3.2.23 \(\int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {a \sin (e+f x)}} \, dx\) [123]

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

[Out]

2*b*(b*tan(f*x+e))^(1/2)/f/(a*sin(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 {b \tan (e+f x)}}{f \sqrt {a \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(3/2)/Sqrt[a*Sin[e + f*x]],x]

[Out]

(2*b*Sqrt[b*Tan[e + f*x]])/(f*Sqrt[a*Sin[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 \frac {(b \tan (e+f x))^{3/2}}{\sqrt {a \sin (e+f x)}} \, dx &=\frac {2 b \sqrt {b \tan (e+f x)}}{f \sqrt {a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 30, normalized size = 1.00 \begin {gather*} \frac {2 b \sqrt {b \tan (e+f x)}}{f \sqrt {a \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^(3/2)/Sqrt[a*Sin[e + f*x]],x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(307\) vs. \(2(26)=52\).
time = 0.35, size = 308, normalized size = 10.27

method	result	size
default	\(\frac {\left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right ) \ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )-\cos \left (f x +e \right ) \ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right )-4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \cos \left (f x +e \right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{2 f \sin \left (f x +e \right )^{3} \sqrt {a \sin \left (f x +e \right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(308\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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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((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(3/2)/sqrt(a*sin(f*x + e)), x)

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Fricas [A]
time = 0.36, size = 49, normalized size = 1.63 \begin {gather*} \frac {2 \, \sqrt {a \sin \left (f x + e\right )} b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a f \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))**(3/2)/(a*sin(f*x+e))**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3007 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(3/2)/sqrt(a*sin(f*x + e)), x)

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Mupad [B]
time = 3.07, size = 39, normalized size = 1.30 \begin {gather*} \frac {2\,b\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{2\,{\cos \left (e+f\,x\right )}^2}}}{f\,\sqrt {a\,\sin \left (e+f\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x))^(3/2)/(a*sin(e + f*x))^(1/2),x)

[Out]

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

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