3.985 \(\int \sec ^{12}(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=179 \[ \frac{a^2 (9 A-2 B) \tan ^9(c+d x)}{99 d}+\frac{4 a^2 (9 A-2 B) \tan ^7(c+d x)}{77 d}+\frac{6 a^2 (9 A-2 B) \tan ^5(c+d x)}{55 d}+\frac{4 a^2 (9 A-2 B) \tan ^3(c+d x)}{33 d}+\frac{a^2 (9 A-2 B) \tan (c+d x)}{11 d}+\frac{a^2 (9 A-2 B) \sec ^9(c+d x)}{99 d}+\frac{(A+B) \sec ^{11}(c+d x) (a \sin (c+d x)+a)^2}{11 d} \]

[Out]

(a^2*(9*A - 2*B)*Sec[c + d*x]^9)/(99*d) + ((A + B)*Sec[c + d*x]^11*(a + a*Sin[c + d*x])^2)/(11*d) + (a^2*(9*A
- 2*B)*Tan[c + d*x])/(11*d) + (4*a^2*(9*A - 2*B)*Tan[c + d*x]^3)/(33*d) + (6*a^2*(9*A - 2*B)*Tan[c + d*x]^5)/(
55*d) + (4*a^2*(9*A - 2*B)*Tan[c + d*x]^7)/(77*d) + (a^2*(9*A - 2*B)*Tan[c + d*x]^9)/(99*d)

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Rubi [A]  time = 0.150762, antiderivative size = 179, 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} \[ \frac{a^2 (9 A-2 B) \tan ^9(c+d x)}{99 d}+\frac{4 a^2 (9 A-2 B) \tan ^7(c+d x)}{77 d}+\frac{6 a^2 (9 A-2 B) \tan ^5(c+d x)}{55 d}+\frac{4 a^2 (9 A-2 B) \tan ^3(c+d x)}{33 d}+\frac{a^2 (9 A-2 B) \tan (c+d x)}{11 d}+\frac{a^2 (9 A-2 B) \sec ^9(c+d x)}{99 d}+\frac{(A+B) \sec ^{11}(c+d x) (a \sin (c+d x)+a)^2}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^12*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(9*A - 2*B)*Sec[c + d*x]^9)/(99*d) + ((A + B)*Sec[c + d*x]^11*(a + a*Sin[c + d*x])^2)/(11*d) + (a^2*(9*A
- 2*B)*Tan[c + d*x])/(11*d) + (4*a^2*(9*A - 2*B)*Tan[c + d*x]^3)/(33*d) + (6*a^2*(9*A - 2*B)*Tan[c + d*x]^5)/(
55*d) + (4*a^2*(9*A - 2*B)*Tan[c + d*x]^7)/(77*d) + (a^2*(9*A - 2*B)*Tan[c + d*x]^9)/(99*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 ^{12}(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac{(A+B) \sec ^{11}(c+d x) (a+a \sin (c+d x))^2}{11 d}+\frac{1}{11} (a (9 A-2 B)) \int \sec ^{10}(c+d x) (a+a \sin (c+d x)) \, dx\\ &=\frac{a^2 (9 A-2 B) \sec ^9(c+d x)}{99 d}+\frac{(A+B) \sec ^{11}(c+d x) (a+a \sin (c+d x))^2}{11 d}+\frac{1}{11} \left (a^2 (9 A-2 B)\right ) \int \sec ^{10}(c+d x) \, dx\\ &=\frac{a^2 (9 A-2 B) \sec ^9(c+d x)}{99 d}+\frac{(A+B) \sec ^{11}(c+d x) (a+a \sin (c+d x))^2}{11 d}-\frac{\left (a^2 (9 A-2 B)\right ) \operatorname{Subst}\left (\int \left (1+4 x^2+6 x^4+4 x^6+x^8\right ) \, dx,x,-\tan (c+d x)\right )}{11 d}\\ &=\frac{a^2 (9 A-2 B) \sec ^9(c+d x)}{99 d}+\frac{(A+B) \sec ^{11}(c+d x) (a+a \sin (c+d x))^2}{11 d}+\frac{a^2 (9 A-2 B) \tan (c+d x)}{11 d}+\frac{4 a^2 (9 A-2 B) \tan ^3(c+d x)}{33 d}+\frac{6 a^2 (9 A-2 B) \tan ^5(c+d x)}{55 d}+\frac{4 a^2 (9 A-2 B) \tan ^7(c+d x)}{77 d}+\frac{a^2 (9 A-2 B) \tan ^9(c+d x)}{99 d}\\ \end{align*}

