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

Optimal. Leaf size=60 \[ \frac{a \tan ^5(c+d x)}{5 d}+\frac{2 a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{b \sec ^6(c+d x)}{6 d} \]

[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)

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Rubi [A]  time = 0.0706185, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3090, 3767, 2606, 30} \[ \frac{a \tan ^5(c+d x)}{5 d}+\frac{2 a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{b \sec ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[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 3090

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 3767

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]

Rule 2606

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 30

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

Rubi steps

\begin{align*} \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx &=\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 \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d}+\frac{b \operatorname{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]  time = 0.147887, size = 53, normalized size = 0.88 \[ \frac{a \left (\frac{1}{5} \tan ^5(c+d x)+\frac{2}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{b \sec ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.078, size = 48, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -a \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) +{\frac{b}{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [A]  time = 1.14306, size = 72, normalized size = 1.2 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 0.472191, size = 147, normalized size = 2.45 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1932, size = 95, normalized size = 1.58 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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