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

Optimal. Leaf size=103 \[ \frac{\sqrt{b} (3 a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 d (a+b)^{3/2}}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{b \cos (c+d x)}{2 a d (a+b) \left (a-b \cos ^2(c+d x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.129107, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3186, 414, 522, 206, 208} \[ \frac{\sqrt{b} (3 a+2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{2 a^2 d (a+b)^{3/2}}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{b \cos (c+d x)}{2 a d (a+b) \left (a-b \cos ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.768188, size = 194, normalized size = 1.88 \[ \frac{\frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{(-a-b)^{3/2}}+\frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{(-a-b)^{3/2}}+2 \left (\frac{a b \cos (c+d x)}{(a+b) (2 a-b \cos (2 (c+d x))+b)}+\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.116, size = 150, normalized size = 1.5 \begin{align*}{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{2\,{a}^{2}d}}-{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{2\,{a}^{2}d}}-{\frac{b\cos \left ( dx+c \right ) }{2\,da \left ( a+b \right ) \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) }}+{\frac{3\,b}{2\,da \left ( a+b \right ) }{\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}}}}+{\frac{{b}^{2}}{{a}^{2}d \left ( a+b \right ) }{\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)/(a+sin(d*x+c)^2*b)^2,x)

[Out]

1/2/d/a^2*ln(-1+cos(d*x+c))-1/2/d/a^2*ln(1+cos(d*x+c))-1/2/d/a*b/(a+b)*cos(d*x+c)/(b*cos(d*x+c)^2-a-b)+3/2/d/a
*b/(a+b)/((a+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/((a+b)*b)^(1/2))+1/d/a^2*b^2/(a+b)/((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)/(a+b*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.31169, size = 1081, normalized size = 10.5 \begin{align*} \left [-\frac{2 \, a b \cos \left (d x + c\right ) -{\left ({\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\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 \,{\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \,{\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left ({\left (a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}, -\frac{a b \cos \left (d x + c\right ) +{\left ({\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \sqrt{-\frac{b}{a + b}} \arctan \left (\sqrt{-\frac{b}{a + b}} \cos \left (d x + c\right )\right ) +{\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left ({\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \,{\left ({\left (a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.16908, size = 332, normalized size = 3.22 \begin{align*} -\frac{\frac{{\left (3 \, a b + 2 \, b^{2}\right )} \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 )}{{\left (a^{3} + a^{2} b\right )} \sqrt{-a b - b^{2}}} - \frac{2 \,{\left (a b - \frac{a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{2 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{{\left (a^{3} + a^{2} b\right )}{\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} + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}} - \frac{\log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((3*a*b + 2*b^2)*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))/((a^
3 + a^2*b)*sqrt(-a*b - b^2)) - 2*(a*b - a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*b^2*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1))/((a^3 + a^2*b)*(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) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)) - log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) +
1))/a^2)/d