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

Optimal. Leaf size=369 \[ \frac{a b \left (-104 a^4 b^2-803 a^2 b^4+8 a^6-256 b^6\right ) \cos (c+d x)}{6 d}+\frac{b \left (2 a^2-7 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^5}{3 d}+\frac{a b \left (2 a^2-13 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{3 d}+\frac{b \left (-72 a^2 b^2+8 a^4-35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{12 d}+\frac{a b \left (-88 a^2 b^2+8 a^4-151 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{12 d}+\frac{b^2 \left (-200 a^4 b^2-866 a^2 b^4+16 a^6-105 b^6\right ) \sin (c+d x) \cos (c+d x)}{24 d}-\frac{\sec (c+d x) \left (5 a b-\left (2 a^2-7 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^6}{3 d}+\frac{35}{8} b^4 x \left (16 a^2 b^2+16 a^4+b^4\right )+\frac{\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^7}{3 d} \]

[Out]

(35*b^4*(16*a^4 + 16*a^2*b^2 + b^4)*x)/8 + (a*b*(8*a^6 - 104*a^4*b^2 - 803*a^2*b^4 - 256*b^6)*Cos[c + d*x])/(6
*d) + (b^2*(16*a^6 - 200*a^4*b^2 - 866*a^2*b^4 - 105*b^6)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + (a*b*(8*a^4 - 88
*a^2*b^2 - 151*b^4)*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(12*d) + (b*(8*a^4 - 72*a^2*b^2 - 35*b^4)*Cos[c + d*x
]*(a + b*Sin[c + d*x])^3)/(12*d) + (a*b*(2*a^2 - 13*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^4)/(3*d) + (b*(2*a^
2 - 7*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^5)/(3*d) + (Sec[c + d*x]^3*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*
x])^7)/(3*d) - (Sec[c + d*x]*(a + b*Sin[c + d*x])^6*(5*a*b - (2*a^2 - 7*b^2)*Sin[c + d*x]))/(3*d)

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Rubi [A]  time = 0.645708, antiderivative size = 369, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2691, 2861, 2753, 2734} \[ \frac{a b \left (-104 a^4 b^2-803 a^2 b^4+8 a^6-256 b^6\right ) \cos (c+d x)}{6 d}+\frac{b \left (2 a^2-7 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^5}{3 d}+\frac{a b \left (2 a^2-13 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{3 d}+\frac{b \left (-72 a^2 b^2+8 a^4-35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{12 d}+\frac{a b \left (-88 a^2 b^2+8 a^4-151 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{12 d}+\frac{b^2 \left (-200 a^4 b^2-866 a^2 b^4+16 a^6-105 b^6\right ) \sin (c+d x) \cos (c+d x)}{24 d}-\frac{\sec (c+d x) \left (5 a b-\left (2 a^2-7 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^6}{3 d}+\frac{35}{8} b^4 x \left (16 a^2 b^2+16 a^4+b^4\right )+\frac{\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^7}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*b^4*(16*a^4 + 16*a^2*b^2 + b^4)*x)/8 + (a*b*(8*a^6 - 104*a^4*b^2 - 803*a^2*b^4 - 256*b^6)*Cos[c + d*x])/(6
*d) + (b^2*(16*a^6 - 200*a^4*b^2 - 866*a^2*b^4 - 105*b^6)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + (a*b*(8*a^4 - 88
*a^2*b^2 - 151*b^4)*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(12*d) + (b*(8*a^4 - 72*a^2*b^2 - 35*b^4)*Cos[c + d*x
]*(a + b*Sin[c + d*x])^3)/(12*d) + (a*b*(2*a^2 - 13*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^4)/(3*d) + (b*(2*a^
2 - 7*b^2)*Cos[c + d*x]*(a + b*Sin[c + d*x])^5)/(3*d) + (Sec[c + d*x]^3*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*
x])^7)/(3*d) - (Sec[c + d*x]*(a + b*Sin[c + d*x])^6*(5*a*b - (2*a^2 - 7*b^2)*Sin[c + d*x]))/(3*d)

