3.837 \(\int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=164 \[ \frac {2 \tan ^7(c+d x)}{7 a^2 d}+\frac {7 \tan ^5(c+d x)}{5 a^2 d}+\frac {3 \tan ^3(c+d x)}{a^2 d}+\frac {5 \tan (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}+\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

[Out]

2*arctanh(cos(d*x+c))/a^2/d-cot(d*x+c)/a^2/d-2*sec(d*x+c)/a^2/d-2/3*sec(d*x+c)^3/a^2/d-2/5*sec(d*x+c)^5/a^2/d-
2/7*sec(d*x+c)^7/a^2/d+5*tan(d*x+c)/a^2/d+3*tan(d*x+c)^3/a^2/d+7/5*tan(d*x+c)^5/a^2/d+2/7*tan(d*x+c)^7/a^2/d

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Rubi [A]  time = 0.33, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2875, 2873, 3767, 2622, 302, 207, 2620, 270} \[ \frac {2 \tan ^7(c+d x)}{7 a^2 d}+\frac {7 \tan ^5(c+d x)}{5 a^2 d}+\frac {3 \tan ^3(c+d x)}{a^2 d}+\frac {5 \tan (c+d x)}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}+\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

(2*ArcTanh[Cos[c + d*x]])/(a^2*d) - Cot[c + d*x]/(a^2*d) - (2*Sec[c + d*x])/(a^2*d) - (2*Sec[c + d*x]^3)/(3*a^
2*d) - (2*Sec[c + d*x]^5)/(5*a^2*d) - (2*Sec[c + d*x]^7)/(7*a^2*d) + (5*Tan[c + d*x])/(a^2*d) + (3*Tan[c + d*x
]^3)/(a^2*d) + (7*Tan[c + d*x]^5)/(5*a^2*d) + (2*Tan[c + d*x]^7)/(7*a^2*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 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 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 ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \csc ^2(c+d x) \sec ^8(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \sec ^8(c+d x)-2 a^2 \csc (c+d x) \sec ^8(c+d x)+a^2 \csc ^2(c+d x) \sec ^8(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \sec ^8(c+d x) \, dx}{a^2}+\frac {\int \csc ^2(c+d x) \sec ^8(c+d x) \, dx}{a^2}-\frac {2 \int \csc (c+d x) \sec ^8(c+d x) \, dx}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^4}{x^2} \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {\operatorname {Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}-\frac {2 \operatorname {Subst}\left (\int \frac {x^8}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {\tan (c+d x)}{a^2 d}+\frac {\tan ^3(c+d x)}{a^2 d}+\frac {3 \tan ^5(c+d x)}{5 a^2 d}+\frac {\tan ^7(c+d x)}{7 a^2 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^2 d}-\frac {2 \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^2 d}\\ &=-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {5 \tan (c+d x)}{a^2 d}+\frac {3 \tan ^3(c+d x)}{a^2 d}+\frac {7 \tan ^5(c+d x)}{5 a^2 d}+\frac {2 \tan ^7(c+d x)}{7 a^2 d}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {5 \tan (c+d x)}{a^2 d}+\frac {3 \tan ^3(c+d x)}{a^2 d}+\frac {7 \tan ^5(c+d x)}{5 a^2 d}+\frac {2 \tan ^7(c+d x)}{7 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 6.10, size = 442, normalized size = 2.70 \[ \frac {16 \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {13 \sin \left (\frac {1}{2} (c+d x)\right )}{384 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{384 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4777 \sin \left (\frac {1}{2} (c+d x)\right )}{13440 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {1}{768 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {997}{26880 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {997 \sin \left (\frac {1}{2} (c+d x)\right )}{13440 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {3}{280 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {3 \sin \left (\frac {1}{2} (c+d x)\right )}{140 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5}-\frac {1}{448 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{224 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^7}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

(16*(-1/32*Cot[(c + d*x)/2]/d + Log[Cos[(c + d*x)/2]]/(8*d) - Log[Sin[(c + d*x)/2]]/(8*d) + 1/(768*d*(Cos[(c +
d*x)/2] - Sin[(c + d*x)/2])^2) + Sin[(c + d*x)/2]/(384*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (13*Sin[(
c + d*x)/2])/(384*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + Sin[(c + d*x)/2]/(224*d*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2])^7) - 1/(448*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6) + (3*Sin[(c + d*x)/2])/(140*d*(Cos[(c + d*
x)/2] + Sin[(c + d*x)/2])^5) - 3/(280*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + (997*Sin[(c + d*x)/2])/(134
40*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 997/(26880*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4777*
Sin[(c + d*x)/2])/(13440*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + Tan[(c + d*x)/2]/(32*d)))/a^2

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fricas [A]  time = 0.48, size = 250, normalized size = 1.52 \[ -\frac {432 \, \cos \left (d x + c\right )^{6} - 660 \, \cos \left (d x + c\right )^{4} + 98 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (2 \, \cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 105 \, {\left (2 \, \cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (327 \, \cos \left (d x + c\right )^{4} - 41 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) + 25}{105 \, {\left (2 \, a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} + {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/105*(432*cos(d*x + c)^6 - 660*cos(d*x + c)^4 + 98*cos(d*x + c)^2 - 105*(2*cos(d*x + c)^5 - 2*cos(d*x + c)^3
+ (cos(d*x + c)^5 - 2*cos(d*x + c)^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 105*(2*cos(d*x + c)^5 - 2*c
os(d*x + c)^3 + (cos(d*x + c)^5 - 2*cos(d*x + c)^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(327*cos(d*
x + c)^4 - 41*cos(d*x + c)^2 - 5)*sin(d*x + c) + 25)/(2*a^2*d*cos(d*x + c)^5 - 2*a^2*d*cos(d*x + c)^3 + (a^2*d
*cos(d*x + c)^5 - 2*a^2*d*cos(d*x + c)^3)*sin(d*x + c))

