3.794 \(\int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=162 \[ \frac {4 \tan ^7(c+d x)}{7 a^3 d}+\frac {13 \tan ^5(c+d x)}{5 a^3 d}+\frac {5 \tan ^3(c+d x)}{a^3 d}+\frac {7 \tan (c+d x)}{a^3 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {4 \sec ^7(c+d x)}{7 a^3 d}-\frac {3 \sec ^5(c+d x)}{5 a^3 d}-\frac {\sec ^3(c+d x)}{a^3 d}-\frac {3 \sec (c+d x)}{a^3 d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[Out]

3*arctanh(cos(d*x+c))/a^3/d-cot(d*x+c)/a^3/d-3*sec(d*x+c)/a^3/d-sec(d*x+c)^3/a^3/d-3/5*sec(d*x+c)^5/a^3/d-4/7*
sec(d*x+c)^7/a^3/d+7*tan(d*x+c)/a^3/d+5*tan(d*x+c)^3/a^3/d+13/5*tan(d*x+c)^5/a^3/d+4/7*tan(d*x+c)^7/a^3/d

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Rubi [A]  time = 0.35, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {2875, 2873, 3767, 2622, 302, 207, 2620, 270, 2606, 30} \[ \frac {4 \tan ^7(c+d x)}{7 a^3 d}+\frac {13 \tan ^5(c+d x)}{5 a^3 d}+\frac {5 \tan ^3(c+d x)}{a^3 d}+\frac {7 \tan (c+d x)}{a^3 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {4 \sec ^7(c+d x)}{7 a^3 d}-\frac {3 \sec ^5(c+d x)}{5 a^3 d}-\frac {\sec ^3(c+d x)}{a^3 d}-\frac {3 \sec (c+d x)}{a^3 d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(a^3*d) - Cot[c + d*x]/(a^3*d) - (3*Sec[c + d*x])/(a^3*d) - Sec[c + d*x]^3/(a^3*d) -
(3*Sec[c + d*x]^5)/(5*a^3*d) - (4*Sec[c + d*x]^7)/(7*a^3*d) + (7*Tan[c + d*x])/(a^3*d) + (5*Tan[c + d*x]^3)/(
a^3*d) + (13*Tan[c + d*x]^5)/(5*a^3*d) + (4*Tan[c + d*x]^7)/(7*a^3*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 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 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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, 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 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(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 2875

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[((g*Cos[e + f*x])^(2*m + 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] && ILtQ[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]

Rubi steps

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

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Mathematica [B]  time = 0.88, size = 351, normalized size = 2.17 \[ \frac {\csc ^3(c+d x) \left (-1316 \sin (c+d x)+3520 \sin (2 (c+d x))-1380 \sin (3 (c+d x))-1056 \sin (4 (c+d x))+176 \sin (5 (c+d x))-440 \cos (2 (c+d x))-2640 \cos (3 (c+d x))+846 \cos (4 (c+d x))+176 \cos (5 (c+d x))-2100 \sin (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+630 \sin (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-1575 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+105 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+14 \cos (c+d x) \left (-105 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+105 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+176\right )+1575 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-105 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+2100 \sin (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-630 \sin (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-966\right )}{140 a^3 d (\sin (c+d x)+1)^3 \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(Csc[c + d*x]^3*(-966 - 440*Cos[2*(c + d*x)] - 2640*Cos[3*(c + d*x)] + 846*Cos[4*(c + d*x)] + 176*Cos[5*(c + d
*x)] - 1575*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 105*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 14*Cos[c + d
*x]*(176 + 105*Log[Cos[(c + d*x)/2]] - 105*Log[Sin[(c + d*x)/2]]) + 1575*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]
] - 105*Cos[5*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 1316*Sin[c + d*x] + 3520*Sin[2*(c + d*x)] + 2100*Log[Cos[(c +
d*x)/2]]*Sin[2*(c + d*x)] - 2100*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] - 1380*Sin[3*(c + d*x)] - 1056*Sin[4*
(c + d*x)] - 630*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] + 630*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)] + 176*Sin
[5*(c + d*x)]))/(140*a^3*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2)*(1 + Sin[c + d*x])^3)

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fricas [A]  time = 0.49, size = 265, normalized size = 1.64 \[ \frac {846 \, \cos \left (d x + c\right )^{4} - 956 \, \cos \left (d x + c\right )^{2} + 105 \, {\left (\cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3} - {\left (3 \, \cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 4 \, \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 105 \, {\left (\cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3} - {\left (3 \, \cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 4 \, \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (176 \, \cos \left (d x + c\right )^{4} - 477 \, \cos \left (d x + c\right )^{2} + 15\right )} \sin \left (d x + c\right ) + 40}{70 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} - 5 \, a^{3} d \cos \left (d x + c\right )^{3} + 4 \, a^{3} d \cos \left (d x + c\right ) - {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/70*(846*cos(d*x + c)^4 - 956*cos(d*x + c)^2 + 105*(cos(d*x + c)^5 - 5*cos(d*x + c)^3 - (3*cos(d*x + c)^3 - 4
*cos(d*x + c))*sin(d*x + c) + 4*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 105*(cos(d*x + c)^5 - 5*cos(d*x +
c)^3 - (3*cos(d*x + c)^3 - 4*cos(d*x + c))*sin(d*x + c) + 4*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(17
6*cos(d*x + c)^4 - 477*cos(d*x + c)^2 + 15)*sin(d*x + c) + 40)/(a^3*d*cos(d*x + c)^5 - 5*a^3*d*cos(d*x + c)^3
+ 4*a^3*d*cos(d*x + c) - (3*a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c))*sin(d*x + c))