Mathematica [A]  time = 0.909396, size = 181, normalized size = 1.01 \[ \frac{a^2 \left (128 (2 B-9 A) \tan ^{11}(c+d x)+35 (18 A+7 B) \sec ^{11}(c+d x)-1155 (9 A-2 B) \tan ^3(c+d x) \sec ^8(c+d x)+1848 (9 A-2 B) \tan ^5(c+d x) \sec ^6(c+d x)-1584 (9 A-2 B) \tan ^7(c+d x) \sec ^4(c+d x)+704 (9 A-2 B) \tan ^9(c+d x) \sec ^2(c+d x)+3465 A \tan (c+d x) \sec ^{10}(c+d x)+385 B \tan ^2(c+d x) \sec ^9(c+d x)\right )}{3465 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^12*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(35*(18*A + 7*B)*Sec[c + d*x]^11 + 3465*A*Sec[c + d*x]^10*Tan[c + d*x] + 385*B*Sec[c + d*x]^9*Tan[c + d*x
]^2 - 1155*(9*A - 2*B)*Sec[c + d*x]^8*Tan[c + d*x]^3 + 1848*(9*A - 2*B)*Sec[c + d*x]^6*Tan[c + d*x]^5 - 1584*(
9*A - 2*B)*Sec[c + d*x]^4*Tan[c + d*x]^7 + 704*(9*A - 2*B)*Sec[c + d*x]^2*Tan[c + d*x]^9 + 128*(-9*A + 2*B)*Ta
n[c + d*x]^11))/(3465*d)

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Maple [B]  time = 0.236, size = 423, normalized size = 2.4 \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)^12*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^2*A*(1/11*sin(d*x+c)^3/cos(d*x+c)^11+8/99*sin(d*x+c)^3/cos(d*x+c)^9+16/231*sin(d*x+c)^3/cos(d*x+c)^7+64
/1155*sin(d*x+c)^3/cos(d*x+c)^5+128/3465*sin(d*x+c)^3/cos(d*x+c)^3)+B*a^2*(1/11*sin(d*x+c)^4/cos(d*x+c)^11+7/9
9*sin(d*x+c)^4/cos(d*x+c)^9+5/99*sin(d*x+c)^4/cos(d*x+c)^7+1/33*sin(d*x+c)^4/cos(d*x+c)^5+1/99*sin(d*x+c)^4/co
s(d*x+c)^3-1/99*sin(d*x+c)^4/cos(d*x+c)-1/99*(2+sin(d*x+c)^2)*cos(d*x+c))+2/11*a^2*A/cos(d*x+c)^11+2*B*a^2*(1/
11*sin(d*x+c)^3/cos(d*x+c)^11+8/99*sin(d*x+c)^3/cos(d*x+c)^9+16/231*sin(d*x+c)^3/cos(d*x+c)^7+64/1155*sin(d*x+
c)^3/cos(d*x+c)^5+128/3465*sin(d*x+c)^3/cos(d*x+c)^3)-a^2*A*(-256/693-1/11*sec(d*x+c)^10-10/99*sec(d*x+c)^8-80
/693*sec(d*x+c)^6-32/231*sec(d*x+c)^4-128/693*sec(d*x+c)^2)*tan(d*x+c)+1/11*B*a^2/cos(d*x+c)^11)

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Maxima [A]  time = 1.04184, size = 321, normalized size = 1.79 \begin{align*} \frac{{\left (315 \, \tan \left (d x + c\right )^{11} + 1540 \, \tan \left (d x + c\right )^{9} + 2970 \, \tan \left (d x + c\right )^{7} + 2772 \, \tan \left (d x + c\right )^{5} + 1155 \, \tan \left (d x + c\right )^{3}\right )} A a^{2} + 5 \,{\left (63 \, \tan \left (d x + c\right )^{11} + 385 \, \tan \left (d x + c\right )^{9} + 990 \, \tan \left (d x + c\right )^{7} + 1386 \, \tan \left (d x + c\right )^{5} + 1155 \, \tan \left (d x + c\right )^{3} + 693 \, \tan \left (d x + c\right )\right )} A a^{2} + 2 \,{\left (315 \, \tan \left (d x + c\right )^{11} + 1540 \, \tan \left (d x + c\right )^{9} + 2970 \, \tan \left (d x + c\right )^{7} + 2772 \, \tan \left (d x + c\right )^{5} + 1155 \, \tan \left (d x + c\right )^{3}\right )} B a^{2} - \frac{35 \,{\left (11 \, \cos \left (d x + c\right )^{2} - 9\right )} B a^{2}}{\cos \left (d x + c\right )^{11}} + \frac{630 \, A a^{2}}{\cos \left (d x + c\right )^{11}} + \frac{315 \, B a^{2}}{\cos \left (d x + c\right )^{11}}}{3465 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/3465*((315*tan(d*x + c)^11 + 1540*tan(d*x + c)^9 + 2970*tan(d*x + c)^7 + 2772*tan(d*x + c)^5 + 1155*tan(d*x
+ c)^3)*A*a^2 + 5*(63*tan(d*x + c)^11 + 385*tan(d*x + c)^9 + 990*tan(d*x + c)^7 + 1386*tan(d*x + c)^5 + 1155*t
an(d*x + c)^3 + 693*tan(d*x + c))*A*a^2 + 2*(315*tan(d*x + c)^11 + 1540*tan(d*x + c)^9 + 2970*tan(d*x + c)^7 +
2772*tan(d*x + c)^5 + 1155*tan(d*x + c)^3)*B*a^2 - 35*(11*cos(d*x + c)^2 - 9)*B*a^2/cos(d*x + c)^11 + 630*A*a
^2/cos(d*x + c)^11 + 315*B*a^2/cos(d*x + c)^11)/d

