3.37 \(\int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=201 \[ \frac{a^2 \tan (c+d x)}{d}-\frac{2 a^2 \cot ^9(c+d x)}{9 d}-\frac{9 a^2 \cot ^7(c+d x)}{7 d}-\frac{16 a^2 \cot ^5(c+d x)}{5 d}-\frac{14 a^2 \cot ^3(c+d x)}{3 d}-\frac{6 a^2 \cot (c+d x)}{d}-\frac{2 a^2 \csc ^9(c+d x)}{9 d}-\frac{2 a^2 \csc ^7(c+d x)}{7 d}-\frac{2 a^2 \csc ^5(c+d x)}{5 d}-\frac{2 a^2 \csc ^3(c+d x)}{3 d}-\frac{2 a^2 \csc (c+d x)}{d}+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(2*a^2*ArcTanh[Sin[c + d*x]])/d - (6*a^2*Cot[c + d*x])/d - (14*a^2*Cot[c + d*x]^3)/(3*d) - (16*a^2*Cot[c + d*x
]^5)/(5*d) - (9*a^2*Cot[c + d*x]^7)/(7*d) - (2*a^2*Cot[c + d*x]^9)/(9*d) - (2*a^2*Csc[c + d*x])/d - (2*a^2*Csc
[c + d*x]^3)/(3*d) - (2*a^2*Csc[c + d*x]^5)/(5*d) - (2*a^2*Csc[c + d*x]^7)/(7*d) - (2*a^2*Csc[c + d*x]^9)/(9*d
) + (a^2*Tan[c + d*x])/d

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Rubi [A]  time = 0.258413, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2873, 3767, 2621, 302, 207, 2620, 270} \[ \frac{a^2 \tan (c+d x)}{d}-\frac{2 a^2 \cot ^9(c+d x)}{9 d}-\frac{9 a^2 \cot ^7(c+d x)}{7 d}-\frac{16 a^2 \cot ^5(c+d x)}{5 d}-\frac{14 a^2 \cot ^3(c+d x)}{3 d}-\frac{6 a^2 \cot (c+d x)}{d}-\frac{2 a^2 \csc ^9(c+d x)}{9 d}-\frac{2 a^2 \csc ^7(c+d x)}{7 d}-\frac{2 a^2 \csc ^5(c+d x)}{5 d}-\frac{2 a^2 \csc ^3(c+d x)}{3 d}-\frac{2 a^2 \csc (c+d x)}{d}+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^10*(a + a*Sec[c + d*x])^2,x]

[Out]

(2*a^2*ArcTanh[Sin[c + d*x]])/d - (6*a^2*Cot[c + d*x])/d - (14*a^2*Cot[c + d*x]^3)/(3*d) - (16*a^2*Cot[c + d*x
]^5)/(5*d) - (9*a^2*Cot[c + d*x]^7)/(7*d) - (2*a^2*Cot[c + d*x]^9)/(9*d) - (2*a^2*Csc[c + d*x])/d - (2*a^2*Csc
[c + d*x]^3)/(3*d) - (2*a^2*Csc[c + d*x]^5)/(5*d) - (2*a^2*Csc[c + d*x]^7)/(7*d) - (2*a^2*Csc[c + d*x]^9)/(9*d
) + (a^2*Tan[c + d*x])/d

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]

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 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 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 207

