3.29 \(\int (a+a \sec (c+d x))^2 \sin ^8(c+d x) \, dx\)

Optimal. Leaf size=199 \[ -\frac {2 a^2 \sin ^7(c+d x)}{7 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \sin (c+d x) \cos ^7(c+d x)}{8 d}-\frac {17 a^2 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac {11 a^2 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac {139 a^2 \sin (c+d x) \cos (c+d x)}{128 d}-\frac {245 a^2 x}{128} \]

[Out]

-245/128*a^2*x+2*a^2*arctanh(sin(d*x+c))/d-2*a^2*sin(d*x+c)/d+139/128*a^2*cos(d*x+c)*sin(d*x+c)/d+11/192*a^2*c
os(d*x+c)^3*sin(d*x+c)/d-17/48*a^2*cos(d*x+c)^5*sin(d*x+c)/d+1/8*a^2*cos(d*x+c)^7*sin(d*x+c)/d-2/3*a^2*sin(d*x
+c)^3/d-2/5*a^2*sin(d*x+c)^5/d-2/7*a^2*sin(d*x+c)^7/d+a^2*tan(d*x+c)/d

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Rubi [A]  time = 0.36, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3872, 2872, 2637, 2635, 8, 2633, 3770, 3767} \[ -\frac {2 a^2 \sin ^7(c+d x)}{7 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \sin (c+d x) \cos ^7(c+d x)}{8 d}-\frac {17 a^2 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac {11 a^2 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac {139 a^2 \sin (c+d x) \cos (c+d x)}{128 d}-\frac {245 a^2 x}{128} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^8,x]

[Out]

(-245*a^2*x)/128 + (2*a^2*ArcTanh[Sin[c + d*x]])/d - (2*a^2*Sin[c + d*x])/d + (139*a^2*Cos[c + d*x]*Sin[c + d*
x])/(128*d) + (11*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(192*d) - (17*a^2*Cos[c + d*x]^5*Sin[c + d*x])/(48*d) + (a^
2*Cos[c + d*x]^7*Sin[c + d*x])/(8*d) - (2*a^2*Sin[c + d*x]^3)/(3*d) - (2*a^2*Sin[c + d*x]^5)/(5*d) - (2*a^2*Si
n[c + d*x]^7)/(7*d) + (a^2*Tan[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2872

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^2 \sin ^8(c+d x) \, dx &=\int (-a-a \cos (c+d x))^2 \sin ^6(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {\int \left (-3 a^{10}-8 a^{10} \cos (c+d x)+2 a^{10} \cos ^2(c+d x)+12 a^{10} \cos ^3(c+d x)+2 a^{10} \cos ^4(c+d x)-8 a^{10} \cos ^5(c+d x)-3 a^{10} \cos ^6(c+d x)+2 a^{10} \cos ^7(c+d x)+a^{10} \cos ^8(c+d x)+2 a^{10} \sec (c+d x)+a^{10} \sec ^2(c+d x)\right ) \, dx}{a^8}\\ &=-3 a^2 x+a^2 \int \cos ^8(c+d x) \, dx+a^2 \int \sec ^2(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^2(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^4(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^7(c+d x) \, dx+\left (2 a^2\right ) \int \sec (c+d x) \, dx-\left (3 a^2\right ) \int \cos ^6(c+d x) \, dx-\left (8 a^2\right ) \int \cos (c+d x) \, dx-\left (8 a^2\right ) \int \cos ^5(c+d x) \, dx+\left (12 a^2\right ) \int \cos ^3(c+d x) \, dx\\ &=-3 a^2 x+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {8 a^2 \sin (c+d x)}{d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{2 d}+\frac {a^2 \cos ^7(c+d x) \sin (c+d x)}{8 d}+\frac {1}{8} \left (7 a^2\right ) \int \cos ^6(c+d x) \, dx+a^2 \int 1 \, dx+\frac {1}{2} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx-\frac {1}{2} \left (5 a^2\right ) \int \cos ^4(c+d x) \, dx-\frac {a^2 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,-\sin (c+d x)\right )}{d}+\frac {\left (8 a^2\right ) \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac {\left (12 a^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=-2 a^2 x+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{4 d}-\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {17 a^2 \cos ^5(c+d x) \sin (c+d x)}{48 d}+\frac {a^2 \cos ^7(c+d x) \sin (c+d x)}{8 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^7(c+d x)}{7 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {1}{48} \left (35 a^2\right ) \int \cos ^4(c+d x) \, dx+\frac {1}{4} \left (3 a^2\right ) \int 1 \, dx-\frac {1}{8} \left (15 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {5 a^2 x}{4}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {13 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {11 a^2 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {17 a^2 \cos ^5(c+d x) \sin (c+d x)}{48 d}+\frac {a^2 \cos ^7(c+d x) \sin (c+d x)}{8 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^7(c+d x)}{7 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {1}{64} \left (35 a^2\right ) \int \cos ^2(c+d x) \, dx-\frac {1}{16} \left (15 a^2\right ) \int 1 \, dx\\ &=-\frac {35 a^2 x}{16}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {139 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {11 a^2 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {17 a^2 \cos ^5(c+d x) \sin (c+d x)}{48 d}+\frac {a^2 \cos ^7(c+d x) \sin (c+d x)}{8 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^7(c+d x)}{7 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {1}{128} \left (35 a^2\right ) \int 1 \, dx\\ &=-\frac {245 a^2 x}{128}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {139 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {11 a^2 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {17 a^2 \cos ^5(c+d x) \sin (c+d x)}{48 d}+\frac {a^2 \cos ^7(c+d x) \sin (c+d x)}{8 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^7(c+d x)}{7 d}+\frac {a^2 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.96, size = 144, normalized size = 0.72 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (30720 \sin ^7(c+d x)+43008 \sin ^5(c+d x)+71680 \sin ^3(c+d x)+215040 \sin (c+d x)-55440 \sin (2 (c+d x))+2520 \sin (4 (c+d x))+560 \sin (6 (c+d x))-105 \sin (8 (c+d x))+37800 \tan ^{-1}(\tan (c+d x))-107520 \tan (c+d x)-215040 \tanh ^{-1}(\sin (c+d x))+168000 c+168000 d x\right )}{430080 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^8,x]

