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

Optimal. Leaf size=201 \[ \frac{1001 a^8 \cos ^5(c+d x)}{10 d}+\frac{143 a^{16} \cos ^7(c+d x)}{2 d \left (a^8-a^8 \sin (c+d x)\right )}+\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac{286 a^{14} \cos ^9(c+d x)}{3 d \left (a^2-a^2 \sin (c+d x)\right )^3}+\frac{26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}-\frac{1001 a^8 \sin (c+d x) \cos ^3(c+d x)}{8 d}-\frac{3003 a^8 \sin (c+d x) \cos (c+d x)}{16 d}-\frac{3003 a^8 x}{16} \]

[Out]

(-3003*a^8*x)/16 + (1001*a^8*Cos[c + d*x]^5)/(10*d) - (3003*a^8*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (1001*a^8*
Cos[c + d*x]^3*Sin[c + d*x])/(8*d) + (2*a^15*Cos[c + d*x]^13)/(d*(a - a*Sin[c + d*x])^7) + (26*a^13*Cos[c + d*
x]^11)/(d*(a - a*Sin[c + d*x])^5) + (286*a^14*Cos[c + d*x]^9)/(3*d*(a^2 - a^2*Sin[c + d*x])^3) + (143*a^16*Cos
[c + d*x]^7)/(2*d*(a^8 - a^8*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.341912, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{1001 a^8 \cos ^5(c+d x)}{10 d}+\frac{143 a^{16} \cos ^7(c+d x)}{2 d \left (a^8-a^8 \sin (c+d x)\right )}+\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac{286 a^{14} \cos ^9(c+d x)}{3 d \left (a^2-a^2 \sin (c+d x)\right )^3}+\frac{26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}-\frac{1001 a^8 \sin (c+d x) \cos ^3(c+d x)}{8 d}-\frac{3003 a^8 \sin (c+d x) \cos (c+d x)}{16 d}-\frac{3003 a^8 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3003*a^8*x)/16 + (1001*a^8*Cos[c + d*x]^5)/(10*d) - (3003*a^8*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (1001*a^8*
Cos[c + d*x]^3*Sin[c + d*x])/(8*d) + (2*a^15*Cos[c + d*x]^13)/(d*(a - a*Sin[c + d*x])^7) + (26*a^13*Cos[c + d*
x]^11)/(d*(a - a*Sin[c + d*x])^5) + (286*a^14*Cos[c + d*x]^9)/(3*d*(a^2 - a^2*Sin[c + d*x])^3) + (143*a^16*Cos
[c + d*x]^7)/(2*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 ^2(c+d x) (a+a \sin (c+d x))^8 \, dx &=a^{16} \int \frac{\cos ^{14}(c+d x)}{(a-a \sin (c+d x))^8} \, dx\\ &=\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}-\left (13 a^{14}\right ) \int \frac{\cos ^{12}(c+d x)}{(a-a \sin (c+d x))^6} \, dx\\ &=\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac{26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}-\left (143 a^{12}\right ) \int \frac{\cos ^{10}(c+d x)}{(a-a \sin (c+d x))^4} \, dx\\ &=\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac{26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}+\frac{286 a^{11} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^3}-\left (429 a^{10}\right ) \int \frac{\cos ^8(c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac{26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}+\frac{286 a^{11} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^3}+\frac{143 a^{10} \cos ^7(c+d x)}{2 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{1}{2} \left (1001 a^9\right ) \int \frac{\cos ^6(c+d x)}{a-a \sin (c+d x)} \, dx\\ &=\frac{1001 a^8 \cos ^5(c+d x)}{10 d}+\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac{26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}+\frac{286 a^{11} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^3}+\frac{143 a^{10} \cos ^7(c+d x)}{2 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{1}{2} \left (1001 a^8\right ) \int \cos ^4(c+d x) \, dx\\ &=\frac{1001 a^8 \cos ^5(c+d x)}{10 d}-\frac{1001 a^8 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac{26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}+\frac{286 a^{11} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^3}+\frac{143 a^{10} \cos ^7(c+d x)}{2 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{1}{8} \left (3003 a^8\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{1001 a^8 \cos ^5(c+d x)}{10 d}-\frac{3003 a^8 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{1001 a^8 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac{26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}+\frac{286 a^{11} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^3}+\frac{143 a^{10} \cos ^7(c+d x)}{2 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac{1}{16} \left (3003 a^8\right ) \int 1 \, dx\\ &=-\frac{3003 a^8 x}{16}+\frac{1001 a^8 \cos ^5(c+d x)}{10 d}-\frac{3003 a^8 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{1001 a^8 \cos ^3(c+d x) \sin (c+d x)}{8 d}+\frac{2 a^{15} \cos ^{13}(c+d x)}{d (a-a \sin (c+d x))^7}+\frac{26 a^{13} \cos ^{11}(c+d x)}{d (a-a \sin (c+d x))^5}+\frac{286 a^{11} \cos ^9(c+d x)}{3 d (a-a \sin (c+d x))^3}+\frac{143 a^{10} \cos ^7(c+d x)}{2 d \left (a^2-a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.0573164, size = 55, normalized size = 0.27 \[ \frac{128 \sqrt{2} a^8 \sqrt{\sin (c+d x)+1} \sec (c+d x) \, _2F_1\left (-\frac{13}{2},-\frac{1}{2};\frac{1}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(128*Sqrt[2]*a^8*Hypergeometric2F1[-13/2, -1/2, 1/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]*Sqrt[1 + Sin[c + d*x]]
)/d

