3.49 \(\int \sec ^4(c+d x) (a+a \sin (c+d x))^8 \, dx\)

Optimal. Leaf size=179 \[ -\frac{385 a^8 \cos ^3(c+d x)}{4 d}-\frac{231 a^{16} \cos ^5(c+d x)}{4 d \left (a^8-a^8 \sin (c+d x)\right )}+\frac{2 a^{15} \cos ^{11}(c+d x)}{3 d (a-a \sin (c+d x))^7}-\frac{66 a^{14} \cos ^7(c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )^3}-\frac{22 a^{13} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^5}+\frac{1155 a^8 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1155 a^8 x}{8} \]

[Out]

(1155*a^8*x)/8 - (385*a^8*Cos[c + d*x]^3)/(4*d) + (1155*a^8*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (2*a^15*Cos[c +
d*x]^11)/(3*d*(a - a*Sin[c + d*x])^7) - (22*a^13*Cos[c + d*x]^9)/(3*d*(a - a*Sin[c + d*x])^5) - (66*a^14*Cos[
c + d*x]^7)/(d*(a^2 - a^2*Sin[c + d*x])^3) - (231*a^16*Cos[c + d*x]^5)/(4*d*(a^8 - a^8*Sin[c + d*x]))

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Rubi [A]  time = 0.318657, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2670, 2680, 2679, 2682, 2635, 8} \[ -\frac{385 a^8 \cos ^3(c+d x)}{4 d}-\frac{231 a^{16} \cos ^5(c+d x)}{4 d \left (a^8-a^8 \sin (c+d x)\right )}+\frac{2 a^{15} \cos ^{11}(c+d x)}{3 d (a-a \sin (c+d x))^7}-\frac{66 a^{14} \cos ^7(c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )^3}-\frac{22 a^{13} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^5}+\frac{1155 a^8 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1155 a^8 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^8,x]

[Out]

(1155*a^8*x)/8 - (385*a^8*Cos[c + d*x]^3)/(4*d) + (1155*a^8*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (2*a^15*Cos[c +
d*x]^11)/(3*d*(a - a*Sin[c + d*x])^7) - (22*a^13*Cos[c + d*x]^9)/(3*d*(a - a*Sin[c + d*x])^5) - (66*a^14*Cos[
c + d*x]^7)/(d*(a^2 - a^2*Sin[c + d*x])^3) - (231*a^16*Cos[c + d*x]^5)/(4*d*(a^8 - a^8*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 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
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] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2679

