3.412 \(\int \sec ^{10}(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=192 \[ \frac {2 a \left (8 a^2-3 b^2\right ) \tan ^5(c+d x)}{105 d}+\frac {4 a \left (8 a^2-3 b^2\right ) \tan ^3(c+d x)}{63 d}+\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x)}{21 d}+\frac {2 b \left (4 a^2-b^2\right ) \sec ^5(c+d x)}{63 d}+\frac {2 \sec ^7(c+d x) (a+b \sin (c+d x)) \left (\left (4 a^2-b^2\right ) \sin (c+d x)+3 a b\right )}{63 d}+\frac {\sec ^9(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{9 d} \]

[Out]

2/63*b*(4*a^2-b^2)*sec(d*x+c)^5/d+1/9*sec(d*x+c)^9*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^2/d+2/63*sec(d*x+c)^7*(a+
b*sin(d*x+c))*(3*a*b+(4*a^2-b^2)*sin(d*x+c))/d+2/21*a*(8*a^2-3*b^2)*tan(d*x+c)/d+4/63*a*(8*a^2-3*b^2)*tan(d*x+
c)^3/d+2/105*a*(8*a^2-3*b^2)*tan(d*x+c)^5/d

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Rubi [A]  time = 0.22, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2691, 2861, 2669, 3767} \[ \frac {2 a \left (8 a^2-3 b^2\right ) \tan ^5(c+d x)}{105 d}+\frac {4 a \left (8 a^2-3 b^2\right ) \tan ^3(c+d x)}{63 d}+\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x)}{21 d}+\frac {2 b \left (4 a^2-b^2\right ) \sec ^5(c+d x)}{63 d}+\frac {2 \sec ^7(c+d x) (a+b \sin (c+d x)) \left (\left (4 a^2-b^2\right ) \sin (c+d x)+3 a b\right )}{63 d}+\frac {\sec ^9(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10*(a + b*Sin[c + d*x])^3,x]

[Out]

(2*b*(4*a^2 - b^2)*Sec[c + d*x]^5)/(63*d) + (Sec[c + d*x]^9*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(9*d)
+ (2*Sec[c + d*x]^7*(a + b*Sin[c + d*x])*(3*a*b + (4*a^2 - b^2)*Sin[c + d*x]))/(63*d) + (2*a*(8*a^2 - 3*b^2)*
Tan[c + d*x])/(21*d) + (4*a*(8*a^2 - 3*b^2)*Tan[c + d*x]^3)/(63*d) + (2*a*(8*a^2 - 3*b^2)*Tan[c + d*x]^5)/(105
*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 2691

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

Rule 2861

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x]))/(f*
g*(p + 1)), x] + Dist[1/(g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(p +
2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x
])

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+b \sin (c+d x))^3 \, dx &=\frac {\sec ^9(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{9 d}-\frac {1}{9} \int \sec ^8(c+d x) (a+b \sin (c+d x)) \left (-8 a^2+2 b^2-6 a b \sin (c+d x)\right ) \, dx\\ &=\frac {\sec ^9(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{9 d}+\frac {2 \sec ^7(c+d x) (a+b \sin (c+d x)) \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{63 d}+\frac {1}{63} \int \sec ^6(c+d x) \left (6 a \left (8 a^2-3 b^2\right )+10 b \left (4 a^2-b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac {2 b \left (4 a^2-b^2\right ) \sec ^5(c+d x)}{63 d}+\frac {\sec ^9(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{9 d}+\frac {2 \sec ^7(c+d x) (a+b \sin (c+d x)) \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{63 d}+\frac {1}{21} \left (2 a \left (8 a^2-3 b^2\right )\right ) \int \sec ^6(c+d x) \, dx\\ &=\frac {2 b \left (4 a^2-b^2\right ) \sec ^5(c+d x)}{63 d}+\frac {\sec ^9(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{9 d}+\frac {2 \sec ^7(c+d x) (a+b \sin (c+d x)) \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{63 d}-\frac {\left (2 a \left (8 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{21 d}\\ &=\frac {2 b \left (4 a^2-b^2\right ) \sec ^5(c+d x)}{63 d}+\frac {\sec ^9(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{9 d}+\frac {2 \sec ^7(c+d x) (a+b \sin (c+d x)) \left (3 a b+\left (4 a^2-b^2\right ) \sin (c+d x)\right )}{63 d}+\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x)}{21 d}+\frac {4 a \left (8 a^2-3 b^2\right ) \tan ^3(c+d x)}{63 d}+\frac {2 a \left (8 a^2-3 b^2\right ) \tan ^5(c+d x)}{105 d}\\ \end {align*}

