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

Optimal. Leaf size=22 \[ \frac{2 (d \tan (a+b x))^{5/2}}{5 b d} \]

[Out]

(2*(d*Tan[a + b*x])^(5/2))/(5*b*d)

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Rubi [A]  time = 0.0432917, antiderivative size = 22, 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} \[ \frac{2 (d \tan (a+b x))^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*Tan[a + b*x])^(5/2))/(5*b*d)

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

Mathematica [A]  time = 0.0529561, size = 22, normalized size = 1. \[ \frac{2 (d \tan (a+b x))^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*Tan[a + b*x])^(5/2))/(5*b*d)

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Maple [A]  time = 0.02, size = 19, normalized size = 0.9 \begin{align*}{\frac{2}{5\,bd} \left ( d\tan \left ( bx+a \right ) \right ) ^{{\frac{5}{2}}}} \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]

2/5*(d*tan(b*x+a))^(5/2)/b/d

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Maxima [A]  time = 0.941616, size = 24, normalized size = 1.09 \begin{align*} \frac{2 \, \left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}}}{5 \, 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]

2/5*(d*tan(b*x + a))^(5/2)/(b*d)

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Fricas [B]  time = 1.71208, size = 111, normalized size = 5.05 \begin{align*} -\frac{2 \,{\left (d \cos \left (b x + a\right )^{2} - d\right )} \sqrt{\frac{d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{5 \, b \cos \left (b x + a\right )^{2}} \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]

-2/5*(d*cos(b*x + a)^2 - d)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*cos(b*x + a)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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]

Timed out

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

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