3.847 \(\int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=200 \[ \frac {4 \tan ^9(c+d x)}{9 a^3 d}+\frac {17 \tan ^7(c+d x)}{7 a^3 d}+\frac {28 \tan ^5(c+d x)}{5 a^3 d}+\frac {22 \tan ^3(c+d x)}{3 a^3 d}+\frac {8 \tan (c+d x)}{a^3 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}-\frac {3 \sec ^7(c+d x)}{7 a^3 d}-\frac {3 \sec ^5(c+d x)}{5 a^3 d}-\frac {\sec ^3(c+d x)}{a^3 d}-\frac {3 \sec (c+d x)}{a^3 d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[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-3/7*
sec(d*x+c)^7/a^3/d-4/9*sec(d*x+c)^9/a^3/d+8*tan(d*x+c)/a^3/d+22/3*tan(d*x+c)^3/a^3/d+28/5*tan(d*x+c)^5/a^3/d+1
7/7*tan(d*x+c)^7/a^3/d+4/9*tan(d*x+c)^9/a^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.39, antiderivative size = 200, 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} \[ \frac {4 \tan ^9(c+d x)}{9 a^3 d}+\frac {17 \tan ^7(c+d x)}{7 a^3 d}+\frac {28 \tan ^5(c+d x)}{5 a^3 d}+\frac {22 \tan ^3(c+d x)}{3 a^3 d}+\frac {8 \tan (c+d x)}{a^3 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}-\frac {3 \sec ^7(c+d x)}{7 a^3 d}-\frac {3 \sec ^5(c+d x)}{5 a^3 d}-\frac {\sec ^3(c+d x)}{a^3 d}-\frac {3 \sec (c+d x)}{a^3 d}+\frac {3 \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(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) - (3*Sec[c + d*x]^7)/(7*a^3*d) - (4*Sec[c + d*x]^9)/(9*a^3*d) + (8*Tan[c + d*x])
/(a^3*d) + (22*Tan[c + d*x]^3)/(3*a^3*d) + (28*Tan[c + d*x]^5)/(5*a^3*d) + (17*Tan[c + d*x]^7)/(7*a^3*d) + (4*
Tan[c + d*x]^9)/(9*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 ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \csc ^2(c+d x) \sec ^{10}(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac {\int \left (3 a^3 \sec ^{10}(c+d x)-3 a^3 \csc (c+d x) \sec ^{10}(c+d x)+a^3 \csc ^2(c+d x) \sec ^{10}(c+d x)-a^3 \sec ^9(c+d x) \tan (c+d x)\right ) \, dx}{a^6}\\ &=\frac {\int \csc ^2(c+d x) \sec ^{10}(c+d x) \, dx}{a^3}-\frac {\int \sec ^9(c+d x) \tan (c+d x) \, dx}{a^3}+\frac {3 \int \sec ^{10}(c+d x) \, dx}{a^3}-\frac {3 \int \csc (c+d x) \sec ^{10}(c+d x) \, dx}{a^3}\\ &=-\frac {\operatorname {Subst}\left (\int x^8 \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^5}{x^2} \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {3 \operatorname {Subst}\left (\int \frac {x^{10}}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac {3 \operatorname {Subst}\left (\int \left (1+4 x^2+6 x^4+4 x^6+x^8\right ) \, dx,x,-\tan (c+d x)\right )}{a^3 d}\\ &=-\frac {\sec ^9(c+d x)}{9 a^3 d}+\frac {3 \tan (c+d x)}{a^3 d}+\frac {4 \tan ^3(c+d x)}{a^3 d}+\frac {18 \tan ^5(c+d x)}{5 a^3 d}+\frac {12 \tan ^7(c+d x)}{7 a^3 d}+\frac {\tan ^9(c+d x)}{3 a^3 d}+\frac {\operatorname {Subst}\left (\int \left (5+\frac {1}{x^2}+10 x^2+10 x^4+5 x^6+x^8\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+x^8+\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 {3 \sec ^7(c+d x)}{7 a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {8 \tan (c+d x)}{a^3 d}+\frac {22 \tan ^3(c+d x)}{3 a^3 d}+\frac {28 \tan ^5(c+d x)}{5 a^3 d}+\frac {17 \tan ^7(c+d x)}{7 a^3 d}+\frac {4 \tan ^9(c+d x)}{9 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 {3 \sec ^7(c+d x)}{7 a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {8 \tan (c+d x)}{a^3 d}+\frac {22 \tan ^3(c+d x)}{3 a^3 d}+\frac {28 \tan ^5(c+d x)}{5 a^3 d}+\frac {17 \tan ^7(c+d x)}{7 a^3 d}+\frac {4 \tan ^9(c+d x)}{9 a^3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.64, size = 230, normalized size = 1.15 \[ \frac {-1935360 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+1935360 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\csc (c+d x) (-707328 \sin (c+d x)+1364182 \sin (2 (c+d x))-1161600 \sin (3 (c+d x))+320984 \sin (4 (c+d x))-329344 \sin (5 (c+d x))-240738 \sin (6 (c+d x))+53248 \sin (7 (c+d x))+1083321 \cos (c+d x)-653248 \cos (2 (c+d x))-601845 \cos (3 (c+d x))+340096 \cos (4 (c+d x))-521599 \cos (5 (c+d x))+259008 \cos (6 (c+d x))+40123 \cos (7 (c+d x))-590976)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^9}}{645120 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