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

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \csc ^{10}(c+d x) (a+a \sec (c+d x))^2 \, dx &=\int (-a-a \cos (c+d x))^2 \csc ^{10}(c+d x) \sec ^2(c+d x) \, dx\\ &=\int \left (a^2 \csc ^{10}(c+d x)+2 a^2 \csc ^{10}(c+d x) \sec (c+d x)+a^2 \csc ^{10}(c+d x) \sec ^2(c+d x)\right ) \, dx\\ &=a^2 \int \csc ^{10}(c+d x) \, dx+a^2 \int \csc ^{10}(c+d x) \sec ^2(c+d x) \, dx+\left (2 a^2\right ) \int \csc ^{10}(c+d x) \sec (c+d x) \, dx\\ &=\frac{a^2 \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^5}{x^{10}} \, dx,x,\tan (c+d x)\right )}{d}-\frac{a^2 \operatorname{Subst}\left (\int \left (1+4 x^2+6 x^4+4 x^6+x^8\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{x^{10}}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac{a^2 \cot (c+d x)}{d}-\frac{4 a^2 \cot ^3(c+d x)}{3 d}-\frac{6 a^2 \cot ^5(c+d x)}{5 d}-\frac{4 a^2 \cot ^7(c+d x)}{7 d}-\frac{a^2 \cot ^9(c+d x)}{9 d}+\frac{a^2 \operatorname{Subst}\left (\int \left (1+\frac{1}{x^{10}}+\frac{5}{x^8}+\frac{10}{x^6}+\frac{10}{x^4}+\frac{5}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \left (1+x^2+x^4+x^6+x^8+\frac{1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac{6 a^2 \cot (c+d x)}{d}-\frac{14 a^2 \cot ^3(c+d x)}{3 d}-\frac{16 a^2 \cot ^5(c+d x)}{5 d}-\frac{9 a^2 \cot ^7(c+d x)}{7 d}-\frac{2 a^2 \cot ^9(c+d x)}{9 d}-\frac{2 a^2 \csc (c+d x)}{d}-\frac{2 a^2 \csc ^3(c+d x)}{3 d}-\frac{2 a^2 \csc ^5(c+d x)}{5 d}-\frac{2 a^2 \csc ^7(c+d x)}{7 d}-\frac{2 a^2 \csc ^9(c+d x)}{9 d}+\frac{a^2 \tan (c+d x)}{d}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{6 a^2 \cot (c+d x)}{d}-\frac{14 a^2 \cot ^3(c+d x)}{3 d}-\frac{16 a^2 \cot ^5(c+d x)}{5 d}-\frac{9 a^2 \cot ^7(c+d x)}{7 d}-\frac{2 a^2 \cot ^9(c+d x)}{9 d}-\frac{2 a^2 \csc (c+d x)}{d}-\frac{2 a^2 \csc ^3(c+d x)}{3 d}-\frac{2 a^2 \csc ^5(c+d x)}{5 d}-\frac{2 a^2 \csc ^7(c+d x)}{7 d}-\frac{2 a^2 \csc ^9(c+d x)}{9 d}+\frac{a^2 \tan (c+d x)}{d}\\ \end{align*}

Mathematica [B]  time = 6.83056, size = 1050, normalized size = 5.22 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[c + d*x]^10*(a + a*Sec[c + d*x])^2,x]

[Out]

(-6899*Cos[c + d*x]^2*Cot[c/2]*Csc[c/2 + (d*x)/2]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(80640*d) - (
193*Cos[c + d*x]^2*Cot[c/2]*Csc[c/2 + (d*x)/2]^4*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(13440*d) - (71*
Cos[c + d*x]^2*Cot[c/2]*Csc[c/2 + (d*x)/2]^6*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(32256*d) - (Cos[c +
d*x]^2*Cot[c/2]*Csc[c/2 + (d*x)/2]^8*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(4608*d) - (Cos[c + d*x]^2*
Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(2*d) + (Cos[c + d*x
]^2*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(2*d) + (123041*
Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(161280*d
) + (6899*Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]^3*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2
])/(80640*d) + (193*Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]^5*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*S
in[(d*x)/2])/(13440*d) + (71*Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]^7*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c +
d*x])^2*Sin[(d*x)/2])/(32256*d) + (Cos[c + d*x]^2*Csc[c/2]*Csc[c/2 + (d*x)/2]^9*Sec[c/2 + (d*x)/2]^4*(a + a*Se
c[c + d*x])^2*Sin[(d*x)/2])/(4608*d) + (803*Cos[c + d*x]^2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(a + a*Sec[c + d*x])^
2*Sin[(d*x)/2])/(7680*d) + (49*Cos[c + d*x]^2*Sec[c/2]*Sec[c/2 + (d*x)/2]^7*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2
])/(7680*d) + (Cos[c + d*x]^2*Sec[c/2]*Sec[c/2 + (d*x)/2]^9*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(2560*d) + (C
os[c + d*x]*Sec[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[d*x])/(4*d) + (49*Cos[c + d*x]^2*Sec[c/2 +
(d*x)/2]^6*(a + a*Sec[c + d*x])^2*Tan[c/2])/(7680*d) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x
])^2*Tan[c/2])/(2560*d)

