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

Optimal. Leaf size=68 \[ \frac{8 a^2 b \sqrt{b \tan (e+f x)}}{3 f \sqrt{a \sin (e+f x)}}-\frac{2 b (a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.104562, 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} \[ \frac{8 a^2 b \sqrt{b \tan (e+f x)}}{3 f \sqrt{a \sin (e+f x)}}-\frac{2 b (a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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))^{3/2} (b \tan (e+f x))^{3/2} \, dx &=-\frac{2 b (a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{3 f}+\frac{1}{3} \left (4 a^2\right ) \int \frac{(b \tan (e+f x))^{3/2}}{\sqrt{a \sin (e+f x)}} \, dx\\ &=\frac{8 a^2 b \sqrt{b \tan (e+f x)}}{3 f \sqrt{a \sin (e+f x)}}-\frac{2 b (a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)}}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.15863, size = 45, normalized size = 0.66 \[ \frac{a^2 b (\cos (2 (e+f x))+7) \sqrt{b \tan (e+f x)}}{3 f \sqrt{a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*b*(7 + Cos[2*(e + f*x)])*Sqrt[b*Tan[e + f*x]])/(3*f*Sqrt[a*Sin[e + f*x]])

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Maple [B]  time = 0.155, size = 492, normalized size = 7.2 \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))^(3/2)*(b*tan(f*x+e))^(3/2),x)

[Out]

1/6/f*(-3*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)/(c
os(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)+3*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)+4*cos(f*x+e)^2-3*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)*(-co
s(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+3*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*(-c
os(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)+12)*cos(f
*x+e)*(a*sin(f*x+e))^(3/2)*(b*sin(f*x+e)/cos(f*x+e))^(3/2)/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{3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.65223, size = 143, normalized size = 2.1 \begin{align*} \frac{2 \,{\left (a b \cos \left (f x + e\right )^{2} + 3 \, a b\right )} \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{3 \, 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))^(3/2)*(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out