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Mathematica [A]  time = 1.52, size = 299, normalized size = 1.56 \[ \frac {\sec ^9(c+d x) \left (2064384 a^3 \sin (c+d x)+1376256 a^3 \sin (3 (c+d x))+589824 a^3 \sin (5 (c+d x))+147456 a^3 \sin (7 (c+d x))+16384 a^3 \sin (9 (c+d x))+3150 b \left (23 b^2-147 a^2\right ) \cos (c+d x)-308700 a^2 b \cos (3 (c+d x))-132300 a^2 b \cos (5 (c+d x))-33075 a^2 b \cos (7 (c+d x))-3675 a^2 b \cos (9 (c+d x))+3440640 a^2 b+3096576 a b^2 \sin (c+d x)-516096 a b^2 \sin (3 (c+d x))-221184 a b^2 \sin (5 (c+d x))-55296 a b^2 \sin (7 (c+d x))-6144 a b^2 \sin (9 (c+d x))-737280 b^3 \cos (2 (c+d x))+48300 b^3 \cos (3 (c+d x))+20700 b^3 \cos (5 (c+d x))+5175 b^3 \cos (7 (c+d x))+575 b^3 \cos (9 (c+d x))+409600 b^3\right )}{10321920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10*(a + b*Sin[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^9*(3440640*a^2*b + 409600*b^3 + 3150*b*(-147*a^2 + 23*b^2)*Cos[c + d*x] - 737280*b^3*Cos[2*(c +
d*x)] - 308700*a^2*b*Cos[3*(c + d*x)] + 48300*b^3*Cos[3*(c + d*x)] - 132300*a^2*b*Cos[5*(c + d*x)] + 20700*b^3
*Cos[5*(c + d*x)] - 33075*a^2*b*Cos[7*(c + d*x)] + 5175*b^3*Cos[7*(c + d*x)] - 3675*a^2*b*Cos[9*(c + d*x)] + 5
75*b^3*Cos[9*(c + d*x)] + 2064384*a^3*Sin[c + d*x] + 3096576*a*b^2*Sin[c + d*x] + 1376256*a^3*Sin[3*(c + d*x)]
- 516096*a*b^2*Sin[3*(c + d*x)] + 589824*a^3*Sin[5*(c + d*x)] - 221184*a*b^2*Sin[5*(c + d*x)] + 147456*a^3*Si
n[7*(c + d*x)] - 55296*a*b^2*Sin[7*(c + d*x)] + 16384*a^3*Sin[9*(c + d*x)] - 6144*a*b^2*Sin[9*(c + d*x)]))/(10
321920*d)

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fricas [A]  time = 0.46, size = 146, normalized size = 0.76 \[ -\frac {45 \, b^{3} \cos \left (d x + c\right )^{2} - 105 \, a^{2} b - 35 \, b^{3} - {\left (16 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{8} + 8 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{6} + 6 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 35 \, a^{3} + 105 \, a b^{2} + 5 \, {\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{315 \, d \cos \left (d x + c\right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/315*(45*b^3*cos(d*x + c)^2 - 105*a^2*b - 35*b^3 - (16*(8*a^3 - 3*a*b^2)*cos(d*x + c)^8 + 8*(8*a^3 - 3*a*b^2
)*cos(d*x + c)^6 + 6*(8*a^3 - 3*a*b^2)*cos(d*x + c)^4 + 35*a^3 + 105*a*b^2 + 5*(8*a^3 - 3*a*b^2)*cos(d*x + c)^
2)*sin(d*x + c))/(d*cos(d*x + c)^9)

