3.2.16 \(\int \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)} \, dx\) [116]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2669} \begin {gather*} -\frac {2 b \sqrt {a \sin (e+f x)}}{f \sqrt {b \tan (e+f x)}} \end {gather*}

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 2669

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*}

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Mathematica [A]
time = 0.15, size = 30, normalized size = 1.00 \begin {gather*} -\frac {2 b \sqrt {a \sin (e+f x)}}{f \sqrt {b \tan (e+f x)}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(294\) vs. \(2(26)=52\).
time = 0.40, size = 295, normalized size = 9.83

method	result	size
risch	\(-\frac {2 i \sqrt {a \sin \left (f x +e \right )}\, \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) f}\)	\(71\)
default	\(-\frac {\left (\cos \left (f x +e \right )-1\right ) \left (-4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )+\ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )-\ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right )-4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \cos \left (f x +e \right ) \sqrt {a \sin \left (f x +e \right )}\, \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}{2 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )^{3}}\)	\(295\)

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,method=_RETURNVERBOSE)

[Out]

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

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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((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 = 0.35, size = 52, normalized size = 1.73 \begin {gather*} -\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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \sin {\left (e + f x \right )}} \sqrt {b \tan {\left (e + f x \right )}}\, dx \end {gather*}

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]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Simplification assuming sageVARa near 0Simplification assuming sageVARf near 0Simplification assuming sageV
ARx near 0S

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Mupad [B]
time = 2.90, size = 60, normalized size = 2.00 \begin {gather*} \frac {\sin \left (2\,e+2\,f\,x\right )\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{2\,{\cos \left (e+f\,x\right )}^2}}}{f\,\left ({\cos \left (e+f\,x\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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