3.984 \(\int \sec ^{10}(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=154 \[ \frac{a^2 (7 A-2 B) \tan ^7(c+d x)}{63 d}+\frac{a^2 (7 A-2 B) \tan ^5(c+d x)}{15 d}+\frac{a^2 (7 A-2 B) \tan ^3(c+d x)}{9 d}+\frac{a^2 (7 A-2 B) \tan (c+d x)}{9 d}+\frac{a^2 (7 A-2 B) \sec ^7(c+d x)}{63 d}+\frac{(A+B) \sec ^9(c+d x) (a \sin (c+d x)+a)^2}{9 d} \]

[Out]

(a^2*(7*A - 2*B)*Sec[c + d*x]^7)/(63*d) + ((A + B)*Sec[c + d*x]^9*(a + a*Sin[c + d*x])^2)/(9*d) + (a^2*(7*A -
2*B)*Tan[c + d*x])/(9*d) + (a^2*(7*A - 2*B)*Tan[c + d*x]^3)/(9*d) + (a^2*(7*A - 2*B)*Tan[c + d*x]^5)/(15*d) +
(a^2*(7*A - 2*B)*Tan[c + d*x]^7)/(63*d)

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Rubi [A]  time = 0.141179, antiderivative size = 154, 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, 2669, 3767} \[ \frac{a^2 (7 A-2 B) \tan ^7(c+d x)}{63 d}+\frac{a^2 (7 A-2 B) \tan ^5(c+d x)}{15 d}+\frac{a^2 (7 A-2 B) \tan ^3(c+d x)}{9 d}+\frac{a^2 (7 A-2 B) \tan (c+d x)}{9 d}+\frac{a^2 (7 A-2 B) \sec ^7(c+d x)}{63 d}+\frac{(A+B) \sec ^9(c+d x) (a \sin (c+d x)+a)^2}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(7*A - 2*B)*Sec[c + d*x]^7)/(63*d) + ((A + B)*Sec[c + d*x]^9*(a + a*Sin[c + d*x])^2)/(9*d) + (a^2*(7*A -
2*B)*Tan[c + d*x])/(9*d) + (a^2*(7*A - 2*B)*Tan[c + d*x]^3)/(9*d) + (a^2*(7*A - 2*B)*Tan[c + d*x]^5)/(15*d) +
(a^2*(7*A - 2*B)*Tan[c + d*x]^7)/(63*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 2669

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

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))^2 (A+B \sin (c+d x)) \, dx &=\frac{(A+B) \sec ^9(c+d x) (a+a \sin (c+d x))^2}{9 d}+\frac{1}{9} (a (7 A-2 B)) \int \sec ^8(c+d x) (a+a \sin (c+d x)) \, dx\\ &=\frac{a^2 (7 A-2 B) \sec ^7(c+d x)}{63 d}+\frac{(A+B) \sec ^9(c+d x) (a+a \sin (c+d x))^2}{9 d}+\frac{1}{9} \left (a^2 (7 A-2 B)\right ) \int \sec ^8(c+d x) \, dx\\ &=\frac{a^2 (7 A-2 B) \sec ^7(c+d x)}{63 d}+\frac{(A+B) \sec ^9(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac{\left (a^2 (7 A-2 B)\right ) \operatorname{Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,-\tan (c+d x)\right )}{9 d}\\ &=\frac{a^2 (7 A-2 B) \sec ^7(c+d x)}{63 d}+\frac{(A+B) \sec ^9(c+d x) (a+a \sin (c+d x))^2}{9 d}+\frac{a^2 (7 A-2 B) \tan (c+d x)}{9 d}+\frac{a^2 (7 A-2 B) \tan ^3(c+d x)}{9 d}+\frac{a^2 (7 A-2 B) \tan ^5(c+d x)}{15 d}+\frac{a^2 (7 A-2 B) \tan ^7(c+d x)}{63 d}\\ \end{align*}

Mathematica [A]  time = 0.448487, size = 156, normalized size = 1.01 \[ \frac{a^2 \left (16 (7 A-2 B) \tan ^9(c+d x)+5 (14 A+5 B) \sec ^9(c+d x)-105 (7 A-2 B) \tan ^3(c+d x) \sec ^6(c+d x)+126 (7 A-2 B) \tan ^5(c+d x) \sec ^4(c+d x)-72 (7 A-2 B) \tan ^7(c+d x) \sec ^2(c+d x)+315 A \tan (c+d x) \sec ^8(c+d x)+45 B \tan ^2(c+d x) \sec ^7(c+d x)\right )}{315 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(5*(14*A + 5*B)*Sec[c + d*x]^9 + 315*A*Sec[c + d*x]^8*Tan[c + d*x] + 45*B*Sec[c + d*x]^7*Tan[c + d*x]^2 -
105*(7*A - 2*B)*Sec[c + d*x]^6*Tan[c + d*x]^3 + 126*(7*A - 2*B)*Sec[c + d*x]^4*Tan[c + d*x]^5 - 72*(7*A - 2*B
)*Sec[c + d*x]^2*Tan[c + d*x]^7 + 16*(7*A - 2*B)*Tan[c + d*x]^9))/(315*d)

