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

Optimal. Leaf size=32 \[ \frac{\left (a \sec ^2(x)\right )^{5/2}}{5 a}-\frac{1}{3} \left (a \sec ^2(x)\right )^{3/2} \]

[Out]

-(a*Sec[x]^2)^(3/2)/3 + (a*Sec[x]^2)^(5/2)/(5*a)

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Rubi [A]  time = 0.0999202, antiderivative size = 32, 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} \[ \frac{\left (a \sec ^2(x)\right )^{5/2}}{5 a}-\frac{1}{3} \left (a \sec ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0495771, size = 22, normalized size = 0.69 \[ \frac{1}{15} \left (3 \sec ^2(x)-5\right ) \left (a \sec ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*Sec[x]^2)^(3/2)*(-5 + 3*Sec[x]^2))/15

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Maple [A]  time = 0.018, size = 29, normalized size = 0.9 \begin{align*}{\frac{1}{5\,a} \left ( a+a \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{1}{3} \left ( a+a \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/a*(a+a*tan(x)^2)^(5/2)-1/3*(a+a*tan(x)^2)^(3/2)

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Maxima [B]  time = 1.93163, size = 755, normalized size = 23.59 \begin{align*} \text{result too large to display} \end{align*}

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]

-8/15*(50*a*cos(4*x)*cos(3*x) + 50*a*sin(4*x)*sin(3*x) + 25*a*sin(3*x)*sin(2*x) + (5*a*cos(7*x) - 2*a*cos(5*x)
+ 5*a*cos(3*x))*cos(10*x) + 5*(5*a*cos(7*x) - 2*a*cos(5*x) + 5*a*cos(3*x))*cos(8*x) + 5*(10*a*cos(6*x) + 10*a
*cos(4*x) + 5*a*cos(2*x) + a)*cos(7*x) - 10*(2*a*cos(5*x) - 5*a*cos(3*x))*cos(6*x) - 2*(10*a*cos(4*x) + 5*a*co
s(2*x) + a)*cos(5*x) + 5*(5*a*cos(2*x) + a)*cos(3*x) + (5*a*sin(7*x) - 2*a*sin(5*x) + 5*a*sin(3*x))*sin(10*x)
+ 5*(5*a*sin(7*x) - 2*a*sin(5*x) + 5*a*sin(3*x))*sin(8*x) + 25*(2*a*sin(6*x) + 2*a*sin(4*x) + a*sin(2*x))*sin(
7*x) - 10*(2*a*sin(5*x) - 5*a*sin(3*x))*sin(6*x) - 10*(2*a*sin(4*x) + a*sin(2*x))*sin(5*x))*sqrt(a)/(2*(5*cos(
8*x) + 10*cos(6*x) + 10*cos(4*x) + 5*cos(2*x) + 1)*cos(10*x) + cos(10*x)^2 + 10*(10*cos(6*x) + 10*cos(4*x) + 5
*cos(2*x) + 1)*cos(8*x) + 25*cos(8*x)^2 + 20*(10*cos(4*x) + 5*cos(2*x) + 1)*cos(6*x) + 100*cos(6*x)^2 + 20*(5*
cos(2*x) + 1)*cos(4*x) + 100*cos(4*x)^2 + 25*cos(2*x)^2 + 10*(sin(8*x) + 2*sin(6*x) + 2*sin(4*x) + sin(2*x))*s
in(10*x) + sin(10*x)^2 + 50*(2*sin(6*x) + 2*sin(4*x) + sin(2*x))*sin(8*x) + 25*sin(8*x)^2 + 100*(2*sin(4*x) +
sin(2*x))*sin(6*x) + 100*sin(6*x)^2 + 100*sin(4*x)^2 + 100*sin(4*x)*sin(2*x) + 25*sin(2*x)^2 + 10*cos(2*x) + 1
)

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Fricas [A]  time = 1.39101, size = 82, normalized size = 2.56 \begin{align*} \frac{1}{15} \,{\left (3 \, a \tan \left (x\right )^{4} + a \tan \left (x\right )^{2} - 2 \, a\right )} \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)^(3/2),x, algorithm="fricas")

[Out]

1/15*(3*a*tan(x)^4 + a*tan(x)^2 - 2*a)*sqrt(a*tan(x)^2 + a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13284, size = 42, normalized size = 1.31 \begin{align*} \frac{3 \,{\left (a \tan \left (x\right )^{2} + a\right )}^{\frac{5}{2}} - 5 \,{\left (a \tan \left (x\right )^{2} + a\right )}^{\frac{3}{2}} a}{15 \, a} \end{align*}

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]

1/15*(3*(a*tan(x)^2 + a)^(5/2) - 5*(a*tan(x)^2 + a)^(3/2)*a)/a