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

Optimal. Leaf size=97 \[ -\frac{8 b \tan (e+f x)}{3 a^3 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{4 b \tan (e+f x)}{3 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\cot (e+f x)}{a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.105595, antiderivative size = 97, 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 = {3663, 271, 192, 191} \[ -\frac{8 b \tan (e+f x)}{3 a^3 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{4 b \tan (e+f x)}{3 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\cot (e+f x)}{a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cot[e + f*x]/(a*f*(a + b*Tan[e + f*x]^2)^(3/2))) - (4*b*Tan[e + f*x])/(3*a^2*f*(a + b*Tan[e + f*x]^2)^(3/2))
- (8*b*Tan[e + f*x])/(3*a^3*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 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 \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x)}{a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=-\frac{\cot (e+f x)}{a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{4 b \tan (e+f x)}{3 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{(8 b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 f}\\ &=-\frac{\cot (e+f x)}{a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{4 b \tan (e+f x)}{3 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{8 b \tan (e+f x)}{3 a^3 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.95503, size = 133, normalized size = 1.37 \[ -\frac{\cot (e+f x) \left (4 \left (3 a^2-8 b^2\right ) \cos (2 (e+f x))+\left (3 a^2-12 a b+8 b^2\right ) \cos (4 (e+f x))+3 \left (3 a^2+4 a b+8 b^2\right )\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{6 \sqrt{2} a^3 f ((a-b) \cos (2 (e+f x))+a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*(3*a^2 + 4*a*b + 8*b^2) + 4*(3*a^2 - 8*b^2)*Cos[2*(e + f*x)] + (3*a^2 - 12*a*b + 8*b^2)*Cos[4*(e + f*x)])
*Cot[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(6*Sqrt[2]*a^3*f*(a + b + (a - b)*Cos[2
*(e + f*x)])^2)

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Maple [A]  time = 0.232, size = 153, normalized size = 1.6 \begin{align*} -{\frac{ \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}ab+8\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{2}+12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab-16\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{2}+8\,{b}^{2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}}{3\,f{a}^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{4}\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{5}{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)^(5/2),x)

[Out]

-1/3/f/a^3/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^4*(3*cos(f*x+e)^4*a^2-12*cos(f*x+e)^4*a*b+8*cos(f*x+e)^4*b^2+12*c
os(f*x+e)^2*a*b-16*cos(f*x+e)^2*b^2+8*b^2)*cos(f*x+e)^5*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(5/2)
/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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 47.7678, size = 358, normalized size = 3.69 \begin{align*} -\frac{{\left ({\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 4 \,{\left (3 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, b^{2} \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}}}}{3 \,{\left (a^{3} b^{2} f +{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b - a^{3} b^{2}\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)^(5/2),x, algorithm="fricas")

[Out]

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

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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*tan(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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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{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*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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