3.1.16 \(\int \frac {1}{\sqrt {b \tan ^4(e+f x)}} \, dx\) [16]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 51, 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 = {3739, 3554, 8} \begin {gather*} -\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 {gather*}

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3554

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 3739

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}{\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] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.04, size = 43, normalized size = 0.84 \begin {gather*} -\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)}{f \sqrt {b \tan ^4(e+f x)}} \end {gather*}

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]))

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 40, normalized size = 0.78

method	result	size
derivativedivides	\(-\frac {\tan \left (f x +e \right ) \left (\arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )+1\right )}{f \sqrt {b \left (\tan ^{4}\left (f x +e \right )\right )}}\)	\(40\)
default	\(-\frac {\tan \left (f x +e \right ) \left (\arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )+1\right )}{f \sqrt {b \left (\tan ^{4}\left (f x +e \right )\right )}}\)	\(40\)
risch	\(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} x}{\sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {2 i \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{\sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} f}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 29, normalized size = 0.57 \begin {gather*} -\frac {\frac {f x + e}{\sqrt {b}} + \frac {1}{\sqrt {b} \tan \left (f x + e\right )}}{f} \end {gather*}

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

________________________________________________________________________________________

Fricas [A]
time = 2.03, size = 42, normalized size = 0.82 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b \tan ^{4}{\left (e + f x \right )}}}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [A]
time = 0.56, size = 48, normalized size = 0.94 \begin {gather*} -\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 {gather*}

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(e + f*x)^4)^(1/2),x)

[Out]

int(1/(b*tan(e + f*x)^4)^(1/2), x)

________________________________________________________________________________________