Rule 2691

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

Rule 2861

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

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 2734

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

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+b \sin (c+d x))^8 \, dx &=\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{3 d}-\frac{1}{3} \int \sec ^2(c+d x) (a+b \sin (c+d x))^6 \left (-2 a^2+7 b^2+5 a b \sin (c+d x)\right ) \, dx\\ &=\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{3 d}-\frac{\sec (c+d x) (a+b \sin (c+d x))^6 \left (5 a b-\left (2 a^2-7 b^2\right ) \sin (c+d x)\right )}{3 d}+\frac{1}{3} \int (a+b \sin (c+d x))^5 \left (30 a b^2-6 b \left (2 a^2-7 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac{b \left (2 a^2-7 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^5}{3 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{3 d}-\frac{\sec (c+d x) (a+b \sin (c+d x))^6 \left (5 a b-\left (2 a^2-7 b^2\right ) \sin (c+d x)\right )}{3 d}+\frac{1}{18} \int (a+b \sin (c+d x))^4 \left (30 b^2 \left (4 a^2+7 b^2\right )-30 a b \left (2 a^2-13 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac{a b \left (2 a^2-13 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{3 d}+\frac{b \left (2 a^2-7 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^5}{3 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{3 d}-\frac{\sec (c+d x) (a+b \sin (c+d x))^6 \left (5 a b-\left (2 a^2-7 b^2\right ) \sin (c+d x)\right )}{3 d}+\frac{1}{90} \int (a+b \sin (c+d x))^3 \left (90 a b^2 \left (4 a^2+29 b^2\right )-30 b \left (8 a^4-72 a^2 b^2-35 b^4\right ) \sin (c+d x)\right ) \, dx\\ &=\frac{b \left (8 a^4-72 a^2 b^2-35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{12 d}+\frac{a b \left (2 a^2-13 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{3 d}+\frac{b \left (2 a^2-7 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^5}{3 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{3 d}-\frac{\sec (c+d x) (a+b \sin (c+d x))^6 \left (5 a b-\left (2 a^2-7 b^2\right ) \sin (c+d x)\right )}{3 d}+\frac{1}{360} \int (a+b \sin (c+d x))^2 \left (90 b^2 \left (8 a^4+188 a^2 b^2+35 b^4\right )-90 a b \left (8 a^4-88 a^2 b^2-151 b^4\right ) \sin (c+d x)\right ) \, dx\\ &=\frac{a b \left (8 a^4-88 a^2 b^2-151 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{12 d}+\frac{b \left (8 a^4-72 a^2 b^2-35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{12 d}+\frac{a b \left (2 a^2-13 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{3 d}+\frac{b \left (2 a^2-7 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^5}{3 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{3 d}-\frac{\sec (c+d x) (a+b \sin (c+d x))^6 \left (5 a b-\left (2 a^2-7 b^2\right ) \sin (c+d x)\right )}{3 d}+\frac{\int (a+b \sin (c+d x)) \left (90 a b^2 \left (8 a^4+740 a^2 b^2+407 b^4\right )-90 b \left (16 a^6-200 a^4 b^2-866 a^2 b^4-105 b^6\right ) \sin (c+d x)\right ) \, dx}{1080}\\ &=\frac{35}{8} b^4 \left (16 a^4+16 a^2 b^2+b^4\right ) x+\frac{a b \left (8 a^6-104 a^4 b^2-803 a^2 b^4-256 b^6\right ) \cos (c+d x)}{6 d}+\frac{b^2 \left (16 a^6-200 a^4 b^2-866 a^2 b^4-105 b^6\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{a b \left (8 a^4-88 a^2 b^2-151 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^2}{12 d}+\frac{b \left (8 a^4-72 a^2 b^2-35 b^4\right ) \cos (c+d x) (a+b \sin (c+d x))^3}{12 d}+\frac{a b \left (2 a^2-13 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^4}{3 d}+\frac{b \left (2 a^2-7 b^2\right ) \cos (c+d x) (a+b \sin (c+d x))^5}{3 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^7}{3 d}-\frac{\sec (c+d x) (a+b \sin (c+d x))^6 \left (5 a b-\left (2 a^2-7 b^2\right ) \sin (c+d x)\right )}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.16431, size = 414, normalized size = 1.12 \[ \frac{\sec ^3(c+d x) \left (5376 a^6 b^2 \sin (c+d x)-1792 a^6 b^2 \sin (3 (c+d x))-17920 a^4 b^4 \sin (3 (c+d x))-6720 a^2 b^6 \sin (c+d x)-14560 a^2 b^6 \sin (3 (c+d x))-672 a^2 b^6 \sin (5 (c+d x))-21504 a^5 b^3 \cos (2 (c+d x))+40320 a^4 b^4 (c+d x) \cos (c+d x)+13440 a^4 b^4 (c+d x) \cos (3 (c+d x))-64512 a^3 b^5 \cos (2 (c+d x))-5376 a^3 b^5 \cos (4 (c+d x))+40320 a^2 b^6 (c+d x) \cos (c+d x)+13440 a^2 b^6 (c+d x) \cos (3 (c+d x))-7168 a^5 b^3-44800 a^3 b^5+2048 a^7 b+384 a^8 \sin (c+d x)+128 a^8 \sin (3 (c+d x))-17472 a b^7 \cos (2 (c+d x))-1920 a b^7 \cos (4 (c+d x))+64 a b^7 \cos (6 (c+d x))-13440 a b^7-525 b^8 \sin (c+d x)-847 b^8 \sin (3 (c+d x))-63 b^8 \sin (5 (c+d x))+3 b^8 \sin (7 (c+d x))+2520 b^8 (c+d x) \cos (c+d x)+840 b^8 (c+d x) \cos (3 (c+d x))\right )}{768 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(2048*a^7*b - 7168*a^5*b^3 - 44800*a^3*b^5 - 13440*a*b^7 + 40320*a^4*b^4*(c + d*x)*Cos[c + d*x
] + 40320*a^2*b^6*(c + d*x)*Cos[c + d*x] + 2520*b^8*(c + d*x)*Cos[c + d*x] - 21504*a^5*b^3*Cos[2*(c + d*x)] -
64512*a^3*b^5*Cos[2*(c + d*x)] - 17472*a*b^7*Cos[2*(c + d*x)] + 13440*a^4*b^4*(c + d*x)*Cos[3*(c + d*x)] + 134
40*a^2*b^6*(c + d*x)*Cos[3*(c + d*x)] + 840*b^8*(c + d*x)*Cos[3*(c + d*x)] - 5376*a^3*b^5*Cos[4*(c + d*x)] - 1
920*a*b^7*Cos[4*(c + d*x)] + 64*a*b^7*Cos[6*(c + d*x)] + 384*a^8*Sin[c + d*x] + 5376*a^6*b^2*Sin[c + d*x] - 67
20*a^2*b^6*Sin[c + d*x] - 525*b^8*Sin[c + d*x] + 128*a^8*Sin[3*(c + d*x)] - 1792*a^6*b^2*Sin[3*(c + d*x)] - 17
920*a^4*b^4*Sin[3*(c + d*x)] - 14560*a^2*b^6*Sin[3*(c + d*x)] - 847*b^8*Sin[3*(c + d*x)] - 672*a^2*b^6*Sin[5*(
c + d*x)] - 63*b^8*Sin[5*(c + d*x)] + 3*b^8*Sin[7*(c + d*x)]))/(768*d)

