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

Optimal. Leaf size=127 \[ -\frac{\left (2 a b+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a^2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}-\frac{b (4 a+3 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{5/2} d (a+b)^{3/2}}-\frac{\cot (c+d x)}{a d \left (a+b \sin ^2(c+d x)\right )} \]

[Out]

-(b*(4*a + 3*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(5/2)*(a + b)^(3/2)*d) - Cot[c + d*x]/(a*d*(a
+ b*Sin[c + d*x]^2)) - ((2*a*b + 3*b^2)*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*(a + b)*d*(a + b*Sin[c + d*x]^2))

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Rubi [A]  time = 0.146421, antiderivative size = 130, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3187, 462, 385, 205} \[ -\frac{\left (2 a^2+4 a b+3 b^2\right ) \tan (c+d x)}{2 a^2 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac{b (4 a+3 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{5/2} d (a+b)^{3/2}}-\frac{\cot (c+d x)}{a d \left ((a+b) \tan ^2(c+d x)+a\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(4*a + 3*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(5/2)*(a + b)^(3/2)*d) - Cot[c + d*x]/(a*d*(a
+ (a + b)*Tan[c + d*x]^2)) - ((2*a^2 + 4*a*b + 3*b^2)*Tan[c + d*x])/(2*a^2*(a + b)*d*(a + (a + b)*Tan[c + d*x
]^2))

Rule 3187

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

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 385

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^2 \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\cot (c+d x)}{a d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{-a-3 b+a x^2}{\left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=-\frac{\cot (c+d x)}{a d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac{\left (2 a^2+4 a b+3 b^2\right ) \tan (c+d x)}{2 a^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac{(b (4 a+3 b)) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{2 a^2 (a+b) d}\\ &=-\frac{b (4 a+3 b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{5/2} (a+b)^{3/2} d}-\frac{\cot (c+d x)}{a d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac{\left (2 a^2+4 a b+3 b^2\right ) \tan (c+d x)}{2 a^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.15663, size = 155, normalized size = 1.22 \[ -\frac{\csc ^4(c+d x) (2 a-b \cos (2 (c+d x))+b) \left (\sqrt{a} \sqrt{a+b} \cot (c+d x) \left (4 a^2-b (2 a+3 b) \cos (2 (c+d x))+6 a b+3 b^2\right )+b (4 a+3 b) (2 a-b \cos (2 (c+d x))+b) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )\right )}{8 a^{5/2} d (a+b)^{3/2} \left (a \csc ^2(c+d x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a + b - b*Cos[2*(c + d*x)])*(b*(4*a + 3*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]]*(2*a + b - b*Cos[2*
(c + d*x)]) + Sqrt[a]*Sqrt[a + b]*(4*a^2 + 6*a*b + 3*b^2 - b*(2*a + 3*b)*Cos[2*(c + d*x)])*Cot[c + d*x])*Csc[c
+ d*x]^4)/(8*a^(5/2)*(a + b)^(3/2)*d*(b + a*Csc[c + d*x]^2)^2)

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Maple [A]  time = 0.118, size = 144, normalized size = 1.1 \begin{align*} -{\frac{{b}^{2}\tan \left ( dx+c \right ) }{2\,{a}^{2}d \left ( a+b \right ) \left ( a \left ( \tan \left ( dx+c \right ) \right ) ^{2}+ \left ( \tan \left ( dx+c \right ) \right ) ^{2}b+a \right ) }}-2\,{\frac{b}{da \left ( a+b \right ) \sqrt{a \left ( a+b \right ) }}\arctan \left ({\frac{ \left ( a+b \right ) \tan \left ( dx+c \right ) }{\sqrt{a \left ( a+b \right ) }}} \right ) }-{\frac{3\,{b}^{2}}{2\,{a}^{2}d \left ( a+b \right ) }\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}}-{\frac{1}{{a}^{2}d\tan \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d/a^2*b^2/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)-2/d/a/(a+b)/(a*(a+b))^(1/2)*arctan((a+b)*tan
(d*x+c)/(a*(a+b))^(1/2))*b-3/2/d/a^2*b^2/(a+b)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-1/d/a^
2/tan(d*x+c)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.95988, size = 1330, normalized size = 10.47 \begin{align*} \left [-\frac{4 \,{\left (2 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} -{\left (4 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3} -{\left (4 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{-a^{2} - a b} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} -{\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) - 4 \,{\left (2 \, a^{4} + 6 \, a^{3} b + 7 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )}{8 \,{\left ({\left (a^{5} b + 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} d\right )} \sin \left (d x + c\right )}, -\frac{2 \,{\left (2 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} +{\left (4 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3} -{\left (4 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{a^{2} + a b} \arctan \left (\frac{{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt{a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \,{\left (2 \, a^{4} + 6 \, a^{3} b + 7 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )}{4 \,{\left ({\left (a^{5} b + 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} d\right )} \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(4*(2*a^3*b + 5*a^2*b^2 + 3*a*b^3)*cos(d*x + c)^3 - (4*a^2*b + 7*a*b^2 + 3*b^3 - (4*a*b^2 + 3*b^3)*cos(d
*x + c)^2)*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2
- 4*((2*a + b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(-a^2 - a*b)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2
*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2))*sin(d*x + c) - 4*(2*a^4 + 6*a^3*b + 7*a^2
*b^2 + 3*a*b^3)*cos(d*x + c))/(((a^5*b + 2*a^4*b^2 + a^3*b^3)*d*cos(d*x + c)^2 - (a^6 + 3*a^5*b + 3*a^4*b^2 +
a^3*b^3)*d)*sin(d*x + c)), -1/4*(2*(2*a^3*b + 5*a^2*b^2 + 3*a*b^3)*cos(d*x + c)^3 + (4*a^2*b + 7*a*b^2 + 3*b^3
- (4*a*b^2 + 3*b^3)*cos(d*x + c)^2)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2 +
a*b)*cos(d*x + c)*sin(d*x + c)))*sin(d*x + c) - 2*(2*a^4 + 6*a^3*b + 7*a^2*b^2 + 3*a*b^3)*cos(d*x + c))/(((a^
5*b + 2*a^4*b^2 + a^3*b^3)*d*cos(d*x + c)^2 - (a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d)*sin(d*x + c))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.20885, size = 242, normalized size = 1.91 \begin{align*} -\frac{\frac{{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a^{2} + a b}}\right )\right )}{\left (4 \, a b + 3 \, b^{2}\right )}}{{\left (a^{3} + a^{2} b\right )} \sqrt{a^{2} + a b}} + \frac{2 \, a^{2} \tan \left (d x + c\right )^{2} + 4 \, a b \tan \left (d x + c\right )^{2} + 3 \, b^{2} \tan \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, a b}{{\left (a \tan \left (d x + c\right )^{3} + b \tan \left (d x + c\right )^{3} + a \tan \left (d x + c\right )\right )}{\left (a^{3} + a^{2} b\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b))
)*(4*a*b + 3*b^2)/((a^3 + a^2*b)*sqrt(a^2 + a*b)) + (2*a^2*tan(d*x + c)^2 + 4*a*b*tan(d*x + c)^2 + 3*b^2*tan(d
*x + c)^2 + 2*a^2 + 2*a*b)/((a*tan(d*x + c)^3 + b*tan(d*x + c)^3 + a*tan(d*x + c))*(a^3 + a^2*b)))/d