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

Optimal. Leaf size=91 \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 f (a+b)^{5/2}}-\frac{3 \cot (e+f x)}{2 f (a+b)^2}+\frac{\cot (e+f x)}{2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

(-3*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*(a + b)^(5/2)*f) - (3*Cot[e + f*x])/(2*(a + b)^2*f)
+ Cot[e + f*x]/(2*(a + b)*f*(a + b + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.0850206, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4132, 290, 325, 205} \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 f (a+b)^{5/2}}-\frac{3 \cot (e+f x)}{2 f (a+b)^2}+\frac{\cot (e+f x)}{2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*(a + b)^(5/2)*f) - (3*Cot[e + f*x])/(2*(a + b)^2*f)
+ Cot[e + f*x]/(2*(a + b)*f*(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 290

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

Rule 325

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] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cot (e+f x)}{2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 (a+b) f}\\ &=-\frac{3 \cot (e+f x)}{2 (a+b)^2 f}+\frac{\cot (e+f x)}{2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a+b)^2 f}\\ &=-\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 (a+b)^{5/2} f}-\frac{3 \cot (e+f x)}{2 (a+b)^2 f}+\frac{\cot (e+f x)}{2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 2.32557, size = 242, normalized size = 2.66 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac{b ((a+2 b) \sin (2 e)-a \sin (2 f x))}{a (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}+2 \csc (e) \sin (f x) \csc (e+f x) (a \cos (2 (e+f x))+a+2 b)+\frac{3 b (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{8 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*((3*b*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin
[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos
[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + 2*(a + 2*b + a*Cos[2*(e + f*x)])*Csc[e]*Csc
[e + f*x]*Sin[f*x] + (b*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(a*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(8*(a +
b)^2*f*(a + b*Sec[e + f*x]^2)^2)

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Maple [A]  time = 0.11, size = 86, normalized size = 1. \begin{align*} -{\frac{1}{f \left ( a+b \right ) ^{2}\tan \left ( fx+e \right ) }}-{\frac{b\tan \left ( fx+e \right ) }{2\,f \left ( a+b \right ) ^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,b}{2\,f \left ( a+b \right ) ^{2}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \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)^2,x)

[Out]

-1/f/(a+b)^2/tan(f*x+e)-1/2/f*b/(a+b)^2*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)-3/2/f*b/(a+b)^2/((a+b)*b)^(1/2)*arctan
(tan(f*x+e)*b/((a+b)*b)^(1/2))

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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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.600493, size = 975, normalized size = 10.71 \begin{align*} \left [-\frac{4 \,{\left (2 \, a - b\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left (a \cos \left (f x + e\right )^{2} + b\right )} \sqrt{-\frac{b}{a + b}} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 12 \, b \cos \left (f x + e\right )}{8 \,{\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 )}, -\frac{2 \,{\left (2 \, a - b\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left (a \cos \left (f x + e\right )^{2} + b\right )} \sqrt{\frac{b}{a + b}} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 6 \, b \cos \left (f x + e\right )}{4 \,{\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 )}\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)^2,x, algorithm="fricas")

[Out]

[-1/8*(4*(2*a - b)*cos(f*x + e)^3 - 3*(a*cos(f*x + e)^2 + b)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f
*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x +
e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) + 12
*b*cos(f*x + e))/(((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)), -1/4*(
2*(2*a - b)*cos(f*x + e)^3 - 3*(a*cos(f*x + e)^2 + b)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b
)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) + 6*b*cos(f*x + e))/(((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(-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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.27012, size = 180, normalized size = 1.98 \begin{align*} -\frac{\frac{3 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )} b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a b + b^{2}}} + \frac{3 \, b \tan \left (f x + e\right )^{2} + 2 \, a + 2 \, b}{{\left (b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )\right )}{\left (a^{2} + 2 \, a b + b^{2}\right )}}}{2 \, f} \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)^2,x, algorithm="giac")

[Out]

-1/2*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*b/((a^2 + 2*a*b + b^2)*
sqrt(a*b + b^2)) + (3*b*tan(f*x + e)^2 + 2*a + 2*b)/((b*tan(f*x + e)^3 + a*tan(f*x + e) + b*tan(f*x + e))*(a^2
+ 2*a*b + b^2)))/f