3.1.96 \(\int (a+a \sin (e+f x))^{3/2} \tan ^2(e+f x) \, dx\) [96]

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

[Out]

7/3*sec(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-2/3*sec(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/f+11/3*a^2*cos(f*x+e)/f/(a+a*s
in(f*x+e))^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2792, 2934, 2725} \begin {gather*} \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} \end {gather*}

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 2725

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]

Rule 2792

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 2934

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]

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*}

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Mathematica [A]
time = 1.98, size = 46, normalized size = 0.52 \begin {gather*} \frac {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} \end {gather*}

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 = 1.05, size = 55, normalized size = 0.62

method	result	size
default	\(-\frac {2 a^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin ^{2}\left (f x +e \right )+4 \sin \left (f x +e \right )-8\right )}{3 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(55\)

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,method=_RETURNVERBOSE)

[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 = 0.51, size = 157, normalized size = 1.78 \begin {gather*} -\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 {gather*}

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 = 0.36, size = 52, normalized size = 0.59 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx \end {gather*}

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]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (82) = 164\).
time = 42.48, size = 222, normalized size = 2.52 \begin {gather*} -\frac {\sqrt {2} {\left (3 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{8} + 60 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{6} + 50 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{4} + 60 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} + 3 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{6 \, {\left (\tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{7} + 3 \, \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{5} + 3 \, \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{3} + \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )\right )} f} \end {gather*}

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]

-1/6*sqrt(2)*(3*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^8 + 60*a*sgn(cos(-1/4*pi
+ 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^6 + 50*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi +
1/4*f*x + 1/4*e)^4 + 60*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 3*a*sgn(cos(-
1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/((tan(-1/8*pi + 1/4*f*x + 1/4*e)^7 + 3*tan(-1/8*pi + 1/4*f*x + 1/4*e)^5 +
3*tan(-1/8*pi + 1/4*f*x + 1/4*e)^3 + tan(-1/8*pi + 1/4*f*x + 1/4*e))*f)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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