3.114 \(\int (a \sin (e+f x))^{5/2} \sqrt{b \tan (e+f x)} \, dx\)

Optimal. Leaf size=68 \[ -\frac{8 a^2 b \sqrt{a \sin (e+f x)}}{5 f \sqrt{b \tan (e+f x)}}-\frac{2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt{b \tan (e+f x)}} \]

[Out]

(-8*a^2*b*Sqrt[a*Sin[e + f*x]])/(5*f*Sqrt[b*Tan[e + f*x]]) - (2*b*(a*Sin[e + f*x])^(5/2))/(5*f*Sqrt[b*Tan[e +
f*x]])

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Rubi [A]  time = 0.0901038, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2598, 2589} \[ -\frac{8 a^2 b \sqrt{a \sin (e+f x)}}{5 f \sqrt{b \tan (e+f x)}}-\frac{2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]

[Out]

(-8*a^2*b*Sqrt[a*Sin[e + f*x]])/(5*f*Sqrt[b*Tan[e + f*x]]) - (2*b*(a*Sin[e + f*x])^(5/2))/(5*f*Sqrt[b*Tan[e +
f*x]])

Rule 2598

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] + Dist[(a^2*(m + n - 1))/m, Int[(a*Sin[e + f*x])^(m - 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && IntegersQ[2
*m, 2*n]

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 (a \sin (e+f x))^{5/2} \sqrt{b \tan (e+f x)} \, dx &=-\frac{2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt{b \tan (e+f x)}}+\frac{1}{5} \left (4 a^2\right ) \int \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)} \, dx\\ &=-\frac{8 a^2 b \sqrt{a \sin (e+f x)}}{5 f \sqrt{b \tan (e+f x)}}-\frac{2 b (a \sin (e+f x))^{5/2}}{5 f \sqrt{b \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.186618, size = 51, normalized size = 0.75 \[ -\frac{a^2 \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)} (\sin (2 (e+f x))+8 \cot (e+f x))}{5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]],x]

[Out]

-(a^2*Sqrt[a*Sin[e + f*x]]*(8*Cot[e + f*x] + Sin[2*(e + f*x)])*Sqrt[b*Tan[e + f*x]])/(5*f)

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Maple [B]  time = 0.303, size = 493, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(f*x+e))^(5/2)*(b*tan(f*x+e))^(1/2),x)

[Out]

-1/10/f*(a*sin(f*x+e))^(5/2)*(5*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*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)-5
*cos(f*x+e)*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)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-4*cos(f*x+e)^3+5*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)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-5*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)*(-cos(f*x+e)/(cos(f*x+e)
+1)^2)^(1/2)+20*cos(f*x+e))*(b*sin(f*x+e)/cos(f*x+e))^(1/2)/sin(f*x+e)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sin \left (f x + e\right )\right )^{\frac{5}{2}} \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))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^(5/2)*sqrt(b*tan(f*x + e)), x)

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Fricas [A]  time = 1.63554, size = 161, normalized size = 2.37 \begin{align*} \frac{2 \,{\left (a^{2} \cos \left (f x + e\right )^{3} - 5 \, a^{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 )}}}{5 \, 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))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/5*(a^2*cos(f*x + e)^3 - 5*a^2*cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b*sin(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))**(5/2)*(b*tan(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [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))^(5/2)*(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out