3.104 \(\int \frac{\tan ^2(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=107 \[ \frac{3 \sec (e+f x) \sqrt{a \sin (e+f x)+a}}{4 a f}-\frac{\sec (e+f x)}{2 f \sqrt{a \sin (e+f x)+a}}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{4 \sqrt{2} \sqrt{a} f} \]

[Out]

(5*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(4*Sqrt[2]*Sqrt[a]*f) - Sec[e + f*x]/(2
*f*Sqrt[a + a*Sin[e + f*x]]) + (3*Sec[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(4*a*f)

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Rubi [A]  time = 0.195246, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2712, 2855, 2649, 206} \[ \frac{3 \sec (e+f x) \sqrt{a \sin (e+f x)+a}}{4 a f}-\frac{\sec (e+f x)}{2 f \sqrt{a \sin (e+f x)+a}}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{4 \sqrt{2} \sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(4*Sqrt[2]*Sqrt[a]*f) - Sec[e + f*x]/(2
*f*Sqrt[a + a*Sin[e + f*x]]) + (3*Sec[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(4*a*f)

Rule 2712

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[(b*(a + b*Sin[e +
f*x])^m)/(a*f*(2*m - 1)*Cos[e + f*x]), x] - Dist[1/(a^2*(2*m - 1)), Int[((a + b*Sin[e + f*x])^(m + 1)*(a*m -
b*(2*m - 1)*Sin[e + f*x]))/Cos[e + f*x]^2, x], x] /; FreeQ[{a, b, e, f}, 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 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^2(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx &=-\frac{\sec (e+f x)}{2 f \sqrt{a+a \sin (e+f x)}}+\frac{\int \sec ^2(e+f x) \sqrt{a+a \sin (e+f x)} \left (-\frac{a}{2}+2 a \sin (e+f x)\right ) \, dx}{2 a^2}\\ &=-\frac{\sec (e+f x)}{2 f \sqrt{a+a \sin (e+f x)}}+\frac{3 \sec (e+f x) \sqrt{a+a \sin (e+f x)}}{4 a f}-\frac{5}{8} \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=-\frac{\sec (e+f x)}{2 f \sqrt{a+a \sin (e+f x)}}+\frac{3 \sec (e+f x) \sqrt{a+a \sin (e+f x)}}{4 a f}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{4 f}\\ &=\frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{4 \sqrt{2} \sqrt{a} f}-\frac{\sec (e+f x)}{2 f \sqrt{a+a \sin (e+f x)}}+\frac{3 \sec (e+f x) \sqrt{a+a \sin (e+f x)}}{4 a f}\\ \end{align*}

Mathematica [C]  time = 0.269125, size = 118, normalized size = 1.1 \[ -\frac{\sec (e+f x) \left (-3 \sin (e+f x)+(5+5 i) (-1)^{3/4} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )-1\right )}{4 f \sqrt{a (\sin (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[e + f*x]*(-1 + (5 + 5*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*
x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 3*Sin[e + f*x]))/(4*f*Sqrt[a*(1 + Sin[e +
f*x])])

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Maple [A]  time = 0.462, size = 130, normalized size = 1.2 \begin{align*}{\frac{1}{8\,f\cos \left ( fx+e \right ) } \left ( \sin \left ( fx+e \right ) \left ( 5\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a\sqrt{a-a\sin \left ( fx+e \right ) }+6\,{a}^{3/2} \right ) +5\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a\sqrt{a-a\sin \left ( fx+e \right ) }+2\,{a}^{3/2} \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sin(f*x+e)*(5*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*(a-a*sin(f*x+e))^(1/2)+6*a^(3
/2))+5*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*(a-a*sin(f*x+e))^(1/2)+2*a^(3/2))/a^(3/2)
/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.54293, size = 549, normalized size = 5.13 \begin{align*} \frac{5 \, \sqrt{2}{\left (\cos \left (f x + e\right ) \sin \left (f x + e\right ) + \cos \left (f x + e\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, \sqrt{a \sin \left (f x + e\right ) + a}{\left (3 \, \sin \left (f x + e\right ) + 1\right )}}{16 \,{\left (a f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a f \cos \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/16*(5*sqrt(2)*(cos(f*x + e)*sin(f*x + e) + cos(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(a*s
in(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x
+ e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*sqrt(a*sin(f*x + e) +
a)*(3*sin(f*x + e) + 1))/(a*f*cos(f*x + e)*sin(f*x + e) + a*f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (e + f x \right )}}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.82308, size = 594, normalized size = 5.55 \begin{align*} -\frac{\frac{5 \, \sqrt{2} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a} + \sqrt{a}\right )}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )} - \frac{4 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a} + \sqrt{a}\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )}^{2} - 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )} \sqrt{a} - a\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )} - \frac{2 \,{\left (3 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )}^{3} +{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )}^{2} \sqrt{a} -{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )} a + a^{\frac{3}{2}}\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )}^{2} + 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )} \sqrt{a} - a\right )}^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-1/4*(5*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a) + sqrt(
a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*f*x + 1/2*e) + 1)) - 4*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x
+ 1/2*e)^2 + a) + sqrt(a))/(((sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2 - 2*(sqrt(a
)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - a)*sgn(tan(1/2*f*x + 1/2*e) + 1)) - 2*(
3*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^3 + (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt
(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(a) - (sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a)
)*a + a^(3/2))/(((sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*f*
x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - a)^2*sgn(tan(1/2*f*x + 1/2*e) + 1)))/f