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

Optimal. Leaf size=187 \[ -\frac {4 \tan ^9(c+d x)}{9 a^3 d}-\frac {15 \tan ^7(c+d x)}{7 a^3 d}-\frac {21 \tan ^5(c+d x)}{5 a^3 d}-\frac {13 \tan ^3(c+d x)}{3 a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}+\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {\sec ^7(c+d x)}{7 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d}+\frac {\sec ^3(c+d x)}{3 a^3 d}+\frac {\sec (c+d x)}{a^3 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[Out]

-arctanh(cos(d*x+c))/a^3/d+sec(d*x+c)/a^3/d+1/3*sec(d*x+c)^3/a^3/d+1/5*sec(d*x+c)^5/a^3/d+1/7*sec(d*x+c)^7/a^3
/d+4/9*sec(d*x+c)^9/a^3/d-3*tan(d*x+c)/a^3/d-13/3*tan(d*x+c)^3/a^3/d-21/5*tan(d*x+c)^5/a^3/d-15/7*tan(d*x+c)^7
/a^3/d-4/9*tan(d*x+c)^9/a^3/d

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Rubi [A]  time = 0.36, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {2875, 2873, 3767, 2622, 302, 207, 2606, 30, 2607, 270} \[ -\frac {4 \tan ^9(c+d x)}{9 a^3 d}-\frac {15 \tan ^7(c+d x)}{7 a^3 d}-\frac {21 \tan ^5(c+d x)}{5 a^3 d}-\frac {13 \tan ^3(c+d x)}{3 a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}+\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {\sec ^7(c+d x)}{7 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d}+\frac {\sec ^3(c+d x)}{3 a^3 d}+\frac {\sec (c+d x)}{a^3 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^3*d)) + Sec[c + d*x]/(a^3*d) + Sec[c + d*x]^3/(3*a^3*d) + Sec[c + d*x]^5/(5*a^3*d)
+ Sec[c + d*x]^7/(7*a^3*d) + (4*Sec[c + d*x]^9)/(9*a^3*d) - (3*Tan[c + d*x])/(a^3*d) - (13*Tan[c + d*x]^3)/(3*
a^3*d) - (21*Tan[c + d*x]^5)/(5*a^3*d) - (15*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 2607

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

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 (c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \csc (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)+a^3 \csc (c+d x) \sec ^{10}(c+d x)+3 a^3 \sec ^9(c+d x) \tan (c+d x)-a^3 \sec ^8(c+d x) \tan ^2(c+d x)\right ) \, dx}{a^6}\\ &=\frac {\int \csc (c+d x) \sec ^{10}(c+d x) \, dx}{a^3}-\frac {\int \sec ^8(c+d x) \tan ^2(c+d x) \, dx}{a^3}-\frac {3 \int \sec ^{10}(c+d x) \, dx}{a^3}+\frac {3 \int \sec ^9(c+d x) \tan (c+d x) \, dx}{a^3}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^{10}}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac {\operatorname {Subst}\left (\int x^2 \left (1+x^2\right )^3 \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac {3 \operatorname {Subst}\left (\int x^8 \, 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)}{3 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 (x^2+3 x^4+3 x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac {\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 {\sec (c+d x)}{a^3 d}+\frac {\sec ^3(c+d x)}{3 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d}+\frac {\sec ^7(c+d x)}{7 a^3 d}+\frac {4 \sec ^9(c+d x)}{9 a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}-\frac {13 \tan ^3(c+d x)}{3 a^3 d}-\frac {21 \tan ^5(c+d x)}{5 a^3 d}-\frac {15 \tan ^7(c+d x)}{7 a^3 d}-\frac {4 \tan ^9(c+d x)}{9 a^3 d}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {\sec (c+d x)}{a^3 d}+\frac {\sec ^3(c+d x)}{3 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d}+\frac {\sec ^7(c+d x)}{7 a^3 d}+\frac {4 \sec ^9(c+d x)}{9 a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}-\frac {13 \tan ^3(c+d x)}{3 a^3 d}-\frac {21 \tan ^5(c+d x)}{5 a^3 d}-\frac {15 \tan ^7(c+d x)}{7 a^3 d}-\frac {4 \tan ^9(c+d x)}{9 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.38, size = 204, normalized size = 1.09 \[ \frac {322560 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-322560 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {196992 \sin (c+d x)-383157 \sin (2 (c+d x))+211648 \sin (3 (c+d x))-170292 \sin (4 (c+d x))+50496 \sin (5 (c+d x))+14191 \sin (6 (c+d x))-510876 \cos (c+d x)+317952 \cos (2 (c+d x))-28382 \cos (3 (c+d x))+20352 \cos (4 (c+d x))+85146 \cos (5 (c+d x))-11776 \cos (6 (c+d x))+357504}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^9}}{322560 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-322560*Log[Cos[(c + d*x)/2]] + 322560*Log[Sin[(c + d*x)/2]] + (357504 - 510876*Cos[c + d*x] + 317952*Cos[2*(
c + d*x)] - 28382*Cos[3*(c + d*x)] + 20352*Cos[4*(c + d*x)] + 85146*Cos[5*(c + d*x)] - 11776*Cos[6*(c + d*x)]
+ 196992*Sin[c + d*x] - 383157*Sin[2*(c + d*x)] + 211648*Sin[3*(c + d*x)] - 170292*Sin[4*(c + d*x)] + 50496*Si
n[5*(c + d*x)] + 14191*Sin[6*(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))/(322560*a^3*d)

