3.1002 \(\int \sec ^{10}(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=140 \[ \frac{a^3 (2 A-B) \tan ^5(c+d x)}{21 d}+\frac{10 a^3 (2 A-B) \tan ^3(c+d x)}{63 d}+\frac{5 a^3 (2 A-B) \tan (c+d x)}{21 d}+\frac{2 (2 A-B) \sec ^7(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{21 d}+\frac{(A+B) \sec ^9(c+d x) (a \sin (c+d x)+a)^3}{9 d} \]

[Out]

((A + B)*Sec[c + d*x]^9*(a + a*Sin[c + d*x])^3)/(9*d) + (2*(2*A - B)*Sec[c + d*x]^7*(a^3 + a^3*Sin[c + d*x]))/
(21*d) + (5*a^3*(2*A - B)*Tan[c + d*x])/(21*d) + (10*a^3*(2*A - B)*Tan[c + d*x]^3)/(63*d) + (a^3*(2*A - B)*Tan
[c + d*x]^5)/(21*d)

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Rubi [A]  time = 0.151401, antiderivative size = 140, 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} \[ \frac{a^3 (2 A-B) \tan ^5(c+d x)}{21 d}+\frac{10 a^3 (2 A-B) \tan ^3(c+d x)}{63 d}+\frac{5 a^3 (2 A-B) \tan (c+d x)}{21 d}+\frac{2 (2 A-B) \sec ^7(c+d x) \left (a^3 \sin (c+d x)+a^3\right )}{21 d}+\frac{(A+B) \sec ^9(c+d x) (a \sin (c+d x)+a)^3}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

((A + B)*Sec[c + d*x]^9*(a + a*Sin[c + d*x])^3)/(9*d) + (2*(2*A - B)*Sec[c + d*x]^7*(a^3 + a^3*Sin[c + d*x]))/
(21*d) + (5*a^3*(2*A - B)*Tan[c + d*x])/(21*d) + (10*a^3*(2*A - B)*Tan[c + d*x]^3)/(63*d) + (a^3*(2*A - B)*Tan
[c + d*x]^5)/(21*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 ^{10}(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac{(A+B) \sec ^9(c+d x) (a+a \sin (c+d x))^3}{9 d}+\frac{1}{3} (a (2 A-B)) \int \sec ^8(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=\frac{(A+B) \sec ^9(c+d x) (a+a \sin (c+d x))^3}{9 d}+\frac{2 (2 A-B) \sec ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{21 d}+\frac{1}{21} \left (5 a^3 (2 A-B)\right ) \int \sec ^6(c+d x) \, dx\\ &=\frac{(A+B) \sec ^9(c+d x) (a+a \sin (c+d x))^3}{9 d}+\frac{2 (2 A-B) \sec ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{21 d}-\frac{\left (5 a^3 (2 A-B)\right ) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{21 d}\\ &=\frac{(A+B) \sec ^9(c+d x) (a+a \sin (c+d x))^3}{9 d}+\frac{2 (2 A-B) \sec ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{21 d}+\frac{5 a^3 (2 A-B) \tan (c+d x)}{21 d}+\frac{10 a^3 (2 A-B) \tan ^3(c+d x)}{63 d}+\frac{a^3 (2 A-B) \tan ^5(c+d x)}{21 d}\\ \end{align*}

Mathematica [A]  time = 0.58123, size = 176, normalized size = 1.26 \[ -\frac{a^3 (27 (B-2 A) \cos (2 (c+d x))+12 (B-2 A) \cos (4 (c+d x))-72 A \sin (c+d x)-4 A \sin (3 (c+d x))+12 A \sin (5 (c+d x))+2 A \cos (6 (c+d x))+36 B \sin (c+d x)+2 B \sin (3 (c+d x))-6 B \sin (5 (c+d x))+B (-\cos (6 (c+d x)))-42 B)}{252 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^9 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

