3.17 \(\int \frac{1}{(b \tan ^4(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=119 \[ -\frac{x \tan ^2(e+f x)}{b \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot (e+f x)}{3 b f \sqrt{b \tan ^4(e+f x)}} \]

[Out]

Cot[e + f*x]/(3*b*f*Sqrt[b*Tan[e + f*x]^4]) - Cot[e + f*x]^3/(5*b*f*Sqrt[b*Tan[e + f*x]^4]) - Tan[e + f*x]/(b*
f*Sqrt[b*Tan[e + f*x]^4]) - (x*Tan[e + f*x]^2)/(b*Sqrt[b*Tan[e + f*x]^4])

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Rubi [A]  time = 0.0446079, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 8} \[ -\frac{x \tan ^2(e+f x)}{b \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot (e+f x)}{3 b f \sqrt{b \tan ^4(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^4)^(-3/2),x]

[Out]

Cot[e + f*x]/(3*b*f*Sqrt[b*Tan[e + f*x]^4]) - Cot[e + f*x]^3/(5*b*f*Sqrt[b*Tan[e + f*x]^4]) - Tan[e + f*x]/(b*
f*Sqrt[b*Tan[e + f*x]^4]) - (x*Tan[e + f*x]^2)/(b*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}{\left (b \tan ^4(e+f x)\right )^{3/2}} \, dx &=\frac{\tan ^2(e+f x) \int \cot ^6(e+f x) \, dx}{b \sqrt{b \tan ^4(e+f x)}}\\ &=-\frac{\cot ^3(e+f x)}{5 b f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan ^2(e+f x) \int \cot ^4(e+f x) \, dx}{b \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot (e+f x)}{3 b f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b f \sqrt{b \tan ^4(e+f x)}}+\frac{\tan ^2(e+f x) \int \cot ^2(e+f x) \, dx}{b \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot (e+f x)}{3 b f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan ^2(e+f x) \int 1 \, dx}{b \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot (e+f x)}{3 b f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b f \sqrt{b \tan ^4(e+f x)}}-\frac{x \tan ^2(e+f x)}{b \sqrt{b \tan ^4(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.0486975, size = 45, normalized size = 0.38 \[ -\frac{\tan (e+f x) \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},-\tan ^2(e+f x)\right )}{5 f \left (b \tan ^4(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^4)^(-3/2),x]

[Out]

-(Hypergeometric2F1[-5/2, 1, -3/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(5*f*(b*Tan[e + f*x]^4)^(3/2))

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Maple [A]  time = 0.02, size = 63, normalized size = 0.5 \begin{align*} -{\frac{\tan \left ( fx+e \right ) \left ( 15\,\arctan \left ( \tan \left ( fx+e \right ) \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{5}+15\, \left ( \tan \left ( fx+e \right ) \right ) ^{4}-5\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}+3 \right ) }{15\,f} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{4} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(f*x+e)^4)^(3/2),x)

[Out]

-1/15/f*tan(f*x+e)*(15*arctan(tan(f*x+e))*tan(f*x+e)^5+15*tan(f*x+e)^4-5*tan(f*x+e)^2+3)/(b*tan(f*x+e)^4)^(3/2
)

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Maxima [A]  time = 1.63716, size = 68, normalized size = 0.57 \begin{align*} -\frac{\frac{15 \,{\left (f x + e\right )}}{b^{\frac{3}{2}}} + \frac{15 \, \tan \left (f x + e\right )^{4} - 5 \, \tan \left (f x + e\right )^{2} + 3}{b^{\frac{3}{2}} \tan \left (f x + e\right )^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^4)^(3/2),x, algorithm="maxima")

[Out]

-1/15*(15*(f*x + e)/b^(3/2) + (15*tan(f*x + e)^4 - 5*tan(f*x + e)^2 + 3)/(b^(3/2)*tan(f*x + e)^5))/f

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Fricas [A]  time = 1.93182, size = 162, normalized size = 1.36 \begin{align*} -\frac{{\left (15 \, f x \tan \left (f x + e\right )^{5} + 15 \, \tan \left (f x + e\right )^{4} - 5 \, \tan \left (f x + e\right )^{2} + 3\right )} \sqrt{b \tan \left (f x + e\right )^{4}}}{15 \, b^{2} f \tan \left (f x + e\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^4)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(15*f*x*tan(f*x + e)^5 + 15*tan(f*x + e)^4 - 5*tan(f*x + e)^2 + 3)*sqrt(b*tan(f*x + e)^4)/(b^2*f*tan(f*x
+ e)^7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)**4)**(3/2),x)

[Out]

Integral((b*tan(e + f*x)**4)**(-3/2), x)

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Giac [A]  time = 1.87698, size = 177, normalized size = 1.49 \begin{align*} -\frac{\frac{480 \,{\left (f x + e\right )}}{\sqrt{b}} - \frac{3 \, b^{\frac{9}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 35 \, b^{\frac{9}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 330 \, b^{\frac{9}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{b^{5}} + \frac{330 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 35 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, \sqrt{b}}{b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5}}}{480 \, b f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^4)^(3/2),x, algorithm="giac")

[Out]

-1/480*(480*(f*x + e)/sqrt(b) - (3*b^(9/2)*tan(1/2*f*x + 1/2*e)^5 - 35*b^(9/2)*tan(1/2*f*x + 1/2*e)^3 + 330*b^
(9/2)*tan(1/2*f*x + 1/2*e))/b^5 + (330*sqrt(b)*tan(1/2*f*x + 1/2*e)^4 - 35*sqrt(b)*tan(1/2*f*x + 1/2*e)^2 + 3*
sqrt(b))/(b*tan(1/2*f*x + 1/2*e)^5))/(b*f)