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

Optimal. Leaf size=66 \[ -\frac {\cot (e+f x)}{2 b f \sqrt {b \tan ^2(e+f x)}}-\frac {\tan (e+f x) \log (\sin (e+f x))}{b f \sqrt {b \tan ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, 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, 3475} \[ -\frac {\cot (e+f x)}{2 b f \sqrt {b \tan ^2(e+f x)}}-\frac {\tan (e+f x) \log (\sin (e+f x))}{b f \sqrt {b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^2)^(-3/2),x]

[Out]

-Cot[e + f*x]/(2*b*f*Sqrt[b*Tan[e + f*x]^2]) - (Log[Sin[e + f*x]]*Tan[e + f*x])/(b*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 )^{3/2}} \, dx &=\frac {\tan (e+f x) \int \cot ^3(e+f x) \, dx}{b \sqrt {b \tan ^2(e+f x)}}\\ &=-\frac {\cot (e+f x)}{2 b f \sqrt {b \tan ^2(e+f x)}}-\frac {\tan (e+f x) \int \cot (e+f x) \, dx}{b \sqrt {b \tan ^2(e+f x)}}\\ &=-\frac {\cot (e+f x)}{2 b f \sqrt {b \tan ^2(e+f x)}}-\frac {\log (\sin (e+f x)) \tan (e+f x)}{b f \sqrt {b \tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 56, normalized size = 0.85 \[ -\frac {\tan ^3(e+f x) \left (\cot ^2(e+f x)+2 \log (\tan (e+f x))+2 \log (\cos (e+f x))\right )}{2 f \left (b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((Cot[e + f*x]^2 + 2*Log[Cos[e + f*x]] + 2*Log[Tan[e + f*x]])*Tan[e + f*x]^3)/(f*(b*Tan[e + f*x]^2)^(3/2)
)

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fricas [A]  time = 0.41, size = 69, normalized size = 1.05 \[ -\frac {\sqrt {b \tan \left (f x + e\right )^{2}} {\left (\log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )^{2} + 1\right )}}{2 \, b^{2} f \tan \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(3/2),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)2/b/f/sqrt(b)/2*(1/8*(4*tan((f*x+exp(1))
/2)^2*sign(-tan((f*x+exp(1))/2)^2+1)-sign(-tan((f*x+exp(1))/2)^2+1))/tan((f*x+exp(1))/2)^2/sign(tan((f*x+exp(1
))/2))-1/8*tan((f*x+exp(1))/2)^2*sign(-tan((f*x+exp(1))/2)^2+1)/sign(tan((f*x+exp(1))/2))-1/2*sign(-tan((f*x+e
xp(1))/2)^2+1)*ln(tan((f*x+exp(1))/2)^2)/sign(tan((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 = 0.31, size = 64, normalized size = 0.97 \[ -\frac {\tan \left (f x +e \right ) \left (2 \ln \left (\tan \left (f x +e \right )\right ) \left (\tan ^{2}\left (f x +e \right )\right )-\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (\tan ^{2}\left (f x +e \right )\right )+1\right )}{2 f \left (b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.99, size = 46, normalized size = 0.70 \[ \frac {\frac {\log \left (\tan \left (f x + e\right )^{2} + 1\right )}{b^{\frac {3}{2}}} - \frac {2 \, \log \left (\tan \left (f x + e\right )\right )}{b^{\frac {3}{2}}} - \frac {1}{b^{\frac {3}{2}} \tan \left (f x + e\right )^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(e + f*x)^2)^(3/2),x)

[Out]

int(1/(b*tan(e + f*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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