3.55 \(\int \csc ^8(c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=192 \[ \frac {3 a^3 \tan (c+d x)}{d}-\frac {4 a^3 \cot ^7(c+d x)}{7 d}-\frac {3 a^3 \cot ^5(c+d x)}{d}-\frac {7 a^3 \cot ^3(c+d x)}{d}-\frac {13 a^3 \cot (c+d x)}{d}-\frac {15 a^3 \csc ^7(c+d x)}{14 d}-\frac {3 a^3 \csc ^5(c+d x)}{2 d}-\frac {5 a^3 \csc ^3(c+d x)}{2 d}-\frac {15 a^3 \csc (c+d x)}{2 d}+\frac {15 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^3 \csc ^7(c+d x) \sec ^2(c+d x)}{2 d} \]

[Out]

15/2*a^3*arctanh(sin(d*x+c))/d-13*a^3*cot(d*x+c)/d-7*a^3*cot(d*x+c)^3/d-3*a^3*cot(d*x+c)^5/d-4/7*a^3*cot(d*x+c
)^7/d-15/2*a^3*csc(d*x+c)/d-5/2*a^3*csc(d*x+c)^3/d-3/2*a^3*csc(d*x+c)^5/d-15/14*a^3*csc(d*x+c)^7/d+1/2*a^3*csc
(d*x+c)^7*sec(d*x+c)^2/d+3*a^3*tan(d*x+c)/d

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Rubi [A]  time = 0.31, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3872, 2873, 3767, 2621, 302, 207, 2620, 270, 288} \[ \frac {3 a^3 \tan (c+d x)}{d}-\frac {4 a^3 \cot ^7(c+d x)}{7 d}-\frac {3 a^3 \cot ^5(c+d x)}{d}-\frac {7 a^3 \cot ^3(c+d x)}{d}-\frac {13 a^3 \cot (c+d x)}{d}-\frac {15 a^3 \csc ^7(c+d x)}{14 d}-\frac {3 a^3 \csc ^5(c+d x)}{2 d}-\frac {5 a^3 \csc ^3(c+d x)}{2 d}-\frac {15 a^3 \csc (c+d x)}{2 d}+\frac {15 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^3 \csc ^7(c+d x) \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^8*(a + a*Sec[c + d*x])^3,x]

[Out]

(15*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (13*a^3*Cot[c + d*x])/d - (7*a^3*Cot[c + d*x]^3)/d - (3*a^3*Cot[c + d*x
]^5)/d - (4*a^3*Cot[c + d*x]^7)/(7*d) - (15*a^3*Csc[c + d*x])/(2*d) - (5*a^3*Csc[c + d*x]^3)/(2*d) - (3*a^3*Cs
c[c + d*x]^5)/(2*d) - (15*a^3*Csc[c + d*x]^7)/(14*d) + (a^3*Csc[c + d*x]^7*Sec[c + d*x]^2)/(2*d) + (3*a^3*Tan[
c + d*x])/d

