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

Optimal. Leaf size=158 \[ -\frac {2 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^3(c+d x)}{a^2 d}-\frac {6 \tan (c+d x)}{a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}+\frac {9 \sec ^5(c+d x)}{10 a^2 d}+\frac {3 \sec ^3(c+d x)}{2 a^2 d}+\frac {9 \sec (c+d x)}{2 a^2 d}-\frac {9 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\csc ^2(c+d x) \sec ^5(c+d x)}{2 a^2 d} \]

[Out]

-9/2*arctanh(cos(d*x+c))/a^2/d+2*cot(d*x+c)/a^2/d+9/2*sec(d*x+c)/a^2/d+3/2*sec(d*x+c)^3/a^2/d+9/10*sec(d*x+c)^
5/a^2/d-1/2*csc(d*x+c)^2*sec(d*x+c)^5/a^2/d-6*tan(d*x+c)/a^2/d-2*tan(d*x+c)^3/a^2/d-2/5*tan(d*x+c)^5/a^2/d

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Rubi [A]  time = 0.35, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2875, 2873, 2622, 302, 207, 2620, 270, 288} \[ -\frac {2 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^3(c+d x)}{a^2 d}-\frac {6 \tan (c+d x)}{a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}+\frac {9 \sec ^5(c+d x)}{10 a^2 d}+\frac {3 \sec ^3(c+d x)}{2 a^2 d}+\frac {9 \sec (c+d x)}{2 a^2 d}-\frac {9 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\csc ^2(c+d x) \sec ^5(c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (2*Cot[c + d*x])/(a^2*d) + (9*Sec[c + d*x])/(2*a^2*d) + (3*Sec[c + d*x]
^3)/(2*a^2*d) + (9*Sec[c + d*x]^5)/(10*a^2*d) - (Csc[c + d*x]^2*Sec[c + d*x]^5)/(2*a^2*d) - (6*Tan[c + d*x])/(
a^2*d) - (2*Tan[c + d*x]^3)/(a^2*d) - (2*Tan[c + d*x]^5)/(5*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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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]

Rubi steps

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

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Mathematica [B]  time = 0.73, size = 328, normalized size = 2.08 \[ -\frac {\csc ^2(c+d x) \sec (c+d x) \left (-432 \sin (c+d x)+744 \sin (2 (c+d x))-176 \sin (3 (c+d x))-372 \sin (4 (c+d x))+128 \sin (5 (c+d x))+176 \cos (2 (c+d x))-651 \cos (3 (c+d x))+332 \cos (4 (c+d x))+93 \cos (5 (c+d x))-720 \sin (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+360 \sin (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-630 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+90 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+18 \cos (c+d x) \left (-30 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+31\right )+630 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-90 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+720 \sin (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-360 \sin (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-348\right )}{320 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/320*(Csc[c + d*x]^2*Sec[c + d*x]*(-348 + 176*Cos[2*(c + d*x)] - 651*Cos[3*(c + d*x)] + 332*Cos[4*(c + d*x)]
+ 93*Cos[5*(c + d*x)] - 630*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 90*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2]
] + 18*Cos[c + d*x]*(31 + 30*Log[Cos[(c + d*x)/2]] - 30*Log[Sin[(c + d*x)/2]]) + 630*Cos[3*(c + d*x)]*Log[Sin[
(c + d*x)/2]] - 90*Cos[5*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 432*Sin[c + d*x] + 744*Sin[2*(c + d*x)] + 720*Log[
Cos[(c + d*x)/2]]*Sin[2*(c + d*x)] - 720*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] - 176*Sin[3*(c + d*x)] - 372*S
in[4*(c + d*x)] - 360*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] + 360*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)] + 12
8*Sin[5*(c + d*x)]))/(a^2*d*(1 + Sin[c + d*x])^2)