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Fricas [A]  time = 1.89079, size = 583, normalized size = 3.26 \begin{align*} -\frac{256 \,{\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{8} - 128 \,{\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 32 \,{\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 16 \,{\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 45 \,{\left (2 \, A - 9 \, B\right )} a^{2} -{\left (128 \,{\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{8} - 192 \,{\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 80 \,{\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 56 \,{\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 45 \,{\left (9 \, A - 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{3465 \,{\left (d \cos \left (d x + c\right )^{9} + 2 \, d \cos \left (d x + c\right )^{7} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{7}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/3465*(256*(9*A - 2*B)*a^2*cos(d*x + c)^8 - 128*(9*A - 2*B)*a^2*cos(d*x + c)^6 - 32*(9*A - 2*B)*a^2*cos(d*x
+ c)^4 - 16*(9*A - 2*B)*a^2*cos(d*x + c)^2 - 45*(2*A - 9*B)*a^2 - (128*(9*A - 2*B)*a^2*cos(d*x + c)^8 - 192*(9
*A - 2*B)*a^2*cos(d*x + c)^6 - 80*(9*A - 2*B)*a^2*cos(d*x + c)^4 - 56*(9*A - 2*B)*a^2*cos(d*x + c)^2 - 45*(9*A
- 2*B)*a^2)*sin(d*x + c))/(d*cos(d*x + c)^9 + 2*d*cos(d*x + c)^7*sin(d*x + c) - 2*d*cos(d*x + c)^7)

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

[Out]

Timed out

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Giac [B]  time = 1.44411, size = 806, normalized size = 4.5 \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)^12*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/443520*(33*(6825*A*a^2*tan(1/2*d*x + 1/2*c)^6 - 2940*B*a^2*tan(1/2*d*x + 1/2*c)^6 + 34965*A*a^2*tan(1/2*d*x
+ 1/2*c)^5 - 13755*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 79800*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 30065*B*a^2*tan(1/2*d*
x + 1/2*c)^4 + 100170*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 36470*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 73017*A*a^2*tan(1/2*
d*x + 1/2*c)^2 - 26166*B*a^2*tan(1/2*d*x + 1/2*c)^2 + 29169*A*a^2*tan(1/2*d*x + 1/2*c) - 10367*B*a^2*tan(1/2*d
*x + 1/2*c) + 5142*A*a^2 - 1901*B*a^2)/(tan(1/2*d*x + 1/2*c) + 1)^7 + (661815*A*a^2*tan(1/2*d*x + 1/2*c)^10 +
97020*B*a^2*tan(1/2*d*x + 1/2*c)^10 - 5083155*A*a^2*tan(1/2*d*x + 1/2*c)^9 - 405405*B*a^2*tan(1/2*d*x + 1/2*c)
^9 + 19355490*A*a^2*tan(1/2*d*x + 1/2*c)^8 + 952875*B*a^2*tan(1/2*d*x + 1/2*c)^8 - 45446940*A*a^2*tan(1/2*d*x
+ 1/2*c)^7 - 1122660*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 72295146*A*a^2*tan(1/2*d*x + 1/2*c)^6 + 557172*B*a^2*tan(1
/2*d*x + 1/2*c)^6 - 80611146*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 563178*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 63771840*A*a
^2*tan(1/2*d*x + 1/2*c)^4 - 1126950*B*a^2*tan(1/2*d*x + 1/2*c)^4 - 35253900*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 955
020*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 13119975*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 406120*B*a^2*tan(1/2*d*x + 1/2*c)^2
- 2978811*A*a^2*tan(1/2*d*x + 1/2*c) + 97163*B*a^2*tan(1/2*d*x + 1/2*c) + 330966*A*a^2 - 13*B*a^2)/(tan(1/2*d
*x + 1/2*c) - 1)^11)/d