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

Optimal. Leaf size=101 \[ -\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{a f}+\frac {5 \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {2} f} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 101, normalized size of antiderivative = 1.00, 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 = {2713, 2855, 2649, 206} \[ -\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{a f}+\frac {5 \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.34, size = 114, normalized size = 1.13 \[ \frac {\sec (e+f x) \sqrt {a (\sin (e+f x)+1)} \left (-2 \sin (e+f x)+(1-i) \sqrt [4]{-1} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {f x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 e+f x)\right )-\sin \left (\frac {1}{4} (2 e+f x)\right )\right )\right )+3\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 169, normalized size = 1.67 \[ \frac {\sqrt {2} \sqrt {a} \cos \left (f x + e\right ) \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, \sqrt {a \sin \left (f x + e\right ) + a} {\left (2 \, \sin \left (f x + e\right ) - 3\right )}}{4 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*sqrt(a)*cos(f*x + e)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(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 - (c
os(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*sqrt(a*sin(f*x + e) + a)*(2*sin(f*x + e) - 3))/(f*cos(f
*x + e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sin \left (f x + e\right ) + a} \tan \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.73, size = 89, normalized size = 0.88 \[ -\frac {\left (1+\sin \left (f x +e \right )\right ) \left (\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {a -a \sin \left (f x +e \right )}+4 a \sin \left (f x +e \right )-6 a \right )}{2 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(1+sin(f*x+e))*(a^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a-a*sin(f*x+e))^(1/2
)+4*a*sin(f*x+e)-6*a)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sin \left (f x + e\right ) + a} \tan \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^2\,\sqrt {a+a\,\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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