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Maple [A]  time = 0.081, size = 326, normalized size = 1.6 \begin{align*} -{\frac{1408\,{a}^{2}\cot \left ( dx+c \right ) }{315\,d}}-{\frac{{a}^{2}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{8}}{9\,d}}-{\frac{8\,{a}^{2}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{6}}{63\,d}}-{\frac{16\,{a}^{2}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{105\,d}}-{\frac{64\,{a}^{2}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{315\,d}}-{\frac{2\,{a}^{2}}{9\,d \left ( \sin \left ( dx+c \right ) \right ) ^{9}}}-{\frac{2\,{a}^{2}}{7\,d \left ( \sin \left ( dx+c \right ) \right ) ^{7}}}-{\frac{2\,{a}^{2}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{2\,{a}^{2}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-2\,{\frac{{a}^{2}}{d\sin \left ( dx+c \right ) }}+2\,{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}}{9\,d \left ( \sin \left ( dx+c \right ) \right ) ^{9}\cos \left ( dx+c \right ) }}-{\frac{10\,{a}^{2}}{63\,d \left ( \sin \left ( dx+c \right ) \right ) ^{7}\cos \left ( dx+c \right ) }}-{\frac{16\,{a}^{2}}{63\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}\cos \left ( dx+c \right ) }}-{\frac{32\,{a}^{2}}{63\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+{\frac{128\,{a}^{2}}{63\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^10*(a+a*sec(d*x+c))^2,x)

[Out]

-1408/315*a^2*cot(d*x+c)/d-1/9/d*a^2*cot(d*x+c)*csc(d*x+c)^8-8/63/d*a^2*cot(d*x+c)*csc(d*x+c)^6-16/105/d*a^2*c
ot(d*x+c)*csc(d*x+c)^4-64/315/d*a^2*cot(d*x+c)*csc(d*x+c)^2-2/9/d*a^2/sin(d*x+c)^9-2/7/d*a^2/sin(d*x+c)^7-2/5/
d*a^2/sin(d*x+c)^5-2/3/d*a^2/sin(d*x+c)^3-2/d*a^2/sin(d*x+c)+2/d*a^2*ln(sec(d*x+c)+tan(d*x+c))-1/9/d*a^2/sin(d
*x+c)^9/cos(d*x+c)-10/63/d*a^2/sin(d*x+c)^7/cos(d*x+c)-16/63/d*a^2/sin(d*x+c)^5/cos(d*x+c)-32/63/d*a^2/sin(d*x
+c)^3/cos(d*x+c)+128/63/d*a^2/sin(d*x+c)/cos(d*x+c)

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Maxima [A]  time = 1.02784, size = 275, normalized size = 1.37 \begin{align*} -\frac{a^{2}{\left (\frac{2 \,{\left (315 \, \sin \left (d x + c\right )^{8} + 105 \, \sin \left (d x + c\right )^{6} + 63 \, \sin \left (d x + c\right )^{4} + 45 \, \sin \left (d x + c\right )^{2} + 35\right )}}{\sin \left (d x + c\right )^{9}} - 315 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 315 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 5 \, a^{2}{\left (\frac{315 \, \tan \left (d x + c\right )^{8} + 210 \, \tan \left (d x + c\right )^{6} + 126 \, \tan \left (d x + c\right )^{4} + 45 \, \tan \left (d x + c\right )^{2} + 7}{\tan \left (d x + c\right )^{9}} - 63 \, \tan \left (d x + c\right )\right )} + \frac{{\left (315 \, \tan \left (d x + c\right )^{8} + 420 \, \tan \left (d x + c\right )^{6} + 378 \, \tan \left (d x + c\right )^{4} + 180 \, \tan \left (d x + c\right )^{2} + 35\right )} a^{2}}{\tan \left (d x + c\right )^{9}}}{315 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^10*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/315*(a^2*(2*(315*sin(d*x + c)^8 + 105*sin(d*x + c)^6 + 63*sin(d*x + c)^4 + 45*sin(d*x + c)^2 + 35)/sin(d*x
+ c)^9 - 315*log(sin(d*x + c) + 1) + 315*log(sin(d*x + c) - 1)) + 5*a^2*((315*tan(d*x + c)^8 + 210*tan(d*x + c
)^6 + 126*tan(d*x + c)^4 + 45*tan(d*x + c)^2 + 7)/tan(d*x + c)^9 - 63*tan(d*x + c)) + (315*tan(d*x + c)^8 + 42
0*tan(d*x + c)^6 + 378*tan(d*x + c)^4 + 180*tan(d*x + c)^2 + 35)*a^2/tan(d*x + c)^9)/d

