3.3.60 \(\int \frac {\sec ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) [260]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.00, 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 = {2687, 32} \begin {gather*} -\frac {2}{b d \sqrt {d \tan (a+b x)}} \end {gather*}

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 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]

Rule 2687

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])

Rubi steps

\begin {align*} \int \frac {\sec ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=\frac {\text {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 = 0.06, size = 20, normalized size = 1.00 \begin {gather*} -\frac {2}{b d \sqrt {d \tan (a+b x)}} \end {gather*}

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]])

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 19, normalized size = 0.95

method	result	size
derivativedivides	\(-\frac {2}{b d \sqrt {d \tan \left (b x +a \right )}}\)	\(19\)
default	\(-\frac {2}{b d \sqrt {d \tan \left (b x +a \right )}}\)	\(19\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 18, normalized size = 0.90 \begin {gather*} -\frac {2}{\sqrt {d \tan \left (b x + a\right )} b d} \end {gather*}

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)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (18) = 36\).
time = 0.38, size = 40, normalized size = 2.00 \begin {gather*} -\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 {gather*}

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))

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{2}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [A]
time = 0.55, size = 18, normalized size = 0.90 \begin {gather*} -\frac {2}{\sqrt {d \tan \left (b x + a\right )} b d} \end {gather*}

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)

________________________________________________________________________________________

Mupad [B]
time = 2.57, size = 51, normalized size = 2.55 \begin {gather*} -\frac {\sin \left (2\,a+2\,b\,x\right )\,\sqrt {\frac {d\,\sin \left (2\,a+2\,b\,x\right )}{\cos \left (2\,a+2\,b\,x\right )+1}}}{b\,d^2\,{\sin \left (a+b\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________