________________________________________________________________________________________

Maple [B]  time = 0.066, size = 389, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{7}+{\frac{7\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{35\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{35\,\sin \left ( dx+c \right ) }{16}} \right ) \cos \left ( dx+c \right ) -{\frac{35\,dx}{16}}-{\frac{35\,c}{16}} \right ) +8\,{a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{\cos \left ( dx+c \right ) }}+ \left ({\frac{16}{5}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{6}+6/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+8/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +28\,{a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+5/4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) \cos \left ( dx+c \right ) -{\frac{15\,dx}{8}}-{\frac{15\,c}{8}} \right ) +56\,{a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}+ \left ( 8/3+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +70\,{a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+3/2\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) -3/2\,dx-3/2\,c \right ) +56\,{a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +28\,{a}^{8} \left ( \tan \left ( dx+c \right ) -dx-c \right ) +8\,{\frac{{a}^{8}}{\cos \left ( dx+c \right ) }}+{a}^{8}\tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sin(d*x+c))^8,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.45832, size = 447, normalized size = 2.22 \begin{align*} \frac{384 \,{\left (\cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3} + \frac{5}{\cos \left (d x + c\right )} + 15 \, \cos \left (d x + c\right )\right )} a^{8} - 4480 \,{\left (\cos \left (d x + c\right )^{3} - \frac{3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{8} - 5 \,{\left (105 \, d x + 105 \, c - \frac{87 \, \tan \left (d x + c\right )^{5} + 136 \, \tan \left (d x + c\right )^{3} + 57 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1} - 48 \, \tan \left (d x + c\right )\right )} a^{8} - 840 \,{\left (15 \, d x + 15 \, c - \frac{9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{8} - 8400 \,{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{8} - 6720 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{8} + 13440 \, a^{8}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 240 \, a^{8} \tan \left (d x + c\right ) + \frac{1920 \, a^{8}}{\cos \left (d x + c\right )}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