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giac [A]  time = 0.25, size = 204, normalized size = 1.24 \[ -\frac {\frac {1680 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {420 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {420 \, {\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {35 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 14\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {7875 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 41055 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 94640 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 119630 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 87507 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 34979 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6122}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/840*(1680*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 420*tan(1/2*d*x + 1/2*c)/a^2 - 420*(4*tan(1/2*d*x + 1/2*c) -
1)/(a^2*tan(1/2*d*x + 1/2*c)) + 35*(15*tan(1/2*d*x + 1/2*c)^2 - 27*tan(1/2*d*x + 1/2*c) + 14)/(a^2*(tan(1/2*d
*x + 1/2*c) - 1)^3) + (7875*tan(1/2*d*x + 1/2*c)^6 + 41055*tan(1/2*d*x + 1/2*c)^5 + 94640*tan(1/2*d*x + 1/2*c)
^4 + 119630*tan(1/2*d*x + 1/2*c)^3 + 87507*tan(1/2*d*x + 1/2*c)^2 + 34979*tan(1/2*d*x + 1/2*c) + 6122)/(a^2*(t
an(1/2*d*x + 1/2*c) + 1)^7))/d

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maple [A]  time = 0.64, size = 266, normalized size = 1.62 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {1}{12 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5}{8 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {4}{7 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {2}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {24}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {7}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {107}{12 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {59}{8 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {75}{8 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c))^2,x)

[Out]

1/2/d/a^2*tan(1/2*d*x+1/2*c)-1/12/d/a^2/(tan(1/2*d*x+1/2*c)-1)^3-1/8/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2-5/8/a^2/d/
(tan(1/2*d*x+1/2*c)-1)-1/2/d/a^2/tan(1/2*d*x+1/2*c)-2/d/a^2*ln(tan(1/2*d*x+1/2*c))-4/7/d/a^2/(tan(1/2*d*x+1/2*
c)+1)^7+2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^6-24/5/a^2/d/(tan(1/2*d*x+1/2*c)+1)^5+7/a^2/d/(tan(1/2*d*x+1/2*c)+1)^4-
107/12/a^2/d/(tan(1/2*d*x+1/2*c)+1)^3+59/8/a^2/d/(tan(1/2*d*x+1/2*c)+1)^2-75/8/a^2/d/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.34, size = 481, normalized size = 2.93 \[ -\frac {\frac {\frac {1828 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3847 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1656 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {12734 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {7952 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {9702 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {12600 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {315 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {5460 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {2205 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + 105}{\frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {14 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {14 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {a^{2} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}} + \frac {420 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {105 \, \sin \left (d x + c\right )}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}}}{210 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/210*((1828*sin(d*x + c)/(cos(d*x + c) + 1) + 3847*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1656*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 - 12734*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 7952*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 +
9702*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 12600*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 315*sin(d*x + c)^8/(co
s(d*x + c) + 1)^8 - 5460*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 2205*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 10
5)/(a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^3/(co
s(d*x + c) + 1)^3 - 8*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 14*a^2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 1
4*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 8*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 3*a^2*sin(d*x + c)^9/(
cos(d*x + c) + 1)^9 - 4*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - a^2*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)
+ 420*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - 105*sin(d*x + c)/(a^2*(cos(d*x + c) + 1)))/d

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mupad [B]  time = 11.24, size = 331, normalized size = 2.02 \[ \frac {21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+52\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-120\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {462\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+\frac {1136\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\frac {12734\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{105}+\frac {552\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{35}-\frac {3847\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{105}-\frac {1828\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}-1}{d\,\left (-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+28\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-28\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*sin(c + d*x)^2*(a + a*sin(c + d*x))^2),x)

[Out]

((552*tan(c/2 + (d*x)/2)^3)/35 - (3847*tan(c/2 + (d*x)/2)^2)/105 - (1828*tan(c/2 + (d*x)/2))/105 + (12734*tan(
c/2 + (d*x)/2)^4)/105 + (1136*tan(c/2 + (d*x)/2)^5)/15 - (462*tan(c/2 + (d*x)/2)^6)/5 - 120*tan(c/2 + (d*x)/2)
^7 + 3*tan(c/2 + (d*x)/2)^8 + 52*tan(c/2 + (d*x)/2)^9 + 21*tan(c/2 + (d*x)/2)^10 - 1)/(d*(8*a^2*tan(c/2 + (d*x
)/2)^2 + 6*a^2*tan(c/2 + (d*x)/2)^3 - 16*a^2*tan(c/2 + (d*x)/2)^4 - 28*a^2*tan(c/2 + (d*x)/2)^5 + 28*a^2*tan(c
/2 + (d*x)/2)^7 + 16*a^2*tan(c/2 + (d*x)/2)^8 - 6*a^2*tan(c/2 + (d*x)/2)^9 - 8*a^2*tan(c/2 + (d*x)/2)^10 - 2*a
^2*tan(c/2 + (d*x)/2)^11 + 2*a^2*tan(c/2 + (d*x)/2))) - (2*log(tan(c/2 + (d*x)/2)))/(a^2*d) + tan(c/2 + (d*x)/
2)/(2*a^2*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)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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