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giac [B]  time = 1.60, size = 473, normalized size = 2.46 \[ -\frac {2 \, {\left (315 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{17} + 945 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{16} - 840 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 1260 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 630 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{14} + 4788 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 1512 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 8820 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} + 1050 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 5112 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 8532 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 3150 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 10658 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4272 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 13230 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 1890 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 5112 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 8532 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1890 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 4788 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1512 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3780 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 270 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 840 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1260 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 90 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, a^{2} b - 10 \, b^{3}\right )}}{315 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{9} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-2/315*(315*a^3*tan(1/2*d*x + 1/2*c)^17 + 945*a^2*b*tan(1/2*d*x + 1/2*c)^16 - 840*a^3*tan(1/2*d*x + 1/2*c)^15
+ 1260*a*b^2*tan(1/2*d*x + 1/2*c)^15 + 630*b^3*tan(1/2*d*x + 1/2*c)^14 + 4788*a^3*tan(1/2*d*x + 1/2*c)^13 + 15
12*a*b^2*tan(1/2*d*x + 1/2*c)^13 + 8820*a^2*b*tan(1/2*d*x + 1/2*c)^12 + 1050*b^3*tan(1/2*d*x + 1/2*c)^12 - 511
2*a^3*tan(1/2*d*x + 1/2*c)^11 + 8532*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 3150*b^3*tan(1/2*d*x + 1/2*c)^10 + 10658*
a^3*tan(1/2*d*x + 1/2*c)^9 + 4272*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 13230*a^2*b*tan(1/2*d*x + 1/2*c)^8 + 1890*b^3
*tan(1/2*d*x + 1/2*c)^8 - 5112*a^3*tan(1/2*d*x + 1/2*c)^7 + 8532*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 1890*b^3*tan(1
/2*d*x + 1/2*c)^6 + 4788*a^3*tan(1/2*d*x + 1/2*c)^5 + 1512*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 3780*a^2*b*tan(1/2*d
*x + 1/2*c)^4 + 270*b^3*tan(1/2*d*x + 1/2*c)^4 - 840*a^3*tan(1/2*d*x + 1/2*c)^3 + 1260*a*b^2*tan(1/2*d*x + 1/2
*c)^3 + 90*b^3*tan(1/2*d*x + 1/2*c)^2 + 315*a^3*tan(1/2*d*x + 1/2*c) + 105*a^2*b - 10*b^3)/((tan(1/2*d*x + 1/2
*c)^2 - 1)^9*d)

