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

Optimal. Leaf size=88 \[ \frac{11 a^2 \cos (e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}-\frac{2 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f}+\frac{7 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{3 f} \]

[Out]

(11*a^2*Cos[e + f*x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) + (7*Sec[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(3*f) - (2*
Sec[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(3*a*f)

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Rubi [A]  time = 0.195731, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2713, 2855, 2646} \[ \frac{11 a^2 \cos (e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}-\frac{2 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f}+\frac{7 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*a^2*Cos[e + f*x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) + (7*Sec[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(3*f) - (2*
Sec[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(3*a*f)

Rule 2713

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> -Simp[(a + b*Sin[e + f
*x])^(m + 1)/(b*f*m*Cos[e + f*x]), x] + Dist[1/(b*m), Int[((a + b*Sin[e + f*x])^m*(b*(m + 1) + a*Sin[e + f*x])
)/Cos[e + f*x]^2, x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !LtQ[m, 0]

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx &=-\frac{2 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{3 a f}+\frac{2 \int \sec ^2(e+f x) (a+a \sin (e+f x))^{3/2} \left (\frac{5 a}{2}+a \sin (e+f x)\right ) \, dx}{3 a}\\ &=\frac{7 \sec (e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}-\frac{2 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{3 a f}-\frac{1}{6} (11 a) \int \sqrt{a+a \sin (e+f x)} \, dx\\ &=\frac{11 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}+\frac{7 \sec (e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}-\frac{2 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{3 a f}\\ \end{align*}

Mathematica [A]  time = 3.99156, size = 46, normalized size = 0.52 \[ \frac{a \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} (-8 \sin (e+f x)+\cos (2 (e+f x))+15)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[e + f*x]*(15 + Cos[2*(e + f*x)] - 8*Sin[e + f*x])*Sqrt[a*(1 + Sin[e + f*x])])/(3*f)

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Maple [A]  time = 0.402, size = 55, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{2}+4\,\sin \left ( fx+e \right ) -8 \right ) }{3\,f\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a^2*(1+sin(f*x+e))*(sin(f*x+e)^2+4*sin(f*x+e)-8)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [A]  time = 1.54691, size = 196, normalized size = 2.23 \begin{align*} -\frac{8 \,{\left (2 \, a^{\frac{3}{2}} - \frac{2 \, a^{\frac{3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{2 \, a^{\frac{3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{2 \, a^{\frac{3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{3 \, f{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/3*(2*a^(3/2) - 2*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 -
2*a^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 2*a^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)/(f*(sin(f*x + e
)/(cos(f*x + e) + 1) - 1)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2))

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Fricas [A]  time = 1.7535, size = 123, normalized size = 1.4 \begin{align*} \frac{2 \,{\left (a \cos \left (f x + e\right )^{2} - 4 \, a \sin \left (f x + e\right ) + 7 \, a\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{3 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a*cos(f*x + e)^2 - 4*a*sin(f*x + e) + 7*a)*sqrt(a*sin(f*x + e) + a)/(f*cos(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+a*sin(f*x+e))**(3/2)*tan(f*x+e)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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