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giac [A]  time = 0.22, size = 187, normalized size = 1.15 \[ -\frac {\frac {840 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {35 \, {\left (12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 17 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{3}} + \frac {3885 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 19880 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 45465 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 57120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 41671 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16632 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2931}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{280 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/280*(840*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 140*tan(1/2*d*x + 1/2*c)/a^3 - 35*(12*tan(1/2*d*x + 1/2*c)^2
- 17*tan(1/2*d*x + 1/2*c) + 4)/((tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c))*a^3) + (3885*tan(1/2*d*x + 1/2
*c)^6 + 19880*tan(1/2*d*x + 1/2*c)^5 + 45465*tan(1/2*d*x + 1/2*c)^4 + 57120*tan(1/2*d*x + 1/2*c)^3 + 41671*tan
(1/2*d*x + 1/2*c)^2 + 16632*tan(1/2*d*x + 1/2*c) + 2931)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

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maple [A]  time = 0.65, size = 224, normalized size = 1.38 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}-\frac {1}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}-\frac {8}{7 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {4}{a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {46}{5 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {13}{a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {31}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {49}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {111}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^3,x)

[Out]

1/2/d/a^3*tan(1/2*d*x+1/2*c)-1/8/a^3/d/(tan(1/2*d*x+1/2*c)-1)-1/2/d/a^3/tan(1/2*d*x+1/2*c)-3/d/a^3*ln(tan(1/2*
d*x+1/2*c))-8/7/a^3/d/(tan(1/2*d*x+1/2*c)+1)^7+4/a^3/d/(tan(1/2*d*x+1/2*c)+1)^6-46/5/a^3/d/(tan(1/2*d*x+1/2*c)
+1)^5+13/a^3/d/(tan(1/2*d*x+1/2*c)+1)^4-31/2/a^3/d/(tan(1/2*d*x+1/2*c)+1)^3+49/4/a^3/d/(tan(1/2*d*x+1/2*c)+1)^
2-111/8/a^3/d/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.34, size = 395, normalized size = 2.44 \[ -\frac {\frac {\frac {934 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3854 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6566 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3556 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3710 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {7070 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {4270 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {1015 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 35}{\frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} + \frac {210 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {35 \, \sin \left (d x + c\right )}{a^{3} {\left (\cos \left (d x + c\right ) + 1\right )}}}{70 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/70*((934*sin(d*x + c)/(cos(d*x + c) + 1) + 3854*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6566*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 + 3556*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3710*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 70
70*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 4270*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 1015*sin(d*x + c)^8/(cos(d
*x + c) + 1)^8 + 35)/(a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 6*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 14*a^3
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 14*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 14*a^3*sin(d*x + c)^6/(cos
(d*x + c) + 1)^6 - 14*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 6*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - a^
3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9) + 210*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - 35*sin(d*x + c)/(a^3*(
cos(d*x + c) + 1)))/d

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mupad [B]  time = 10.84, size = 274, normalized size = 1.69 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d}-\frac {-29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-122\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-202\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-106\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {508\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {938\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}+\frac {3854\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}+\frac {934\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+1}{d\,\left (-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*sin(c + d*x)^2*(a + a*sin(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)/(2*a^3*d) - ((934*tan(c/2 + (d*x)/2))/35 + (3854*tan(c/2 + (d*x)/2)^2)/35 + (938*tan(c/2 +
(d*x)/2)^3)/5 + (508*tan(c/2 + (d*x)/2)^4)/5 - 106*tan(c/2 + (d*x)/2)^5 - 202*tan(c/2 + (d*x)/2)^6 - 122*tan(c
/2 + (d*x)/2)^7 - 29*tan(c/2 + (d*x)/2)^8 + 1)/(d*(12*a^3*tan(c/2 + (d*x)/2)^2 + 28*a^3*tan(c/2 + (d*x)/2)^3 +
28*a^3*tan(c/2 + (d*x)/2)^4 - 28*a^3*tan(c/2 + (d*x)/2)^6 - 28*a^3*tan(c/2 + (d*x)/2)^7 - 12*a^3*tan(c/2 + (d
*x)/2)^8 - 2*a^3*tan(c/2 + (d*x)/2)^9 + 2*a^3*tan(c/2 + (d*x)/2))) - (3*log(tan(c/2 + (d*x)/2)))/(a^3*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)**2*sec(d*x+c)**2/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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