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

Optimal. Leaf size=30 \[ \frac {1}{3 \left (a \sec ^2(x)\right )^{3/2}}-\frac {1}{a \sqrt {a \sec ^2(x)}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 30, normalized size of antiderivative = 1.00, 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} \[ \frac {1}{3 \left (a \sec ^2(x)\right )^{3/2}}-\frac {1}{a \sqrt {a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(3*(a*Sec[x]^2)^(3/2)) - 1/(a*Sqrt[a*Sec[x]^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])

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 \frac {\tan ^3(x)}{\left (a+a \tan ^2(x)\right )^{3/2}} \, dx &=\int \frac {\tan ^3(x)}{\left (a \sec ^2(x)\right )^{3/2}} \, dx\\ &=\frac {1}{2} a \operatorname {Subst}\left (\int \frac {-1+x}{(a x)^{5/2}} \, dx,x,\sec ^2(x)\right )\\ &=\frac {1}{2} a \operatorname {Subst}\left (\int \left (-\frac {1}{(a x)^{5/2}}+\frac {1}{a (a x)^{3/2}}\right ) \, dx,x,\sec ^2(x)\right )\\ &=\frac {1}{3 \left (a \sec ^2(x)\right )^{3/2}}-\frac {1}{a \sqrt {a \sec ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.77 \[ \frac {\cos (2 x)-5}{6 a \sqrt {a \sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5 + Cos[2*x])/(6*a*Sqrt[a*Sec[x]^2])

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fricas [A]  time = 0.41, size = 43, normalized size = 1.43 \[ -\frac {\sqrt {a \tan \relax (x)^{2} + a} {\left (3 \, \tan \relax (x)^{2} + 2\right )}}{3 \, {\left (a^{2} \tan \relax (x)^{4} + 2 \, a^{2} \tan \relax (x)^{2} + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(a*tan(x)^2 + a)*(3*tan(x)^2 + 2)/(a^2*tan(x)^4 + 2*a^2*tan(x)^2 + a^2)

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giac [A]  time = 0.33, size = 26, normalized size = 0.87 \[ -\frac {3 \, a \tan \relax (x)^{2} + 2 \, a}{3 \, {\left (a \tan \relax (x)^{2} + a\right )}^{\frac {3}{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.17, size = 29, normalized size = 0.97 \[ -\frac {1}{a \sqrt {a +a \left (\tan ^{2}\relax (x )\right )}}+\frac {1}{3 \left (a +a \left (\tan ^{2}\relax (x )\right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 38, normalized size = 1.27 \[ \frac {{\left (\sin \relax (x)^{2} + 2\right )} {\left (\sin \relax (x) + 1\right )}^{\frac {3}{2}} {\left (-\sin \relax (x) + 1\right )}^{\frac {3}{2}}}{3 \, {\left (a^{\frac {3}{2}} \sin \relax (x)^{2} - a^{\frac {3}{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sin(x)^2 + 2)*(sin(x) + 1)^(3/2)*(-sin(x) + 1)^(3/2)/(a^(3/2)*sin(x)^2 - a^(3/2))

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mupad [B]  time = 11.75, size = 29, normalized size = 0.97 \[ -\frac {\left ({\mathrm {tan}\relax (x)}^2+\frac {2}{3}\right )\,\sqrt {a\,{\mathrm {tan}\relax (x)}^2+a}}{a^2\,{\left ({\mathrm {tan}\relax (x)}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.57, size = 36, normalized size = 1.20 \[ \begin {cases} \frac {\frac {a}{3 \left (a \tan ^{2}{\relax (x )} + a\right )^{\frac {3}{2}}} - \frac {1}{\sqrt {a \tan ^{2}{\relax (x )} + a}}}{a} & \text {for}\: a \neq 0 \\\tilde {\infty } \tan ^{4}{\relax (x )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((a/(3*(a*tan(x)**2 + a)**(3/2)) - 1/sqrt(a*tan(x)**2 + a))/a, Ne(a, 0)), (zoo*tan(x)**4, True))

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