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 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]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 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 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 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 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 \csc ^8(c+d x) (a+a \sec (c+d x))^3 \, dx &=-\int (-a-a \cos (c+d x))^3 \csc ^8(c+d x) \sec ^3(c+d x) \, dx\\ &=\int \left (a^3 \csc ^8(c+d x)+3 a^3 \csc ^8(c+d x) \sec (c+d x)+3 a^3 \csc ^8(c+d x) \sec ^2(c+d x)+a^3 \csc ^8(c+d x) \sec ^3(c+d x)\right ) \, dx\\ &=a^3 \int \csc ^8(c+d x) \, dx+a^3 \int \csc ^8(c+d x) \sec ^3(c+d x) \, dx+\left (3 a^3\right ) \int \csc ^8(c+d x) \sec (c+d x) \, dx+\left (3 a^3\right ) \int \csc ^8(c+d x) \sec ^2(c+d x) \, dx\\ &=-\frac {a^3 \operatorname {Subst}\left (\int \frac {x^{10}}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{d}-\frac {a^3 \operatorname {Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {x^8}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}+\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^4}{x^8} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {a^3 \cot (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{d}-\frac {3 a^3 \cot ^5(c+d x)}{5 d}-\frac {a^3 \cot ^7(c+d x)}{7 d}+\frac {a^3 \csc ^7(c+d x) \sec ^2(c+d x)}{2 d}+\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (1+\frac {1}{x^8}+\frac {4}{x^6}+\frac {6}{x^4}+\frac {4}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (1+x^2+x^4+x^6+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{d}-\frac {\left (9 a^3\right ) \operatorname {Subst}\left (\int \frac {x^8}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{2 d}\\ &=-\frac {13 a^3 \cot (c+d x)}{d}-\frac {7 a^3 \cot ^3(c+d x)}{d}-\frac {3 a^3 \cot ^5(c+d x)}{d}-\frac {4 a^3 \cot ^7(c+d x)}{7 d}-\frac {3 a^3 \csc (c+d x)}{d}-\frac {a^3 \csc ^3(c+d x)}{d}-\frac {3 a^3 \csc ^5(c+d x)}{5 d}-\frac {3 a^3 \csc ^7(c+d x)}{7 d}+\frac {a^3 \csc ^7(c+d x) \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}-\frac {\left (9 a^3\right ) \operatorname {Subst}\left (\int \left (1+x^2+x^4+x^6+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{2 d}\\ &=\frac {3 a^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {13 a^3 \cot (c+d x)}{d}-\frac {7 a^3 \cot ^3(c+d x)}{d}-\frac {3 a^3 \cot ^5(c+d x)}{d}-\frac {4 a^3 \cot ^7(c+d x)}{7 d}-\frac {15 a^3 \csc (c+d x)}{2 d}-\frac {5 a^3 \csc ^3(c+d x)}{2 d}-\frac {3 a^3 \csc ^5(c+d x)}{2 d}-\frac {15 a^3 \csc ^7(c+d x)}{14 d}+\frac {a^3 \csc ^7(c+d x) \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}-\frac {\left (9 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{2 d}\\ &=\frac {15 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {13 a^3 \cot (c+d x)}{d}-\frac {7 a^3 \cot ^3(c+d x)}{d}-\frac {3 a^3 \cot ^5(c+d x)}{d}-\frac {4 a^3 \cot ^7(c+d x)}{7 d}-\frac {15 a^3 \csc (c+d x)}{2 d}-\frac {5 a^3 \csc ^3(c+d x)}{2 d}-\frac {3 a^3 \csc ^5(c+d x)}{2 d}-\frac {15 a^3 \csc ^7(c+d x)}{14 d}+\frac {a^3 \csc ^7(c+d x) \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 1.25, size = 430, normalized size = 2.24 \[ \frac {a^3 \cos (c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1)^3 \left (-8 \csc (2 c) (2776 \sin (c-d x)-6080 \sin (c+d x)+8816 \sin (2 (c+d x))-7904 \sin (3 (c+d x))+4864 \sin (4 (c+d x))-1824 \sin (5 (c+d x))+304 \sin (6 (c+d x))-9580 \sin (2 c+d x)-10024 \sin (3 c+d x)+13891 \sin (c+2 d x)+7720 \sin (2 (c+2 d x))+13891 \sin (3 c+2 d x)+10080 \sin (4 c+2 d x)-10060 \sin (c+3 d x)-12454 \sin (2 c+3 d x)-12454 \sin (4 c+3 d x)-6580 \sin (5 c+3 d x)+7664 \sin (3 c+4 d x)+7664 \sin (5 c+4 d x)+2520 \sin (6 c+4 d x)-3420 \sin (3 c+5 d x)-2874 \sin (4 c+5 d x)-2874 \sin (6 c+5 d x)-420 \sin (7 c+5 d x)+640 \sin (4 c+6 d x)+479 \sin (5 c+6 d x)+479 \sin (7 c+6 d x)+5264 \sin (2 c)-9580 \sin (d x)+8480 \sin (2 d x)) \csc (c+d x) \csc ^6\left (\frac {1}{2} (c+d x)\right )-860160 \cos ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+860160 \cos ^2(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{917504 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^8*(a + a*Sec[c + d*x])^3,x]

[Out]

(a^3*Cos[c + d*x]*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(-860160*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] - Sin[(
c + d*x)/2]] + 860160*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 8*Csc[2*c]*Csc[(c + d*x)/2]^6*
Csc[c + d*x]*(5264*Sin[2*c] - 9580*Sin[d*x] + 8480*Sin[2*d*x] + 2776*Sin[c - d*x] - 6080*Sin[c + d*x] + 8816*S
in[2*(c + d*x)] - 7904*Sin[3*(c + d*x)] + 4864*Sin[4*(c + d*x)] - 1824*Sin[5*(c + d*x)] + 304*Sin[6*(c + d*x)]
- 9580*Sin[2*c + d*x] - 10024*Sin[3*c + d*x] + 13891*Sin[c + 2*d*x] + 7720*Sin[2*(c + 2*d*x)] + 13891*Sin[3*c
+ 2*d*x] + 10080*Sin[4*c + 2*d*x] - 10060*Sin[c + 3*d*x] - 12454*Sin[2*c + 3*d*x] - 12454*Sin[4*c + 3*d*x] -
6580*Sin[5*c + 3*d*x] + 7664*Sin[3*c + 4*d*x] + 7664*Sin[5*c + 4*d*x] + 2520*Sin[6*c + 4*d*x] - 3420*Sin[3*c +
5*d*x] - 2874*Sin[4*c + 5*d*x] - 2874*Sin[6*c + 5*d*x] - 420*Sin[7*c + 5*d*x] + 640*Sin[4*c + 6*d*x] + 479*Si
n[5*c + 6*d*x] + 479*Sin[7*c + 6*d*x])))/(917504*d)