(1935360*Log[Cos[(c + d*x)/2]] - 1935360*Log[Sin[(c + d*x)/2]] + (Csc[c + d*x]*(-590976 + 1083321*Cos[c + d*x]
- 653248*Cos[2*(c + d*x)] - 601845*Cos[3*(c + d*x)] + 340096*Cos[4*(c + d*x)] - 521599*Cos[5*(c + d*x)] + 259
008*Cos[6*(c + d*x)] + 40123*Cos[7*(c + d*x)] - 707328*Sin[c + d*x] + 1364182*Sin[2*(c + d*x)] - 1161600*Sin[3
*(c + d*x)] + 320984*Sin[4*(c + d*x)] - 329344*Sin[5*(c + d*x)] - 240738*Sin[6*(c + d*x)] + 53248*Sin[7*(c + d
*x)]))/((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^9))/(645120*a^3*d)

________________________________________________________________________________________

fricas [A]  time = 0.49, size = 297, normalized size = 1.48 \[ \frac {8094 \, \cos \left (d x + c\right )^{6} - 9484 \, \cos \left (d x + c\right )^{4} + 620 \, \cos \left (d x + c\right )^{2} + 945 \, {\left (\cos \left (d x + c\right )^{7} - 5 \, \cos \left (d x + c\right )^{5} + 4 \, \cos \left (d x + c\right )^{3} - {\left (3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 945 \, {\left (\cos \left (d x + c\right )^{7} - 5 \, \cos \left (d x + c\right )^{5} + 4 \, \cos \left (d x + c\right )^{3} - {\left (3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (1664 \, \cos \left (d x + c\right )^{6} - 4653 \, \cos \left (d x + c\right )^{4} + 285 \, \cos \left (d x + c\right )^{2} + 35\right )} \sin \left (d x + c\right ) + 140}{630 \, {\left (a^{3} d \cos \left (d x + c\right )^{7} - 5 \, a^{3} d \cos \left (d x + c\right )^{5} + 4 \, a^{3} d \cos \left (d x + c\right )^{3} - {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

1/630*(8094*cos(d*x + c)^6 - 9484*cos(d*x + c)^4 + 620*cos(d*x + c)^2 + 945*(cos(d*x + c)^7 - 5*cos(d*x + c)^5
+ 4*cos(d*x + c)^3 - (3*cos(d*x + c)^5 - 4*cos(d*x + c)^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 945*(c
os(d*x + c)^7 - 5*cos(d*x + c)^5 + 4*cos(d*x + c)^3 - (3*cos(d*x + c)^5 - 4*cos(d*x + c)^3)*sin(d*x + c))*log(
-1/2*cos(d*x + c) + 1/2) + 2*(1664*cos(d*x + c)^6 - 4653*cos(d*x + c)^4 + 285*cos(d*x + c)^2 + 35)*sin(d*x + c
) + 140)/(a^3*d*cos(d*x + c)^7 - 5*a^3*d*cos(d*x + c)^5 + 4*a^3*d*cos(d*x + c)^3 - (3*a^3*d*cos(d*x + c)^5 - 4
*a^3*d*cos(d*x + c)^3)*sin(d*x + c))

________________________________________________________________________________________

giac [A]  time = 0.28, size = 230, normalized size = 1.15 \[ -\frac {\frac {30240 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {5040 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {5040 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {105 \, {\left (33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 31\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {157815 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 1093680 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3488940 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6524280 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 7788186 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6052704 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2995596 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 864504 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 113591}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{9}}}{10080 \, d} \]

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))^3,x, algorithm="giac")

[Out]

-1/10080*(30240*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 5040*tan(1/2*d*x + 1/2*c)/a^3 - 5040*(6*tan(1/2*d*x + 1/2
*c) - 1)/(a^3*tan(1/2*d*x + 1/2*c)) + 105*(33*tan(1/2*d*x + 1/2*c)^2 - 60*tan(1/2*d*x + 1/2*c) + 31)/(a^3*(tan
(1/2*d*x + 1/2*c) - 1)^3) + (157815*tan(1/2*d*x + 1/2*c)^8 + 1093680*tan(1/2*d*x + 1/2*c)^7 + 3488940*tan(1/2*
d*x + 1/2*c)^6 + 6524280*tan(1/2*d*x + 1/2*c)^5 + 7788186*tan(1/2*d*x + 1/2*c)^4 + 6052704*tan(1/2*d*x + 1/2*c
)^3 + 2995596*tan(1/2*d*x + 1/2*c)^2 + 864504*tan(1/2*d*x + 1/2*c) + 113591)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^9
))/d

________________________________________________________________________________________

maple [A]  time = 0.80, size = 308, normalized size = 1.54 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}-\frac {1}{24 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {11}{32 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}-\frac {8}{9 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {4}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {76}{7 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {58}{3 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {267}{10 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {111}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {25}{a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {67}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {501}{32 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c))^3,x)