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fricas [A]  time = 0.47, size = 260, normalized size = 1.65 \[ -\frac {166 \, \cos \left (d x + c\right )^{4} - 144 \, \cos \left (d x + c\right )^{2} + 45 \, {\left (\cos \left (d x + c\right )^{5} - 3 \, \cos \left (d x + c\right )^{3} - 2 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 45 \, {\left (\cos \left (d x + c\right )^{5} - 3 \, \cos \left (d x + c\right )^{3} - 2 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 4 \, {\left (32 \, \cos \left (d x + c\right )^{4} - 35 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) - 12}{20 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 3 \, a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right ) - 2 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - a^{2} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(166*cos(d*x + c)^4 - 144*cos(d*x + c)^2 + 45*(cos(d*x + c)^5 - 3*cos(d*x + c)^3 - 2*(cos(d*x + c)^3 - c
os(d*x + c))*sin(d*x + c) + 2*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 45*(cos(d*x + c)^5 - 3*cos(d*x + c)^
3 - 2*(cos(d*x + c)^3 - cos(d*x + c))*sin(d*x + c) + 2*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 4*(32*cos(
d*x + c)^4 - 35*cos(d*x + c)^2 - 2)*sin(d*x + c) - 12)/(a^2*d*cos(d*x + c)^5 - 3*a^2*d*cos(d*x + c)^3 + 2*a^2*
d*cos(d*x + c) - 2*(a^2*d*cos(d*x + c)^3 - a^2*d*cos(d*x + c))*sin(d*x + c))

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giac [A]  time = 0.23, size = 187, normalized size = 1.18 \[ \frac {\frac {180 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {5 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{4}} - \frac {10}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {5 \, {\left (54 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {2 \, {\left (245 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 870 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 810 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 211\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{40 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(180*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 5*(a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c))/a^4
- 10/(a^2*(tan(1/2*d*x + 1/2*c) - 1)) - 5*(54*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(1
/2*d*x + 1/2*c)^2) + 2*(245*tan(1/2*d*x + 1/2*c)^4 + 870*tan(1/2*d*x + 1/2*c)^3 + 1240*tan(1/2*d*x + 1/2*c)^2
+ 810*tan(1/2*d*x + 1/2*c) + 211)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d

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maple [A]  time = 0.62, size = 219, normalized size = 1.39 \[ -\frac {1}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2} d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2}}-\frac {1}{8 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {1}{d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {9 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}+\frac {4}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {11}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {49}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^2/d/(tan(1/2*d*x+1/2*c)-1)+1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^2*tan(1/2*d*x+1/2*c)-1/8/a^2/d/tan(1/2*
d*x+1/2*c)^2+1/d/a^2/tan(1/2*d*x+1/2*c)+9/2/d/a^2*ln(tan(1/2*d*x+1/2*c))+4/5/a^2/d/(tan(1/2*d*x+1/2*c)+1)^5-2/
a^2/d/(tan(1/2*d*x+1/2*c)+1)^4+5/a^2/d/(tan(1/2*d*x+1/2*c)+1)^3-11/2/a^2/d/(tan(1/2*d*x+1/2*c)+1)^2+49/4/a^2/d
/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.34, size = 354, normalized size = 2.24 \[ \frac {\frac {\frac {20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {567 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1448 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {985 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {820 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {1355 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {520 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 5}{\frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {5 \, {\left (\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{a^{2}} + \frac {180 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{40 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*((20*sin(d*x + c)/(cos(d*x + c) + 1) + 567*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1448*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 985*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 820*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1355*si
n(d*x + c)^6/(cos(d*x + c) + 1)^6 - 520*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 5)/(a^2*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 4*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 5*a^2*si
n(d*x + c)^6/(cos(d*x + c) + 1)^6 - 4*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a^2*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8) - 5*(8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^2 + 180*log(sin(d*x
+ c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 10.09, size = 191, normalized size = 1.21 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}+\frac {9\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {271\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{8}-\frac {41\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+\frac {197\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{8}+\frac {181\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}+\frac {567\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{40}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {1}{8}\right )}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^2*d) + (9*log(tan(c/2 + (d*x)/2)))/(2*a^2*d) - tan(c/2 + (d*x)/2)/(a^2*d) - (cot(c/2
+ (d*x)/2)^2*(tan(c/2 + (d*x)/2)/2 + (567*tan(c/2 + (d*x)/2)^2)/40 + (181*tan(c/2 + (d*x)/2)^3)/5 + (197*tan(
c/2 + (d*x)/2)^4)/8 - (41*tan(c/2 + (d*x)/2)^5)/2 - (271*tan(c/2 + (d*x)/2)^6)/8 - 13*tan(c/2 + (d*x)/2)^7 - 1
/8))/(a^2*d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2) + 1)^5)

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

[Out]

Timed out

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