3.54 \(\int \csc ^6(c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=165 \[ -\frac {a^6 \tan (c+d x) \sec (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac {11 a^5 \tan (c+d x) \sec (c+d x)}{15 d (a-a \cos (c+d x))^2}+\frac {152 a^3 \tan (c+d x)}{15 d}+\frac {13 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {13 a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac {76 a^6 \tan (c+d x) \sec (c+d x)}{15 d \left (a^3-a^3 \cos (c+d x)\right )} \]

[Out]

13/2*a^3*arctanh(sin(d*x+c))/d+152/15*a^3*tan(d*x+c)/d+13/2*a^3*sec(d*x+c)*tan(d*x+c)/d-1/5*a^6*sec(d*x+c)*tan
(d*x+c)/d/(a-a*cos(d*x+c))^3-11/15*a^5*sec(d*x+c)*tan(d*x+c)/d/(a-a*cos(d*x+c))^2-76/15*a^6*sec(d*x+c)*tan(d*x
+c)/d/(a^3-a^3*cos(d*x+c))

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Rubi [A]  time = 0.44, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3872, 2869, 2766, 2978, 2748, 3768, 3770, 3767, 8} \[ \frac {152 a^3 \tan (c+d x)}{15 d}+\frac {13 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {13 a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac {76 a^6 \tan (c+d x) \sec (c+d x)}{15 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac {a^6 \tan (c+d x) \sec (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac {11 a^5 \tan (c+d x) \sec (c+d x)}{15 d (a-a \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6*(a + a*Sec[c + d*x])^3,x]

[Out]

(13*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (152*a^3*Tan[c + d*x])/(15*d) + (13*a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d
) - (a^6*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a - a*Cos[c + d*x])^3) - (11*a^5*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a
- a*Cos[c + d*x])^2) - (76*a^6*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a^3 - a^3*Cos[c + d*x]))

Rule 8

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2869

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

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
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 3768

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

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 \csc ^6(c+d x) (a+a \sec (c+d x))^3 \, dx &=-\int (-a-a \cos (c+d x))^3 \csc ^6(c+d x) \sec ^3(c+d x) \, dx\\ &=-\left (a^6 \int \frac {\sec ^3(c+d x)}{(-a+a \cos (c+d x))^3} \, dx\right )\\ &=-\frac {a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac {1}{5} a^4 \int \frac {(-7 a-4 a \cos (c+d x)) \sec ^3(c+d x)}{(-a+a \cos (c+d x))^2} \, dx\\ &=-\frac {a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac {11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac {1}{15} a^2 \int \frac {\left (43 a^2+33 a^2 \cos (c+d x)\right ) \sec ^3(c+d x)}{-a+a \cos (c+d x)} \, dx\\ &=-\frac {a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac {11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac {76 a^4 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))}-\frac {1}{15} \int \left (-195 a^3-152 a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=-\frac {a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac {11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac {76 a^4 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))}+\frac {1}{15} \left (152 a^3\right ) \int \sec ^2(c+d x) \, dx+\left (13 a^3\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {13 a^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac {11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac {76 a^4 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))}+\frac {1}{2} \left (13 a^3\right ) \int \sec (c+d x) \, dx-\frac {\left (152 a^3\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {13 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {152 a^3 \tan (c+d x)}{15 d}+\frac {13 a^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac {11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac {76 a^4 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 1.22, size = 353, normalized size = 2.14 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (24960 \cos ^2(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\csc \left (\frac {c}{2}\right ) \sec (c) \left (4329 \sin \left (c-\frac {d x}{2}\right )-1989 \sin \left (c+\frac {d x}{2}\right )-3575 \sin \left (2 c+\frac {d x}{2}\right )+475 \sin \left (c+\frac {3 d x}{2}\right )+2005 \sin \left (2 c+\frac {3 d x}{2}\right )+2275 \sin \left (3 c+\frac {3 d x}{2}\right )-2673 \sin \left (c+\frac {5 d x}{2}\right )+105 \sin \left (2 c+\frac {5 d x}{2}\right )-1593 \sin \left (3 c+\frac {5 d x}{2}\right )-975 \sin \left (4 c+\frac {5 d x}{2}\right )+1325 \sin \left (2 c+\frac {7 d x}{2}\right )-255 \sin \left (3 c+\frac {7 d x}{2}\right )+875 \sin \left (4 c+\frac {7 d x}{2}\right )+195 \sin \left (5 c+\frac {7 d x}{2}\right )-304 \sin \left (3 c+\frac {9 d x}{2}\right )+90 \sin \left (4 c+\frac {9 d x}{2}\right )-214 \sin \left (5 c+\frac {9 d x}{2}\right )-1235 \sin \left (\frac {d x}{2}\right )+3805 \sin \left (\frac {3 d x}{2}\right )\right ) \csc ^5\left (\frac {1}{2} (c+d x)\right )\right )}{30720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^6*(a + a*Sec[c + d*x])^3,x]

