3.1.96 \(\int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [96]

Optimal. Leaf size=43 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{\sqrt {a+b} f} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 43, 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 = {4219, 385, 213} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f \sqrt {a+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 213

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

Rule 385

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 4219

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^
n)^p/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 86, normalized size = 2.00 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right ) \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x)}{\sqrt {2} \sqrt {a+b} f \sqrt {a+b \sec ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(279\) vs. \(2(37)=74\).
time = 0.15, size = 280, normalized size = 6.51

method result size
default \(\frac {\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\ln \left (-\frac {2 \left (\cos \left (f x +e \right )-1\right ) \left (\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {a +b}+\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}-\cos \left (f x +e \right ) a +b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a +b}}\right )+\ln \left (-\frac {4 \left (\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {a +b}+\cos \left (f x +e \right ) a +\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}+b \right )}{\cos \left (f x +e \right )-1}\right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )}{2 f \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \sqrt {a +b}}\) \(280\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.69, size = 148, normalized size = 3.44 \begin {gather*} \left [\frac {\log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \, \sqrt {a + b} f}, \frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a + b}\right )}{{\left (a + b\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (37) = 74\).
time = 0.79, size = 305, normalized size = 7.09 \begin {gather*} -\frac {\frac {\log \left ({\left | -\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} + \sqrt {a + b} \right |}\right )}{\sqrt {a + b}} - \frac {\log \left ({\left | -\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} - \sqrt {a + b} \right |}\right )}{\sqrt {a + b}} - \frac {\log \left ({\left | -{\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )} {\left (a + b\right )} + \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{\sqrt {a + b}}}{2 \, f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(log(abs(-sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 + sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 -
 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b) + sqrt(a + b)))/sqrt(a + b) - log(abs(-sqrt(
a + b)*tan(1/2*f*x + 1/2*e)^2 + sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1
/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b) - sqrt(a + b)))/sqrt(a + b) - log(abs(-(sqrt(a + b)*tan(1/2*f*x
+ 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1
/2*f*x + 1/2*e)^2 + a + b))*(a + b) + sqrt(a + b)*(a - b)))/sqrt(a + b))/(f*sgn(cos(f*x + e)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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