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

Rule 2682

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

Rule 2635

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+a \sin (c+d x))^8 \, dx &=a^{16} \int \frac{\cos ^{12}(c+d x)}{(a-a \sin (c+d x))^8} \, dx\\ &=\frac{2 a^{15} \cos ^{11}(c+d x)}{3 d (a-a \sin (c+d x))^7}-\frac{1}{3} \left (11 a^{14}\right ) \int \frac{\cos ^{10}(c+d x)}{(a-a \sin (c+d x))^6} \, dx\\ &=\frac{2 a^{15} \cos ^{11}(c+d x)}{3 d (a-a \sin (c+d x))^7}-\frac{22 a^{13} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^5}+\left (33 a^{12}\right ) \int \frac{\cos ^8(c+d x)}{(a-a \sin (c+d x))^4} \, dx\\ &=\frac{2 a^{15} \cos ^{11}(c+d x)}{3 d (a-a \sin (c+d x))^7}-\frac{22 a^{13} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^5}-\frac{66 a^{11} \cos ^7(c+d x)}{d (a-a \sin (c+d x))^3}+\left (231 a^{10}\right ) \int \frac{\cos ^6(c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac{2 a^{15} \cos ^{11}(c+d x)}{3 d (a-a \sin (c+d x))^7}-\frac{22 a^{13} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^5}-\frac{66 a^{11} \cos ^7(c+d x)}{d (a-a \sin (c+d x))^3}-\frac{231 a^{10} \cos ^5(c+d x)}{4 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{1}{4} \left (1155 a^9\right ) \int \frac{\cos ^4(c+d x)}{a-a \sin (c+d x)} \, dx\\ &=-\frac{385 a^8 \cos ^3(c+d x)}{4 d}+\frac{2 a^{15} \cos ^{11}(c+d x)}{3 d (a-a \sin (c+d x))^7}-\frac{22 a^{13} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^5}-\frac{66 a^{11} \cos ^7(c+d x)}{d (a-a \sin (c+d x))^3}-\frac{231 a^{10} \cos ^5(c+d x)}{4 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{1}{4} \left (1155 a^8\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{385 a^8 \cos ^3(c+d x)}{4 d}+\frac{1155 a^8 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{2 a^{15} \cos ^{11}(c+d x)}{3 d (a-a \sin (c+d x))^7}-\frac{22 a^{13} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^5}-\frac{66 a^{11} \cos ^7(c+d x)}{d (a-a \sin (c+d x))^3}-\frac{231 a^{10} \cos ^5(c+d x)}{4 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{1}{8} \left (1155 a^8\right ) \int 1 \, dx\\ &=\frac{1155 a^8 x}{8}-\frac{385 a^8 \cos ^3(c+d x)}{4 d}+\frac{1155 a^8 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{2 a^{15} \cos ^{11}(c+d x)}{3 d (a-a \sin (c+d x))^7}-\frac{22 a^{13} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^5}-\frac{66 a^{11} \cos ^7(c+d x)}{d (a-a \sin (c+d x))^3}-\frac{231 a^{10} \cos ^5(c+d x)}{4 d \left (a^2-a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.0528785, size = 59, normalized size = 0.33 \[ \frac{64 \sqrt{2} a^8 (\sin (c+d x)+1)^{3/2} \sec ^3(c+d x) \, _2F_1\left (-\frac{11}{2},-\frac{3}{2};-\frac{1}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^8,x]

[Out]

(64*Sqrt[2]*a^8*Hypergeometric2F1[-11/2, -3/2, -1/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^3*(1 + Sin[c + d*x])^(
3/2))/(3*d)

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Maple [B]  time = 0.11, size = 478, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^8,x)

[Out]

1/d*(a^8*(1/3*sin(d*x+c)^9/cos(d*x+c)^3-2*sin(d*x+c)^9/cos(d*x+c)-2*(sin(d*x+c)^7+7/6*sin(d*x+c)^5+35/24*sin(d
*x+c)^3+35/16*sin(d*x+c))*cos(d*x+c)+35/8*d*x+35/8*c)+8*a^8*(1/3*sin(d*x+c)^8/cos(d*x+c)^3-5/3*sin(d*x+c)^8/co
s(d*x+c)-5/3*(16/5+sin(d*x+c)^6+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c))+28*a^8*(1/3*sin(d*x+c)^7/cos(d*
x+c)^3-4/3*sin(d*x+c)^7/cos(d*x+c)-4/3*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/2*d*x+5/2*
c)+56*a^8*(1/3*sin(d*x+c)^6/cos(d*x+c)^3-sin(d*x+c)^6/cos(d*x+c)-(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c
))+70*a^8*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c)+56*a^8*(1/3*sin(d*x+c)^4/cos(d*x+c)^3-1/3*sin(d*x+c)^4/cos(d*x+c
)-1/3*(2+sin(d*x+c)^2)*cos(d*x+c))+28/3*a^8*sin(d*x+c)^3/cos(d*x+c)^3+8/3*a^8/cos(d*x+c)^3-a^8*(-2/3-1/3*sec(d
*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.46647, size = 420, normalized size = 2.35 \begin{align*} \frac{224 \, a^{8} \tan \left (d x + c\right )^{3} + 64 \,{\left (\cos \left (d x + c\right )^{3} - \frac{9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} a^{8} +{\left (8 \, \tan \left (d x + c\right )^{3} + 105 \, d x + 105 \, c - \frac{3 \,{\left (13 \, \tan \left (d x + c\right )^{3} + 11 \, \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 72 \, \tan \left (d x + c\right )\right )} a^{8} + 112 \,{\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac{3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a^{8} + 560 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{8} + 8 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{8} - 448 \, a^{8}{\left (\frac{6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} - \frac{448 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{8}}{\cos \left (d x + c\right )^{3}} + \frac{64 \, a^{8}}{\cos \left (d x + c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

