3.228 \(\int \sec ^2(e+f x) \sqrt {d \tan (e+f x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {2 (d \tan (e+f x))^{3/2}}{3 d f} \]

[Out]

2/3*(d*tan(f*x+e))^(3/2)/d/f

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Rubi [A]  time = 0.04, antiderivative size = 22, 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 = {2607, 32} \[ \frac {2 (d \tan (e+f x))^{3/2}}{3 d f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2*Sqrt[d*Tan[e + f*x]],x]

[Out]

(2*(d*Tan[e + f*x])^(3/2))/(3*d*f)

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 22, normalized size = 1.00 \[ \frac {2 (d \tan (e+f x))^{3/2}}{3 d f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2*Sqrt[d*Tan[e + f*x]],x]

[Out]

(2*(d*Tan[e + f*x])^(3/2))/(3*d*f)

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fricas [B]  time = 0.42, size = 37, normalized size = 1.68 \[ \frac {2 \, \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{3 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/3*sqrt(d*sin(f*x + e)/cos(f*x + e))*sin(f*x + e)/(f*cos(f*x + e))

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giac [A]  time = 0.49, size = 23, normalized size = 1.05 \[ \frac {2 \, \sqrt {d \tan \left (f x + e\right )} \tan \left (f x + e\right )}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

2/3*sqrt(d*tan(f*x + e))*tan(f*x + e)/f

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maple [A]  time = 0.15, size = 19, normalized size = 0.86 \[ \frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^2*(d*tan(f*x+e))^(1/2),x)

[Out]

2/3*(d*tan(f*x+e))^(3/2)/d/f

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maxima [A]  time = 0.46, size = 18, normalized size = 0.82 \[ \frac {2 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{3 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2/3*(d*tan(f*x + e))^(3/2)/(d*f)

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mupad [B]  time = 2.57, size = 53, normalized size = 2.41 \[ \frac {2\,\sin \left (2\,e+2\,f\,x\right )\,\sqrt {\frac {d\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}}{3\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(1/2)/cos(e + f*x)^2,x)

[Out]

(2*sin(2*e + 2*f*x)*((d*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2))/(3*f*(cos(2*e + 2*f*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d \tan {\left (e + f x \right )}} \sec ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**2*(d*tan(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(d*tan(e + f*x))*sec(e + f*x)**2, x)

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