3.42 \(\int \csc (x) \sqrt{A^2+B^2 \sin ^2(x)} \, dx\)

Optimal. Leaf size=49 \[ -B \tan ^{-1}\left (\frac{B \cos (x)}{\sqrt{A^2+B^2 \sin ^2(x)}}\right )-A \tanh ^{-1}\left (\frac{A \cos (x)}{\sqrt{A^2+B^2 \sin ^2(x)}}\right ) \]

[Out]

-(B*ArcTan[(B*Cos[x])/Sqrt[A^2 + B^2*Sin[x]^2]]) - A*ArcTanh[(A*Cos[x])/Sqrt[A^2 + B^2*Sin[x]^2]]

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Rubi [A]  time = 0.0846305, antiderivative size = 57, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3186, 402, 217, 203, 377, 206} \[ -B \tan ^{-1}\left (\frac{B \cos (x)}{\sqrt{A^2-B^2 \cos ^2(x)+B^2}}\right )-A \tanh ^{-1}\left (\frac{A \cos (x)}{\sqrt{A^2-B^2 \cos ^2(x)+B^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]*Sqrt[A^2 + B^2*Sin[x]^2],x]

[Out]

-(B*ArcTan[(B*Cos[x])/Sqrt[A^2 + B^2 - B^2*Cos[x]^2]]) - A*ArcTanh[(A*Cos[x])/Sqrt[A^2 + B^2 - B^2*Cos[x]^2]]

Rule 3186

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

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

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

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 206

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

Rubi steps

\begin{align*} \int \csc (x) \sqrt{A^2+B^2 \sin ^2(x)} \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{A^2+B^2-B^2 x^2}}{1-x^2} \, dx,x,\cos (x)\right )\\ &=-\left (A^2 \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{A^2+B^2-B^2 x^2}} \, dx,x,\cos (x)\right )\right )-B^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{A^2+B^2-B^2 x^2}} \, dx,x,\cos (x)\right )\\ &=-\left (A^2 \operatorname{Subst}\left (\int \frac{1}{1-A^2 x^2} \, dx,x,\frac{\cos (x)}{\sqrt{A^2+B^2-B^2 \cos ^2(x)}}\right )\right )-B^2 \operatorname{Subst}\left (\int \frac{1}{1+B^2 x^2} \, dx,x,\frac{\cos (x)}{\sqrt{A^2+B^2-B^2 \cos ^2(x)}}\right )\\ &=-B \tan ^{-1}\left (\frac{B \cos (x)}{\sqrt{A^2+B^2-B^2 \cos ^2(x)}}\right )-A \tanh ^{-1}\left (\frac{A \cos (x)}{\sqrt{A^2+B^2-B^2 \cos ^2(x)}}\right )\\ \end{align*}

Mathematica [B]  time = 0.106308, size = 99, normalized size = 2.02 \[ \sqrt{-B^2} \log \left (\sqrt{2 A^2-B^2 \cos (2 x)+B^2}+\sqrt{2} \sqrt{-B^2} \cos (x)\right )-\sqrt{A^2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{A^2} \cos (x)}{\sqrt{2 A^2-B^2 \cos (2 x)+B^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]*Sqrt[A^2 + B^2*Sin[x]^2],x]

[Out]

-(Sqrt[A^2]*ArcTanh[(Sqrt[2]*Sqrt[A^2]*Cos[x])/Sqrt[2*A^2 + B^2 - B^2*Cos[2*x]]]) + Sqrt[-B^2]*Log[Sqrt[2]*Sqr
t[-B^2]*Cos[x] + Sqrt[2*A^2 + B^2 - B^2*Cos[2*x]]]

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Maple [C]  time = 0.096, size = 149, normalized size = 3. \begin{align*} -{\frac{1}{2\,\cos \left ( x \right ) }\sqrt{ \left ({A}^{2}+{B}^{2} \left ( \sin \left ( x \right ) \right ) ^{2} \right ) \left ( \cos \left ( x \right ) \right ) ^{2}} \left ( A{\it csgn} \left ( A \right ) \ln \left ( -{\frac{1}{ \left ( \sin \left ( x \right ) \right ) ^{2}} \left ({A}^{2} \left ( \sin \left ( x \right ) \right ) ^{2}-{B}^{2} \left ( \sin \left ( x \right ) \right ) ^{2}-2\,{\it csgn} \left ( A \right ) A\sqrt{ \left ({A}^{2}+{B}^{2} \left ( \sin \left ( x \right ) \right ) ^{2} \right ) \left ( \cos \left ( x \right ) \right ) ^{2}}-2\,{A}^{2} \right ) } \right ) -B{\it csgn} \left ( B \right ) \arctan \left ({\frac{{\it csgn} \left ( B \right ) \left ( 2\,{B}^{2} \left ( \sin \left ( x \right ) \right ) ^{2}+{A}^{2}-{B}^{2} \right ) }{2\,B}{\frac{1}{\sqrt{ \left ({A}^{2}+{B}^{2} \left ( \sin \left ( x \right ) \right ) ^{2} \right ) \left ( \cos \left ( x \right ) \right ) ^{2}}}}} \right ) \right ){\frac{1}{\sqrt{{A}^{2}+{B}^{2} \left ( \sin \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A^2+B^2*sin(x)^2)^(1/2)/sin(x),x)

