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

Optimal. Leaf size=54 \[ -\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{f (a+b)^{3/2}}-\frac {\cot (e+f x)}{f (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4132, 325, 205} \[ -\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{f (a+b)^{3/2}}-\frac {\cot (e+f x)}{f (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.64, size = 189, normalized size = 3.50 \[ \frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt {a+b} \csc (e) \sin (f x) \sqrt {b (\cos (e)-i \sin (e))^4} \csc (e+f x)+b (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )\right )}{2 f (a+b)^{3/2} \sqrt {b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 271, normalized size = 5.02 \[ \left [\frac {\sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 4 \, \cos \left (f x + e\right )}{4 \, {\left (a + b\right )} f \sin \left (f x + e\right )}, \frac {\sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )}{2 \, {\left (a + b\right )} f \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/4*(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) - 4*cos(f*x + e))/((a + b)*f*sin(f*x + e)), 1/2*(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) -
2*cos(f*x + e))/((a + b)*f*sin(f*x + e))]

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giac [A]  time = 0.75, size = 74, normalized size = 1.37 \[ -\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b}{\sqrt {a b + b^{2}} {\left (a + b\right )}} + \frac {1}{{\left (a + b\right )} \tan \left (f x + e\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.83, size = 54, normalized size = 1.00 \[ -\frac {b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \left (a +b \right ) \sqrt {\left (a +b \right ) b}}-\frac {1}{f \left (a +b \right ) \tan \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 50, normalized size = 0.93 \[ -\frac {\frac {b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} {\left (a + b\right )}} + \frac {1}{{\left (a + b\right )} \tan \left (f x + e\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

-(b*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*(a + b)) + 1/((a + b)*tan(f*x + e)))/f

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mupad [B]  time = 4.28, size = 46, normalized size = 0.85 \[ -\frac {\mathrm {cot}\left (e+f\,x\right )}{f\,\left (a+b\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )}{f\,{\left (a+b\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^2*(a + b/cos(e + f*x)^2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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