3.26 \(\int \sec ^8(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=82 \[ \frac{a^2 \tan ^5(c+d x)}{7 d}+\frac{10 a^2 \tan ^3(c+d x)}{21 d}+\frac{5 a^2 \tan (c+d x)}{7 d}+\frac{2 \sec ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{7 d} \]

[Out]

(2*Sec[c + d*x]^7*(a^2 + a^2*Sin[c + d*x]))/(7*d) + (5*a^2*Tan[c + d*x])/(7*d) + (10*a^2*Tan[c + d*x]^3)/(21*d
) + (a^2*Tan[c + d*x]^5)/(7*d)

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Rubi [A]  time = 0.058504, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2676, 3767} \[ \frac{a^2 \tan ^5(c+d x)}{7 d}+\frac{10 a^2 \tan ^3(c+d x)}{21 d}+\frac{5 a^2 \tan (c+d x)}{7 d}+\frac{2 \sec ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + a*Sin[c + d*x])^2,x]

[Out]

(2*Sec[c + d*x]^7*(a^2 + a^2*Sin[c + d*x]))/(7*d) + (5*a^2*Tan[c + d*x])/(7*d) + (10*a^2*Tan[c + d*x]^3)/(21*d
) + (a^2*Tan[c + d*x]^5)/(7*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 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))^2 \, dx &=\frac{2 \sec ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{7 d}+\frac{1}{7} \left (5 a^2\right ) \int \sec ^6(c+d x) \, dx\\ &=\frac{2 \sec ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{7 d}-\frac{\left (5 a^2\right ) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{7 d}\\ &=\frac{2 \sec ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{7 d}+\frac{5 a^2 \tan (c+d x)}{7 d}+\frac{10 a^2 \tan ^3(c+d x)}{21 d}+\frac{a^2 \tan ^5(c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.0234543, size = 110, normalized size = 1.34 \[ -\frac{8 a^2 \tan ^7(c+d x)}{21 d}+\frac{2 a^2 \sec ^7(c+d x)}{7 d}-\frac{5 a^2 \tan ^3(c+d x) \sec ^4(c+d x)}{3 d}+\frac{4 a^2 \tan ^5(c+d x) \sec ^2(c+d x)}{3 d}+\frac{a^2 \tan (c+d x) \sec ^6(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + a*Sin[c + d*x])^2,x]

[Out]

(2*a^2*Sec[c + d*x]^7)/(7*d) + (a^2*Sec[c + d*x]^6*Tan[c + d*x])/d - (5*a^2*Sec[c + d*x]^4*Tan[c + d*x]^3)/(3*
d) + (4*a^2*Sec[c + d*x]^2*Tan[c + d*x]^5)/(3*d) - (8*a^2*Tan[c + d*x]^7)/(21*d)

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Maple [A]  time = 0.112, size = 121, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{7\, \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{2\,{a}^{2}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}-{a}^{2} \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))^2,x)

[Out]

1/d*(a^2*(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)+2/7*a^
2/cos(d*x+c)^7-a^2*(-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 = 0.974856, size = 132, normalized size = 1.61 \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^{2} + 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^{2} + \frac{30 \, a^{2}}{\cos \left (d x + c\right )^{7}}}{105 \, 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))^2,x, algorithm="maxima")

[Out]

1/105*((15*tan(d*x + c)^7 + 42*tan(d*x + c)^5 + 35*tan(d*x + c)^3)*a^2 + 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^2 + 30*a^2/cos(d*x + c)^7)/d

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Fricas [A]  time = 1.68832, size = 278, normalized size = 3.39 \begin{align*} -\frac{16 \, a^{2} \cos \left (d x + c\right )^{4} - 8 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} -{\left (8 \, a^{2} \cos \left (d x + c\right )^{4} - 12 \, a^{2} \cos \left (d x + c\right )^{2} - 5 \, a^{2}\right )} \sin \left (d x + c\right )}{21 \,{\left (d \cos \left (d x + c\right )^{5} + 2 \, d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{3}\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))^2,x, algorithm="fricas")

[Out]

-1/21*(16*a^2*cos(d*x + c)^4 - 8*a^2*cos(d*x + c)^2 - 2*a^2 - (8*a^2*cos(d*x + c)^4 - 12*a^2*cos(d*x + c)^2 -
5*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5 + 2*d*cos(d*x + c)^3*sin(d*x + c) - 2*d*cos(d*x + c)^3)

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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))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.18891, size = 231, normalized size = 2.82 \begin{align*} -\frac{\frac{7 \,{\left (9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}} + \frac{273 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 1155 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2450 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2870 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2037 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 791 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 152 \, a^{2}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{7}}}{168 \, 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))^2,x, algorithm="giac")

[Out]

-1/168*(7*(9*a^2*tan(1/2*d*x + 1/2*c)^2 + 15*a^2*tan(1/2*d*x + 1/2*c) + 8*a^2)/(tan(1/2*d*x + 1/2*c) + 1)^3 +
(273*a^2*tan(1/2*d*x + 1/2*c)^6 - 1155*a^2*tan(1/2*d*x + 1/2*c)^5 + 2450*a^2*tan(1/2*d*x + 1/2*c)^4 - 2870*a^2
*tan(1/2*d*x + 1/2*c)^3 + 2037*a^2*tan(1/2*d*x + 1/2*c)^2 - 791*a^2*tan(1/2*d*x + 1/2*c) + 152*a^2)/(tan(1/2*d
*x + 1/2*c) - 1)^7)/d