiderivative size = 44 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {b \sec ^4(c+d x)}{4 d}+\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \]

[Out]

1/4*b*sec(d*x+c)^4/d+a*tan(d*x+c)/d+1/3*a*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, 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} \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {b \sec ^4(c+d x)}{4 d} \]

[In]

Int[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^4)/(4*d) + (a*Tan[c + d*x])/d + (a*Tan[c + d*x]^3)/(3*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 ^4(c+d x)+b \sec ^4(c+d x) \tan (c+d x)\right ) \, dx \\ & = a \int \sec ^4(c+d x) \, dx+b \int \sec ^4(c+d x) \tan (c+d x) \, dx \\ & = -\frac {a \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}+\frac {b \text {Subst}\left (\int x^3 \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {b \sec ^4(c+d x)}{4 d}+\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {b \sec ^4(c+d x)}{4 d}+\frac {a \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \]

[In]

Integrate[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^4)/(4*d) + (a*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {b}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(38\)
default	\(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {b}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(38\)
parts	\(-\frac {a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b \sec \left (d x +c \right )^{4}}{4 d}\)	\(40\)
risch	\(\frac {4 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+4 b \,{\mathrm e}^{4 i \left (d x +c \right )}+\frac {16 i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}+\frac {4 i a}{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}\)	\(62\)
parallelrisch	\(-\frac {2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b -\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a}{3}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}{3}-b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}\)	\(112\)
norman	\(\frac {\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\)	\(167\)

[In]

int(sec(d*x+c)^5*(cos(d*x+c)*a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+1/4*b/cos(d*x+c)^4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.02 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {4 \, {\left (2 \, a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 3 \, b}{12 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(4*(2*a*cos(d*x + c)^3 + a*cos(d*x + c))*sin(d*x + c) + 3*b)/(d*cos(d*x + c)^4)

Sympy [F]

\[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\int \left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**5*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Integral((a*cos(c + d*x) + b*sin(c + d*x))*sec(c + d*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a + \frac {3 \, b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{12 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a + 3*b/(sin(d*x + c)^2 - 1)^2)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {3 \, b \tan \left (d x + c\right )^{4} + 4 \, a \tan \left (d x + c\right )^{3} + 6 \, b \tan \left (d x + c\right )^{2} + 12 \, a \tan \left (d x + c\right )}{12 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/12*(3*b*tan(d*x + c)^4 + 4*a*tan(d*x + c)^3 + 6*b*tan(d*x + c)^2 + 12*a*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 21.96 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx=\frac {\frac {b}{4}+\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{3}+\frac {a\,\sin \left (4\,c+4\,d\,x\right )}{12}}{d\,{\cos \left (c+d\,x\right )}^4} \]

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))/cos(c + d*x)^5,x)

[Out]

(b/4 + (a*sin(2*c + 2*d*x))/3 + (a*sin(4*c + 4*d*x))/12)/(d*cos(c + d*x)^4)