3.16 \(\int \frac{1}{\sqrt{b \tan ^4(e+f x)}} \, dx\)

Optimal. Leaf size=51 \[ -\frac{x \tan ^2(e+f x)}{\sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{f \sqrt{b \tan ^4(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0211575, antiderivative size = 51, 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} \[ -\frac{x \tan ^2(e+f x)}{\sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{f \sqrt{b \tan ^4(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Tan[e + f*x]^4],x]

[Out]

-(Tan[e + f*x]/(f*Sqrt[b*Tan[e + f*x]^4])) - (x*Tan[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 \frac{1}{\sqrt{b \tan ^4(e+f x)}} \, dx &=\frac{\tan ^2(e+f x) \int \cot ^2(e+f x) \, dx}{\sqrt{b \tan ^4(e+f x)}}\\ &=-\frac{\tan (e+f x)}{f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan ^2(e+f x) \int 1 \, dx}{\sqrt{b \tan ^4(e+f x)}}\\ &=-\frac{\tan (e+f x)}{f \sqrt{b \tan ^4(e+f x)}}-\frac{x \tan ^2(e+f x)}{\sqrt{b \tan ^4(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.0503223, size = 43, normalized size = 0.84 \[ -\frac{\tan (e+f x) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-\tan ^2(e+f x)\right )}{f \sqrt{b \tan ^4(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Tan[e + f*x]^4],x]

[Out]

-((Hypergeometric2F1[-1/2, 1, 1/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(f*Sqrt[b*Tan[e + f*x]^4]))

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Maple [A]  time = 0.026, size = 40, normalized size = 0.8 \begin{align*} -{\frac{\tan \left ( fx+e \right ) \left ( \arctan \left ( \tan \left ( fx+e \right ) \right ) \tan \left ( fx+e \right ) +1 \right ) }{f}{\frac{1}{\sqrt{b \left ( \tan \left ( fx+e \right ) \right ) ^{4}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.53725, size = 36, normalized size = 0.71 \begin{align*} -\frac{\frac{f x + e}{\sqrt{b}} + \frac{1}{\sqrt{b} \tan \left (f x + e\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((f*x + e)/sqrt(b) + 1/(sqrt(b)*tan(f*x + e)))/f

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Fricas [A]  time = 2.01841, size = 93, normalized size = 1.82 \begin{align*} -\frac{\sqrt{b \tan \left (f x + e\right )^{4}}{\left (f x \tan \left (f x + e\right ) + 1\right )}}{b f \tan \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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) + 1)/(b*f*tan(f*x + e)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \tan ^{4}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b*tan(e + f*x)**4), x)

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Giac [A]  time = 1.53148, size = 65, normalized size = 1.27 \begin{align*} -\frac{\frac{2 \,{\left (f x + e\right )}}{\sqrt{b}} - \frac{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{b}} + \frac{1}{\sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(f*x + e)/sqrt(b) - tan(1/2*f*x + 1/2*e)/sqrt(b) + 1/(sqrt(b)*tan(1/2*f*x + 1/2*e)))/f