3.38 \(\int \sec ^6(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=92 \[ \frac{2 a^6 \cos (c+d x)}{15 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac{a^6 \cos (c+d x)}{5 d (a-a \sin (c+d x))^3}+\frac{2 a^5 \cos (c+d x)}{15 d (a-a \sin (c+d x))^2} \]

[Out]

(a^6*Cos[c + d*x])/(5*d*(a - a*Sin[c + d*x])^3) + (2*a^5*Cos[c + d*x])/(15*d*(a - a*Sin[c + d*x])^2) + (2*a^6*
Cos[c + d*x])/(15*d*(a^3 - a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.0934226, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2670, 2650, 2648} \[ \frac{2 a^6 \cos (c+d x)}{15 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac{a^6 \cos (c+d x)}{5 d (a-a \sin (c+d x))^3}+\frac{2 a^5 \cos (c+d x)}{15 d (a-a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^6*Cos[c + d*x])/(5*d*(a - a*Sin[c + d*x])^3) + (2*a^5*Cos[c + d*x])/(15*d*(a - a*Sin[c + d*x])^2) + (2*a^6*
Cos[c + d*x])/(15*d*(a^3 - a^3*Sin[c + d*x]))

Rule 2670

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a/g)^
(2*m), Int[(g*Cos[e + f*x])^(2*m + p)/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 -
b^2, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 \, dx &=a^6 \int \frac{1}{(a-a \sin (c+d x))^3} \, dx\\ &=\frac{a^6 \cos (c+d x)}{5 d (a-a \sin (c+d x))^3}+\frac{1}{5} \left (2 a^5\right ) \int \frac{1}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac{a^6 \cos (c+d x)}{5 d (a-a \sin (c+d x))^3}+\frac{2 a^5 \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{1}{15} \left (2 a^4\right ) \int \frac{1}{a-a \sin (c+d x)} \, dx\\ &=\frac{a^6 \cos (c+d x)}{5 d (a-a \sin (c+d x))^3}+\frac{2 a^5 \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{2 a^4 \cos (c+d x)}{15 d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.0196873, size = 110, normalized size = 1.2 \[ \frac{2 a^3 \tan ^5(c+d x)}{15 d}+\frac{7 a^3 \sec ^5(c+d x)}{15 d}+\frac{a^3 \tan ^2(c+d x) \sec ^3(c+d x)}{3 d}-\frac{a^3 \tan ^3(c+d x) \sec ^2(c+d x)}{3 d}+\frac{a^3 \tan (c+d x) \sec ^4(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^3,x]

[Out]

(7*a^3*Sec[c + d*x]^5)/(15*d) + (a^3*Sec[c + d*x]^4*Tan[c + d*x])/d + (a^3*Sec[c + d*x]^3*Tan[c + d*x]^2)/(3*d
) - (a^3*Sec[c + d*x]^2*Tan[c + d*x]^3)/(3*d) + (2*a^3*Tan[c + d*x]^5)/(15*d)

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Maple [A]  time = 0.075, size = 171, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{15}} \right ) +3\,{a}^{3} \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{3\,{a}^{3}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{a}^{3} \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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(1/5*sin(d*x+c)^4/cos(d*x+c)^5+1/15*sin(d*x+c)^4/cos(d*x+c)^3-1/15*sin(d*x+c)^4/cos(d*x+c)-1/15*(2+si
n(d*x+c)^2)*cos(d*x+c))+3*a^3*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)+3/5*a^3/cos(d*x+c
)^5-a^3*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 0.971747, size = 139, normalized size = 1.51 \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^{3} + 3 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a^{3} - \frac{{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} a^{3}}{\cos \left (d x + c\right )^{5}} + \frac{9 \, a^{3}}{\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+a*sin(d*x+c))^3,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 + 3*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*a
^3 - (5*cos(d*x + c)^2 - 3)*a^3/cos(d*x + c)^5 + 9*a^3/cos(d*x + c)^5)/d

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Fricas [A]  time = 1.60439, size = 367, normalized size = 3.99 \begin{align*} \frac{2 \, a^{3} \cos \left (d x + c\right )^{3} - 4 \, a^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{3} \cos \left (d x + c\right ) - 3 \, a^{3} +{\left (2 \, a^{3} \cos \left (d x + c\right )^{2} + 6 \, a^{3} \cos \left (d x + c\right ) - 3 \, a^{3}\right )} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{3} + 3 \, d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) -{\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(2*a^3*cos(d*x + c)^3 - 4*a^3*cos(d*x + c)^2 - 9*a^3*cos(d*x + c) - 3*a^3 + (2*a^3*cos(d*x + c)^2 + 6*a^3
*cos(d*x + c) - 3*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3 + 3*d*cos(d*x + c)^2 - 2*d*cos(d*x + c) - (d*cos(d*x +
c)^2 - 2*d*cos(d*x + c) - 4*d)*sin(d*x + c) - 4*d)

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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+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.15781, size = 116, normalized size = 1.26 \begin{align*} -\frac{2 \,{\left (15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 20 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7 \, a^{3}\right )}}{15 \, d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-2/15*(15*a^3*tan(1/2*d*x + 1/2*c)^4 - 30*a^3*tan(1/2*d*x + 1/2*c)^3 + 40*a^3*tan(1/2*d*x + 1/2*c)^2 - 20*a^3*
tan(1/2*d*x + 1/2*c) + 7*a^3)/(d*(tan(1/2*d*x + 1/2*c) - 1)^5)