3.994 \(\int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=162 \[ \frac {a^7 (A+B)}{16 d (a-a \sin (c+d x))^4}+\frac {a^6 A}{12 d (a-a \sin (c+d x))^3}+\frac {a^5 (3 A-B)}{32 d (a-a \sin (c+d x))^2}+\frac {a^4 (2 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^4 (A-B)}{32 d (a \sin (c+d x)+a)}+\frac {a^3 (5 A-3 B) \tanh ^{-1}(\sin (c+d x))}{32 d} \]

[Out]

1/32*a^3*(5*A-3*B)*arctanh(sin(d*x+c))/d+1/16*a^7*(A+B)/d/(a-a*sin(d*x+c))^4+1/12*a^6*A/d/(a-a*sin(d*x+c))^3+1
/32*a^5*(3*A-B)/d/(a-a*sin(d*x+c))^2+1/16*a^4*(2*A-B)/d/(a-a*sin(d*x+c))-1/32*a^4*(A-B)/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.19, antiderivative size = 162, 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 = {2836, 77, 206} \[ \frac {a^7 (A+B)}{16 d (a-a \sin (c+d x))^4}+\frac {a^5 (3 A-B)}{32 d (a-a \sin (c+d x))^2}+\frac {a^4 (2 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^4 (A-B)}{32 d (a \sin (c+d x)+a)}+\frac {a^3 (5 A-3 B) \tanh ^{-1}(\sin (c+d x))}{32 d}+\frac {a^6 A}{12 d (a-a \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(5*A - 3*B)*ArcTanh[Sin[c + d*x]])/(32*d) + (a^7*(A + B))/(16*d*(a - a*Sin[c + d*x])^4) + (a^6*A)/(12*d*(
a - a*Sin[c + d*x])^3) + (a^5*(3*A - B))/(32*d*(a - a*Sin[c + d*x])^2) + (a^4*(2*A - B))/(16*d*(a - a*Sin[c +
d*x])) - (a^4*(A - B))/(32*d*(a + a*Sin[c + d*x]))

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

\begin {align*} \int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac {a^9 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^5 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^9 \operatorname {Subst}\left (\int \left (\frac {A+B}{4 a^2 (a-x)^5}+\frac {A}{4 a^3 (a-x)^4}+\frac {3 A-B}{16 a^4 (a-x)^3}+\frac {2 A-B}{16 a^5 (a-x)^2}+\frac {A-B}{32 a^5 (a+x)^2}+\frac {5 A-3 B}{32 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 (A+B)}{16 d (a-a \sin (c+d x))^4}+\frac {a^6 A}{12 d (a-a \sin (c+d x))^3}+\frac {a^5 (3 A-B)}{32 d (a-a \sin (c+d x))^2}+\frac {a^4 (2 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^4 (A-B)}{32 d (a+a \sin (c+d x))}+\frac {\left (a^4 (5 A-3 B)\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{32 d}\\ &=\frac {a^3 (5 A-3 B) \tanh ^{-1}(\sin (c+d x))}{32 d}+\frac {a^7 (A+B)}{16 d (a-a \sin (c+d x))^4}+\frac {a^6 A}{12 d (a-a \sin (c+d x))^3}+\frac {a^5 (3 A-B)}{32 d (a-a \sin (c+d x))^2}+\frac {a^4 (2 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^4 (A-B)}{32 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 151, normalized size = 0.93 \[ \frac {a^9 \left (\frac {(5 A-3 B) \tanh ^{-1}(\sin (c+d x))}{32 a^6}+\frac {2 A-B}{16 a^5 (a-a \sin (c+d x))}-\frac {A-B}{32 a^5 (a \sin (c+d x)+a)}+\frac {3 A-B}{32 a^4 (a-a \sin (c+d x))^2}+\frac {A}{12 a^3 (a-a \sin (c+d x))^3}+\frac {A+B}{16 a^2 (a-a \sin (c+d x))^4}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^9*(((5*A - 3*B)*ArcTanh[Sin[c + d*x]])/(32*a^6) + (A + B)/(16*a^2*(a - a*Sin[c + d*x])^4) + A/(12*a^3*(a -
a*Sin[c + d*x])^3) + (3*A - B)/(32*a^4*(a - a*Sin[c + d*x])^2) + (2*A - B)/(16*a^5*(a - a*Sin[c + d*x])) - (A
- B)/(32*a^5*(a + a*Sin[c + d*x]))))/d