[Out]

1/2/d/a^3*tan(1/2*d*x+1/2*c)-1/24/d/a^3/(tan(1/2*d*x+1/2*c)-1)^3-1/16/d/a^3/(tan(1/2*d*x+1/2*c)-1)^2-11/32/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))-8/9/d/a^3/(tan(1/2*d*x+1
/2*c)+1)^9+4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^8-76/7/a^3/d/(tan(1/2*d*x+1/2*c)+1)^7+58/3/a^3/d/(tan(1/2*d*x+1/2*c)
+1)^6-267/10/a^3/d/(tan(1/2*d*x+1/2*c)+1)^5+111/4/a^3/d/(tan(1/2*d*x+1/2*c)+1)^4-25/a^3/d/(tan(1/2*d*x+1/2*c)+
1)^3+67/4/a^3/d/(tan(1/2*d*x+1/2*c)+1)^2-501/32/a^3/d/(tan(1/2*d*x+1/2*c)+1)

________________________________________________________________________________________

maxima [B]  time = 0.35, size = 567, normalized size = 2.84 \[ -\frac {\frac {\frac {8786 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35076 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {43062 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {41753 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {152172 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {99072 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {93324 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {157689 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {44730 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {50820 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {42210 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {10395 \, \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + 315}{\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 {12 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {27 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {36 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {36 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {27 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {12 \, a^{3} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} - \frac {a^{3} \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}}} + \frac {1890 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {315 \, \sin \left (d x + c\right )}{a^{3} {\left (\cos \left (d x + c\right ) + 1\right )}}}{630 \, 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))^3,x, algorithm="maxima")

[Out]

-1/630*((8786*sin(d*x + c)/(cos(d*x + c) + 1) + 35076*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 43062*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3 - 41753*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 152172*sin(d*x + c)^5/(cos(d*x + c) + 1)
^5 - 99072*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 93324*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 157689*sin(d*x +
c)^8/(cos(d*x + c) + 1)^8 + 44730*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 50820*sin(d*x + c)^10/(cos(d*x + c) +
1)^10 - 42210*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 10395*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + 315)/(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 + 12*a^3*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 2*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 27*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 36*a^3*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6 + 36*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 27*a^3*sin(d*x + c)^9/(cos(d
*x + c) + 1)^9 - 2*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 12*a^3*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 -
6*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - a^3*sin(d*x + c)^13/(cos(d*x + c) + 1)^13) + 1890*log(sin(d*x +
c)/(cos(d*x + c) + 1))/a^3 - 315*sin(d*x + c)/(a^3*(cos(d*x + c) + 1)))/d

________________________________________________________________________________________

mupad [B]  time = 11.32, size = 390, normalized size = 1.95 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {-33\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-134\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-\frac {484\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{3}+142\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {2503\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{5}+\frac {4444\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{15}-\frac {11008\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{35}-\frac {16908\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{35}-\frac {41753\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{315}+\frac {14354\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{105}+\frac {11692\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{105}+\frac {8786\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{315}+1}{d\,\left (-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-24\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+54\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+72\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-72\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-54\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

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))^3),x)

[Out]

tan(c/2 + (d*x)/2)/(2*a^3*d) - (3*log(tan(c/2 + (d*x)/2)))/(a^3*d) - ((8786*tan(c/2 + (d*x)/2))/315 + (11692*t
an(c/2 + (d*x)/2)^2)/105 + (14354*tan(c/2 + (d*x)/2)^3)/105 - (41753*tan(c/2 + (d*x)/2)^4)/315 - (16908*tan(c/
2 + (d*x)/2)^5)/35 - (11008*tan(c/2 + (d*x)/2)^6)/35 + (4444*tan(c/2 + (d*x)/2)^7)/15 + (2503*tan(c/2 + (d*x)/
2)^8)/5 + 142*tan(c/2 + (d*x)/2)^9 - (484*tan(c/2 + (d*x)/2)^10)/3 - 134*tan(c/2 + (d*x)/2)^11 - 33*tan(c/2 +
(d*x)/2)^12 + 1)/(d*(12*a^3*tan(c/2 + (d*x)/2)^2 + 24*a^3*tan(c/2 + (d*x)/2)^3 + 4*a^3*tan(c/2 + (d*x)/2)^4 -
54*a^3*tan(c/2 + (d*x)/2)^5 - 72*a^3*tan(c/2 + (d*x)/2)^6 + 72*a^3*tan(c/2 + (d*x)/2)^8 + 54*a^3*tan(c/2 + (d*
x)/2)^9 - 4*a^3*tan(c/2 + (d*x)/2)^10 - 24*a^3*tan(c/2 + (d*x)/2)^11 - 12*a^3*tan(c/2 + (d*x)/2)^12 - 2*a^3*ta
n(c/2 + (d*x)/2)^13 + 2*a^3*tan(c/2 + (d*x)/2)))

________________________________________________________________________________________

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))**3,x)

[Out]

Timed out

________________________________________________________________________________________