3.119 \(\int \frac{\csc (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=42 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{\sqrt{a} f} \]

[Out]

-(ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]]/(Sqrt[a]*f))

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Rubi [A]  time = 0.0683289, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3664, 377, 207} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]]/(Sqrt[a]*f))

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 2.34934, size = 226, normalized size = 5.38 \[ -\frac{\cos (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\tanh ^{-1}\left (\frac{a-(a-2 b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )+\tanh ^{-1}\left (\frac{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt{a} \sqrt{a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )}}\right )\right )}{2 \sqrt{a} f \sqrt{\sec ^4\left (\frac{1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((ArcTanh[(a - (a - 2*b)*Tan[(e + f*x)/2]^2)/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^
2)^2])] + ArcTanh[(2*b + a*(-1 + Tan[(e + f*x)/2]^2))/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e +
f*x)/2]^2)^2])])*Cos[e + f*x]*Sec[(e + f*x)/2]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(2*S
qrt[a]*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])

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Maple [B]  time = 0.21, size = 351, normalized size = 8.4 \begin{align*}{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{2\,f\cos \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) -1 \right ) }\sqrt{{\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 ) +1 \right ) ^{2}}}} \left ( \ln \left ( -2\,{\frac{\cos \left ( fx+e \right ) -1}{\sqrt{a} \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( \cos \left ( fx+e \right ) \sqrt{{\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 ) +1 \right ) ^{2}}}}\sqrt{a}-\cos \left ( fx+e \right ) a+b\cos \left ( fx+e \right ) +\sqrt{{\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 ) +1 \right ) ^{2}}}}\sqrt{a}+b \right ) } \right ) +\ln \left ( -4\,{\frac{1}{\cos \left ( fx+e \right ) -1} \left ( \cos \left ( fx+e \right ) \sqrt{{\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 ) +1 \right ) ^{2}}}}\sqrt{a}+\cos \left ( fx+e \right ) a-b\cos \left ( fx+e \right ) +\sqrt{{\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 ) +1 \right ) ^{2}}}}\sqrt{a}+b \right ) } \right ) \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{\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}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

1/2/f/a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*(ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*
x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+
e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)+ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos
(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(c
os(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1)))*sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2
)^(1/2)/cos(f*x+e)/(cos(f*x+e)-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{\sqrt{b \tan \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.45442, size = 346, normalized size = 8.24 \begin{align*} \left [\frac{\log \left (-\frac{2 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt{a} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \, \sqrt{a} f}, \frac{\sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right )}{a f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)
+ a + b)/(cos(f*x + e)^2 - 1))/(sqrt(a)*f), sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*
x + e)^2)*cos(f*x + e)/a)/(a*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc{\left (e + f x \right )}}{\sqrt{a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{\sqrt{b \tan \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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