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

Optimal. Leaf size=30 \[ -\frac{2 b \sqrt{a \sin (e+f x)}}{f \sqrt{b \tan (e+f x)}} \]

[Out]

(-2*b*Sqrt[a*Sin[e + f*x]])/(f*Sqrt[b*Tan[e + f*x]])

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Rubi [A]  time = 0.041555, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2589} \[ -\frac{2 b \sqrt{a \sin (e+f x)}}{f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]],x]

[Out]

(-2*b*Sqrt[a*Sin[e + f*x]])/(f*Sqrt[b*Tan[e + f*x]])

Rule 2589

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

Rubi steps

\begin{align*} \int \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)} \, dx &=-\frac{2 b \sqrt{a \sin (e+f x)}}{f \sqrt{b \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.129498, size = 30, normalized size = 1. \[ -\frac{2 b \sqrt{a \sin (e+f x)}}{f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]],x]

[Out]

(-2*b*Sqrt[a*Sin[e + f*x]])/(f*Sqrt[b*Tan[e + f*x]])

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Maple [B]  time = 0.197, size = 295, normalized size = 9.8 \begin{align*}{\frac{ \left ( \cos \left ( fx+e \right ) -1 \right ) \cos \left ( fx+e \right ) }{2\,f \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( 4\,\cos \left ( fx+e \right ) \sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+\ln \left ( -2\,{\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 ) -\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 ) +4\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}} \right ) \sqrt{a\sin \left ( fx+e \right ) }\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \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((a*sin(f*x+e))^(1/2)*(b*tan(f*x+e))^(1/2),x)

[Out]

1/2/f*(cos(f*x+e)-1)*(4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+ln(-2*(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)/(cos(f*x+e)+1)^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)-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)/(cos(f*x+e)+1)^2)^(1/2)+2*cos(f*
x+e)-1)/sin(f*x+e)^2)+4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*cos(f*x+e)*(a*sin(f*x+e))^(1/2)*(b*sin(f*x+e)/co
s(f*x+e))^(1/2)/sin(f*x+e)^3/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((a*sin(f*x+e))^(1/2)*(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.5758, size = 120, normalized size = 4. \begin{align*} -\frac{2 \, \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 )}{f \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)/(f*sin(f*x + e))

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

[Out]

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