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fricas [A]  time = 0.62, size = 278, normalized size = 1.45 \[ -\frac {320 \, a^{3} \cos \left (d x + c\right )^{6} - 750 \, a^{3} \cos \left (d x + c\right )^{5} + 170 \, a^{3} \cos \left (d x + c\right )^{4} + 720 \, a^{3} \cos \left (d x + c\right )^{3} - 520 \, a^{3} \cos \left (d x + c\right )^{2} + 42 \, a^{3} \cos \left (d x + c\right ) + 14 \, a^{3} - 105 \, {\left (a^{3} \cos \left (d x + c\right )^{5} - 3 \, a^{3} \cos \left (d x + c\right )^{4} + 3 \, a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 105 \, {\left (a^{3} \cos \left (d x + c\right )^{5} - 3 \, a^{3} \cos \left (d x + c\right )^{4} + 3 \, a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{28 \, {\left (d \cos \left (d x + c\right )^{5} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^8*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/28*(320*a^3*cos(d*x + c)^6 - 750*a^3*cos(d*x + c)^5 + 170*a^3*cos(d*x + c)^4 + 720*a^3*cos(d*x + c)^3 - 520
*a^3*cos(d*x + c)^2 + 42*a^3*cos(d*x + c) + 14*a^3 - 105*(a^3*cos(d*x + c)^5 - 3*a^3*cos(d*x + c)^4 + 3*a^3*co
s(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(sin(d*x + c) + 1)*sin(d*x + c) + 105*(a^3*cos(d*x + c)^5 - 3*a^3*cos(d*
x + c)^4 + 3*a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(-sin(d*x + c) + 1)*sin(d*x + c))/((d*cos(d*x + c)^5
- 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^3 - d*cos(d*x + c)^2)*sin(d*x + c))

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giac [A]  time = 0.69, size = 169, normalized size = 0.88 \[ \frac {840 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 840 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {112 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac {1050 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 112 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 14 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}}}{112 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^8*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/112*(840*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 840*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 7*a^3*tan(1/2
*d*x + 1/2*c) - 112*(5*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2
- (1050*a^3*tan(1/2*d*x + 1/2*c)^6 + 112*a^3*tan(1/2*d*x + 1/2*c)^4 + 14*a^3*tan(1/2*d*x + 1/2*c)^2 + a^3)/ta
n(1/2*d*x + 1/2*c)^7)/d

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maple [B]  time = 1.12, size = 360, normalized size = 1.88 \[ -\frac {80 a^{3} \cot \left (d x +c \right )}{7 d}-\frac {a^{3} \cot \left (d x +c \right ) \left (\csc ^{6}\left (d x +c \right )\right )}{7 d}-\frac {6 a^{3} \cot \left (d x +c \right ) \left (\csc ^{4}\left (d x +c \right )\right )}{35 d}-\frac {8 a^{3} \cot \left (d x +c \right ) \left (\csc ^{2}\left (d x +c \right )\right )}{35 d}-\frac {3 a^{3}}{7 d \sin \left (d x +c \right )^{7}}-\frac {3 a^{3}}{5 d \sin \left (d x +c \right )^{5}}-\frac {a^{3}}{d \sin \left (d x +c \right )^{3}}-\frac {15 a^{3}}{2 d \sin \left (d x +c \right )}+\frac {15 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}-\frac {3 a^{3}}{7 d \sin \left (d x +c \right )^{7} \cos \left (d x +c \right )}-\frac {24 a^{3}}{35 d \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )}-\frac {48 a^{3}}{35 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {192 a^{3}}{35 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {a^{3}}{7 d \sin \left (d x +c \right )^{7} \cos \left (d x +c \right )^{2}}-\frac {9 a^{3}}{35 d \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{2}}-\frac {3 a^{3}}{5 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {3 a^{3}}{2 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^8*(a+a*sec(d*x+c))^3,x)

[Out]

