3.268 \(\int \tan (x) (a+a \tan ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=14 \[ \frac {1}{3} \left (a \sec ^2(x)\right )^{3/2} \]

[Out]

1/3*(a*sec(x)^2)^(3/2)

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Rubi [A]  time = 0.05, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3657, 4124, 32} \[ \frac {1}{3} \left (a \sec ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]*(a + a*Tan[x]^2)^(3/2),x]

[Out]

(a*Sec[x]^2)^(3/2)/3

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \[ \frac {1}{3} \left (a \sec ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[x]^2)^(3/2)/3

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fricas [A]  time = 0.45, size = 12, normalized size = 0.86 \[ \frac {1}{3} \, {\left (a \tan \relax (x)^{2} + a\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*tan(x)^2 + a)^(3/2)

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giac [A]  time = 0.34, size = 12, normalized size = 0.86 \[ \frac {1}{3} \, {\left (a \tan \relax (x)^{2} + a\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*tan(x)^2 + a)^(3/2)

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maple [A]  time = 0.08, size = 13, normalized size = 0.93 \[ \frac {\left (a +a \left (\tan ^{2}\relax (x )\right )\right )^{\frac {3}{2}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \tan \relax (x)^{2} + a\right )}^{\frac {3}{2}} \tan \relax (x)\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.18, size = 11, normalized size = 0.79 \[ \frac {a^{3/2}}{3\,{\left ({\cos \relax (x)}^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)*(a + a*tan(x)^2)^(3/2),x)

[Out]

a^(3/2)/(3*(cos(x)^2)^(3/2))

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sympy [A]  time = 1.94, size = 12, normalized size = 0.86 \[ \frac {\left (a \tan ^{2}{\relax (x )} + a\right )^{\frac {3}{2}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*tan(x)**2 + a)**(3/2)/3

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