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

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 = 0.05, 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 = {2589} \[ \frac {2 b \sqrt {b \tan (e+f x)}}{f \sqrt {a \sin (e+f x)}} \]

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 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 \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 \[ \frac {2 b \sqrt {b \tan (e+f x)}}{f \sqrt {a \sin (e+f x)}} \]

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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fricas [A]  time = 0.74, size = 45, normalized size = 1.50 \[ \frac {2 \, \sqrt {a \sin \left (f x + e\right )} b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a f \sin \left (f x + e\right )} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {a \sin \left (f x + e\right )}}\,{d x} \]

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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maple [B]  time = 0.53, size = 308, normalized size = 10.27 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (\cos \left (f x +e \right ) \ln \left (-\frac {2 \left (2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )^{2}}\right )-\cos \left (f x +e \right ) \ln \left (-\frac {2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \cos \left (f x +e \right )-1}{\sin \left (f x +e \right )^{2}}\right )+4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) \cos \left (f x +e \right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{2 f \sqrt {a \sin \left (f x +e \right )}\, \sin \left (f x +e \right )^{3} \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}} \]

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)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {a \sin \left (f x + e\right )}}\,{d x} \]

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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