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fricas [B]  time = 0.75, size = 353, normalized size = 2.18 \[ \frac {6 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 26 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 12 \, {\left (3 \, A - 5 \, B\right )} a^{3} + 3 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - {\left ({\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - {\left ({\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, A - 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{192 \, {\left (3 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/192*(6*(5*A - 3*B)*a^3*cos(d*x + c)^4 - 26*(5*A - 3*B)*a^3*cos(d*x + c)^2 + 12*(3*A - 5*B)*a^3 + 3*(3*(5*A -
3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*a^3*cos(d*x + c)^2 - ((5*A - 3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*
a^3*cos(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) - 3*(3*(5*A - 3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*
a^3*cos(d*x + c)^2 - ((5*A - 3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*a^3*cos(d*x + c)^2)*sin(d*x + c))*log(-si
n(d*x + c) + 1) + 6*(3*(5*A - 3*B)*a^3*cos(d*x + c)^2 - 2*(5*A - 3*B)*a^3)*sin(d*x + c))/(3*d*cos(d*x + c)^4 -
4*d*cos(d*x + c)^2 - (d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2)*sin(d*x + c))

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giac [A]  time = 0.32, size = 237, normalized size = 1.46 \[ \frac {12 \, {\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 12 \, {\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {12 \, {\left (5 \, A a^{3} \sin \left (d x + c\right ) - 3 \, B a^{3} \sin \left (d x + c\right ) + 7 \, A a^{3} - 5 \, B a^{3}\right )}}{\sin \left (d x + c\right ) + 1} + \frac {125 \, A a^{3} \sin \left (d x + c\right )^{4} - 75 \, B a^{3} \sin \left (d x + c\right )^{4} - 596 \, A a^{3} \sin \left (d x + c\right )^{3} + 348 \, B a^{3} \sin \left (d x + c\right )^{3} + 1110 \, A a^{3} \sin \left (d x + c\right )^{2} - 618 \, B a^{3} \sin \left (d x + c\right )^{2} - 996 \, A a^{3} \sin \left (d x + c\right ) + 492 \, B a^{3} \sin \left (d x + c\right ) + 405 \, A a^{3} - 99 \, B a^{3}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{4}}}{768 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/768*(12*(5*A*a^3 - 3*B*a^3)*log(abs(sin(d*x + c) + 1)) - 12*(5*A*a^3 - 3*B*a^3)*log(abs(sin(d*x + c) - 1)) -
12*(5*A*a^3*sin(d*x + c) - 3*B*a^3*sin(d*x + c) + 7*A*a^3 - 5*B*a^3)/(sin(d*x + c) + 1) + (125*A*a^3*sin(d*x
+ c)^4 - 75*B*a^3*sin(d*x + c)^4 - 596*A*a^3*sin(d*x + c)^3 + 348*B*a^3*sin(d*x + c)^3 + 1110*A*a^3*sin(d*x +
c)^2 - 618*B*a^3*sin(d*x + c)^2 - 996*A*a^3*sin(d*x + c) + 492*B*a^3*sin(d*x + c) + 405*A*a^3 - 99*B*a^3)/(sin
(d*x + c) - 1)^4)/d

