3.1.54 \(\int \frac {\csc ^3(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\) [54]

Optimal. Leaf size=78 \[ -\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac {5 \sec (c+d x)}{2 a^2 d}+\frac {5 \sec ^3(c+d x)}{6 a^2 d}-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3254, 2702, 294, 308, 213} \begin {gather*} \frac {5 \sec ^3(c+d x)}{6 a^2 d}+\frac {5 \sec (c+d x)}{2 a^2 d}-\frac {5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 294

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 308

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 2702

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 3254

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

Rubi steps

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

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(208\) vs. \(2(78)=156\).
time = 0.32, size = 208, normalized size = 2.67 \begin {gather*} \frac {2 \csc ^8(c+d x) \left (22-40 \cos (2 (c+d x))+13 \cos (3 (c+d x))-30 \cos (4 (c+d x))+13 \cos (5 (c+d x))+15 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+15 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-15 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-15 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (-26-30 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+30 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3 a^2 d \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Csc[c + d*x]^8*(22 - 40*Cos[2*(c + d*x)] + 13*Cos[3*(c + d*x)] - 30*Cos[4*(c + d*x)] + 13*Cos[5*(c + d*x)]
+ 15*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 15*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 15*Cos[3*(c + d*x)]*
Log[Sin[(c + d*x)/2]] - 15*Cos[5*(c + d*x)]*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(-26 - 30*Log[Cos[(c + d*x)/2
]] + 30*Log[Sin[(c + d*x)/2]])))/(3*a^2*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2)^3)

________________________________________________________________________________________

Maple [A]
time = 0.31, size = 75, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {\frac {1}{4+4 \cos \left (d x +c \right )}-\frac {5 \ln \left (1+\cos \left (d x +c \right )\right )}{4}+\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {2}{\cos \left (d x +c \right )}+\frac {1}{4 \cos \left (d x +c \right )-4}+\frac {5 \ln \left (\cos \left (d x +c \right )-1\right )}{4}}{d \,a^{2}}\)	\(75\)
default	\(\frac {\frac {1}{4+4 \cos \left (d x +c \right )}-\frac {5 \ln \left (1+\cos \left (d x +c \right )\right )}{4}+\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {2}{\cos \left (d x +c \right )}+\frac {1}{4 \cos \left (d x +c \right )-4}+\frac {5 \ln \left (\cos \left (d x +c \right )-1\right )}{4}}{d \,a^{2}}\)	\(75\)
risch	\(\frac {15 \,{\mathrm e}^{9 i \left (d x +c \right )}+20 \,{\mathrm e}^{7 i \left (d x +c \right )}-22 \,{\mathrm e}^{5 i \left (d x +c \right )}+20 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 a^{2} d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 a^{2} d}\)	\(132\)
norman	\(\frac {\frac {1}{8 a d}+\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {75 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}-\frac {65 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}-\frac {55 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}}{a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3/(a-a*sin(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(1/4/(1+cos(d*x+c))-5/4*ln(1+cos(d*x+c))+1/3/cos(d*x+c)^3+2/cos(d*x+c)+1/4/(cos(d*x+c)-1)+5/4*ln(cos(d
*x+c)-1))

________________________________________________________________________________________

Maxima [A]
time = 0.34, size = 86, normalized size = 1.10 \begin {gather*} \frac {\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 2\right )}}{a^{2} \cos \left (d x + c\right )^{5} - a^{2} \cos \left (d x + c\right )^{3}} - \frac {15 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {15 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 - 2)/(a^2*cos(d*x + c)^5 - a^2*cos(d*x + c)^3) - 15*log(cos(d*x
+ c) + 1)/a^2 + 15*log(cos(d*x + c) - 1)/a^2)/d

________________________________________________________________________________________

Fricas [A]
time = 0.41, size = 118, normalized size = 1.51 \begin {gather*} \frac {30 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(30*cos(d*x + c)^4 - 20*cos(d*x + c)^2 - 15*(cos(d*x + c)^5 - cos(d*x + c)^3)*log(1/2*cos(d*x + c) + 1/2)
+ 15*(cos(d*x + c)^5 - cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) - 4)/(a^2*d*cos(d*x + c)^5 - a^2*d*cos(d*
x + c)^3)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\csc ^{3}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} - 2 \sin ^{2}{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(c + d*x)**3/(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1), x)/a**2

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (70) = 140\).
time = 0.44, size = 175, normalized size = 2.24 \begin {gather*} -\frac {\frac {3 \, {\left (\frac {10 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {30 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}} + \frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {16 \, {\left (\frac {12 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {9 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 7\right )}}{a^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(10*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)*(cos(d*x + c) + 1)/(a^2*(cos(d*x + c) - 1)) - 30*log(a
bs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + 3*(cos(d*x + c) - 1)/(a^2*(cos(d*x + c) + 1)) - 16*(12*(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 7)/(a^2*((cos(d*x + c) - 1)/
(cos(d*x + c) + 1) + 1)^3))/d

________________________________________________________________________________________

Mupad [B]
time = 13.76, size = 70, normalized size = 0.90 \begin {gather*} \frac {-\frac {5\,{\cos \left (c+d\,x\right )}^4}{2}+\frac {5\,{\cos \left (c+d\,x\right )}^2}{3}+\frac {1}{3}}{d\,\left (a^2\,{\cos \left (c+d\,x\right )}^3-a^2\,{\cos \left (c+d\,x\right )}^5\right )}-\frac {5\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{2\,a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*cos(c + d*x)^2)/3 - (5*cos(c + d*x)^4)/2 + 1/3)/(d*(a^2*cos(c + d*x)^3 - a^2*cos(c + d*x)^5)) - (5*atanh(c
os(c + d*x)))/(2*a^2*d)

________________________________________________________________________________________