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

Optimal. Leaf size=68 \[ -\frac{2 b \tan (e+f x)}{f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{\cot (e+f x)}{f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}} \]

[Out]

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

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Rubi [A]  time = 0.0914931, antiderivative size = 68, 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 = {4132, 271, 191} \[ -\frac{2 b \tan (e+f x)}{f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{\cot (e+f x)}{f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 1.77019, size = 76, normalized size = 1.12 \[ -\frac{\csc (e+f x) \sec ^3(e+f x) (a \cos (2 (e+f x))+a+2 b) ((a-b) \cos (2 (e+f x))+a+3 b)}{4 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(a + 3*b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]*Sec[e + f*x]^3)/(4*(a + b)^
2*f*(a + b*Sec[e + f*x]^2)^(3/2))

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Maple [A]  time = 0.282, size = 89, normalized size = 1.3 \begin{align*} -{\frac{ \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{f \left ( a+b \right ) ^{2} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}\sin \left ( fx+e \right ) } \left ({\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \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*sec(f*x+e)^2)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.85312, size = 238, normalized size = 3.5 \begin{align*} -\frac{{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{{\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} f\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*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

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

[Out]

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