[Out]

-1/430080*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(168000*c + 168000*d*x + 37800*ArcTan[Tan[c + d*x]] - 2
15040*ArcTanh[Sin[c + d*x]] + 215040*Sin[c + d*x] + 71680*Sin[c + d*x]^3 + 43008*Sin[c + d*x]^5 + 30720*Sin[c
+ d*x]^7 - 55440*Sin[2*(c + d*x)] + 2520*Sin[4*(c + d*x)] + 560*Sin[6*(c + d*x)] - 105*Sin[8*(c + d*x)] - 1075
20*Tan[c + d*x]))/d

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fricas [A]  time = 0.71, size = 185, normalized size = 0.93 \[ -\frac {25725 \, a^{2} d x \cos \left (d x + c\right ) - 13440 \, a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + 13440 \, a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (1680 \, a^{2} \cos \left (d x + c\right )^{8} + 3840 \, a^{2} \cos \left (d x + c\right )^{7} - 4760 \, a^{2} \cos \left (d x + c\right )^{6} - 16896 \, a^{2} \cos \left (d x + c\right )^{5} + 770 \, a^{2} \cos \left (d x + c\right )^{4} + 31232 \, a^{2} \cos \left (d x + c\right )^{3} + 14595 \, a^{2} \cos \left (d x + c\right )^{2} - 45056 \, a^{2} \cos \left (d x + c\right ) + 13440 \, a^{2}\right )} \sin \left (d x + c\right )}{13440 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^8,x, algorithm="fricas")

[Out]

-1/13440*(25725*a^2*d*x*cos(d*x + c) - 13440*a^2*cos(d*x + c)*log(sin(d*x + c) + 1) + 13440*a^2*cos(d*x + c)*l
og(-sin(d*x + c) + 1) - (1680*a^2*cos(d*x + c)^8 + 3840*a^2*cos(d*x + c)^7 - 4760*a^2*cos(d*x + c)^6 - 16896*a
^2*cos(d*x + c)^5 + 770*a^2*cos(d*x + c)^4 + 31232*a^2*cos(d*x + c)^3 + 14595*a^2*cos(d*x + c)^2 - 45056*a^2*c
os(d*x + c) + 13440*a^2)*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 0.79, size = 225, normalized size = 1.13 \[ -\frac {25725 \, {\left (d x + c\right )} a^{2} - 26880 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 26880 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {26880 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (39165 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{15} + 300265 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 989261 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1791073 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1814943 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 670131 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 147735 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 14595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{8}}}{13440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^8,x, algorithm="giac")

[Out]

-1/13440*(25725*(d*x + c)*a^2 - 26880*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 26880*a^2*log(abs(tan(1/2*d*x +
1/2*c) - 1)) + 26880*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(39165*a^2*tan(1/2*d*x + 1/2*c
)^15 + 300265*a^2*tan(1/2*d*x + 1/2*c)^13 + 989261*a^2*tan(1/2*d*x + 1/2*c)^11 + 1791073*a^2*tan(1/2*d*x + 1/2
*c)^9 + 1814943*a^2*tan(1/2*d*x + 1/2*c)^7 + 670131*a^2*tan(1/2*d*x + 1/2*c)^5 + 147735*a^2*tan(1/2*d*x + 1/2*
c)^3 + 14595*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^8)/d