[Out]

-1/30720*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^2*(24960*Cos[c + d*x]^2*(Log[Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Csc[c/2]*Csc[(c + d*x)/2]^5*Sec[c]*(-1235
*Sin[(d*x)/2] + 3805*Sin[(3*d*x)/2] + 4329*Sin[c - (d*x)/2] - 1989*Sin[c + (d*x)/2] - 3575*Sin[2*c + (d*x)/2]
+ 475*Sin[c + (3*d*x)/2] + 2005*Sin[2*c + (3*d*x)/2] + 2275*Sin[3*c + (3*d*x)/2] - 2673*Sin[c + (5*d*x)/2] + 1
05*Sin[2*c + (5*d*x)/2] - 1593*Sin[3*c + (5*d*x)/2] - 975*Sin[4*c + (5*d*x)/2] + 1325*Sin[2*c + (7*d*x)/2] - 2
55*Sin[3*c + (7*d*x)/2] + 875*Sin[4*c + (7*d*x)/2] + 195*Sin[5*c + (7*d*x)/2] - 304*Sin[3*c + (9*d*x)/2] + 90*
Sin[4*c + (9*d*x)/2] - 214*Sin[5*c + (9*d*x)/2])))/d

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fricas [A]  time = 0.74, size = 225, normalized size = 1.36 \[ -\frac {608 \, a^{3} \cos \left (d x + c\right )^{5} - 826 \, a^{3} \cos \left (d x + c\right )^{4} - 476 \, a^{3} \cos \left (d x + c\right )^{3} + 868 \, a^{3} \cos \left (d x + c\right )^{2} - 120 \, a^{3} \cos \left (d x + c\right ) - 30 \, a^{3} - 195 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 195 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(608*a^3*cos(d*x + c)^5 - 826*a^3*cos(d*x + c)^4 - 476*a^3*cos(d*x + c)^3 + 868*a^3*cos(d*x + c)^2 - 120
*a^3*cos(d*x + c) - 30*a^3 - 195*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^3 + a^3*cos(d*x + c)^2)*log(sin(d*x
+ c) + 1)*sin(d*x + c) + 195*(a^3*cos(d*x + c)^4 - 2*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)^4 - 2*d*cos(d*x + c)^3 + d*cos(d*x + c)^2)*sin(d*x + c))

