3.15 \(\int \sqrt{b \tan ^4(e+f x)} \, dx\)

Optimal. Leaf size=50 \[ \frac{\cot (e+f x) \sqrt{b \tan ^4(e+f x)}}{f}-x \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)} \]

[Out]

(Cot[e + f*x]*Sqrt[b*Tan[e + f*x]^4])/f - x*Cot[e + f*x]^2*Sqrt[b*Tan[e + f*x]^4]

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Rubi [A]  time = 0.0207209, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 8} \[ \frac{\cot (e+f x) \sqrt{b \tan ^4(e+f x)}}{f}-x \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[e + f*x]^4],x]

[Out]

(Cot[e + f*x]*Sqrt[b*Tan[e + f*x]^4])/f - x*Cot[e + f*x]^2*Sqrt[b*Tan[e + f*x]^4]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \sqrt{b \tan ^4(e+f x)} \, dx &=\left (\cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}\right ) \int \tan ^2(e+f x) \, dx\\ &=\frac{\cot (e+f x) \sqrt{b \tan ^4(e+f x)}}{f}-\left (\cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}\right ) \int 1 \, dx\\ &=\frac{\cot (e+f x) \sqrt{b \tan ^4(e+f x)}}{f}-x \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}\\ \end{align*}

Mathematica [A]  time = 0.0914212, size = 41, normalized size = 0.82 \[ -\frac{\cot (e+f x) \sqrt{b \tan ^4(e+f x)} \left (\tan ^{-1}(\tan (e+f x)) \cot (e+f x)-1\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tan[e + f*x]^4],x]

[Out]

-((Cot[e + f*x]*(-1 + ArcTan[Tan[e + f*x]]*Cot[e + f*x])*Sqrt[b*Tan[e + f*x]^4])/f)

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Maple [A]  time = 0.021, size = 42, normalized size = 0.8 \begin{align*} -{\frac{-\tan \left ( fx+e \right ) +\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f \left ( \tan \left ( fx+e \right ) \right ) ^{2}}\sqrt{b \left ( \tan \left ( fx+e \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^4)^(1/2),x)

[Out]

-1/f*(b*tan(f*x+e)^4)^(1/2)*(-tan(f*x+e)+arctan(tan(f*x+e)))/tan(f*x+e)^2

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Maxima [A]  time = 1.63842, size = 35, normalized size = 0.7 \begin{align*} -\frac{{\left (f x + e\right )} \sqrt{b} - \sqrt{b} \tan \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(1/2),x, algorithm="maxima")

[Out]

-((f*x + e)*sqrt(b) - sqrt(b)*tan(f*x + e))/f

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Fricas [A]  time = 2.07455, size = 88, normalized size = 1.76 \begin{align*} -\frac{\sqrt{b \tan \left (f x + e\right )^{4}}{\left (f x - \tan \left (f x + e\right )\right )}}{f \tan \left (f x + e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(b*tan(f*x + e)^4)*(f*x - tan(f*x + e))/(f*tan(f*x + e)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan ^{4}{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**4)**(1/2),x)

[Out]

Integral(sqrt(b*tan(e + f*x)**4), x)

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Giac [B]  time = 1.40617, size = 338, normalized size = 6.76 \begin{align*} \frac{{\left (\pi - 4 \, f x \tan \left (f x\right ) \tan \left (e\right ) - \pi \mathrm{sgn}\left (2 \, \tan \left (f x\right )^{2} \tan \left (e\right ) + 2 \, \tan \left (f x\right ) \tan \left (e\right )^{2} - 2 \, \tan \left (f x\right ) - 2 \, \tan \left (e\right )\right ) \tan \left (f x\right ) \tan \left (e\right ) - \pi \tan \left (f x\right ) \tan \left (e\right ) + 2 \, \arctan \left (\frac{\tan \left (f x\right ) \tan \left (e\right ) - 1}{\tan \left (f x\right ) + \tan \left (e\right )}\right ) \tan \left (f x\right ) \tan \left (e\right ) + 2 \, \arctan \left (\frac{\tan \left (f x\right ) + \tan \left (e\right )}{\tan \left (f x\right ) \tan \left (e\right ) - 1}\right ) \tan \left (f x\right ) \tan \left (e\right ) + 4 \, f x + \pi \mathrm{sgn}\left (2 \, \tan \left (f x\right )^{2} \tan \left (e\right ) + 2 \, \tan \left (f x\right ) \tan \left (e\right )^{2} - 2 \, \tan \left (f x\right ) - 2 \, \tan \left (e\right )\right ) - 2 \, \arctan \left (\frac{\tan \left (f x\right ) \tan \left (e\right ) - 1}{\tan \left (f x\right ) + \tan \left (e\right )}\right ) - 2 \, \arctan \left (\frac{\tan \left (f x\right ) + \tan \left (e\right )}{\tan \left (f x\right ) \tan \left (e\right ) - 1}\right ) - 4 \, \tan \left (f x\right ) - 4 \, \tan \left (e\right )\right )} \sqrt{b}}{4 \,{\left (f \tan \left (f x\right ) \tan \left (e\right ) - f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(1/2),x, algorithm="giac")

[Out]

1/4*(pi - 4*f*x*tan(f*x)*tan(e) - pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e))*ta
n(f*x)*tan(e) - pi*tan(f*x)*tan(e) + 2*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x) + tan(e)))*tan(f*x)*tan(e) + 2*a
rctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e) - 1))*tan(f*x)*tan(e) + 4*f*x + pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(
f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e)) - 2*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x) + tan(e))) - 2*arctan((tan(f
*x) + tan(e))/(tan(f*x)*tan(e) - 1)) - 4*tan(f*x) - 4*tan(e))*sqrt(b)/(f*tan(f*x)*tan(e) - f)