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]

1/63*a^2*(7*A-2*B)*sec(d*x+c)^7/d+1/9*(A+B)*sec(d*x+c)^9*(a+a*sin(d*x+c))^2/d+1/9*a^2*(7*A-2*B)*tan(d*x+c)/d+1
/9*a^2*(7*A-2*B)*tan(d*x+c)^3/d+1/15*a^2*(7*A-2*B)*tan(d*x+c)^5/d+1/63*a^2*(7*A-2*B)*tan(d*x+c)^7/d

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Rubi [A]  time = 0.14, antiderivative size = 154, 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} \[ \frac {a^2 (7 A-2 B) \tan ^7(c+d x)}{63 d}+\frac {a^2 (7 A-2 B) \tan ^5(c+d x)}{15 d}+\frac {a^2 (7 A-2 B) \tan ^3(c+d x)}{9 d}+\frac {a^2 (7 A-2 B) \tan (c+d x)}{9 d}+\frac {a^2 (7 A-2 B) \sec ^7(c+d x)}{63 d}+\frac {(A+B) \sec ^9(c+d x) (a \sin (c+d x)+a)^2}{9 d} \]

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 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 ^{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*}

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Mathematica [A]  time = 0.45, 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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fricas [A]  time = 0.69, size = 197, normalized size = 1.28 \[ -\frac {32 \, {\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 16 \, {\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 4 \, {\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 7 \, {\left (2 \, A - 7 \, B\right )} a^{2} - {\left (16 \, {\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 24 \, {\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 10 \, {\left (7 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 7 \, {\left (7 \, A - 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{7} + 2 \, d \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{5}\right )}} \]

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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giac [B]  time = 0.25, size = 461, normalized size = 2.99 \[ -\frac {\frac {21 \, {\left (435 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 225 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1470 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 690 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2060 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 940 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1330 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 590 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 353 \, A a^{2} - 163 \, B a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}} + \frac {31185 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 4725 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 185220 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 11340 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 546840 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 15120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 961380 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3780 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1101618 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24318 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 828492 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33852 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 404208 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 19368 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 116172 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6732 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16373 \, A a^{2} - 223 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{9}}}{20160 \, d} \]

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

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maple [B]  time = 0.71, size = 359, normalized size = 2.33 \[ \frac {a^{2} A \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {5 \left (\sin ^{4}\left (d x +c \right )\right )}{63 \cos \left (d x +c \right )^{7}}+\frac {\sin ^{4}\left (d x +c \right )}{21 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{63 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{63 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{63}\right )+\frac {2 a^{2} A}{9 \cos \left (d x +c \right )^{9}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )-a^{2} A \left (-\frac {128}{315}-\frac {\left (\sec ^{8}\left (d x +c \right )\right )}{9}-\frac {8 \left (\sec ^{6}\left (d x +c \right )\right )}{63}-\frac {16 \left (\sec ^{4}\left (d x +c \right )\right )}{105}-\frac {64 \left (\sec ^{2}\left (d x +c \right )\right )}{315}\right ) \tan \left (d x +c \right )+\frac {B \,a^{2}}{9 \cos \left (d x +c \right )^{9}}}{d} \]

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 = 0.47, size = 207, normalized size = 1.34 \[ \frac {{\left (35 \, \tan \left (d x + c\right )^{9} + 180 \, \tan \left (d x + c\right )^{7} + 378 \, \tan \left (d x + c\right )^{5} + 420 \, \tan \left (d x + c\right )^{3} + 315 \, \tan \left (d x + c\right )\right )} A a^{2} + {\left (35 \, \tan \left (d x + c\right )^{9} + 135 \, \tan \left (d x + c\right )^{7} + 189 \, \tan \left (d x + c\right )^{5} + 105 \, \tan \left (d x + c\right )^{3}\right )} A a^{2} + 2 \, {\left (35 \, \tan \left (d x + c\right )^{9} + 135 \, \tan \left (d x + c\right )^{7} + 189 \, \tan \left (d x + c\right )^{5} + 105 \, \tan \left (d x + c\right )^{3}\right )} B a^{2} - \frac {5 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 7\right )} B a^{2}}{\cos \left (d x + c\right )^{9}} + \frac {70 \, A a^{2}}{\cos \left (d x + c\right )^{9}} + \frac {35 \, B a^{2}}{\cos \left (d x + c\right )^{9}}}{315 \, d} \]