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Maple [B]  time = 0.131, size = 359, normalized size = 2.3 \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)^10*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^2*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+16/31
5*sin(d*x+c)^3/cos(d*x+c)^3)+B*a^2*(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)
)+2/9*a^2*A/cos(d*x+c)^9+2*B*a^2*(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^2*A*(-128/315-1/9*sec(d*x+c)^8-8/63*sec(d*x+c)^6-16/105*s
ec(d*x+c)^4-64/315*sec(d*x+c)^2)*tan(d*x+c)+1/9*B*a^2/cos(d*x+c)^9)

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Maxima [A]  time = 1.04463, size = 279, normalized size = 1.81 \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^{2} +{\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^{2} + 2 \,{\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^{2} - \frac{5 \,{\left (9 \, \cos \left (d x + c\right )^{2} - 7\right )} B a^{2}}{\cos \left (d x + c\right )^{9}} + \frac{70 \, A a^{2}}{\cos \left (d x + c\right )^{9}} + \frac{35 \, B a^{2}}{\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))^2*(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^2 + (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^2 + 2*(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^2 - 5*(9*cos(d*x + c)^2 - 7)*B*a^
2/cos(d*x + c)^9 + 70*A*a^2/cos(d*x + c)^9 + 35*B*a^2/cos(d*x + c)^9)/d

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Fricas [A]  time = 1.84669, size = 475, normalized size = 3.08 \begin{align*} -\frac{32 \,{\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 16 \,{\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 4 \,{\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 7 \,{\left (2 \, A - 7 \, B\right )} a^{2} -{\left (16 \,{\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 24 \,{\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 10 \,{\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 7 \,{\left (7 \, A - 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{315 \,{\left (d \cos \left (d x + c\right )^{7} + 2 \, d \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{5}\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))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/315*(32*(7*A - 2*B)*a^2*cos(d*x + c)^6 - 16*(7*A - 2*B)*a^2*cos(d*x + c)^4 - 4*(7*A - 2*B)*a^2*cos(d*x + c)
^2 - 7*(2*A - 7*B)*a^2 - (16*(7*A - 2*B)*a^2*cos(d*x + c)^6 - 24*(7*A - 2*B)*a^2*cos(d*x + c)^4 - 10*(7*A - 2*
B)*a^2*cos(d*x + c)^2 - 7*(7*A - 2*B)*a^2)*sin(d*x + c))/(d*cos(d*x + c)^7 + 2*d*cos(d*x + c)^5*sin(d*x + c) -
2*d*cos(d*x + c)^5)

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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))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.31404, size = 622, normalized size = 4.04 \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)^10*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/20160*(21*(435*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 225*B*a^2*tan(1/2*d*x + 1/2*c)^4 + 1470*A*a^2*tan(1/2*d*x + 1
/2*c)^3 - 690*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 2060*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 940*B*a^2*tan(1/2*d*x + 1/2*c
)^2 + 1330*A*a^2*tan(1/2*d*x + 1/2*c) - 590*B*a^2*tan(1/2*d*x + 1/2*c) + 353*A*a^2 - 163*B*a^2)/(tan(1/2*d*x +
1/2*c) + 1)^5 + (31185*A*a^2*tan(1/2*d*x + 1/2*c)^8 + 4725*B*a^2*tan(1/2*d*x + 1/2*c)^8 - 185220*A*a^2*tan(1/
2*d*x + 1/2*c)^7 - 11340*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 546840*A*a^2*tan(1/2*d*x + 1/2*c)^6 + 15120*B*a^2*tan(
1/2*d*x + 1/2*c)^6 - 961380*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 3780*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 1101618*A*a^2*t
an(1/2*d*x + 1/2*c)^4 - 24318*B*a^2*tan(1/2*d*x + 1/2*c)^4 - 828492*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 33852*B*a^2
*tan(1/2*d*x + 1/2*c)^3 + 404208*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 19368*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 116172*A*
a^2*tan(1/2*d*x + 1/2*c) + 6732*B*a^2*tan(1/2*d*x + 1/2*c) + 16373*A*a^2 - 223*B*a^2)/(tan(1/2*d*x + 1/2*c) -
1)^9)/d