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fricas [A]  time = 0.48, size = 250, normalized size = 1.34 \[ \frac {736 \, \cos \left (d x + c\right )^{6} - 1422 \, \cos \left (d x + c\right )^{4} - 510 \, \cos \left (d x + c\right )^{2} - 315 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 315 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (789 \, \cos \left (d x + c\right )^{4} + 235 \, \cos \left (d x + c\right )^{2} + 35\right )} \sin \left (d x + c\right ) - 140}{630 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} + {\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/630*(736*cos(d*x + c)^6 - 1422*cos(d*x + c)^4 - 510*cos(d*x + c)^2 - 315*(3*cos(d*x + c)^5 - 4*cos(d*x + c)^
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) + 315*(3*cos(d*x + c)^5 - 4*
cos(d*x + c)^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*(789*cos(d
*x + c)^4 + 235*cos(d*x + c)^2 + 35)*sin(d*x + c) - 140)/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3 + (a
^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3)*sin(d*x + c))

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giac [A]  time = 0.24, size = 187, normalized size = 1.00 \[ \frac {\frac {10080 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {105 \, {\left (27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {63315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 412020 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1273440 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 2324700 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2731302 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2097228 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1032552 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 297828 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 40127}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{9}}}{10080 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/10080*(10080*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 105*(27*tan(1/2*d*x + 1/2*c)^2 - 48*tan(1/2*d*x + 1/2*c) +
25)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) + (63315*tan(1/2*d*x + 1/2*c)^8 + 412020*tan(1/2*d*x + 1/2*c)^7 + 1273
440*tan(1/2*d*x + 1/2*c)^6 + 2324700*tan(1/2*d*x + 1/2*c)^5 + 2731302*tan(1/2*d*x + 1/2*c)^4 + 2097228*tan(1/2
*d*x + 1/2*c)^3 + 1032552*tan(1/2*d*x + 1/2*c)^2 + 297828*tan(1/2*d*x + 1/2*c) + 40127)/(a^3*(tan(1/2*d*x + 1/
2*c) + 1)^9))/d

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maple [A]  time = 0.79, size = 271, normalized size = 1.45 \[ -\frac {1}{24 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {9}{32 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {8}{9 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}-\frac {4}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}+\frac {72}{7 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {52}{3 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {219}{10 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {83}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {193}{12 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {75}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {201}{32 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)^2-9/32/a^3/d/(tan(1/2*d*x+1/2*c)-1)+1/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+72/7/a^3/d/(ta
n(1/2*d*x+1/2*c)+1)^7-52/3/a^3/d/(tan(1/2*d*x+1/2*c)+1)^6+219/10/a^3/d/(tan(1/2*d*x+1/2*c)+1)^5-83/4/a^3/d/(ta
n(1/2*d*x+1/2*c)+1)^4+193/12/a^3/d/(tan(1/2*d*x+1/2*c)+1)^3-75/8/a^3/d/(tan(1/2*d*x+1/2*c)+1)^2+201/32/a^3/d/(
tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.34, size = 508, normalized size = 2.72 \[ \frac {\frac {2 \, {\left (\frac {3063 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4866 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1289 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {11736 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {10566 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5292 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {13482 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {6300 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {2625 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {3150 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {945 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} + 668\right )}}{a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, 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 )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {27 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {12 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} + \frac {315 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/315*(2*(3063*sin(d*x + c)/(cos(d*x + c) + 1) + 4866*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1289*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 - 11736*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 10566*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 5292*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 13482*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 6300*sin(d*x + c)^8/
(cos(d*x + c) + 1)^8 - 2625*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 3150*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 -
945*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 668)/(a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 27*a^3*sin(d*x + c)^4/(cos(d*x + c
) + 1)^4 - 36*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 36*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 27*a^3*si
n(d*x + c)^8/(cos(d*x + c) + 1)^8 - 2*a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 12*a^3*sin(d*x + c)^10/(cos(d*
x + c) + 1)^10 - 6*a^3*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) + 31
5*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 12.56, size = 195, normalized size = 1.04 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-\frac {50\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{3}+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {428\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5}+\frac {168\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}-\frac {2348\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{35}-\frac {2608\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{35}-\frac {2578\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{315}+\frac {3244\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{105}+\frac {2042\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}+\frac {1336}{315}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*sin(c + d*x)*(a + a*sin(c + d*x))^3),x)

[Out]

log(tan(c/2 + (d*x)/2))/(a^3*d) - ((2042*tan(c/2 + (d*x)/2))/105 + (3244*tan(c/2 + (d*x)/2)^2)/105 - (2578*tan
(c/2 + (d*x)/2)^3)/315 - (2608*tan(c/2 + (d*x)/2)^4)/35 - (2348*tan(c/2 + (d*x)/2)^5)/35 + (168*tan(c/2 + (d*x
)/2)^6)/5 + (428*tan(c/2 + (d*x)/2)^7)/5 + 40*tan(c/2 + (d*x)/2)^8 - (50*tan(c/2 + (d*x)/2)^9)/3 - 20*tan(c/2
+ (d*x)/2)^10 - 6*tan(c/2 + (d*x)/2)^11 + 1336/315)/(a^3*d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)
^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)*sec(d*x+c)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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