[Out]

-1/2*((A^2+B^2*sin(x)^2)*cos(x)^2)^(1/2)*(A*csgn(A)*ln(-(A^2*sin(x)^2-B^2*sin(x)^2-2*csgn(A)*A*((A^2+B^2*sin(x
)^2)*cos(x)^2)^(1/2)-2*A^2)/sin(x)^2)-B*csgn(B)*arctan(1/2*csgn(B)/B*(2*B^2*sin(x)^2+A^2-B^2)/((A^2+B^2*sin(x)
^2)*cos(x)^2)^(1/2)))/cos(x)/(A^2+B^2*sin(x)^2)^(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((A^2+B^2*sin(x)^2)^(1/2)/sin(x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.99838, size = 635, normalized size = 12.96 \begin{align*} \frac{1}{2} \, B \arctan \left (-\frac{{\left (A^{4} + 2 \, A^{2} B^{2} + B^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) - 2 \,{\left (2 \, B^{3} \cos \left (x\right )^{3} -{\left (A^{2} B + B^{3}\right )} \cos \left (x\right )\right )} \sqrt{-B^{2} \cos \left (x\right )^{2} + A^{2} + B^{2}}}{4 \, B^{4} \cos \left (x\right )^{4} + A^{4} + 2 \, A^{2} B^{2} + B^{4} -{\left (A^{4} + 6 \, A^{2} B^{2} + 5 \, B^{4}\right )} \cos \left (x\right )^{2}}\right ) - \frac{1}{2} \, B \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right )}\right ) - \frac{1}{2} \, A \log \left (-B^{2} \cos \left (x\right )^{2} + A B \cos \left (x\right ) \sin \left (x\right ) + A^{2} + B^{2} + \sqrt{-B^{2} \cos \left (x\right )^{2} + A^{2} + B^{2}}{\left (A \cos \left (x\right ) + B \sin \left (x\right )\right )}\right ) + \frac{1}{2} \, A \log \left (-B^{2} \cos \left (x\right )^{2} - A B \cos \left (x\right ) \sin \left (x\right ) + A^{2} + B^{2} - \sqrt{-B^{2} \cos \left (x\right )^{2} + A^{2} + B^{2}}{\left (A \cos \left (x\right ) - B \sin \left (x\right )\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A^2+B^2*sin(x)^2)^(1/2)/sin(x),x, algorithm="fricas")

[Out]

1/2*B*arctan(-((A^4 + 2*A^2*B^2 + B^4)*cos(x)*sin(x) - 2*(2*B^3*cos(x)^3 - (A^2*B + B^3)*cos(x))*sqrt(-B^2*cos
(x)^2 + A^2 + B^2))/(4*B^4*cos(x)^4 + A^4 + 2*A^2*B^2 + B^4 - (A^4 + 6*A^2*B^2 + 5*B^4)*cos(x)^2)) - 1/2*B*arc
tan(sin(x)/cos(x)) - 1/2*A*log(-B^2*cos(x)^2 + A*B*cos(x)*sin(x) + A^2 + B^2 + sqrt(-B^2*cos(x)^2 + A^2 + B^2)
*(A*cos(x) + B*sin(x))) + 1/2*A*log(-B^2*cos(x)^2 - A*B*cos(x)*sin(x) + A^2 + B^2 - sqrt(-B^2*cos(x)^2 + A^2 +
B^2)*(A*cos(x) - B*sin(x)))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{A^{2} + B^{2} \sin ^{2}{\left (x \right )}}}{\sin{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A**2+B**2*sin(x)**2)**(1/2)/sin(x),x)

[Out]

Integral(sqrt(A**2 + B**2*sin(x)**2)/sin(x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{B^{2} \sin \left (x\right )^{2} + A^{2}}}{\sin \left (x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A^2+B^2*sin(x)^2)^(1/2)/sin(x),x, algorithm="giac")

[Out]

integrate(sqrt(B^2*sin(x)^2 + A^2)/sin(x), x)