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

Optimal. Leaf size=85 \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a+b}}-\frac{(a-2 b) \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

-((a - 2*b)*ArcTanh[Cos[c + d*x]])/(2*a^2*d) - (b^(3/2)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]])/(a^2*Sqrt
[a + b]*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*d)

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Rubi [A]  time = 0.116098, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3186, 414, 522, 206, 208} \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a+b}}-\frac{(a-2 b) \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - 2*b)*ArcTanh[Cos[c + d*x]])/(2*a^2*d) - (b^(3/2)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]])/(a^2*Sqrt
[a + b]*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*d)

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 206

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

Rule 208

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

Rubi steps

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

Mathematica [C]  time = 2.31014, size = 224, normalized size = 2.64 \[ -\frac{\csc ^2(c+d x) (2 a-b \cos (2 (c+d x))+b) \left (-8 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )-8 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )+\sqrt{-a-b} \left (4 (a-2 b) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+a \csc ^2\left (\frac{1}{2} (c+d x)\right )-a \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )}{16 a^2 d \sqrt{-a-b} \left (a \csc ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a + b - b*Cos[2*(c + d*x)])*Csc[c + d*x]^2*(-8*b^(3/2)*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt
[-a - b]] - 8*b^(3/2)*ArcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]] + Sqrt[-a - b]*(a*Csc[(c + d
*x)/2]^2 + 4*(a - 2*b)*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]) - a*Sec[(c + d*x)/2]^2)))/(16*a^2*Sqrt[
-a - b]*d*(b + a*Csc[c + d*x]^2))

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Maple [A]  time = 0.122, size = 142, normalized size = 1.7 \begin{align*}{\frac{1}{4\,da \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{4\,da}}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) b}{2\,{a}^{2}d}}+{\frac{1}{4\,da \left ( 1+\cos \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{4\,da}}+{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) b}{2\,{a}^{2}d}}-{\frac{{b}^{2}}{{a}^{2}d}{\it Artanh} \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94234, size = 824, normalized size = 9.69 \begin{align*} \left [\frac{2 \,{\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt{\frac{b}{a + b}} \log \left (-\frac{b \cos \left (d x + c\right )^{2} - 2 \,{\left (a + b\right )} \sqrt{\frac{b}{a + b}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) + 2 \, a \cos \left (d x + c\right ) -{\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}}, \frac{4 \,{\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt{-\frac{b}{a + b}} \arctan \left (\sqrt{-\frac{b}{a + b}} \cos \left (d x + c\right )\right ) + 2 \, a \cos \left (d x + c\right ) -{\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left ({\left (a - 2 \, b\right )} \cos \left (d x + c\right )^{2} - a + 2 \, b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(b*cos(d*x + c)^2 - b)*sqrt(b/(a + b))*log(-(b*cos(d*x + c)^2 - 2*(a + b)*sqrt(b/(a + b))*cos(d*x + c)
+ a + b)/(b*cos(d*x + c)^2 - a - b)) + 2*a*cos(d*x + c) - ((a - 2*b)*cos(d*x + c)^2 - a + 2*b)*log(1/2*cos(d*
x + c) + 1/2) + ((a - 2*b)*cos(d*x + c)^2 - a + 2*b)*log(-1/2*cos(d*x + c) + 1/2))/(a^2*d*cos(d*x + c)^2 - a^2
*d), 1/4*(4*(b*cos(d*x + c)^2 - b)*sqrt(-b/(a + b))*arctan(sqrt(-b/(a + b))*cos(d*x + c)) + 2*a*cos(d*x + c) -
((a - 2*b)*cos(d*x + c)^2 - a + 2*b)*log(1/2*cos(d*x + c) + 1/2) + ((a - 2*b)*cos(d*x + c)^2 - a + 2*b)*log(-
1/2*cos(d*x + c) + 1/2))/(a^2*d*cos(d*x + c)^2 - a^2*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.20978, size = 265, normalized size = 3.12 \begin{align*} \frac{\frac{8 \, b^{2} \arctan \left (\frac{b \cos \left (d x + c\right ) + a + b}{\sqrt{-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} a^{2}} + \frac{2 \,{\left (a - 2 \, b\right )} \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{{\left (a - \frac{2 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{4 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}} - \frac{\cos \left (d x + c\right ) - 1}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*b^2*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))/(sqrt(-a*b - b^
2)*a^2) + 2*(a - 2*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + (a - 2*a*(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) + 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(a^2*(cos(d*x + c) - 1)) - (cos
(d*x + c) - 1)/(a*(cos(d*x + c) + 1)))/d