3.155 \(\int \frac{\csc (e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=79 \[ \frac{b \cos (e+f x)}{a f (a+b) \sqrt{a-b \cos ^2(e+f x)+b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{a^{3/2} f} \]

[Out]

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

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Rubi [A]  time = 0.0938341, antiderivative size = 79, normalized size of antiderivative = 1., 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 = {3186, 382, 377, 206} \[ \frac{b \cos (e+f x)}{a f (a+b) \sqrt{a-b \cos ^2(e+f x)+b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]]/(a^(3/2)*f)) + (b*Cos[e + f*x])/(a*(a + b)*f*
Sqrt[a + b - b*Cos[e + f*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 382

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[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 377

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

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])

Rubi steps

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

Mathematica [A]  time = 0.338754, size = 93, normalized size = 1.18 \[ \frac{\frac{\sqrt{2} \sqrt{a} b \cos (e+f x)}{(a+b) \sqrt{2 a-b \cos (2 (e+f x))+b}}-\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \cos (e+f x)}{\sqrt{2 a-b \cos (2 (e+f x))+b}}\right )}{a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 2.155, size = 165, normalized size = 2.1 \begin{align*}{\frac{1}{f\cos \left ( fx+e \right ) }\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( -{\frac{1}{2}\ln \left ({\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,a+ \left ( -a+b \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{a}\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ){a}^{-{\frac{3}{2}}}}+{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a \left ( a+b \right ) }{\frac{1}{\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \right ){\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)*(-1/2/a^(3/2)*ln((2*a+(-a+b)*sin(f*x+e)^2+2*a^(1/2)*(-(-b*sin(f*x+e)
^2-a)*cos(f*x+e)^2)^(1/2))/sin(f*x+e)^2)+1/a*b*cos(f*x+e)^2/(a+b)/(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2))/c
os(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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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(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.28678, size = 1008, normalized size = 12.76 \begin{align*} \left [-\frac{4 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) -{\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sqrt{a} \log \left (\frac{2 \,{\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} +{\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right )}{4 \,{\left ({\left (a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}, -\frac{2 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) -{\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sqrt{-a} \arctan \left (-\frac{{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a}}{2 \,{\left (a b \cos \left (f x + e\right )^{3} -{\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right )}{2 \,{\left ({\left (a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*b*cos(f*x + e) - ((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*s
qrt(a)*log(2*((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^2 - 4*((a - b)*cos(f*x
+ e)^3 + (a + b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) + a^2 + 2*a*b + b^2)/(cos(f*x + e)^4 -
2*cos(f*x + e)^2 + 1)))/((a^3*b + a^2*b^2)*f*cos(f*x + e)^2 - (a^4 + 2*a^3*b + a^2*b^2)*f), -1/2*(2*sqrt(-b*c
os(f*x + e)^2 + a + b)*a*b*cos(f*x + e) - ((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sqrt(-a)*arctan(-1/
2*((a - b)*cos(f*x + e)^2 + a + b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/(a*b*cos(f*x + e)^3 - (a^2 + a*b)*
cos(f*x + e))))/((a^3*b + a^2*b^2)*f*cos(f*x + e)^2 - (a^4 + 2*a^3*b + a^2*b^2)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Integral(csc(e + f*x)/(a + b*sin(e + f*x)**2)**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(csc(f*x + e)/(b*sin(f*x + e)^2 + a)^(3/2), x)