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giac [A]  time = 0.40, size = 141, normalized size = 0.85 \[ \frac {390 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 390 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {60 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac {465 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(390*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 390*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 60*(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 - (465*a^3*tan(1/2*d*x + 1/2*c
)^4 + 40*a^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^3)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 1.05, size = 274, normalized size = 1.66 \[ -\frac {152 a^{3} \cot \left (d x +c \right )}{15 d}-\frac {a^{3} \cot \left (d x +c \right ) \left (\csc ^{4}\left (d x +c \right )\right )}{5 d}-\frac {4 a^{3} \cot \left (d x +c \right ) \left (\csc ^{2}\left (d x +c \right )\right )}{15 d}-\frac {3 a^{3}}{5 d \sin \left (d x +c \right )^{5}}-\frac {a^{3}}{d \sin \left (d x +c \right )^{3}}-\frac {13 a^{3}}{2 d \sin \left (d x +c \right )}+\frac {13 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}-\frac {3 a^{3}}{5 d \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )}-\frac {6 a^{3}}{5 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {24 a^{3}}{5 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {a^{3}}{5 d \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{2}}-\frac {7 a^{3}}{15 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {7 a^{3}}{6 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^6*(a+a*sec(d*x+c))^3,x)

[Out]

-152/15*a^3*cot(d*x+c)/d-1/5/d*a^3*cot(d*x+c)*csc(d*x+c)^4-4/15/d*a^3*cot(d*x+c)*csc(d*x+c)^2-3/5/d*a^3/sin(d*
x+c)^5-1/d*a^3/sin(d*x+c)^3-13/2/d*a^3/sin(d*x+c)+13/2/d*a^3*ln(sec(d*x+c)+tan(d*x+c))-3/5/d*a^3/sin(d*x+c)^5/
cos(d*x+c)-6/5/d*a^3/sin(d*x+c)^3/cos(d*x+c)+24/5/d*a^3/sin(d*x+c)/cos(d*x+c)-1/5/d*a^3/sin(d*x+c)^5/cos(d*x+c
)^2-7/15/d*a^3/sin(d*x+c)^3/cos(d*x+c)^2+7/6/d*a^3/sin(d*x+c)/cos(d*x+c)^2

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maxima [A]  time = 0.49, size = 228, normalized size = 1.38 \[ -\frac {a^{3} {\left (\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{6} - 70 \, \sin \left (d x + c\right )^{4} - 14 \, \sin \left (d x + c\right )^{2} - 6\right )}}{\sin \left (d x + c\right )^{7} - \sin \left (d x + c\right )^{5}} - 105 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 5 \, \sin \left (d x + c\right )^{2} + 3\right )}}{\sin \left (d x + c\right )^{5}} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, a^{3} {\left (\frac {15 \, \tan \left (d x + c\right )^{4} + 5 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{5}} - 5 \, \tan \left (d x + c\right )\right )} + \frac {4 \, {\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{3}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(a^3*(2*(105*sin(d*x + c)^6 - 70*sin(d*x + c)^4 - 14*sin(d*x + c)^2 - 6)/(sin(d*x + c)^7 - sin(d*x + c)^
5) - 105*log(sin(d*x + c) + 1) + 105*log(sin(d*x + c) - 1)) + 6*a^3*(2*(15*sin(d*x + c)^4 + 5*sin(d*x + c)^2 +
3)/sin(d*x + c)^5 - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) + 36*a^3*((15*tan(d*x + c)^4 + 5*tan
(d*x + c)^2 + 1)/tan(d*x + c)^5 - 5*tan(d*x + c)) + 4*(15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*a^3/tan(d*x
+ c)^5)/d

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mupad [B]  time = 4.91, size = 136, normalized size = 0.82 \[ \frac {13\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {51\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {262\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {388\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+\frac {34\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {a^3}{5}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^3/sin(c + d*x)^6,x)

[Out]

(13*a^3*atanh(tan(c/2 + (d*x)/2)))/d - ((34*a^3*tan(c/2 + (d*x)/2)^2)/15 + (388*a^3*tan(c/2 + (d*x)/2)^4)/15 -
(262*a^3*tan(c/2 + (d*x)/2)^6)/3 + 51*a^3*tan(c/2 + (d*x)/2)^8 + a^3/5)/(d*(4*tan(c/2 + (d*x)/2)^5 - 8*tan(c/
2 + (d*x)/2)^7 + 4*tan(c/2 + (d*x)/2)^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(csc(d*x+c)**6*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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