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]

-3/2*cot(f*x+e)/(a+b)^2/f-3/2*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^(1/2)/(a+b)^(5/2)/f+1/2*cot(f*x+e)/(a+b
)/f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.09, antiderivative size = 91, normalized size of antiderivative = 1.00, 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 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 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 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)}{\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*}

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Mathematica [C]  time = 2.22, 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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fricas [B]  time = 0.64, size = 407, normalized size = 4.47 \[ \left [-\frac {4 \, {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 12 \, b \cos \left (f x + e\right )}{8 \, {\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 6 \, b \cos \left (f x + e\right )}{4 \, {\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]

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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giac [A]  time = 0.47, size = 133, normalized size = 1.46 \[ -\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b + b^{2}}} + \frac {3 \, b \tan \left (f x + e\right )^{2} + 2 \, a + 2 \, b}{{\left (b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )\right )} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{2 \, f} \]

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

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maple [A]  time = 1.05, size = 86, normalized size = 0.95 \[ -\frac {b \tan \left (f x +e \right )}{2 f \left (a +b \right )^{2} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \left (a +b \right )^{2} \sqrt {\left (a +b \right ) b}}-\frac {1}{f \left (a +b \right )^{2} \tan \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 117, normalized size = 1.29 \[ -\frac {\frac {3 \, b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {3 \, b \tan \left (f x + e\right )^{2} + 2 \, a + 2 \, b}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.40, size = 91, normalized size = 1.00 \[ -\frac {\frac {1}{a+b}+\frac {3\,b\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,{\left (a+b\right )}^2}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (a+b\right )\,\mathrm {tan}\left (e+f\,x\right )\right )}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^2+2\,a\,b+b^2\right )}{{\left (a+b\right )}^{5/2}}\right )}{2\,f\,{\left (a+b\right )}^{5/2}} \]

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)^2),x)

[Out]

- (1/(a + b) + (3*b*tan(e + f*x)^2)/(2*(a + b)^2))/(f*(b*tan(e + f*x)^3 + tan(e + f*x)*(a + b))) - (3*b^(1/2)*
atan((b^(1/2)*tan(e + f*x)*(2*a*b + a^2 + b^2))/(a + b)^(5/2)))/(2*f*(a + b)^(5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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