1/24*(224*a^8*tan(d*x + c)^3 + 64*(cos(d*x + c)^3 - (9*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*a^
8 + (8*tan(d*x + c)^3 + 105*d*x + 105*c - 3*(13*tan(d*x + c)^3 + 11*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x
+ c)^2 + 1) - 72*tan(d*x + c))*a^8 + 112*(2*tan(d*x + c)^3 + 15*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 +
1) - 12*tan(d*x + c))*a^8 + 560*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^8 + 8*(tan(d*x + c)^3 + 3*ta
n(d*x + c))*a^8 - 448*a^8*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c)) - 448*(3*cos(d*x + c)^2 - 1
)*a^8/cos(d*x + c)^3 + 64*a^8/cos(d*x + c)^3)/d

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Fricas [A]  time = 1.71977, size = 639, normalized size = 3.57 \begin{align*} -\frac{6 \, a^{8} \cos \left (d x + c\right )^{6} - 52 \, a^{8} \cos \left (d x + c\right )^{5} - 317 \, a^{8} \cos \left (d x + c\right )^{4} + 1286 \, a^{8} \cos \left (d x + c\right )^{3} + 6930 \, a^{8} d x + 512 \, a^{8} -{\left (3465 \, a^{8} d x + 5641 \, a^{8}\right )} \cos \left (d x + c\right )^{2} +{\left (3465 \, a^{8} d x - 6674 \, a^{8}\right )} \cos \left (d x + c\right ) -{\left (6 \, a^{8} \cos \left (d x + c\right )^{5} + 58 \, a^{8} \cos \left (d x + c\right )^{4} - 259 \, a^{8} \cos \left (d x + c\right )^{3} + 6930 \, a^{8} d x - 1545 \, a^{8} \cos \left (d x + c\right )^{2} - 512 \, a^{8} +{\left (3465 \, a^{8} d x - 7186 \, a^{8}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \,{\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) +{\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^8,x, algorithm="fricas")

[Out]

-1/24*(6*a^8*cos(d*x + c)^6 - 52*a^8*cos(d*x + c)^5 - 317*a^8*cos(d*x + c)^4 + 1286*a^8*cos(d*x + c)^3 + 6930*
a^8*d*x + 512*a^8 - (3465*a^8*d*x + 5641*a^8)*cos(d*x + c)^2 + (3465*a^8*d*x - 6674*a^8)*cos(d*x + c) - (6*a^8
*cos(d*x + c)^5 + 58*a^8*cos(d*x + c)^4 - 259*a^8*cos(d*x + c)^3 + 6930*a^8*d*x - 1545*a^8*cos(d*x + c)^2 - 51
2*a^8 + (3465*a^8*d*x - 7186*a^8)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x
+ c) + 2*d)*sin(d*x + c) - 2*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)**4*(a+a*sin(d*x+c))**8,x)

[Out]

Timed out

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Giac [A]  time = 1.23457, size = 270, normalized size = 1.51 \begin{align*} \frac{3465 \,{\left (d x + c\right )} a^{8} + \frac{1024 \,{\left (6 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7 \, a^{8}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{2 \,{\left (369 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1728 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 393 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 5568 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 393 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5696 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 369 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1856 \, a^{8}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^8,x, algorithm="giac")

[Out]

1/24*(3465*(d*x + c)*a^8 + 1024*(6*a^8*tan(1/2*d*x + 1/2*c)^2 - 15*a^8*tan(1/2*d*x + 1/2*c) + 7*a^8)/(tan(1/2*
d*x + 1/2*c) - 1)^3 + 2*(369*a^8*tan(1/2*d*x + 1/2*c)^7 - 1728*a^8*tan(1/2*d*x + 1/2*c)^6 + 393*a^8*tan(1/2*d*
x + 1/2*c)^5 - 5568*a^8*tan(1/2*d*x + 1/2*c)^4 - 393*a^8*tan(1/2*d*x + 1/2*c)^3 - 5696*a^8*tan(1/2*d*x + 1/2*c
)^2 - 369*a^8*tan(1/2*d*x + 1/2*c) - 1856*a^8)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d