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]

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

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Rubi [A]  time = 0.04, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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 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
]

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*}

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Mathematica [A]  time = 0.06, 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]

-1/3*((-3 + Sin[c + d*x]^2)*Tan[c + d*x])/(a*d*Sqrt[a*Sec[c + d*x]^2])

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fricas [A]  time = 0.43, size = 70, normalized size = 1.21 \[ \frac {\sqrt {a \tan \left (d x + c\right )^{2} + a} {\left (2 \, \tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )}}{3 \, {\left (a^{2} d \tan \left (d x + c\right )^{4} + 2 \, a^{2} d \tan \left (d x + c\right )^{2} + a^{2} d\right )}} \]

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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giac [A]  time = 4.02, size = 81, normalized size = 1.40 \[ -\frac {2 \, {\left (3 \, \sqrt {a} {\left (\frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2} - 4 \, \sqrt {a}\right )}}{3 \, a^{2} d {\left (\frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )} \]

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))

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maple [A]  time = 0.34, size = 57, normalized size = 0.98 \[ \frac {a \left (\frac {\tan \left (d x +c \right )}{3 a \left (a +a \left (\tan ^{2}\left (d x +c \right )\right )\right )^{\frac {3}{2}}}+\frac {2 \tan \left (d x +c \right )}{3 a^{2} \sqrt {a +a \left (\tan ^{2}\left (d x +c \right )\right )}}\right )}{d} \]

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 = 0.71, size = 26, normalized size = 0.45 \[ \frac {\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (d x + c\right )}{12 \, a^{\frac {3}{2}} d} \]

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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mupad [B]  time = 11.71, size = 35, normalized size = 0.60 \[ \frac {\frac {2\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+\mathrm {tan}\left (c+d\,x\right )}{d\,{\left (a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tan(c + d*x)^2)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \tan ^{2}{\left (c + d x \right )} + a\right )^{\frac {3}{2}}}\, dx \]

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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