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maple [A]  time = 0.70, size = 210, normalized size = 1.06 \[ \frac {7 a^{2} \left (\sin ^{7}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{8 d}+\frac {49 a^{2} \cos \left (d x +c \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{48 d}+\frac {245 a^{2} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{192 d}+\frac {245 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{128 d}-\frac {245 a^{2} x}{128}-\frac {245 a^{2} c}{128 d}-\frac {2 a^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{7 d}-\frac {2 a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{5 d}-\frac {2 a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 a^{2} \sin \left (d x +c \right )}{d}+\frac {2 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} \left (\sin ^{9}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*sin(d*x+c)^8,x)

[Out]

7/8/d*a^2*sin(d*x+c)^7*cos(d*x+c)+49/48/d*a^2*cos(d*x+c)*sin(d*x+c)^5+245/192/d*a^2*cos(d*x+c)*sin(d*x+c)^3+24
5/128*a^2*cos(d*x+c)*sin(d*x+c)/d-245/128*a^2*x-245/128/d*a^2*c-2/7*a^2*sin(d*x+c)^7/d-2/5*a^2*sin(d*x+c)^5/d-
2/3*a^2*sin(d*x+c)^3/d-2*a^2*sin(d*x+c)/d+2/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^2*sin(d*x+c)^9/cos(d*x+c)

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maxima [A]  time = 0.44, size = 215, normalized size = 1.08 \[ -\frac {1024 \, {\left (30 \, \sin \left (d x + c\right )^{7} + 42 \, \sin \left (d x + c\right )^{5} + 70 \, \sin \left (d x + c\right )^{3} - 105 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 210 \, \sin \left (d x + c\right )\right )} a^{2} - 35 \, {\left (128 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 840 \, d x + 840 \, c + 3 \, \sin \left (8 \, d x + 8 \, c\right ) + 168 \, \sin \left (4 \, d x + 4 \, c\right ) - 768 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} + 2240 \, {\left (105 \, d x + 105 \, c - \frac {87 \, \tan \left (d x + c\right )^{5} + 136 \, \tan \left (d x + c\right )^{3} + 57 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1} - 48 \, \tan \left (d x + c\right )\right )} a^{2}}{107520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^8,x, algorithm="maxima")

[Out]

-1/107520*(1024*(30*sin(d*x + c)^7 + 42*sin(d*x + c)^5 + 70*sin(d*x + c)^3 - 105*log(sin(d*x + c) + 1) + 105*l
og(sin(d*x + c) - 1) + 210*sin(d*x + c))*a^2 - 35*(128*sin(2*d*x + 2*c)^3 + 840*d*x + 840*c + 3*sin(8*d*x + 8*
c) + 168*sin(4*d*x + 4*c) - 768*sin(2*d*x + 2*c))*a^2 + 2240*(105*d*x + 105*c - (87*tan(d*x + c)^5 + 136*tan(d
*x + c)^3 + 57*tan(d*x + c))/(tan(d*x + c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1) - 48*tan(d*x + c))*a^2
)/d

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mupad [B]  time = 2.54, size = 293, normalized size = 1.47 \[ \frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {245\,a^2\,x}{128}+\frac {\frac {501\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{64}+\frac {2633\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{48}+\frac {38047\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{240}+\frac {388613\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{1680}+\frac {13781\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{96}-\frac {32681\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{560}-\frac {1739\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{80}-\frac {61\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{16}-\frac {11\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^8*(a + a/cos(c + d*x))^2,x)

[Out]

(4*a^2*atanh(tan(c/2 + (d*x)/2)))/d - (245*a^2*x)/128 + ((13781*a^2*tan(c/2 + (d*x)/2)^9)/96 - (1739*a^2*tan(c
/2 + (d*x)/2)^5)/80 - (32681*a^2*tan(c/2 + (d*x)/2)^7)/560 - (61*a^2*tan(c/2 + (d*x)/2)^3)/16 + (388613*a^2*ta
n(c/2 + (d*x)/2)^11)/1680 + (38047*a^2*tan(c/2 + (d*x)/2)^13)/240 + (2633*a^2*tan(c/2 + (d*x)/2)^15)/48 + (501
*a^2*tan(c/2 + (d*x)/2)^17)/64 - (11*a^2*tan(c/2 + (d*x)/2))/64)/(d*(7*tan(c/2 + (d*x)/2)^2 + 20*tan(c/2 + (d*
x)/2)^4 + 28*tan(c/2 + (d*x)/2)^6 + 14*tan(c/2 + (d*x)/2)^8 - 14*tan(c/2 + (d*x)/2)^10 - 28*tan(c/2 + (d*x)/2)
^12 - 20*tan(c/2 + (d*x)/2)^14 - 7*tan(c/2 + (d*x)/2)^16 - tan(c/2 + (d*x)/2)^18 + 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((a+a*sec(d*x+c))**2*sin(d*x+c)**8,x)

[Out]

Timed out

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