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Maple [A]  time = 0.134, size = 495, normalized size = 1.3 \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+b*sin(d*x+c))^8,x)

[Out]

1/d*(-a^8*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+8/3*a^7*b/cos(d*x+c)^3+28/3*a^6*b^2*sin(d*x+c)^3/cos(d*x+c)^3+56*
a^5*b^3*(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))+70*a^4*b^4
*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c)+56*a^3*b^5*(1/3*sin(d*x+c)^6/cos(d*x+c)^3-sin(d*x+c)^6/cos(d*x+c)-(8/3+si
n(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c))+28*a^2*b^6*(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)+8*a*b^7*(1/3*sin(d*x+c)^8/cos(d
*x+c)^3-5/3*sin(d*x+c)^8/cos(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))+b^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+3
5/16*sin(d*x+c))*cos(d*x+c)+35/8*d*x+35/8*c))

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Maxima [A]  time = 1.47095, size = 443, normalized size = 1.2 \begin{align*} \frac{224 \, a^{6} b^{2} \tan \left (d x + c\right )^{3} + 8 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \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^{4} b^{4} + 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^{2} b^{6} + 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 b^{7} +{\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 )} b^{8} - 448 \, a^{3} b^{5}{\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^{5} b^{3}}{\cos \left (d x + c\right )^{3}} + \frac{64 \, a^{7} b}{\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+b*sin(d*x+c))^8,x, algorithm="maxima")

[Out]

1/24*(224*a^6*b^2*tan(d*x + c)^3 + 8*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^8 + 560*(tan(d*x + c)^3 + 3*d*x + 3*c
- 3*tan(d*x + c))*a^4*b^4 + 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^2*b^6 + 64*(cos(d*x + c)^3 - (9*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*a*b^7 + (
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))*b^8 - 448*a^3*b^5*((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^5*b^3/cos(d*x + c)^3 + 64*a^7*b/cos(d*x + c)^3)/d

