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]

1/99*a^2*(9*A-2*B)*sec(d*x+c)^9/d+1/11*(A+B)*sec(d*x+c)^11*(a+a*sin(d*x+c))^2/d+1/11*a^2*(9*A-2*B)*tan(d*x+c)/
d+4/33*a^2*(9*A-2*B)*tan(d*x+c)^3/d+6/55*a^2*(9*A-2*B)*tan(d*x+c)^5/d+4/77*a^2*(9*A-2*B)*tan(d*x+c)^7/d+1/99*a
^2*(9*A-2*B)*tan(d*x+c)^9/d

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Rubi [A]  time = 0.15, antiderivative size = 179, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [A]  time = 0.92, 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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fricas [A]  time = 0.77, size = 237, normalized size = 1.32 \[ -\frac {256 \, {\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{8} - 128 \, {\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 32 \, {\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 16 \, {\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 45 \, {\left (2 \, A - 9 \, B\right )} a^{2} - {\left (128 \, {\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{8} - 192 \, {\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 80 \, {\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 56 \, {\left (9 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 45 \, {\left (9 \, A - 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{9} + 2 \, d \cos \left (d x + c\right )^{7} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{7}\right )}} \]

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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giac [B]  time = 0.30, size = 597, normalized size = 3.34 \[ -\frac {\frac {33 \, {\left (6825 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 2940 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 34965 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 13755 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 79800 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30065 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 100170 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36470 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 73017 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 26166 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 29169 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 10367 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5142 \, A a^{2} - 1901 \, B a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}} + \frac {661815 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 97020 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 5083155 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 405405 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 19355490 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 952875 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 45446940 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1122660 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72295146 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 557172 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 80611146 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 563178 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63771840 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1126950 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 35253900 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 955020 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 13119975 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 406120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2978811 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 97163 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 330966 \, A a^{2} - 13 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{11}}}{443520 \, d} \]

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

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maple [B]  time = 0.81, size = 423, normalized size = 2.36 \[ \frac {a^{2} A \left (\frac {\sin ^{3}\left (d x +c \right )}{11 \cos \left (d x +c \right )^{11}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{99 \cos \left (d x +c \right )^{9}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{231 \cos \left (d x +c \right )^{7}}+\frac {64 \left (\sin ^{3}\left (d x +c \right )\right )}{1155 \cos \left (d x +c \right )^{5}}+\frac {128 \left (\sin ^{3}\left (d x +c \right )\right )}{3465 \cos \left (d x +c \right )^{3}}\right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{11 \cos \left (d x +c \right )^{11}}+\frac {7 \left (\sin ^{4}\left (d x +c \right )\right )}{99 \cos \left (d x +c \right )^{9}}+\frac {5 \left (\sin ^{4}\left (d x +c \right )\right )}{99 \cos \left (d x +c \right )^{7}}+\frac {\sin ^{4}\left (d x +c \right )}{33 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{99 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{99 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{99}\right )+\frac {2 a^{2} A}{11 \cos \left (d x +c \right )^{11}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{11 \cos \left (d x +c \right )^{11}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{99 \cos \left (d x +c \right )^{9}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{231 \cos \left (d x +c \right )^{7}}+\frac {64 \left (\sin ^{3}\left (d x +c \right )\right )}{1155 \cos \left (d x +c \right )^{5}}+\frac {128 \left (\sin ^{3}\left (d x +c \right )\right )}{3465 \cos \left (d x +c \right )^{3}}\right )-a^{2} A \left (-\frac {256}{693}-\frac {\left (\sec ^{10}\left (d x +c \right )\right )}{11}-\frac {10 \left (\sec ^{8}\left (d x +c \right )\right )}{99}-\frac {80 \left (\sec ^{6}\left (d x +c \right )\right )}{693}-\frac {32 \left (\sec ^{4}\left (d x +c \right )\right )}{231}-\frac {128 \left (\sec ^{2}\left (d x +c \right )\right )}{693}\right ) \tan \left (d x +c \right )+\frac {B \,a^{2}}{11 \cos \left (d x +c \right )^{11}}}{d} \]

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 = 0.71, size = 238, normalized size = 1.33 \[ \frac {{\left (315 \, \tan \left (d x + c\right )^{11} + 1540 \, \tan \left (d x + c\right )^{9} + 2970 \, \tan \left (d x + c\right )^{7} + 2772 \, \tan \left (d x + c\right )^{5} + 1155 \, \tan \left (d x + c\right )^{3}\right )} A a^{2} + 5 \, {\left (63 \, \tan \left (d x + c\right )^{11} + 385 \, \tan \left (d x + c\right )^{9} + 990 \, \tan \left (d x + c\right )^{7} + 1386 \, \tan \left (d x + c\right )^{5} + 1155 \, \tan \left (d x + c\right )^{3} + 693 \, \tan \left (d x + c\right )\right )} A a^{2} + 2 \, {\left (315 \, \tan \left (d x + c\right )^{11} + 1540 \, \tan \left (d x + c\right )^{9} + 2970 \, \tan \left (d x + c\right )^{7} + 2772 \, \tan \left (d x + c\right )^{5} + 1155 \, \tan \left (d x + c\right )^{3}\right )} B a^{2} - \frac {35 \, {\left (11 \, \cos \left (d x + c\right )^{2} - 9\right )} B a^{2}}{\cos \left (d x + c\right )^{11}} + \frac {630 \, A a^{2}}{\cos \left (d x + c\right )^{11}} + \frac {315 \, B a^{2}}{\cos \left (d x + c\right )^{11}}}{3465 \, d} \]

