3.131 \(\int \frac{1}{\sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \, dx\)

Optimal. Leaf size=106 \[ \frac{\sqrt{a \sin (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{\sqrt{a \sin (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

[Out]

(ArcTan[Sqrt[Cos[e + f*x]]]*Sqrt[a*Sin[e + f*x]])/(a*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) - (ArcTanh[Sqr
t[Cos[e + f*x]]]*Sqrt[a*Sin[e + f*x]])/(a*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])

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Rubi [A]  time = 0.0829008, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2601, 12, 2565, 329, 298, 203, 206} \[ \frac{\sqrt{a \sin (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{\sqrt{a \sin (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]]),x]

[Out]

(ArcTan[Sqrt[Cos[e + f*x]]]*Sqrt[a*Sin[e + f*x]])/(a*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) - (ArcTanh[Sqr
t[Cos[e + f*x]]]*Sqrt[a*Sin[e + f*x]])/(a*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]])

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[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 \frac{1}{\sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \, dx &=\frac{\sqrt{a \sin (e+f x)} \int \frac{\sqrt{\cos (e+f x)} \csc (e+f x)}{a} \, dx}{\sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=\frac{\sqrt{a \sin (e+f x)} \int \sqrt{\cos (e+f x)} \csc (e+f x) \, dx}{a \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{\sqrt{a \sin (e+f x)} \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-x^2} \, dx,x,\cos (e+f x)\right )}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{\left (2 \sqrt{a \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-x^4} \, dx,x,\sqrt{\cos (e+f x)}\right )}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{\sqrt{a \sin (e+f x)} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\cos (e+f x)}\right )}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}+\frac{\sqrt{a \sin (e+f x)} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\cos (e+f x)}\right )}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=\frac{\tan ^{-1}\left (\sqrt{\cos (e+f x)}\right ) \sqrt{a \sin (e+f x)}}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{\tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right ) \sqrt{a \sin (e+f x)}}{a f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.144943, size = 80, normalized size = 0.75 \[ \frac{\sin (2 (e+f x)) \left (\tan ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-\tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right )}{2 f \cos ^2(e+f x)^{3/4} \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]]),x]

[Out]

((ArcTan[(Cos[e + f*x]^2)^(1/4)] - ArcTanh[(Cos[e + f*x]^2)^(1/4)])*Sin[2*(e + f*x)])/(2*f*(Cos[e + f*x]^2)^(3
/4)*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])

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Maple [A]  time = 0.119, size = 177, normalized size = 1.7 \begin{align*} -{\frac{\cos \left ( fx+e \right ) -1}{2\,f\sin \left ( fx+e \right ) } \left ( \ln \left ( -{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 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}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+2\,\cos \left ( fx+e \right ) -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{a\sin \left ( fx+e \right ) }}}{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}}{\frac{1}{\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sin \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(a*sin(f*x + e))*sqrt(b*tan(f*x + e))), x)

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Fricas [B]  time = 3.72479, size = 1072, normalized size = 10.11 \begin{align*} \left [\frac{2 \, \sqrt{-a b} \arctan \left (\frac{2 \, \sqrt{-a b} \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (a b \cos \left (f x + e\right ) + a b\right )} \sin \left (f x + e\right )}\right ) - \sqrt{-a b} \log \left (-\frac{a b \cos \left (f x + e\right )^{3} - 5 \, a b \cos \left (f x + e\right )^{2} + 4 \, \sqrt{-a b} \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, a b \cos \left (f x + e\right ) + a b}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, a b f}, -\frac{2 \, \sqrt{a b} \arctan \left (\frac{2 \, \sqrt{a b} \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (a b \cos \left (f x + e\right ) - a b\right )} \sin \left (f x + e\right )}\right ) - \sqrt{a b} \log \left (\frac{4 \, \sqrt{a b}{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} -{\left (a b \cos \left (f x + e\right )^{2} + 6 \, a b \cos \left (f x + e\right ) + a b\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}\right )}{4 \, a b f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(-a*b)*arctan(2*sqrt(-a*b)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)/((a
*b*cos(f*x + e) + a*b)*sin(f*x + e))) - sqrt(-a*b)*log(-(a*b*cos(f*x + e)^3 - 5*a*b*cos(f*x + e)^2 + 4*sqrt(-a
*b)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 5*a*b*cos(f*x + e) + a*
b)/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + 3*cos(f*x + e) + 1)))/(a*b*f), -1/4*(2*sqrt(a*b)*arctan(2*sqrt(a*b)*sq
rt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)/((a*b*cos(f*x + e) - a*b)*sin(f*x + e))) - s
qrt(a*b)*log((4*sqrt(a*b)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e
)) - (a*b*cos(f*x + e)^2 + 6*a*b*cos(f*x + e) + a*b)*sin(f*x + e))/((cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*sin(
f*x + e))))/(a*b*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sin \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(a*sin(f*x + e))*sqrt(b*tan(f*x + e))), x)