3.260 \(\int \frac{\sec ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=20 \[ -\frac{2}{b d \sqrt{d \tan (a+b x)}} \]

[Out]

-2/(b*d*Sqrt[d*Tan[a + b*x]])

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Rubi [A]  time = 0.0434227, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2607, 32} \[ -\frac{2}{b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(b*d*Sqrt[d*Tan[a + b*x]])

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

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]

Rubi steps

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

Mathematica [A]  time = 0.0584714, size = 20, normalized size = 1. \[ -\frac{2}{b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(b*d*Sqrt[d*Tan[a + b*x]])

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Maple [A]  time = 0.02, size = 19, normalized size = 1. \begin{align*} -2\,{\frac{1}{bd\sqrt{d\tan \left ( bx+a \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.944843, size = 24, normalized size = 1.2 \begin{align*} -\frac{2}{\sqrt{d \tan \left (b x + a\right )} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(sqrt(d*tan(b*x + a))*b*d)

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Fricas [B]  time = 1.66253, size = 97, normalized size = 4.85 \begin{align*} -\frac{2 \, \sqrt{\frac{d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )}{b d^{2} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(d*sin(b*x + a)/cos(b*x + a))*cos(b*x + a)/(b*d^2*sin(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(a + b*x)**2/(d*tan(a + b*x))**(3/2), x)

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Giac [A]  time = 1.34494, size = 24, normalized size = 1.2 \begin{align*} -\frac{2}{\sqrt{d \tan \left (b x + a\right )} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(sqrt(d*tan(b*x + a))*b*d)