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Fricas [A]  time = 2.91977, size = 633, normalized size = 1.72 \begin{align*} \frac{64 \, a b^{7} \cos \left (d x + c\right )^{6} + 64 \, a^{7} b + 448 \, a^{5} b^{3} + 448 \, a^{3} b^{5} + 64 \, a b^{7} + 105 \,{\left (16 \, a^{4} b^{4} + 16 \, a^{2} b^{6} + b^{8}\right )} d x \cos \left (d x + c\right )^{3} - 192 \,{\left (7 \, a^{3} b^{5} + 3 \, a b^{7}\right )} \cos \left (d x + c\right )^{4} - 192 \,{\left (7 \, a^{5} b^{3} + 14 \, a^{3} b^{5} + 3 \, a b^{7}\right )} \cos \left (d x + c\right )^{2} +{\left (6 \, b^{8} \cos \left (d x + c\right )^{6} + 8 \, a^{8} + 224 \, a^{6} b^{2} + 560 \, a^{4} b^{4} + 224 \, a^{2} b^{6} + 8 \, b^{8} - 3 \,{\left (112 \, a^{2} b^{6} + 13 \, b^{8}\right )} \cos \left (d x + c\right )^{4} + 16 \,{\left (a^{8} - 14 \, a^{6} b^{2} - 140 \, a^{4} b^{4} - 98 \, a^{2} b^{6} - 5 \, b^{8}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(64*a*b^7*cos(d*x + c)^6 + 64*a^7*b + 448*a^5*b^3 + 448*a^3*b^5 + 64*a*b^7 + 105*(16*a^4*b^4 + 16*a^2*b^6
+ b^8)*d*x*cos(d*x + c)^3 - 192*(7*a^3*b^5 + 3*a*b^7)*cos(d*x + c)^4 - 192*(7*a^5*b^3 + 14*a^3*b^5 + 3*a*b^7)
*cos(d*x + c)^2 + (6*b^8*cos(d*x + c)^6 + 8*a^8 + 224*a^6*b^2 + 560*a^4*b^4 + 224*a^2*b^6 + 8*b^8 - 3*(112*a^2
*b^6 + 13*b^8)*cos(d*x + c)^4 + 16*(a^8 - 14*a^6*b^2 - 140*a^4*b^4 - 98*a^2*b^6 - 5*b^8)*cos(d*x + c)^2)*sin(d
*x + c))/(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)**4*(a+b*sin(d*x+c))**8,x)

[Out]

Timed out

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Giac [A]  time = 1.18347, size = 923, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(105*(16*a^4*b^4 + 16*a^2*b^6 + b^8)*(d*x + c) - 16*(3*a^8*tan(1/2*d*x + 1/2*c)^5 - 210*a^4*b^4*tan(1/2*d
*x + 1/2*c)^5 - 168*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 - 9*b^8*tan(1/2*d*x + 1/2*c)^5 + 24*a^7*b*tan(1/2*d*x + 1/2
*c)^4 - 168*a^3*b^5*tan(1/2*d*x + 1/2*c)^4 - 48*a*b^7*tan(1/2*d*x + 1/2*c)^4 - 2*a^8*tan(1/2*d*x + 1/2*c)^3 +
112*a^6*b^2*tan(1/2*d*x + 1/2*c)^3 + 700*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 + 448*a^2*b^6*tan(1/2*d*x + 1/2*c)^3 +
22*b^8*tan(1/2*d*x + 1/2*c)^3 + 336*a^5*b^3*tan(1/2*d*x + 1/2*c)^2 + 672*a^3*b^5*tan(1/2*d*x + 1/2*c)^2 + 144
*a*b^7*tan(1/2*d*x + 1/2*c)^2 + 3*a^8*tan(1/2*d*x + 1/2*c) - 210*a^4*b^4*tan(1/2*d*x + 1/2*c) - 168*a^2*b^6*ta
n(1/2*d*x + 1/2*c) - 9*b^8*tan(1/2*d*x + 1/2*c) + 8*a^7*b - 112*a^5*b^3 - 280*a^3*b^5 - 64*a*b^7)/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^3 + 2*(336*a^2*b^6*tan(1/2*d*x + 1/2*c)^7 + 33*b^8*tan(1/2*d*x + 1/2*c)^7 - 1344*a^3*b^5*tan(
1/2*d*x + 1/2*c)^6 - 384*a*b^7*tan(1/2*d*x + 1/2*c)^6 + 336*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 + 57*b^8*tan(1/2*d*
x + 1/2*c)^5 - 4032*a^3*b^5*tan(1/2*d*x + 1/2*c)^4 - 1536*a*b^7*tan(1/2*d*x + 1/2*c)^4 - 336*a^2*b^6*tan(1/2*d
*x + 1/2*c)^3 - 57*b^8*tan(1/2*d*x + 1/2*c)^3 - 4032*a^3*b^5*tan(1/2*d*x + 1/2*c)^2 - 1664*a*b^7*tan(1/2*d*x +
1/2*c)^2 - 336*a^2*b^6*tan(1/2*d*x + 1/2*c) - 33*b^8*tan(1/2*d*x + 1/2*c) - 1344*a^3*b^5 - 512*a*b^7)/(tan(1/
2*d*x + 1/2*c)^2 + 1)^4)/d