3.1.40 \(\int \frac {\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) [40]

Optimal. Leaf size=58 \[ -\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac {3 \sec (c+d x)}{2 a d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3254, 2702, 294, 327, 213} \begin {gather*} \frac {3 \sec (c+d x)}{2 a d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTanh[Cos[c + d*x]])/(2*a*d) + (3*Sec[c + d*x])/(2*a*d) - (Csc[c + d*x]^2*Sec[c + d*x])/(2*a*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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(146\) vs. \(2(58)=116\).
time = 0.18, size = 146, normalized size = 2.52 \begin {gather*} \frac {\csc ^4(c+d x) \left (2-6 \cos (2 (c+d x))+2 \cos (3 (c+d x))+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 (-2-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{2 a d \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^4*(2 - 6*Cos[2*(c + d*x)] + 2*Cos[3*(c + d*x)] + 3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 3*Co
s[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(-2 - 3*Log[Cos[(c + d*x)/2]] + 3*Log[Sin[(c + d*x)/2]])))
/(2*a*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2))

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Maple [A]
time = 0.28, size = 63, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {\frac {1}{4+4 \cos \left (d x +c \right )}-\frac {3 \ln \left (1+\cos \left (d x +c \right )\right )}{4}+\frac {1}{4 \cos \left (d x +c \right )-4}+\frac {3 \ln \left (\cos \left (d x +c \right )-1\right )}{4}+\frac {1}{\cos \left (d x +c \right )}}{d a}\)	\(63\)
default	\(\frac {\frac {1}{4+4 \cos \left (d x +c \right )}-\frac {3 \ln \left (1+\cos \left (d x +c \right )\right )}{4}+\frac {1}{4 \cos \left (d x +c \right )-4}+\frac {3 \ln \left (\cos \left (d x +c \right )-1\right )}{4}+\frac {1}{\cos \left (d x +c \right )}}{d a}\)	\(63\)
norman	\(\frac {\frac {1}{8 a d}+\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {9 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}\)	\(94\)
risch	\(\frac {3 \,{\mathrm e}^{5 i \left (d x +c \right )}-2 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 a d}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 70, normalized size = 1.21 \begin {gather*} \frac {\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )} - \frac {3 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{4 \, 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.43, size = 98, normalized size = 1.69 \begin {gather*} \frac {6 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{3} - \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 )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4}{4 \, {\left (a d \cos \left (d x + c\right )^{3} - a d \cos \left (d x + c\right )\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),x, algorithm="fricas")

[Out]

1/4*(6*cos(d*x + c)^2 - 3*(cos(d*x + c)^3 - cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 3*(cos(d*x + c)^3 - co
s(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 4)/(a*d*cos(d*x + c)^3 - a*d*cos(d*x + c))

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (52) = 104\).
time = 0.50, size = 149, normalized size = 2.57 \begin {gather*} \frac {\frac {6 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} + \frac {\frac {14 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + \frac {{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}} - \frac {\cos \left (d x + c\right ) - 1}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, 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),x, algorithm="giac")

[Out]

1/8*(6*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a + (14*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3*(co
s(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1)/(a*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + (cos(d*x + c) - 1)^2/(
cos(d*x + c) + 1)^2)) - (cos(d*x + c) - 1)/(a*(cos(d*x + c) + 1)))/d

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Mupad [B]
time = 0.09, size = 55, normalized size = 0.95 \begin {gather*} -\frac {\frac {3\,{\cos \left (c+d\,x\right )}^2}{2}-1}{d\,\left (a\,\cos \left (c+d\,x\right )-a\,{\cos \left (c+d\,x\right )}^3\right )}-\frac {3\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{2\,a\,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)),x)

[Out]

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

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