3.284 \(\int \frac{1}{(a+a \tan ^2(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=58 \[ \frac{2 \tan (c+d x)}{3 a d \sqrt{a \sec ^2(c+d x)}}+\frac{\tan (c+d x)}{3 d \left (a \sec ^2(c+d x)\right )^{3/2}} \]

[Out]

Tan[c + d*x]/(3*d*(a*Sec[c + d*x]^2)^(3/2)) + (2*Tan[c + d*x])/(3*a*d*Sqrt[a*Sec[c + d*x]^2])

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Rubi [A]  time = 0.0351514, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3657, 4122, 192, 191} \[ \frac{2 \tan (c+d x)}{3 a d \sqrt{a \sec ^2(c+d x)}}+\frac{\tan (c+d x)}{3 d \left (a \sec ^2(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tan[c + d*x]^2)^(-3/2),x]

[Out]

Tan[c + d*x]/(3*d*(a*Sec[c + d*x]^2)^(3/2)) + (2*Tan[c + d*x])/(3*a*d*Sqrt[a*Sec[c + d*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 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.0606538, size = 40, normalized size = 0.69 \[ -\frac{\left (\sin ^2(c+d x)-3\right ) \tan (c+d x)}{3 a d \sqrt{a \sec ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-3 + Sin[c + d*x]^2)*Tan[c + d*x])/(3*a*d*Sqrt[a*Sec[c + d*x]^2])

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Maple [A]  time = 0.019, size = 57, normalized size = 1. \begin{align*}{\frac{a}{d} \left ({\frac{\tan \left ( dx+c \right ) }{3\,a} \left ( a+a \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,\tan \left ( dx+c \right ) }{3\,{a}^{2}}{\frac{1}{\sqrt{a+a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.95516, size = 35, normalized size = 0.6 \begin{align*} \frac{\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (d x + c\right )}{12 \, a^{\frac{3}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(sin(3*d*x + 3*c) + 9*sin(d*x + c))/(a^(3/2)*d)

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Fricas [A]  time = 1.54072, size = 167, normalized size = 2.88 \begin{align*} \frac{\sqrt{a \tan \left (d x + c\right )^{2} + a}{\left (2 \, \tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )}}{3 \,{\left (a^{2} d \tan \left (d x + c\right )^{4} + 2 \, a^{2} d \tan \left (d x + c\right )^{2} + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*tan(c + d*x)**2 + a)**(-3/2), x)

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Giac [A]  time = 1.66812, size = 109, normalized size = 1.88 \begin{align*} -\frac{2 \,{\left (3 \, \sqrt{a}{\left (\frac{1}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{2} - 4 \, \sqrt{a}\right )}}{3 \, a^{2} d{\left (\frac{1}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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