-(a^3*(-42*B + 27*(-2*A + B)*Cos[2*(c + d*x)] + 12*(-2*A + B)*Cos[4*(c + d*x)] + 2*A*Cos[6*(c + d*x)] - B*Cos[
6*(c + d*x)] - 72*A*Sin[c + d*x] + 36*B*Sin[c + d*x] - 4*A*Sin[3*(c + d*x)] + 2*B*Sin[3*(c + d*x)] + 12*A*Sin[
5*(c + d*x)] - 6*B*Sin[5*(c + d*x)]))/(252*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^9*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2])^3)

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Maple [B]  time = 0.138, size = 535, 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)^10*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(1/9*sin(d*x+c)^4/cos(d*x+c)^9+5/63*sin(d*x+c)^4/cos(d*x+c)^7+1/21*sin(d*x+c)^4/cos(d*x+c)^5+1/63*s
in(d*x+c)^4/cos(d*x+c)^3-1/63*sin(d*x+c)^4/cos(d*x+c)-1/63*(2+sin(d*x+c)^2)*cos(d*x+c))+B*a^3*(1/9*sin(d*x+c)^
5/cos(d*x+c)^9+4/63*sin(d*x+c)^5/cos(d*x+c)^7+8/315*sin(d*x+c)^5/cos(d*x+c)^5)+3*a^3*A*(1/9*sin(d*x+c)^3/cos(d
*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x+c)^3/cos(d*x+c)^5+16/315*sin(d*x+c)^3/cos(d*x+c)^3)+3*B*a
^3*(1/9*sin(d*x+c)^4/cos(d*x+c)^9+5/63*sin(d*x+c)^4/cos(d*x+c)^7+1/21*sin(d*x+c)^4/cos(d*x+c)^5+1/63*sin(d*x+c
)^4/cos(d*x+c)^3-1/63*sin(d*x+c)^4/cos(d*x+c)-1/63*(2+sin(d*x+c)^2)*cos(d*x+c))+1/3*a^3*A/cos(d*x+c)^9+3*B*a^3
*(1/9*sin(d*x+c)^3/cos(d*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x+c)^3/cos(d*x+c)^5+16/315*sin(d*x+
c)^3/cos(d*x+c)^3)-a^3*A*(-128/315-1/9*sec(d*x+c)^8-8/63*sec(d*x+c)^6-16/105*sec(d*x+c)^4-64/315*sec(d*x+c)^2)
*tan(d*x+c)+1/9*B*a^3/cos(d*x+c)^9)

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Maxima [B]  time = 1.0801, size = 365, normalized size = 2.61 \begin{align*} \frac{{\left (35 \, \tan \left (d x + c\right )^{9} + 180 \, \tan \left (d x + c\right )^{7} + 378 \, \tan \left (d x + c\right )^{5} + 420 \, \tan \left (d x + c\right )^{3} + 315 \, \tan \left (d x + c\right )\right )} A a^{3} + 3 \,{\left (35 \, \tan \left (d x + c\right )^{9} + 135 \, \tan \left (d x + c\right )^{7} + 189 \, \tan \left (d x + c\right )^{5} + 105 \, \tan \left (d x + c\right )^{3}\right )} A a^{3} + 3 \,{\left (35 \, \tan \left (d x + c\right )^{9} + 135 \, \tan \left (d x + c\right )^{7} + 189 \, \tan \left (d x + c\right )^{5} + 105 \, \tan \left (d x + c\right )^{3}\right )} B a^{3} +{\left (35 \, \tan \left (d x + c\right )^{9} + 90 \, \tan \left (d x + c\right )^{7} + 63 \, \tan \left (d x + c\right )^{5}\right )} B a^{3} - \frac{5 \,{\left (9 \, \cos \left (d x + c\right )^{2} - 7\right )} A a^{3}}{\cos \left (d x + c\right )^{9}} - \frac{15 \,{\left (9 \, \cos \left (d x + c\right )^{2} - 7\right )} B a^{3}}{\cos \left (d x + c\right )^{9}} + \frac{105 \, A a^{3}}{\cos \left (d x + c\right )^{9}} + \frac{35 \, B a^{3}}{\cos \left (d x + c\right )^{9}}}{315 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/315*((35*tan(d*x + c)^9 + 180*tan(d*x + c)^7 + 378*tan(d*x + c)^5 + 420*tan(d*x + c)^3 + 315*tan(d*x + c))*A
*a^3 + 3*(35*tan(d*x + c)^9 + 135*tan(d*x + c)^7 + 189*tan(d*x + c)^5 + 105*tan(d*x + c)^3)*A*a^3 + 3*(35*tan(
d*x + c)^9 + 135*tan(d*x + c)^7 + 189*tan(d*x + c)^5 + 105*tan(d*x + c)^3)*B*a^3 + (35*tan(d*x + c)^9 + 90*tan
(d*x + c)^7 + 63*tan(d*x + c)^5)*B*a^3 - 5*(9*cos(d*x + c)^2 - 7)*A*a^3/cos(d*x + c)^9 - 15*(9*cos(d*x + c)^2
- 7)*B*a^3/cos(d*x + c)^9 + 105*A*a^3/cos(d*x + c)^9 + 35*B*a^3/cos(d*x + c)^9)/d