-80/7*a^3*cot(d*x+c)/d-1/7/d*a^3*cot(d*x+c)*csc(d*x+c)^6-6/35/d*a^3*cot(d*x+c)*csc(d*x+c)^4-8/35/d*a^3*cot(d*x
+c)*csc(d*x+c)^2-3/7/d*a^3/sin(d*x+c)^7-3/5/d*a^3/sin(d*x+c)^5-1/d*a^3/sin(d*x+c)^3-15/2/d*a^3/sin(d*x+c)+15/2
/d*a^3*ln(sec(d*x+c)+tan(d*x+c))-3/7/d*a^3/sin(d*x+c)^7/cos(d*x+c)-24/35/d*a^3/sin(d*x+c)^5/cos(d*x+c)-48/35/d
*a^3/sin(d*x+c)^3/cos(d*x+c)+192/35/d*a^3/sin(d*x+c)/cos(d*x+c)-1/7/d*a^3/sin(d*x+c)^7/cos(d*x+c)^2-9/35/d*a^3
/sin(d*x+c)^5/cos(d*x+c)^2-3/5/d*a^3/sin(d*x+c)^3/cos(d*x+c)^2+3/2/d*a^3/sin(d*x+c)/cos(d*x+c)^2

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maxima [A]  time = 0.37, size = 268, normalized size = 1.40 \[ -\frac {a^{3} {\left (\frac {2 \, {\left (315 \, \sin \left (d x + c\right )^{8} - 210 \, \sin \left (d x + c\right )^{6} - 42 \, \sin \left (d x + c\right )^{4} - 18 \, \sin \left (d x + c\right )^{2} - 10\right )}}{\sin \left (d x + c\right )^{9} - \sin \left (d x + c\right )^{7}} - 315 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 315 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{3} {\left (\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{6} + 35 \, \sin \left (d x + c\right )^{4} + 21 \, \sin \left (d x + c\right )^{2} + 15\right )}}{\sin \left (d x + c\right )^{7}} - 105 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} {\left (\frac {140 \, \tan \left (d x + c\right )^{6} + 70 \, \tan \left (d x + c\right )^{4} + 28 \, \tan \left (d x + c\right )^{2} + 5}{\tan \left (d x + c\right )^{7}} - 35 \, \tan \left (d x + c\right )\right )} + \frac {4 \, {\left (35 \, \tan \left (d x + c\right )^{6} + 35 \, \tan \left (d x + c\right )^{4} + 21 \, \tan \left (d x + c\right )^{2} + 5\right )} a^{3}}{\tan \left (d x + c\right )^{7}}}{140 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^8*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/140*(a^3*(2*(315*sin(d*x + c)^8 - 210*sin(d*x + c)^6 - 42*sin(d*x + c)^4 - 18*sin(d*x + c)^2 - 10)/(sin(d*x
+ c)^9 - sin(d*x + c)^7) - 315*log(sin(d*x + c) + 1) + 315*log(sin(d*x + c) - 1)) + 2*a^3*(2*(105*sin(d*x + c
)^6 + 35*sin(d*x + c)^4 + 21*sin(d*x + c)^2 + 15)/sin(d*x + c)^7 - 105*log(sin(d*x + c) + 1) + 105*log(sin(d*x
+ c) - 1)) + 12*a^3*((140*tan(d*x + c)^6 + 70*tan(d*x + c)^4 + 28*tan(d*x + c)^2 + 5)/tan(d*x + c)^7 - 35*tan
(d*x + c)) + 4*(35*tan(d*x + c)^6 + 35*tan(d*x + c)^4 + 21*tan(d*x + c)^2 + 5)*a^3/tan(d*x + c)^7)/d

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mupad [B]  time = 2.90, size = 169, normalized size = 0.88 \[ \frac {15\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {230\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-396\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+120\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {85\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}+\frac {12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{7}+\frac {a^3}{7}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\right )}-\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^3/sin(c + d*x)^8,x)

[Out]

(15*a^3*atanh(tan(c/2 + (d*x)/2)))/d - ((12*a^3*tan(c/2 + (d*x)/2)^2)/7 + (85*a^3*tan(c/2 + (d*x)/2)^4)/7 + 12
0*a^3*tan(c/2 + (d*x)/2)^6 - 396*a^3*tan(c/2 + (d*x)/2)^8 + 230*a^3*tan(c/2 + (d*x)/2)^10 + a^3/7)/(d*(16*tan(
c/2 + (d*x)/2)^7 - 32*tan(c/2 + (d*x)/2)^9 + 16*tan(c/2 + (d*x)/2)^11)) - (a^3*tan(c/2 + (d*x)/2))/(16*d)

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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(csc(d*x+c)**8*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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