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Fricas [B]  time = 1.94829, size = 1026, normalized size = 5.1 \begin{align*} -\frac{1408 \, a^{2} \cos \left (d x + c\right )^{8} - 2186 \, a^{2} \cos \left (d x + c\right )^{7} - 3372 \, a^{2} \cos \left (d x + c\right )^{6} + 6200 \, a^{2} \cos \left (d x + c\right )^{5} + 2060 \, a^{2} \cos \left (d x + c\right )^{4} - 5784 \, a^{2} \cos \left (d x + c\right )^{3} + 268 \, a^{2} \cos \left (d x + c\right )^{2} + 1756 \, a^{2} \cos \left (d x + c\right ) - 315 \,{\left (a^{2} \cos \left (d x + c\right )^{7} - 2 \, a^{2} \cos \left (d x + c\right )^{6} - a^{2} \cos \left (d x + c\right )^{5} + 4 \, a^{2} \cos \left (d x + c\right )^{4} - a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 315 \,{\left (a^{2} \cos \left (d x + c\right )^{7} - 2 \, a^{2} \cos \left (d x + c\right )^{6} - a^{2} \cos \left (d x + c\right )^{5} + 4 \, a^{2} \cos \left (d x + c\right )^{4} - a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 315 \, a^{2}}{315 \,{\left (d \cos \left (d x + c\right )^{7} - 2 \, d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{5} + 4 \, d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{3} - 2 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^10*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/315*(1408*a^2*cos(d*x + c)^8 - 2186*a^2*cos(d*x + c)^7 - 3372*a^2*cos(d*x + c)^6 + 6200*a^2*cos(d*x + c)^5
+ 2060*a^2*cos(d*x + c)^4 - 5784*a^2*cos(d*x + c)^3 + 268*a^2*cos(d*x + c)^2 + 1756*a^2*cos(d*x + c) - 315*(a^
2*cos(d*x + c)^7 - 2*a^2*cos(d*x + c)^6 - a^2*cos(d*x + c)^5 + 4*a^2*cos(d*x + c)^4 - a^2*cos(d*x + c)^3 - 2*a
^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*log(sin(d*x + c) + 1)*sin(d*x + c) + 315*(a^2*cos(d*x + c)^7 - 2*a^2*cos
(d*x + c)^6 - a^2*cos(d*x + c)^5 + 4*a^2*cos(d*x + c)^4 - a^2*cos(d*x + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2*cos(
d*x + c))*log(-sin(d*x + c) + 1)*sin(d*x + c) - 315*a^2)/((d*cos(d*x + c)^7 - 2*d*cos(d*x + c)^6 - d*cos(d*x +
c)^5 + 4*d*cos(d*x + c)^4 - d*cos(d*x + c)^3 - 2*d*cos(d*x + c)^2 + d*cos(d*x + c))*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**10*(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.38362, size = 270, normalized size = 1.34 \begin{align*} \frac{63 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1155 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 80640 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 80640 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 17955 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{80640 \, 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{139545 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 19635 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 3591 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 495 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 35 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9}}}{40320 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^10*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/40320*(63*a^2*tan(1/2*d*x + 1/2*c)^5 + 1155*a^2*tan(1/2*d*x + 1/2*c)^3 + 80640*a^2*log(abs(tan(1/2*d*x + 1/2
*c) + 1)) - 80640*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 17955*a^2*tan(1/2*d*x + 1/2*c) - 80640*a^2*tan(1/2*
d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (139545*a^2*tan(1/2*d*x + 1/2*c)^8 + 19635*a^2*tan(1/2*d*x + 1/2*c
)^6 + 3591*a^2*tan(1/2*d*x + 1/2*c)^4 + 495*a^2*tan(1/2*d*x + 1/2*c)^2 + 35*a^2)/tan(1/2*d*x + 1/2*c)^9)/d