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maple [A]  time = 0.38, size = 265, normalized size = 1.38 \[ \frac {-a^{3} \left (-\frac {128}{315}-\frac {\left (\sec ^{8}\left (d x +c \right )\right )}{9}-\frac {8 \left (\sec ^{6}\left (d x +c \right )\right )}{63}-\frac {16 \left (\sec ^{4}\left (d x +c \right )\right )}{105}-\frac {64 \left (\sec ^{2}\left (d x +c \right )\right )}{315}\right ) \tan \left (d x +c \right )+\frac {a^{2} b}{3 \cos \left (d x +c \right )^{9}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {5 \left (\sin ^{4}\left (d x +c \right )\right )}{63 \cos \left (d x +c \right )^{7}}+\frac {\sin ^{4}\left (d x +c \right )}{21 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{63 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{63 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{63}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^10*(a+b*sin(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.34, size = 145, normalized size = 0.76 \[ \frac {{\left (35 \, \tan \left (d x + c\right )^{9} + 180 \, \tan \left (d x + c\right )^{7} + 378 \, \tan \left (d x + c\right )^{5} + 420 \, \tan \left (d x + c\right )^{3} + 315 \, \tan \left (d x + c\right )\right )} a^{3} + 3 \, {\left (35 \, \tan \left (d x + c\right )^{9} + 135 \, \tan \left (d x + c\right )^{7} + 189 \, \tan \left (d x + c\right )^{5} + 105 \, \tan \left (d x + c\right )^{3}\right )} a b^{2} - \frac {5 \, {\left (9 \, \cos \left (d x + c\right )^{2} - 7\right )} b^{3}}{\cos \left (d x + c\right )^{9}} + \frac {105 \, a^{2} b}{\cos \left (d x + c\right )^{9}}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a+b*sin(d*x+c))^3,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
^3 + 3*(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*b^2 - 5*(9*cos(d*x
+ c)^2 - 7)*b^3/cos(d*x + c)^9 + 105*a^2*b/cos(d*x + c)^9)/d

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mupad [B]  time = 6.13, size = 275, normalized size = 1.43 \[ \frac {b^3}{9\,d\,{\cos \left (c+d\,x\right )}^9}-\frac {b^3}{7\,d\,{\cos \left (c+d\,x\right )}^7}+\frac {a^2\,b}{3\,d\,{\cos \left (c+d\,x\right )}^9}+\frac {128\,a^3\,\sin \left (c+d\,x\right )}{315\,d\,\cos \left (c+d\,x\right )}+\frac {64\,a^3\,\sin \left (c+d\,x\right )}{315\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {16\,a^3\,\sin \left (c+d\,x\right )}{105\,d\,{\cos \left (c+d\,x\right )}^5}+\frac {8\,a^3\,\sin \left (c+d\,x\right )}{63\,d\,{\cos \left (c+d\,x\right )}^7}+\frac {a^3\,\sin \left (c+d\,x\right )}{9\,d\,{\cos \left (c+d\,x\right )}^9}-\frac {16\,a\,b^2\,\sin \left (c+d\,x\right )}{105\,d\,\cos \left (c+d\,x\right )}-\frac {8\,a\,b^2\,\sin \left (c+d\,x\right )}{105\,d\,{\cos \left (c+d\,x\right )}^3}-\frac {2\,a\,b^2\,\sin \left (c+d\,x\right )}{35\,d\,{\cos \left (c+d\,x\right )}^5}-\frac {a\,b^2\,\sin \left (c+d\,x\right )}{21\,d\,{\cos \left (c+d\,x\right )}^7}+\frac {a\,b^2\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^3/cos(c + d*x)^10,x)

[Out]

b^3/(9*d*cos(c + d*x)^9) - b^3/(7*d*cos(c + d*x)^7) + (a^2*b)/(3*d*cos(c + d*x)^9) + (128*a^3*sin(c + d*x))/(3
15*d*cos(c + d*x)) + (64*a^3*sin(c + d*x))/(315*d*cos(c + d*x)^3) + (16*a^3*sin(c + d*x))/(105*d*cos(c + d*x)^
5) + (8*a^3*sin(c + d*x))/(63*d*cos(c + d*x)^7) + (a^3*sin(c + d*x))/(9*d*cos(c + d*x)^9) - (16*a*b^2*sin(c +
d*x))/(105*d*cos(c + d*x)) - (8*a*b^2*sin(c + d*x))/(105*d*cos(c + d*x)^3) - (2*a*b^2*sin(c + d*x))/(35*d*cos(
c + d*x)^5) - (a*b^2*sin(c + d*x))/(21*d*cos(c + d*x)^7) + (a*b^2*sin(c + d*x))/(3*d*cos(c + d*x)^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+b*sin(d*x+c))**3,x)

[Out]

Timed out

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