3.1001 \(\int \sec ^8(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=115 \[ \frac{a^3 (4 A-3 B) \tan ^3(c+d x)}{35 d}+\frac{3 a^3 (4 A-3 B) \tan (c+d x)}{35 d}+\frac{2 (4 A-3 B) \sec ^5(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{35 d}+\frac{(A+B) \sec ^7(c+d x) (a \sin (c+d x)+a)^3}{7 d} \]

[Out]

((A + B)*Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3)/(7*d) + (2*(4*A - 3*B)*Sec[c + d*x]^5*(a^3 + a^3*Sin[c + d*x])
)/(35*d) + (3*a^3*(4*A - 3*B)*Tan[c + d*x])/(35*d) + (a^3*(4*A - 3*B)*Tan[c + d*x]^3)/(35*d)

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Rubi [A]  time = 0.145609, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {2855, 2676, 3767} \[ \frac{a^3 (4 A-3 B) \tan ^3(c+d x)}{35 d}+\frac{3 a^3 (4 A-3 B) \tan (c+d x)}{35 d}+\frac{2 (4 A-3 B) \sec ^5(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{35 d}+\frac{(A+B) \sec ^7(c+d x) (a \sin (c+d x)+a)^3}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + B)*Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3)/(7*d) + (2*(4*A - 3*B)*Sec[c + d*x]^5*(a^3 + a^3*Sin[c + d*x])
)/(35*d) + (3*a^3*(4*A - 3*B)*Tan[c + d*x])/(35*d) + (a^3*(4*A - 3*B)*Tan[c + d*x]^3)/(35*d)

Rule 2855

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

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

Mathematica [A]  time = 0.456818, size = 135, normalized size = 1.17 \[ \frac{a^3 (14 (4 A-3 B) \cos (2 (c+d x))+(3 B-4 A) \cos (4 (c+d x))+56 A \sin (c+d x)-24 A \sin (3 (c+d x))-42 B \sin (c+d x)+18 B \sin (3 (c+d x))+35 B)}{140 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^7 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(35*B + 14*(4*A - 3*B)*Cos[2*(c + d*x)] + (-4*A + 3*B)*Cos[4*(c + d*x)] + 56*A*Sin[c + d*x] - 42*B*Sin[c
+ d*x] - 24*A*Sin[3*(c + d*x)] + 18*B*Sin[3*(c + d*x)]))/(140*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^7*(Cos[(
c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [B]  time = 0.123, size = 435, normalized size = 3.8 \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)^8*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(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*s
in(d*x+c)^4/cos(d*x+c)-1/35*(2+sin(d*x+c)^2)*cos(d*x+c))+B*a^3*(1/7*sin(d*x+c)^5/cos(d*x+c)^7+2/35*sin(d*x+c)^
5/cos(d*x+c)^5)+3*a^3*A*(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*B*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))+3/7*a^3*A/cos(d*x+c)^7+3*B*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)-a^3*A*(-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)+1/7*B*a^3/cos(d*x+c)^7)

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Maxima [B]  time = 1.05954, size = 308, normalized size = 2.68 \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 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 a^{3} +{\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 )} B a^{3} +{\left (5 \, \tan \left (d x + c\right )^{7} + 7 \, \tan \left (d x + c\right )^{5}\right )} B a^{3} - \frac{{\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} A a^{3}}{\cos \left (d x + c\right )^{7}} - \frac{3 \,{\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} B a^{3}}{\cos \left (d x + c\right )^{7}} + \frac{15 \, A a^{3}}{\cos \left (d x + c\right )^{7}} + \frac{5 \, B 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*(A+B*sin(d*x+c)),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*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*a^3 + (15*tan(d*x + c)^7 + 42*tan(d*x + c)^5 + 35*tan(d*x + c)^3)*B
*a^3 + (5*tan(d*x + c)^7 + 7*tan(d*x + c)^5)*B*a^3 - (7*cos(d*x + c)^2 - 5)*A*a^3/cos(d*x + c)^7 - 3*(7*cos(d*
x + c)^2 - 5)*B*a^3/cos(d*x + c)^7 + 15*A*a^3/cos(d*x + c)^7 + 5*B*a^3/cos(d*x + c)^7)/d

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Fricas [A]  time = 1.66649, size = 350, normalized size = 3.04 \begin{align*} \frac{2 \,{\left (4 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 9 \,{\left (4 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \,{\left (3 \, A - 4 \, B\right )} a^{3} +{\left (6 \,{\left (4 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - 5 \,{\left (4 \, A - 3 \, B\right )} 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*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/35*(2*(4*A - 3*B)*a^3*cos(d*x + c)^4 - 9*(4*A - 3*B)*a^3*cos(d*x + c)^2 + 5*(3*A - 4*B)*a^3 + (6*(4*A - 3*B)
*a^3*cos(d*x + c)^2 - 5*(4*A - 3*B)*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*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.36736, size = 351, normalized size = 3.05 \begin{align*} -\frac{\frac{35 \,{\left (A a^{3} - B a^{3}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1} + \frac{525 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 35 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 1960 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 280 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4025 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 665 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 4480 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1120 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3143 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 791 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1176 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 392 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 243 \, A a^{3} - 51 \, B 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*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/280*(35*(A*a^3 - B*a^3)/(tan(1/2*d*x + 1/2*c) + 1) + (525*A*a^3*tan(1/2*d*x + 1/2*c)^6 + 35*B*a^3*tan(1/2*d
*x + 1/2*c)^6 - 1960*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 280*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 4025*A*a^3*tan(1/2*d*x
+ 1/2*c)^4 - 665*B*a^3*tan(1/2*d*x + 1/2*c)^4 - 4480*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 1120*B*a^3*tan(1/2*d*x + 1
/2*c)^3 + 3143*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 791*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 1176*A*a^3*tan(1/2*d*x + 1/2*
c) + 392*B*a^3*tan(1/2*d*x + 1/2*c) + 243*A*a^3 - 51*B*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^7)/d