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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mupad [B]  time = 12.95, size = 370, normalized size = 2.40 \[ -\frac {a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {455\,A\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{32}-\frac {1575\,A\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{32}-35\,A\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )+7\,A\,\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )-\frac {259\,A\,\cos \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{32}+\frac {35\,A\,\cos \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{32}-45\,B\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1755\,B\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{64}-\frac {1115\,B\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{64}+10\,B\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )-2\,B\,\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )+\frac {103\,B\,\cos \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{64}+\frac {25\,B\,\cos \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{64}-\frac {623\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+77\,A\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )-\frac {441\,A\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{8}+\frac {175\,A\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{8}-\frac {35\,A\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{8}+\frac {21\,A\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{8}+\frac {7\,A\,\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{4}+\frac {131\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {49\,B\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{8}+\frac {27\,B\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{16}+\frac {125\,B\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{16}-\frac {25\,B\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{16}+\frac {33\,B\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )}{16}-\frac {B\,\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )}{2}\right )}{20160\,d\,{\cos \left (\frac {c}{2}-\frac {\pi }{4}+\frac {d\,x}{2}\right )}^5\,{\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}^9} \]

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)^10,x)

[Out]

-(a^2*cos(c/2 + (d*x)/2)*((455*A*cos((5*c)/2 + (5*d*x)/2))/32 - (1575*A*cos((3*c)/2 + (3*d*x)/2))/32 - 35*A*co
s((7*c)/2 + (7*d*x)/2) + 7*A*cos((9*c)/2 + (9*d*x)/2) - (259*A*cos((11*c)/2 + (11*d*x)/2))/32 + (35*A*cos((13*
c)/2 + (13*d*x)/2))/32 - 45*B*cos(c/2 + (d*x)/2) + (1755*B*cos((3*c)/2 + (3*d*x)/2))/64 - (1115*B*cos((5*c)/2
+ (5*d*x)/2))/64 + 10*B*cos((7*c)/2 + (7*d*x)/2) - 2*B*cos((9*c)/2 + (9*d*x)/2) + (103*B*cos((11*c)/2 + (11*d*
x)/2))/64 + (25*B*cos((13*c)/2 + (13*d*x)/2))/64 - (623*A*sin(c/2 + (d*x)/2))/4 + 77*A*sin((3*c)/2 + (3*d*x)/2
) - (441*A*sin((5*c)/2 + (5*d*x)/2))/8 + (175*A*sin((7*c)/2 + (7*d*x)/2))/8 - (35*A*sin((9*c)/2 + (9*d*x)/2))/
8 + (21*A*sin((11*c)/2 + (11*d*x)/2))/8 + (7*A*sin((13*c)/2 + (13*d*x)/2))/4 + (131*B*sin(c/2 + (d*x)/2))/8 +
(49*B*sin((3*c)/2 + (3*d*x)/2))/8 + (27*B*sin((5*c)/2 + (5*d*x)/2))/16 + (125*B*sin((7*c)/2 + (7*d*x)/2))/16 -
(25*B*sin((9*c)/2 + (9*d*x)/2))/16 + (33*B*sin((11*c)/2 + (11*d*x)/2))/16 - (B*sin((13*c)/2 + (13*d*x)/2))/2)
)/(20160*d*cos(c/2 - pi/4 + (d*x)/2)^5*cos(c/2 + pi/4 + (d*x)/2)^9)

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

[Out]

Timed out

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