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

Optimal. Leaf size=151 \[ -\frac {4 \tan ^7(c+d x)}{7 a^3 d}-\frac {11 \tan ^5(c+d x)}{5 a^3 d}-\frac {10 \tan ^3(c+d x)}{3 a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}+\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d}+\frac {\sec ^3(c+d x)}{3 a^3 d}+\frac {\sec (c+d x)}{a^3 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[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+4/7*sec(d*x+c)^7/a^3
/d-3*tan(d*x+c)/a^3/d-10/3*tan(d*x+c)^3/a^3/d-11/5*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 = 0.29, antiderivative size = 151, 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} \[ -\frac {4 \tan ^7(c+d x)}{7 a^3 d}-\frac {11 \tan ^5(c+d x)}{5 a^3 d}-\frac {10 \tan ^3(c+d x)}{3 a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}+\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\sec ^5(c+d x)}{5 a^3 d}+\frac {\sec ^3(c+d x)}{3 a^3 d}+\frac {\sec (c+d x)}{a^3 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]*Sec[c + d*x]^2)/(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)
+ (4*Sec[c + d*x]^7)/(7*a^3*d) - (3*Tan[c + d*x])/(a^3*d) - (10*Tan[c + d*x]^3)/(3*a^3*d) - (11*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 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 ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \csc (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)+a^3 \csc (c+d x) \sec ^8(c+d x)+3 a^3 \sec ^7(c+d x) \tan (c+d x)-a^3 \sec ^6(c+d x) \tan ^2(c+d x)\right ) \, dx}{a^6}\\ &=\frac {\int \csc (c+d x) \sec ^8(c+d x) \, dx}{a^3}-\frac {\int \sec ^6(c+d x) \tan ^2(c+d x) \, dx}{a^3}-\frac {3 \int \sec ^8(c+d x) \, dx}{a^3}+\frac {3 \int \sec ^7(c+d x) \tan (c+d x) \, dx}{a^3}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^8}{-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 )^2 \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac {3 \operatorname {Subst}\left (\int x^6 \, 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 {3 \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 (x^2+2 x^4+x^6\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac {\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 {\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 {4 \sec ^7(c+d x)}{7 a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}-\frac {10 \tan ^3(c+d x)}{3 a^3 d}-\frac {11 \tan ^5(c+d x)}{5 a^3 d}-\frac {4 \tan ^7(c+d x)}{7 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 {4 \sec ^7(c+d x)}{7 a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}-\frac {10 \tan ^3(c+d x)}{3 a^3 d}-\frac {11 \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 = 0.43, size = 341, normalized size = 2.26 \[ \frac {\frac {105 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-2281 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5+353 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4-706 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3+162 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2-324 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-\frac {120 \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}-840 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6+840 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6+60}{840 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60 - (120*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - 324*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]) + 162*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 706*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2])^3 + 353*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 2281*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2])^5 - 840*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6 + 840*Log[Sin[(c + d*x)/2
]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6 + (105*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)/(C
os[(c + d*x)/2] - Sin[(c + d*x)/2]))/(840*d*(a + a*Sin[c + d*x])^3)

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fricas [A]  time = 0.46, size = 218, normalized size = 1.44 \[ \frac {272 \, \cos \left (d x + c\right )^{4} - 594 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (3 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 105 \, {\left (3 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 6 \, {\left (101 \, \cos \left (d x + c\right )^{2} + 15\right )} \sin \left (d x + c\right ) - 120}{210 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(272*cos(d*x + c)^4 - 594*cos(d*x + c)^2 - 105*(3*cos(d*x + c)^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*(3*cos(d*x + c)^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) - 6*(101*cos(d*x + c)^2 + 15)*sin(d*x + c) -
120)/(3*a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c) + (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 = 0.27, size = 135, normalized size = 0.89 \[ \frac {\frac {840 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {105}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} + \frac {5145 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 24360 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 54005 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 66080 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 47691 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 18872 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3431}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(840*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 105/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) + (5145*tan(1/2*d*x + 1/2
*c)^6 + 24360*tan(1/2*d*x + 1/2*c)^5 + 54005*tan(1/2*d*x + 1/2*c)^4 + 66080*tan(1/2*d*x + 1/2*c)^3 + 47691*tan
(1/2*d*x + 1/2*c)^2 + 18872*tan(1/2*d*x + 1/2*c) + 3431)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

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maple [A]  time = 0.65, size = 187, normalized size = 1.24 \[ -\frac {1}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {8}{7 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4}{a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {42}{5 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {11}{a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {67}{6 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {31}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {49}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/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))+8/7/a^3/d/(tan(1/2*d*x+1/2*c)+1)^7-4/a^3/d/(t
an(1/2*d*x+1/2*c)+1)^6+42/5/a^3/d/(tan(1/2*d*x+1/2*c)+1)^5-11/a^3/d/(tan(1/2*d*x+1/2*c)+1)^4+67/6/a^3/d/(tan(1
/2*d*x+1/2*c)+1)^3-31/4/a^3/d/(tan(1/2*d*x+1/2*c)+1)^2+49/8/a^3/d/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.35, size = 336, normalized size = 2.23 \[ \frac {\frac {2 \, {\left (\frac {1011 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {1939 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1379 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {525 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {1715 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {1155 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {315 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 221\right )}}{a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {105 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(2*(1011*sin(d*x + c)/(cos(d*x + c) + 1) + 1939*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1379*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 - 525*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1715*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 -
1155*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 315*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 221)/(a^3 + 6*a^3*sin(d*x
+ c)/(cos(d*x + c) + 1) + 14*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)^5/(cos(d*x + c) + 1)^5 - 14*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6*a^3*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 - a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 105*log(sin(d*x + c)/(cos(d*x + c) +
1))/a^3)/d

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mupad [B]  time = 11.33, size = 143, normalized size = 0.95 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {98\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {394\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{15}+\frac {554\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {674\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+\frac {442}{105}}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(c/2 + (d*x)/2))/(a^3*d) - ((674*tan(c/2 + (d*x)/2))/35 + (554*tan(c/2 + (d*x)/2)^2)/15 + (394*tan(c/2
+ (d*x)/2)^3)/15 - 10*tan(c/2 + (d*x)/2)^4 - (98*tan(c/2 + (d*x)/2)^5)/3 - 22*tan(c/2 + (d*x)/2)^6 - 6*tan(c/2
+ (d*x)/2)^7 + 442/105)/(a^3*d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2) + 1)^7)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\csc {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(c + d*x)*sec(c + d*x)**2/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

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