3.40 \(\int \sec ^8(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=99 \[ \frac{3 a^3 \tan ^5(c+d x)}{35 d}+\frac{2 a^3 \tan ^3(c+d x)}{7 d}+\frac{3 a^3 \tan (c+d x)}{7 d}+\frac{3 a^3 \sec ^5(c+d x)}{35 d}+\frac{2 a \sec ^7(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

[Out]

(3*a^3*Sec[c + d*x]^5)/(35*d) + (2*a*Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2)/(7*d) + (3*a^3*Tan[c + d*x])/(7*d)
+ (2*a^3*Tan[c + d*x]^3)/(7*d) + (3*a^3*Tan[c + d*x]^5)/(35*d)

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Rubi [A]  time = 0.083488, antiderivative size = 99, 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 = {2676, 2669, 3767} \[ \frac{3 a^3 \tan ^5(c+d x)}{35 d}+\frac{2 a^3 \tan ^3(c+d x)}{7 d}+\frac{3 a^3 \tan (c+d x)}{7 d}+\frac{3 a^3 \sec ^5(c+d x)}{35 d}+\frac{2 a \sec ^7(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + a*Sin[c + d*x])^3,x]

[Out]

(3*a^3*Sec[c + d*x]^5)/(35*d) + (2*a*Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2)/(7*d) + (3*a^3*Tan[c + d*x])/(7*d)
+ (2*a^3*Tan[c + d*x]^3)/(7*d) + (3*a^3*Tan[c + d*x]^5)/(35*d)

Rule 2676

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

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 ^8(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{2 a \sec ^7(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac{1}{7} \left (3 a^2\right ) \int \sec ^6(c+d x) (a+a \sin (c+d x)) \, dx\\ &=\frac{3 a^3 \sec ^5(c+d x)}{35 d}+\frac{2 a \sec ^7(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac{1}{7} \left (3 a^3\right ) \int \sec ^6(c+d x) \, dx\\ &=\frac{3 a^3 \sec ^5(c+d x)}{35 d}+\frac{2 a \sec ^7(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{7 d}\\ &=\frac{3 a^3 \sec ^5(c+d x)}{35 d}+\frac{2 a \sec ^7(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac{3 a^3 \tan (c+d x)}{7 d}+\frac{2 a^3 \tan ^3(c+d x)}{7 d}+\frac{3 a^3 \tan ^5(c+d x)}{35 d}\\ \end{align*}

Mathematica [A]  time = 0.0114133, size = 134, normalized size = 1.35 \[ -\frac{8 a^3 \tan ^7(c+d x)}{35 d}+\frac{13 a^3 \sec ^7(c+d x)}{35 d}+\frac{a^3 \tan ^2(c+d x) \sec ^5(c+d x)}{5 d}-\frac{a^3 \tan ^3(c+d x) \sec ^4(c+d x)}{d}+\frac{4 a^3 \tan ^5(c+d x) \sec ^2(c+d x)}{5 d}+\frac{a^3 \tan (c+d x) \sec ^6(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + a*Sin[c + d*x])^3,x]

[Out]

(13*a^3*Sec[c + d*x]^7)/(35*d) + (a^3*Sec[c + d*x]^6*Tan[c + d*x])/d + (a^3*Sec[c + d*x]^5*Tan[c + d*x]^2)/(5*
d) - (a^3*Sec[c + d*x]^4*Tan[c + d*x]^3)/d + (4*a^3*Sec[c + d*x]^2*Tan[c + d*x]^5)/(5*d) - (8*a^3*Tan[c + d*x]
^7)/(35*d)

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Maple [B]  time = 0.082, size = 217, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{35}} \right ) +3\,{a}^{3} \left ( 1/7\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{3\,{a}^{3}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}-{a}^{3} \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{35}} \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)^8*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*(a^3*(1/7*sin(d*x+c)^4/cos(d*x+c)^7+3/35*sin(d*x+c)^4/cos(d*x+c)^5+1/35*sin(d*x+c)^4/cos(d*x+c)^3-1/35*sin
(d*x+c)^4/cos(d*x+c)-1/35*(2+sin(d*x+c)^2)*cos(d*x+c))+3*a^3*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/
cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)+3/7*a^3/cos(d*x+c)^7-a^3*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c
)^4-8/35*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 0.955943, size = 165, normalized size = 1.67 \begin{align*} \frac{{\left (15 \, \tan \left (d x + c\right )^{7} + 42 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3}\right )} a^{3} +{\left (5 \, \tan \left (d x + c\right )^{7} + 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} a^{3} - \frac{{\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} a^{3}}{\cos \left (d x + c\right )^{7}} + \frac{15 \, a^{3}}{\cos \left (d x + c\right )^{7}}}{35 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/35*((15*tan(d*x + c)^7 + 42*tan(d*x + c)^5 + 35*tan(d*x + c)^3)*a^3 + (5*tan(d*x + c)^7 + 21*tan(d*x + c)^5
+ 35*tan(d*x + c)^3 + 35*tan(d*x + c))*a^3 - (7*cos(d*x + c)^2 - 5)*a^3/cos(d*x + c)^7 + 15*a^3/cos(d*x + c)^7
)/d

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Fricas [A]  time = 1.65027, size = 274, normalized size = 2.77 \begin{align*} \frac{8 \, a^{3} \cos \left (d x + c\right )^{4} - 36 \, a^{3} \cos \left (d x + c\right )^{2} + 15 \, a^{3} + 4 \,{\left (6 \, a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3}\right )} \sin \left (d x + c\right )}{35 \,{\left (3 \, d \cos \left (d x + c\right )^{3} - 4 \, d \cos \left (d x + c\right ) -{\left (d \cos \left (d x + c\right )^{3} - 4 \, d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/35*(8*a^3*cos(d*x + c)^4 - 36*a^3*cos(d*x + c)^2 + 15*a^3 + 4*(6*a^3*cos(d*x + c)^2 - 5*a^3)*sin(d*x + c))/(
3*d*cos(d*x + c)^3 - 4*d*cos(d*x + c) - (d*cos(d*x + c)^3 - 4*d*cos(d*x + c))*sin(d*x + c))

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

[Out]

Timed out

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Giac [A]  time = 1.15263, size = 186, normalized size = 1.88 \begin{align*} -\frac{\frac{35 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1} + \frac{525 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 1960 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4025 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 4480 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3143 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1176 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 243 \, a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{7}}}{280 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/280*(35*a^3/(tan(1/2*d*x + 1/2*c) + 1) + (525*a^3*tan(1/2*d*x + 1/2*c)^6 - 1960*a^3*tan(1/2*d*x + 1/2*c)^5
+ 4025*a^3*tan(1/2*d*x + 1/2*c)^4 - 4480*a^3*tan(1/2*d*x + 1/2*c)^3 + 3143*a^3*tan(1/2*d*x + 1/2*c)^2 - 1176*a
^3*tan(1/2*d*x + 1/2*c) + 243*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^7)/d