3.273 \(\int \frac{\tan ^3(x)}{\sqrt{a+a \tan ^2(x)}} \, dx\)

Optimal. Leaf size=25 \[ \frac{\sqrt{a \sec ^2(x)}}{a}+\frac{1}{\sqrt{a \sec ^2(x)}} \]

[Out]

1/Sqrt[a*Sec[x]^2] + Sqrt[a*Sec[x]^2]/a

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Rubi [A]  time = 0.0922876, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3657, 4124, 43} \[ \frac{\sqrt{a \sec ^2(x)}}{a}+\frac{1}{\sqrt{a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^3/Sqrt[a + a*Tan[x]^2],x]

[Out]

1/Sqrt[a*Sec[x]^2] + Sqrt[a*Sec[x]^2]/a

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 4124

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Dist[b/(2*f), Subst[In
t[(-1 + x)^((m - 1)/2)*(b*x)^(p - 1), x], x, Sec[e + f*x]^2], x] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p] &&
IntegerQ[(m - 1)/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^3(x)}{\sqrt{a+a \tan ^2(x)}} \, dx &=\int \frac{\tan ^3(x)}{\sqrt{a \sec ^2(x)}} \, dx\\ &=\frac{1}{2} a \operatorname{Subst}\left (\int \frac{-1+x}{(a x)^{3/2}} \, dx,x,\sec ^2(x)\right )\\ &=\frac{1}{2} a \operatorname{Subst}\left (\int \left (-\frac{1}{(a x)^{3/2}}+\frac{1}{a \sqrt{a x}}\right ) \, dx,x,\sec ^2(x)\right )\\ &=\frac{1}{\sqrt{a \sec ^2(x)}}+\frac{\sqrt{a \sec ^2(x)}}{a}\\ \end{align*}

Mathematica [A]  time = 0.0238331, size = 17, normalized size = 0.68 \[ \frac{\sec ^2(x)+1}{\sqrt{a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^3/Sqrt[a + a*Tan[x]^2],x]

[Out]

(1 + Sec[x]^2)/Sqrt[a*Sec[x]^2]

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Maple [A]  time = 0.03, size = 26, normalized size = 1. \begin{align*}{\frac{1}{a}\sqrt{a+a \left ( \tan \left ( x \right ) \right ) ^{2}}}+{\frac{1}{\sqrt{a+a \left ( \tan \left ( x \right ) \right ) ^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^3/(a+a*tan(x)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.13729, size = 50, normalized size = 2. \begin{align*} \frac{{\left (\sin \left (x\right )^{2} - 2\right )} \sqrt{\sin \left (x\right ) + 1} \sqrt{-\sin \left (x\right ) + 1}}{\sqrt{a} \sin \left (x\right )^{2} - \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+a*tan(x)^2)^(1/2),x, algorithm="maxima")

[Out]

(sin(x)^2 - 2)*sqrt(sin(x) + 1)*sqrt(-sin(x) + 1)/(sqrt(a)*sin(x)^2 - sqrt(a))

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Fricas [A]  time = 1.33972, size = 50, normalized size = 2. \begin{align*} \frac{\tan \left (x\right )^{2} + 2}{\sqrt{a \tan \left (x\right )^{2} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+a*tan(x)^2)^(1/2),x, algorithm="fricas")

[Out]

(tan(x)^2 + 2)/sqrt(a*tan(x)^2 + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\left (x \right )}}{\sqrt{a \left (\tan ^{2}{\left (x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**3/(a+a*tan(x)**2)**(1/2),x)

[Out]

Integral(tan(x)**3/sqrt(a*(tan(x)**2 + 1)), x)

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Giac [A]  time = 1.10193, size = 36, normalized size = 1.44 \begin{align*} \frac{\sqrt{a \tan \left (x\right )^{2} + a} + \frac{a}{\sqrt{a \tan \left (x\right )^{2} + a}}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+a*tan(x)^2)^(1/2),x, algorithm="giac")

[Out]

(sqrt(a*tan(x)^2 + a) + a/sqrt(a*tan(x)^2 + a))/a