3.375 \(\int \csc (e+f x) \sqrt{b \sec (e+f x)} \, dx\)

Optimal. Leaf size=58 \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f} \]

[Out]

(Sqrt[b]*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/f - (Sqrt[b]*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/f

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Rubi [A]  time = 0.0458734, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2622, 329, 298, 203, 206} \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]*Sqrt[b*Sec[e + f*x]],x]

[Out]

(Sqrt[b]*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/f - (Sqrt[b]*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/f

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 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 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 (e+f x) \sqrt{b \sec (e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{-1+\frac{x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{b f}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{-1+\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{b f}\\ &=-\frac{b \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{f}\\ &=\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{f}\\ \end{align*}

Mathematica [A]  time = 0.34326, size = 73, normalized size = 1.26 \[ \frac{\sqrt{b \sec (e+f x)} \left (\log \left (1-\sqrt{\sec (e+f x)}\right )-\log \left (\sqrt{\sec (e+f x)}+1\right )+2 \tan ^{-1}\left (\sqrt{\sec (e+f x)}\right )\right )}{2 f \sqrt{\sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]*Sqrt[b*Sec[e + f*x]],x]

[Out]

((2*ArcTan[Sqrt[Sec[e + f*x]]] + Log[1 - Sqrt[Sec[e + f*x]]] - Log[1 + Sqrt[Sec[e + f*x]]])*Sqrt[b*Sec[e + f*x
]])/(2*f*Sqrt[Sec[e + f*x]])

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Maple [B]  time = 0.122, size = 169, normalized size = 2.9 \begin{align*}{\frac{\cos \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }{2\,f \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{b}{\cos \left ( fx+e \right ) }}} \left ( \ln \left ( -2\,{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\cos \left ( fx+e \right ) -2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}-1 \right ) } \right ) -\arctan \left ({\frac{1}{2}{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}}} \right ) \right ){\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(b/cos(f*x+e))^(1/2)*cos(f*x+e)*(ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2
+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2
)^(1/2)))*(-1+cos(f*x+e))/sin(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^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(csc(f*x+e)*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.65422, size = 660, normalized size = 11.38 \begin{align*} \left [\frac{2 \, \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{b}{\cos \left (f x + e\right )}}{\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + \sqrt{-b} \log \left (\frac{b \cos \left (f x + e\right )^{2} - 4 \,{\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt{-b} \sqrt{\frac{b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, f}, -\frac{2 \, \sqrt{b} \arctan \left (\frac{\sqrt{\frac{b}{\cos \left (f x + e\right )}}{\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt{b}}\right ) - \sqrt{b} \log \left (\frac{b \cos \left (f x + e\right )^{2} - 4 \,{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{b} \sqrt{\frac{b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(-b)*arctan(1/2*sqrt(-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b) + sqrt(-b)*log((b*cos(f*x + e)
^2 - 4*(cos(f*x + e)^2 - cos(f*x + e))*sqrt(-b)*sqrt(b/cos(f*x + e)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 +
2*cos(f*x + e) + 1)))/f, -1/4*(2*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f*x + e) - 1)/sqrt(b)) - sqrt(b
)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(b)*sqrt(b/cos(f*x + e)) + 6*b*cos(f*x + e) +
b)/(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)))/f]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15376, size = 86, normalized size = 1.48 \begin{align*} \frac{b^{2}{\left (\frac{\arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{-b}}\right )}{\sqrt{-b} b} - \frac{\arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{b}}\right )}{b^{\frac{3}{2}}}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*(arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b) - arctan(sqrt(b*cos(f*x + e))/sqrt(b))/b^(3/2))*sgn(co
s(f*x + e))/f