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

Optimal. Leaf size=93 \[ \frac {\tan ^3(c+d x)}{3 a d}+\frac {2 \tan (c+d x)}{a d}-\frac {\cot (c+d x)}{a d}-\frac {\sec ^3(c+d x)}{3 a d}-\frac {\sec (c+d x)}{a d}+\frac {\tanh ^{-1}(\cos (c+d x))}{a d} \]

[Out]

arctanh(cos(d*x+c))/a/d-cot(d*x+c)/a/d-sec(d*x+c)/a/d-1/3*sec(d*x+c)^3/a/d+2*tan(d*x+c)/a/d+1/3*tan(d*x+c)^3/a
/d

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Rubi [A]  time = 0.19, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2839, 2620, 270, 2622, 302, 207} \[ \frac {\tan ^3(c+d x)}{3 a d}+\frac {2 \tan (c+d x)}{a d}-\frac {\cot (c+d x)}{a d}-\frac {\sec ^3(c+d x)}{3 a d}-\frac {\sec (c+d x)}{a d}+\frac {\tanh ^{-1}(\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Cos[c + d*x]]/(a*d) - Cot[c + d*x]/(a*d) - Sec[c + d*x]/(a*d) - Sec[c + d*x]^3/(3*a*d) + (2*Tan[c + d*
x])/(a*d) + Tan[c + d*x]^3/(3*a*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 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rubi steps

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

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Mathematica [B]  time = 0.63, size = 245, normalized size = 2.63 \[ -\frac {\csc ^3(c+d x) \left (4 \sin (c+d x)-16 \sin (2 (c+d x))+8 \sin (3 (c+d x))+10 \cos (2 (c+d x))+8 \cos (3 (c+d x))+6 \sin (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-3 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8\right )-6 \sin (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2\right )}{3 a d (\sin (c+d x)+1) \left (\csc \left (\frac {1}{2} (c+d x)\right )-\sec \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Csc[c + d*x]^3*(2 + 10*Cos[2*(c + d*x)] + 8*Cos[3*(c + d*x)] + 3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]]
- 3*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(-8 - 3*Log[Cos[(c + d*x)/2]] + 3*Log[Sin[(c + d*x)/
2]]) + 4*Sin[c + d*x] - 16*Sin[2*(c + d*x)] - 6*Log[Cos[(c + d*x)/2]]*Sin[2*(c + d*x)] + 6*Log[Sin[(c + d*x)/2
]]*Sin[2*(c + d*x)] + 8*Sin[3*(c + d*x)]))/(a*d*(Csc[(c + d*x)/2] - Sec[(c + d*x)/2])*(Csc[(c + d*x)/2] + Sec[
(c + d*x)/2])*(1 + Sin[c + d*x]))

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fricas [A]  time = 0.47, size = 162, normalized size = 1.74 \[ \frac {10 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (8 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 4}{6 \, {\left (a d \cos \left (d x + c\right )^{3} - a d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(10*cos(d*x + c)^2 + 3*(cos(d*x + c)^3 - cos(d*x + c)*sin(d*x + c) - cos(d*x + c))*log(1/2*cos(d*x + c) +
1/2) - 3*(cos(d*x + c)^3 - cos(d*x + c)*sin(d*x + c) - cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(8*cos(d
*x + c)^2 - 1)*sin(d*x + c) - 4)/(a*d*cos(d*x + c)^3 - a*d*cos(d*x + c)*sin(d*x + c) - a*d*cos(d*x + c))

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giac [A]  time = 0.18, size = 133, normalized size = 1.43 \[ -\frac {\frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a} + \frac {21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 19}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*log(abs(tan(1/2*d*x + 1/2*c)))/a - 3*tan(1/2*d*x + 1/2*c)/a - 3*(tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*
x + 1/2*c) + 1)/((tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c))*a) + (21*tan(1/2*d*x + 1/2*c)^2 + 36*tan(1/2*
d*x + 1/2*c) + 19)/(a*(tan(1/2*d*x + 1/2*c) + 1)^3))/d

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maple [A]  time = 0.43, size = 139, normalized size = 1.49 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {1}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {7}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a/d*tan(1/2*d*x+1/2*c)-1/2/a/d/(tan(1/2*d*x+1/2*c)-1)-1/2/a/d/tan(1/2*d*x+1/2*c)-1/a/d*ln(tan(1/2*d*x+1/2*
c))-2/3/a/d/(tan(1/2*d*x+1/2*c)+1)^3+1/a/d/(tan(1/2*d*x+1/2*c)+1)^2-7/2/a/d/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.33, size = 215, normalized size = 2.31 \[ -\frac {\frac {\frac {22 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 3}{\frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((22*sin(d*x + c)/(cos(d*x + c) + 1) + 8*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 30*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 27*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 - 2*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) +
6*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - 3*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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mupad [B]  time = 9.11, size = 150, normalized size = 1.61 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {22\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+1}{d\,\left (-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)/(2*a*d) - ((22*tan(c/2 + (d*x)/2))/3 + (8*tan(c/2 + (d*x)/2)^2)/3 - 10*tan(c/2 + (d*x)/2)^3
- 9*tan(c/2 + (d*x)/2)^4 + 1)/(d*(2*a*tan(c/2 + (d*x)/2) + 4*a*tan(c/2 + (d*x)/2)^2 - 4*a*tan(c/2 + (d*x)/2)^
4 - 2*a*tan(c/2 + (d*x)/2)^5)) - log(tan(c/2 + (d*x)/2))/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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