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Fricas [A]  time = 1.94768, size = 436, normalized size = 3.11 \begin{align*} \frac{8 \,{\left (2 \, A - B\right )} a^{3} \cos \left (d x + c\right )^{6} - 36 \,{\left (2 \, A - B\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \,{\left (2 \, A - B\right )} a^{3} \cos \left (d x + c\right )^{2} + 7 \,{\left (A - 2 \, B\right )} a^{3} +{\left (24 \,{\left (2 \, A - B\right )} a^{3} \cos \left (d x + c\right )^{4} - 20 \,{\left (2 \, A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 7 \,{\left (2 \, A - B\right )} a^{3}\right )} \sin \left (d x + c\right )}{63 \,{\left (3 \, d \cos \left (d x + c\right )^{5} - 4 \, d \cos \left (d x + c\right )^{3} -{\left (d \cos \left (d x + c\right )^{5} - 4 \, d \cos \left (d x + c\right )^{3}\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)^10*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/63*(8*(2*A - B)*a^3*cos(d*x + c)^6 - 36*(2*A - B)*a^3*cos(d*x + c)^4 + 15*(2*A - B)*a^3*cos(d*x + c)^2 + 7*(
A - 2*B)*a^3 + (24*(2*A - B)*a^3*cos(d*x + c)^4 - 20*(2*A - B)*a^3*cos(d*x + c)^2 - 7*(2*A - B)*a^3)*sin(d*x +
c))/(3*d*cos(d*x + c)^5 - 4*d*cos(d*x + c)^3 - (d*cos(d*x + c)^5 - 4*d*cos(d*x + c)^3)*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)**10*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.31763, size = 531, normalized size = 3.79 \begin{align*} -\frac{\frac{21 \,{\left (21 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 36 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 19 \, A a^{3} - 13 \, B a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}} + \frac{3591 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 315 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 19656 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 756 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 56196 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 4200 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 95760 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 11340 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 107730 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 14994 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 79464 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 13356 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 38484 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6768 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 10944 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2196 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1615 \, A a^{3} - 209 \, B a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{9}}}{2016 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/2016*(21*(21*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 15*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 36*A*a^3*tan(1/2*d*x + 1/2*c)
- 24*B*a^3*tan(1/2*d*x + 1/2*c) + 19*A*a^3 - 13*B*a^3)/(tan(1/2*d*x + 1/2*c) + 1)^3 + (3591*A*a^3*tan(1/2*d*x
+ 1/2*c)^8 + 315*B*a^3*tan(1/2*d*x + 1/2*c)^8 - 19656*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 756*B*a^3*tan(1/2*d*x +
1/2*c)^7 + 56196*A*a^3*tan(1/2*d*x + 1/2*c)^6 - 4200*B*a^3*tan(1/2*d*x + 1/2*c)^6 - 95760*A*a^3*tan(1/2*d*x +
1/2*c)^5 + 11340*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 107730*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 14994*B*a^3*tan(1/2*d*x
+ 1/2*c)^4 - 79464*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 13356*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 38484*A*a^3*tan(1/2*d*x
+ 1/2*c)^2 - 6768*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 10944*A*a^3*tan(1/2*d*x + 1/2*c) + 2196*B*a^3*tan(1/2*d*x +
1/2*c) + 1615*A*a^3 - 209*B*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^9)/d