3.18 \(\int \frac{1}{(b \tan ^4(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=183 \[ -\frac{x \tan ^2(e+f x)}{b^2 \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}} \]

[Out]

Cot[e + f*x]/(3*b^2*f*Sqrt[b*Tan[e + f*x]^4]) - Cot[e + f*x]^3/(5*b^2*f*Sqrt[b*Tan[e + f*x]^4]) + Cot[e + f*x]
^5/(7*b^2*f*Sqrt[b*Tan[e + f*x]^4]) - Cot[e + f*x]^7/(9*b^2*f*Sqrt[b*Tan[e + f*x]^4]) - Tan[e + f*x]/(b^2*f*Sq
rt[b*Tan[e + f*x]^4]) - (x*Tan[e + f*x]^2)/(b^2*Sqrt[b*Tan[e + f*x]^4])

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Rubi [A]  time = 0.0646674, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 8} \[ -\frac{x \tan ^2(e+f x)}{b^2 \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^4)^(-5/2),x]

[Out]

Cot[e + f*x]/(3*b^2*f*Sqrt[b*Tan[e + f*x]^4]) - Cot[e + f*x]^3/(5*b^2*f*Sqrt[b*Tan[e + f*x]^4]) + Cot[e + f*x]
^5/(7*b^2*f*Sqrt[b*Tan[e + f*x]^4]) - Cot[e + f*x]^7/(9*b^2*f*Sqrt[b*Tan[e + f*x]^4]) - Tan[e + f*x]/(b^2*f*Sq
rt[b*Tan[e + f*x]^4]) - (x*Tan[e + f*x]^2)/(b^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}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx &=\frac{\tan ^2(e+f x) \int \cot ^{10}(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan ^2(e+f x) \int \cot ^8(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\tan ^2(e+f x) \int \cot ^6(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan ^2(e+f x) \int \cot ^4(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\tan ^2(e+f x) \int \cot ^2(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan ^2(e+f x) \int 1 \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{x \tan ^2(e+f x)}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.0326329, size = 45, normalized size = 0.25 \[ -\frac{\tan (e+f x) \text{Hypergeometric2F1}\left (-\frac{9}{2},1,-\frac{7}{2},-\tan ^2(e+f x)\right )}{9 f \left (b \tan ^4(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Hypergeometric2F1[-9/2, 1, -7/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(9*f*(b*Tan[e + f*x]^4)^(5/2))

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Maple [A]  time = 0.021, size = 83, normalized size = 0.5 \begin{align*} -{\frac{\tan \left ( fx+e \right ) \left ( 315\,\arctan \left ( \tan \left ( fx+e \right ) \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{9}+315\, \left ( \tan \left ( fx+e \right ) \right ) ^{8}-105\, \left ( \tan \left ( fx+e \right ) \right ) ^{6}+63\, \left ( \tan \left ( fx+e \right ) \right ) ^{4}-45\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}+35 \right ) }{315\,f} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{4} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(f*x+e)^4)^(5/2),x)

[Out]

-1/315/f*tan(f*x+e)*(315*arctan(tan(f*x+e))*tan(f*x+e)^9+315*tan(f*x+e)^8-105*tan(f*x+e)^6+63*tan(f*x+e)^4-45*
tan(f*x+e)^2+35)/(b*tan(f*x+e)^4)^(5/2)

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Maxima [A]  time = 1.6532, size = 95, normalized size = 0.52 \begin{align*} -\frac{\frac{315 \,{\left (f x + e\right )}}{b^{\frac{5}{2}}} + \frac{315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35}{b^{\frac{5}{2}} \tan \left (f x + e\right )^{9}}}{315 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^4)^(5/2),x, algorithm="maxima")

[Out]

-1/315*(315*(f*x + e)/b^(5/2) + (315*tan(f*x + e)^8 - 105*tan(f*x + e)^6 + 63*tan(f*x + e)^4 - 45*tan(f*x + e)
^2 + 35)/(b^(5/2)*tan(f*x + e)^9))/f

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Fricas [A]  time = 1.92304, size = 225, normalized size = 1.23 \begin{align*} -\frac{{\left (315 \, f x \tan \left (f x + e\right )^{9} + 315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35\right )} \sqrt{b \tan \left (f x + e\right )^{4}}}{315 \, b^{3} f \tan \left (f x + e\right )^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^4)^(5/2),x, algorithm="fricas")

[Out]

-1/315*(315*f*x*tan(f*x + e)^9 + 315*tan(f*x + e)^8 - 105*tan(f*x + e)^6 + 63*tan(f*x + e)^4 - 45*tan(f*x + e)
^2 + 35)*sqrt(b*tan(f*x + e)^4)/(b^3*f*tan(f*x + e)^11)

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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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)**4)**(5/2),x)

[Out]

Integral((b*tan(e + f*x)**4)**(-5/2), x)

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Giac [A]  time = 2.79148, size = 265, normalized size = 1.45 \begin{align*} -\frac{\frac{161280 \,{\left (f x + e\right )}}{b^{\frac{5}{2}}} + \frac{121590 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 18480 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 3528 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 495 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 35 \, \sqrt{b}}{b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9}} - \frac{35 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} - 495 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 3528 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 18480 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 121590 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{b^{27}}}{161280 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^4)^(5/2),x, algorithm="giac")

[Out]

-1/161280*(161280*(f*x + e)/b^(5/2) + (121590*sqrt(b)*tan(1/2*f*x + 1/2*e)^8 - 18480*sqrt(b)*tan(1/2*f*x + 1/2
*e)^6 + 3528*sqrt(b)*tan(1/2*f*x + 1/2*e)^4 - 495*sqrt(b)*tan(1/2*f*x + 1/2*e)^2 + 35*sqrt(b))/(b^3*tan(1/2*f*
x + 1/2*e)^9) - (35*b^(49/2)*tan(1/2*f*x + 1/2*e)^9 - 495*b^(49/2)*tan(1/2*f*x + 1/2*e)^7 + 3528*b^(49/2)*tan(
1/2*f*x + 1/2*e)^5 - 18480*b^(49/2)*tan(1/2*f*x + 1/2*e)^3 + 121590*b^(49/2)*tan(1/2*f*x + 1/2*e))/b^27)/f