3.129 \(\int \frac{\csc ^2(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=106 \[ -\frac{8 b \tan (e+f x)}{3 f (a+b)^3 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{4 b \tan (e+f x)}{3 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{\cot (e+f x)}{f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[Out]

-(Cot[e + f*x]/((a + b)*f*(a + b + b*Tan[e + f*x]^2)^(3/2))) - (4*b*Tan[e + f*x])/(3*(a + b)^2*f*(a + b + b*Ta
n[e + f*x]^2)^(3/2)) - (8*b*Tan[e + f*x])/(3*(a + b)^3*f*Sqrt[a + b + b*Tan[e + f*x]^2])

________________________________________________________________________________________

Rubi [A]  time = 0.108249, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {4132, 271, 192, 191} \[ -\frac{8 b \tan (e+f x)}{3 f (a+b)^3 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{4 b \tan (e+f x)}{3 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{\cot (e+f x)}{f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-(Cot[e + f*x]/((a + b)*f*(a + b + b*Tan[e + f*x]^2)^(3/2))) - (4*b*Tan[e + f*x])/(3*(a + b)^2*f*(a + b + b*Ta
n[e + f*x]^2)^(3/2)) - (8*b*Tan[e + f*x])/(3*(a + b)^3*f*Sqrt[a + b + b*Tan[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 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -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 )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x)}{(a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{(a+b) f}\\ &=-\frac{\cot (e+f x)}{(a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{4 b \tan (e+f x)}{3 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(8 b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 (a+b)^2 f}\\ &=-\frac{\cot (e+f x)}{(a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{4 b \tan (e+f x)}{3 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{8 b \tan (e+f x)}{3 (a+b)^3 f \sqrt{a+b+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.09656, size = 108, normalized size = 1.02 \[ -\frac{\tan ^3(e+f x) \sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-6 \left (a^2-b^2\right ) \csc ^2(e+f x)+3 a^2+3 (a+b)^2 \csc ^4(e+f x)-6 a b-b^2\right )}{6 f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(3*a^2 - 6*a*b - b^2 - 6*(a^2 - b^2)*Csc[e + f*x]^2 + 3*(a + b)^2*Csc[e + f*x
]^4)*Sec[e + f*x]^2*Tan[e + f*x]^3)/(6*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.403, size = 146, normalized size = 1.4 \begin{align*} -{\frac{ \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}-6\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}ab- \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{2}+12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab-4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{2}+8\,{b}^{2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}}{3\,f \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ( a+b \right ) \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{4}\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{5}{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)^(5/2),x)

[Out]

-1/3/f/(a^2+2*a*b+b^2)/(a+b)/(b+a*cos(f*x+e)^2)^4*(3*cos(f*x+e)^4*a^2-6*cos(f*x+e)^4*a*b-cos(f*x+e)^4*b^2+12*c
os(f*x+e)^2*a*b-4*cos(f*x+e)^2*b^2+8*b^2)*cos(f*x+e)^5*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(5/2)/sin(f*x+e)

________________________________________________________________________________________

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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.40064, size = 429, normalized size = 4.05 \begin{align*} -\frac{{\left ({\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{5} + 4 \,{\left (3 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, b^{2} \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}}}}{3 \,{\left ({\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\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)^(5/2),x, algorithm="fricas")

[Out]

-1/3*((3*a^2 - 6*a*b - b^2)*cos(f*x + e)^5 + 4*(3*a*b - b^2)*cos(f*x + e)^3 + 8*b^2*cos(f*x + e))*sqrt((a*cos(
f*x + e)^2 + b)/cos(f*x + e)^2)/(((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*f*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^
2 + 3*a^2*b^3 + a*b^4)*f*cos(f*x + e)^2 + (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f)*sin(f*x + e))

________________________________________________________________________________________

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(f*x+e)**2/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^2/(b*sec(f*x + e)^2 + a)^(5/2), x)