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

Optimal. Leaf size=97 \[ -\frac {\cot ^3(e+f x)}{4 b^2 f \sqrt {b \tan ^2(e+f x)}}+\frac {\cot (e+f x)}{2 b^2 f \sqrt {b \tan ^2(e+f x)}}+\frac {\tan (e+f x) \log (\sin (e+f x))}{b^2 f \sqrt {b \tan ^2(e+f x)}} \]

[Out]

1/2*cot(f*x+e)/b^2/f/(b*tan(f*x+e)^2)^(1/2)-1/4*cot(f*x+e)^3/b^2/f/(b*tan(f*x+e)^2)^(1/2)+ln(sin(f*x+e))*tan(f
*x+e)/b^2/f/(b*tan(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3658, 3473, 3475} \[ -\frac {\cot ^3(e+f x)}{4 b^2 f \sqrt {b \tan ^2(e+f x)}}+\frac {\cot (e+f x)}{2 b^2 f \sqrt {b \tan ^2(e+f x)}}+\frac {\tan (e+f x) \log (\sin (e+f x))}{b^2 f \sqrt {b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^2)^(-5/2),x]

[Out]

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

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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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
]])

Rubi steps

\begin {align*} \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\tan (e+f x) \int \cot ^5(e+f x) \, dx}{b^2 \sqrt {b \tan ^2(e+f x)}}\\ &=-\frac {\cot ^3(e+f x)}{4 b^2 f \sqrt {b \tan ^2(e+f x)}}-\frac {\tan (e+f x) \int \cot ^3(e+f x) \, dx}{b^2 \sqrt {b \tan ^2(e+f x)}}\\ &=\frac {\cot (e+f x)}{2 b^2 f \sqrt {b \tan ^2(e+f x)}}-\frac {\cot ^3(e+f x)}{4 b^2 f \sqrt {b \tan ^2(e+f x)}}+\frac {\tan (e+f x) \int \cot (e+f x) \, dx}{b^2 \sqrt {b \tan ^2(e+f x)}}\\ &=\frac {\cot (e+f x)}{2 b^2 f \sqrt {b \tan ^2(e+f x)}}-\frac {\cot ^3(e+f x)}{4 b^2 f \sqrt {b \tan ^2(e+f x)}}+\frac {\log (\sin (e+f x)) \tan (e+f x)}{b^2 f \sqrt {b \tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 68, normalized size = 0.70 \[ \frac {\tan ^5(e+f x) \left (-\cot ^4(e+f x)+2 \cot ^2(e+f x)+4 \log (\tan (e+f x))+4 \log (\cos (e+f x))\right )}{4 f \left (b \tan ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 82, normalized size = 0.85 \[ \frac {{\left (2 \, \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} - 1\right )} \sqrt {b \tan \left (f x + e\right )^{2}}}{4 \, b^{3} f \tan \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f/sqrt(b)/b^2/2*
(1/1024*(192*tan((f*x+exp(1))/2)^2*sign(tan((f*x+exp(1))/2))*sign(-tan((f*x+exp(1))/2)^2+1)-16*tan((f*x+exp(1)
)/2)^4*sign(tan((f*x+exp(1))/2))*sign(-tan((f*x+exp(1))/2)^2+1))+1/64*(12*tan((f*x+exp(1))/2)^2*sign(-tan((f*x
+exp(1))/2)^2+1)-48*tan((f*x+exp(1))/2)^4*sign(-tan((f*x+exp(1))/2)^2+1)-sign(-tan((f*x+exp(1))/2)^2+1))/tan((
f*x+exp(1))/2)^4/sign(tan((f*x+exp(1))/2))+1/2*sign(-tan((f*x+exp(1))/2)^2+1)*ln(tan((f*x+exp(1))/2)^2)/sign(t
an((f*x+exp(1))/2))-sign(-tan((f*x+exp(1))/2)^2+1)*ln(tan((f*x+exp(1))/2)^2+1)/sign(tan((f*x+exp(1))/2)))

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maple [A]  time = 0.35, size = 74, normalized size = 0.76 \[ \frac {\tan \left (f x +e \right ) \left (4 \ln \left (\tan \left (f x +e \right )\right ) \left (\tan ^{4}\left (f x +e \right )\right )-2 \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (\tan ^{4}\left (f x +e \right )\right )+2 \left (\tan ^{2}\left (f x +e \right )\right )-1\right )}{4 f \left (b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.21, size = 66, normalized size = 0.68 \[ -\frac {\frac {2 \, \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{b^{\frac {5}{2}}} - \frac {4 \, \log \left (\tan \left (f x + e\right )\right )}{b^{\frac {5}{2}}} - \frac {2 \, \sqrt {b} \tan \left (f x + e\right )^{2} - \sqrt {b}}{b^{3} \tan \left (f x + e\right )^{4}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(e + f*x)^2)^(5/2),x)

[Out]

int(1/(b*tan(e + f*x)^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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