3.103 \(\int \frac{\csc ^2(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx\)

Optimal. Leaf size=33 \[ -\frac{\cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{f (a+b)} \]

[Out]

-((Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/((a + b)*f))

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Rubi [A]  time = 0.0710493, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {4132, 264} \[ -\frac{\cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{f (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^2/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

-((Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/((a + b)*f))

Rule 4132

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

Rule 264

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

Rubi steps

\begin{align*} \int \frac{\csc ^2(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{(a+b) f}\\ \end{align*}

Mathematica [A]  time = 0.112694, size = 55, normalized size = 1.67 \[ -\frac{\csc (e+f x) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)}{2 f (a+b) \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^2/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

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

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Maple [A]  time = 0.322, size = 48, normalized size = 1.5 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{f \left ( a+b \right ) \sin \left ( fx+e \right ) }\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

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

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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)^2/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.529402, size = 113, normalized size = 3.42 \begin{align*} -\frac{\sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a + b\right )} f \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/((a + b)*f*sin(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**2/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Integral(csc(e + f*x)**2/sqrt(a + b*sec(e + f*x)**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 )^{2}}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^2/sqrt(b*sec(f*x + e)^2 + a), x)