3.259 \(\int \tan ^2(x) \sqrt{a+a \tan ^2(x)} \, dx\)

Optimal. Leaf size=36 \[ \frac{1}{2} \tan (x) \sqrt{a \sec ^2(x)}-\frac{1}{2} \cos (x) \sqrt{a \sec ^2(x)} \tanh ^{-1}(\sin (x)) \]

[Out]

-(ArcTanh[Sin[x]]*Cos[x]*Sqrt[a*Sec[x]^2])/2 + (Sqrt[a*Sec[x]^2]*Tan[x])/2

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Rubi [A]  time = 0.0926689, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {3657, 4125, 2611, 3770} \[ \frac{1}{2} \tan (x) \sqrt{a \sec ^2(x)}-\frac{1}{2} \cos (x) \sqrt{a \sec ^2(x)} \tanh ^{-1}(\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^2*Sqrt[a + a*Tan[x]^2],x]

[Out]

-(ArcTanh[Sin[x]]*Cos[x]*Sqrt[a*Sec[x]^2])/2 + (Sqrt[a*Sec[x]^2]*Tan[x])/2

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4125

Int[(u_.)*((b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sec[e + f*x]^n)^FracPart[p])/(Sec[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sec[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0490207, size = 24, normalized size = 0.67 \[ \frac{1}{2} \sqrt{a \sec ^2(x)} \left (\tan (x)-\cos (x) \tanh ^{-1}(\sin (x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*Sec[x]^2]*(-(ArcTanh[Sin[x]]*Cos[x]) + Tan[x]))/2

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Maple [A]  time = 0.029, size = 39, normalized size = 1.1 \begin{align*}{\frac{\tan \left ( x \right ) }{2}\sqrt{a+a \left ( \tan \left ( x \right ) \right ) ^{2}}}-{\frac{1}{2}\sqrt{a}\ln \left ( \sqrt{a}\tan \left ( x \right ) +\sqrt{a+a \left ( \tan \left ( x \right ) \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a+a*tan(x)^2)^(1/2)*tan(x)-1/2*a^(1/2)*ln(a^(1/2)*tan(x)+(a+a*tan(x)^2)^(1/2))

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Maxima [B]  time = 1.87742, size = 398, normalized size = 11.06 \begin{align*} \frac{{\left (4 \,{\left (\sin \left (3 \, x\right ) - \sin \left (x\right )\right )} \cos \left (4 \, x\right ) -{\left (2 \,{\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) + 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \sin \left (x\right ) + 1\right ) +{\left (2 \,{\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) + 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) + 1\right ) - 4 \,{\left (\cos \left (3 \, x\right ) - \cos \left (x\right )\right )} \sin \left (4 \, x\right ) + 4 \,{\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \sin \left (3 \, x\right ) - 8 \, \cos \left (3 \, x\right ) \sin \left (2 \, x\right ) + 8 \, \cos \left (x\right ) \sin \left (2 \, x\right ) - 8 \, \cos \left (2 \, x\right ) \sin \left (x\right ) - 4 \, \sin \left (x\right )\right )} \sqrt{a}}{4 \,{\left (2 \,{\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.41434, size = 147, normalized size = 4.08 \begin{align*} \frac{1}{4} \, \sqrt{a} \log \left (2 \, a \tan \left (x\right )^{2} - 2 \, \sqrt{a \tan \left (x\right )^{2} + a} \sqrt{a} \tan \left (x\right ) + a\right ) + \frac{1}{2} \, \sqrt{a \tan \left (x\right )^{2} + a} \tan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(a)*log(2*a*tan(x)^2 - 2*sqrt(a*tan(x)^2 + a)*sqrt(a)*tan(x) + a) + 1/2*sqrt(a*tan(x)^2 + a)*tan(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17488, size = 54, normalized size = 1.5 \begin{align*} \frac{1}{2} \, \sqrt{a} \log \left ({\left | -\sqrt{a} \tan \left (x\right ) + \sqrt{a \tan \left (x\right )^{2} + a} \right |}\right ) + \frac{1}{2} \, \sqrt{a \tan \left (x\right )^{2} + a} \tan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(a)*log(abs(-sqrt(a)*tan(x) + sqrt(a*tan(x)^2 + a))) + 1/2*sqrt(a*tan(x)^2 + a)*tan(x)