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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mupad [B]  time = 13.96, size = 466, normalized size = 2.60 \[ -\frac {a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {8127\,A\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{64}-\frac {24255\,A\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{64}-\frac {21357\,A\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{64}+\frac {5229\,A\,\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{64}-\frac {8379\,A\,\cos \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{64}+\frac {1467\,A\,\cos \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{64}-\frac {2619\,A\,\cos \left (\frac {15\,c}{2}+\frac {15\,d\,x}{2}\right )}{128}+\frac {315\,A\,\cos \left (\frac {17\,c}{2}+\frac {17\,d\,x}{2}\right )}{128}-385\,B\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {30415\,B\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{128}-\frac {23247\,B\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{128}+\frac {12957\,B\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{128}-\frac {5789\,B\,\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{128}+\frac {3339\,B\,\cos \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{128}-\frac {267\,B\,\cos \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{128}+\frac {779\,B\,\cos \left (\frac {15\,c}{2}+\frac {15\,d\,x}{2}\right )}{256}+\frac {245\,B\,\cos \left (\frac {17\,c}{2}+\frac {17\,d\,x}{2}\right )}{256}-\frac {47889\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{32}+\frac {25713\,A\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{32}-\frac {21303\,A\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{32}+\frac {9207\,A\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{32}-\frac {4797\,A\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{32}+\frac {1917\,A\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{32}-\frac {27\,A\,\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{32}+\frac {171\,A\,\sin \left (\frac {15\,c}{2}+\frac {15\,d\,x}{2}\right )}{32}+\frac {9\,A\,\sin \left (\frac {17\,c}{2}+\frac {17\,d\,x}{2}\right )}{2}+\frac {7809\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}+\frac {2047\,B\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{64}+\frac {1383\,B\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{64}+\frac {3993\,B\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{64}-\frac {563\,B\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{64}+\frac {1843\,B\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{64}-\frac {373\,B\,\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{64}+\frac {309\,B\,\sin \left (\frac {15\,c}{2}+\frac {15\,d\,x}{2}\right )}{64}-B\,\sin \left (\frac {17\,c}{2}+\frac {17\,d\,x}{2}\right )\right )}{887040\,d\,{\cos \left (\frac {c}{2}-\frac {\pi }{4}+\frac {d\,x}{2}\right )}^7\,{\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^12,x)

[Out]

-(a^2*cos(c/2 + (d*x)/2)*((8127*A*cos((5*c)/2 + (5*d*x)/2))/64 - (24255*A*cos((3*c)/2 + (3*d*x)/2))/64 - (2135
7*A*cos((7*c)/2 + (7*d*x)/2))/64 + (5229*A*cos((9*c)/2 + (9*d*x)/2))/64 - (8379*A*cos((11*c)/2 + (11*d*x)/2))/
64 + (1467*A*cos((13*c)/2 + (13*d*x)/2))/64 - (2619*A*cos((15*c)/2 + (15*d*x)/2))/128 + (315*A*cos((17*c)/2 +
(17*d*x)/2))/128 - 385*B*cos(c/2 + (d*x)/2) + (30415*B*cos((3*c)/2 + (3*d*x)/2))/128 - (23247*B*cos((5*c)/2 +
(5*d*x)/2))/128 + (12957*B*cos((7*c)/2 + (7*d*x)/2))/128 - (5789*B*cos((9*c)/2 + (9*d*x)/2))/128 + (3339*B*cos
((11*c)/2 + (11*d*x)/2))/128 - (267*B*cos((13*c)/2 + (13*d*x)/2))/128 + (779*B*cos((15*c)/2 + (15*d*x)/2))/256
+ (245*B*cos((17*c)/2 + (17*d*x)/2))/256 - (47889*A*sin(c/2 + (d*x)/2))/32 + (25713*A*sin((3*c)/2 + (3*d*x)/2
))/32 - (21303*A*sin((5*c)/2 + (5*d*x)/2))/32 + (9207*A*sin((7*c)/2 + (7*d*x)/2))/32 - (4797*A*sin((9*c)/2 + (
9*d*x)/2))/32 + (1917*A*sin((11*c)/2 + (11*d*x)/2))/32 - (27*A*sin((13*c)/2 + (13*d*x)/2))/32 + (171*A*sin((15
*c)/2 + (15*d*x)/2))/32 + (9*A*sin((17*c)/2 + (17*d*x)/2))/2 + (7809*B*sin(c/2 + (d*x)/2))/64 + (2047*B*sin((3
*c)/2 + (3*d*x)/2))/64 + (1383*B*sin((5*c)/2 + (5*d*x)/2))/64 + (3993*B*sin((7*c)/2 + (7*d*x)/2))/64 - (563*B*
sin((9*c)/2 + (9*d*x)/2))/64 + (1843*B*sin((11*c)/2 + (11*d*x)/2))/64 - (373*B*sin((13*c)/2 + (13*d*x)/2))/64
+ (309*B*sin((15*c)/2 + (15*d*x)/2))/64 - B*sin((17*c)/2 + (17*d*x)/2)))/(887040*d*cos(c/2 - pi/4 + (d*x)/2)^7
*cos(c/2 + pi/4 + (d*x)/2)^11)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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