3.386 \(\int \sec ^6(c+d x) (a+b \sin (c+d x)) \, dx\)

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

[Out]

(b*Sec[c + d*x]^5)/(5*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 = 0.0400507, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2669, 3767} \[ \frac{a \tan ^5(c+d x)}{5 d}+\frac{2 a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{b \sec ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + b*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^5)/(5*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 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 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]

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a+b \sin (c+d x)) \, dx &=\frac{b \sec ^5(c+d x)}{5 d}+a \int \sec ^6(c+d x) \, dx\\ &=\frac{b \sec ^5(c+d x)}{5 d}-\frac{a \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{b \sec ^5(c+d x)}{5 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 = 0.179222, size = 53, normalized size = 0.88 \[ \frac{a \left (\frac{1}{5} \tan ^5(c+d x)+\frac{2}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{b \sec ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + b*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^5)/(5*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 = 0.032, 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}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+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)+1/5*b/cos(d*x+c)^5)

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Maxima [A]  time = 0.951036, size = 65, normalized size = 1.08 \begin{align*} \frac{{\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{3 \, b}{\cos \left (d x + c\right )^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/15*((3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a + 3*b/cos(d*x + c)^5)/d

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Fricas [A]  time = 2.09528, size = 127, normalized size = 2.12 \begin{align*} \frac{{\left (8 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 3 \, a\right )} \sin \left (d x + c\right ) + 3 \, b}{15 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/15*((8*a*cos(d*x + c)^4 + 4*a*cos(d*x + c)^2 + 3*a)*sin(d*x + c) + 3*b)/(d*cos(d*x + c)^5)

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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)**6*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.10191, size = 162, normalized size = 2.7 \begin{align*} -\frac{2 \,{\left (15 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 15 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 20 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 58 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 30 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 20 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, b\right )}}{15 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-2/15*(15*a*tan(1/2*d*x + 1/2*c)^9 + 15*b*tan(1/2*d*x + 1/2*c)^8 - 20*a*tan(1/2*d*x + 1/2*c)^7 + 58*a*tan(1/2*
d*x + 1/2*c)^5 + 30*b*tan(1/2*d*x + 1/2*c)^4 - 20*a*tan(1/2*d*x + 1/2*c)^3 + 15*a*tan(1/2*d*x + 1/2*c) + 3*b)/
((tan(1/2*d*x + 1/2*c)^2 - 1)^5*d)