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maple [B]  time = 0.65, size = 669, normalized size = 4.13 \[ -\frac {B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{128 d}-\frac {3 B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{32 d}+\frac {15 a^{3} A \sin \left (d x +c \right )}{128 d}+\frac {5 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{32 d}+\frac {B \,a^{3}}{8 d \cos \left (d x +c \right )^{8}}+\frac {3 a^{3} A}{8 d \cos \left (d x +c \right )^{8}}+\frac {a^{3} A \left (\sin ^{4}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}}+\frac {3 B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}}+\frac {B \,a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{6}}+\frac {5 B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{6}}-\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{128 d \cos \left (d x +c \right )^{2}}+\frac {35 a^{3} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{128 d}+\frac {35 a^{3} A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{192 d}+\frac {a^{3} A \left (\sin ^{4}\left (d x +c \right )\right )}{12 d \cos \left (d x +c \right )^{6}}+\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{64 d \cos \left (d x +c \right )^{4}}+\frac {15 a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{64 d \cos \left (d x +c \right )^{4}}+\frac {B \,a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {15 B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{64 d \cos \left (d x +c \right )^{4}}+\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{6}}+\frac {5 a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{6}}+\frac {7 a^{3} A \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{48 d}+\frac {3 a^{3} B \sin \left (d x +c \right )}{32 d}+\frac {a^{3} A \left (\sin ^{4}\left (d x +c \right )\right )}{24 d \cos \left (d x +c \right )^{4}}+\frac {15 a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{128 d \cos \left (d x +c \right )^{2}}+\frac {15 B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{128 d \cos \left (d x +c \right )^{2}}+\frac {3 B \,a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}}+\frac {a^{3} A \tan \left (d x +c \right ) \left (\sec ^{7}\left (d x +c \right )\right )}{8 d}+\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}}+\frac {3 a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/64/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^4+15/64/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*B*a^3*sin(d*x+c)^4/cos(d*
x+c)^4+15/64/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)^4+35/192/d*a^3*A*tan(d*x+c)*sec(d*x+c)^3-1/128/d*B*a^3*sin(d*x+c)
^3-3/32/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+15/128/d*a^3*A*sin(d*x+c)+5/32/d*a^3*A*ln(sec(d*x+c)+tan(d*x+c))+1/1
6/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^6+5/16/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^6+1/4/d*B*a^3*sin(d*x+c)^4/cos(d*x+c)
^6+5/16/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)^6+7/48/d*a^3*A*tan(d*x+c)*sec(d*x+c)^5+1/8/d*B*a^3/cos(d*x+c)^8+3/8/d*
a^3*A/cos(d*x+c)^8-1/128/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^2+15/128/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^2+15/128/d*B
*a^3*sin(d*x+c)^3/cos(d*x+c)^2+35/128/d*a^3*A*sec(d*x+c)*tan(d*x+c)+1/24/d*a^3*A*sin(d*x+c)^4/cos(d*x+c)^4+1/8
/d*a^3*A*sin(d*x+c)^4/cos(d*x+c)^8+3/8/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)^8+1/8/d*a^3*A*tan(d*x+c)*sec(d*x+c)^7+1
/8/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^8+3/8/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^8+3/8/d*B*a^3*sin(d*x+c)^4/cos(d*x+c)
^8+1/12/d*a^3*A*sin(d*x+c)^4/cos(d*x+c)^6+3/32*a^3*B*sin(d*x+c)/d

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maxima [A]  time = 0.32, size = 185, normalized size = 1.14 \[ \frac {3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{4} - 9 \, {\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} + 7 \, {\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{2} + 3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right ) - 32 \, A a^{3}\right )}}{\sin \left (d x + c\right )^{5} - 3 \, \sin \left (d x + c\right )^{4} + 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right )^{2} - 3 \, \sin \left (d x + c\right ) + 1}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/192*(3*(5*A - 3*B)*a^3*log(sin(d*x + c) + 1) - 3*(5*A - 3*B)*a^3*log(sin(d*x + c) - 1) - 2*(3*(5*A - 3*B)*a^
3*sin(d*x + c)^4 - 9*(5*A - 3*B)*a^3*sin(d*x + c)^3 + 7*(5*A - 3*B)*a^3*sin(d*x + c)^2 + 3*(5*A - 3*B)*a^3*sin
(d*x + c) - 32*A*a^3)/(sin(d*x + c)^5 - 3*sin(d*x + c)^4 + 2*sin(d*x + c)^3 + 2*sin(d*x + c)^2 - 3*sin(d*x + c
) + 1))/d

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mupad [B]  time = 9.18, size = 172, normalized size = 1.06 \[ \frac {a^3\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (5\,A-3\,B\right )}{32\,d}-\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {5\,A\,a^3}{32}-\frac {3\,B\,a^3}{32}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {15\,A\,a^3}{32}-\frac {9\,B\,a^3}{32}\right )+{\sin \left (c+d\,x\right )}^2\,\left (\frac {35\,A\,a^3}{96}-\frac {7\,B\,a^3}{32}\right )-\frac {A\,a^3}{3}+\sin \left (c+d\,x\right )\,\left (\frac {5\,A\,a^3}{32}-\frac {3\,B\,a^3}{32}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^5-3\,{\sin \left (c+d\,x\right )}^4+2\,{\sin \left (c+d\,x\right )}^3+2\,{\sin \left (c+d\,x\right )}^2-3\,\sin \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^9,x)

[Out]

(a^3*atanh(sin(c + d*x))*(5*A - 3*B))/(32*d) - (sin(c + d*x)^4*((5*A*a^3)/32 - (3*B*a^3)/32) - sin(c + d*x)^3*
((15*A*a^3)/32 - (9*B*a^3)/32) + sin(c + d*x)^2*((35*A*a^3)/96 - (7*B*a^3)/32) - (A*a^3)/3 + sin(c + d*x)*((5*
A*a^3)/32 - (3*B*a^3)/32))/(d*(2*sin(c + d*x)^2 - 3*sin(c + d*x) + 2*sin(c + d*x)^3 - 3*sin(c + d*x)^4 + sin(c
+ d*x)^5 + 1))

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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)**9*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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