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

Optimal. Leaf size=62 \[ -\frac{2 b \tan (e+f x)}{a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot (e+f x)}{a f \sqrt{a+b \tan ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0979908, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3663, 271, 191} \[ -\frac{2 b \tan (e+f x)}{a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot (e+f x)}{a f \sqrt{a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3663

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

Rule 271

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

Rule 191

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

Rubi steps

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

Mathematica [A]  time = 0.652551, size = 74, normalized size = 1.19 \[ -\frac{\csc (e+f x) \sec (e+f x) ((a-2 b) \cos (2 (e+f x))+a+2 b)}{\sqrt{2} a^2 f \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.17, size = 109, normalized size = 1.8 \begin{align*} -{\frac{ \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+2\,b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{f{a}^{2} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}\sin \left ( fx+e \right ) } \left ({\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f/a^2/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*(a*cos(f*x+e)^2-2*cos(f*x+e)^2*b+2*b)*cos(f*x+e)^3*((a*cos(f*x+e)
^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)/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*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.8659, size = 211, normalized size = 3.4 \begin{align*} -\frac{{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{{\left (a^{2} b f +{\left (a^{3} - a^{2} b\right )} f \cos \left (f x + e\right )^{2}\right )} \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*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

-((a - 2*b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^2*b*f + (
a^3 - a^2*b)*f*cos(f*x + e)^2)*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 )}}{\left (a + b \tan ^{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)**2/(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral(csc(e + f*x)**2/(a + b*tan(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 )^{2}}{{\left (b \tan \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)^2/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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