1/240*(384*(cos(d*x + c)^5 - 5*cos(d*x + c)^3 + 5/cos(d*x + c) + 15*cos(d*x + c))*a^8 - 4480*(cos(d*x + c)^3 -
3/cos(d*x + c) - 6*cos(d*x + c))*a^8 - 5*(105*d*x + 105*c - (87*tan(d*x + c)^5 + 136*tan(d*x + c)^3 + 57*tan(
d*x + c))/(tan(d*x + c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1) - 48*tan(d*x + c))*a^8 - 840*(15*d*x + 15
*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1) - 8*tan(d*x + c))*a^8 - 8400*
(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*a^8 - 6720*(d*x + c - tan(d*x + c))*a^8 + 1
3440*a^8*(1/cos(d*x + c) + cos(d*x + c)) + 240*a^8*tan(d*x + c) + 1920*a^8/cos(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.87544, size = 632, normalized size = 3.14 \begin{align*} \frac{40 \, a^{8} \cos \left (d x + c\right )^{7} + 384 \, a^{8} \cos \left (d x + c\right )^{6} - 1526 \, a^{8} \cos \left (d x + c\right )^{5} - 6400 \, a^{8} \cos \left (d x + c\right )^{4} + 11865 \, a^{8} \cos \left (d x + c\right )^{3} - 45045 \, a^{8} d x + 46080 \, a^{8} \cos \left (d x + c\right )^{2} + 30720 \, a^{8} - 15 \,{\left (3003 \, a^{8} d x - 4027 \, a^{8}\right )} \cos \left (d x + c\right ) +{\left (40 \, a^{8} \cos \left (d x + c\right )^{6} - 344 \, a^{8} \cos \left (d x + c\right )^{5} - 1870 \, a^{8} \cos \left (d x + c\right )^{4} + 4530 \, a^{8} \cos \left (d x + c\right )^{3} + 45045 \, a^{8} d x + 16395 \, a^{8} \cos \left (d x + c\right )^{2} - 29685 \, a^{8} \cos \left (d x + c\right ) + 30720 \, a^{8}\right )} \sin \left (d x + c\right )}{240 \,{\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^8,x, algorithm="fricas")

[Out]

1/240*(40*a^8*cos(d*x + c)^7 + 384*a^8*cos(d*x + c)^6 - 1526*a^8*cos(d*x + c)^5 - 6400*a^8*cos(d*x + c)^4 + 11
865*a^8*cos(d*x + c)^3 - 45045*a^8*d*x + 46080*a^8*cos(d*x + c)^2 + 30720*a^8 - 15*(3003*a^8*d*x - 4027*a^8)*c
os(d*x + c) + (40*a^8*cos(d*x + c)^6 - 344*a^8*cos(d*x + c)^5 - 1870*a^8*cos(d*x + c)^4 + 4530*a^8*cos(d*x + c
)^3 + 45045*a^8*d*x + 16395*a^8*cos(d*x + c)^2 - 29685*a^8*cos(d*x + c) + 30720*a^8)*sin(d*x + c))/(d*cos(d*x
+ c) - d*sin(d*x + c) + d)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.21141, size = 312, normalized size = 1.55 \begin{align*} -\frac{45045 \,{\left (d x + c\right )} a^{8} + \frac{61440 \, a^{8}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1} + \frac{2 \,{\left (14565 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 28800 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 50855 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 174720 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 36930 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 400640 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 36930 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 426240 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 50855 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 211584 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 14565 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 40064 \, a^{8}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^8,x, algorithm="giac")

[Out]

-1/240*(45045*(d*x + c)*a^8 + 61440*a^8/(tan(1/2*d*x + 1/2*c) - 1) + 2*(14565*a^8*tan(1/2*d*x + 1/2*c)^11 - 28
800*a^8*tan(1/2*d*x + 1/2*c)^10 + 50855*a^8*tan(1/2*d*x + 1/2*c)^9 - 174720*a^8*tan(1/2*d*x + 1/2*c)^8 + 36930
*a^8*tan(1/2*d*x + 1/2*c)^7 - 400640*a^8*tan(1/2*d*x + 1/2*c)^6 - 36930*a^8*tan(1/2*d*x + 1/2*c)^5 - 426240*a^
8*tan(1/2*d*x + 1/2*c)^4 - 50855*a^8*tan(1/2*d*x + 1/2*c)^3 - 211584*a^8*tan(1/2*d*x + 1/2*c)^2 - 14565*a^8*ta
n(1/2*d*x + 1/2*c) - 40064*a^8)/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d