3.1.22 \(\int \frac {1}{\sqrt {b \tan ^n(e+f x)}} \, dx\) [22]

Optimal. Leaf size=62 \[ \frac {2 \, _2F_1\left (1,\frac {2-n}{4};\frac {6-n}{4};-\tan ^2(e+f x)\right ) \tan (e+f x)}{f (2-n) \sqrt {b \tan ^n(e+f x)}} \]

[Out]

2*hypergeom([1, 1/2-1/4*n],[3/2-1/4*n],-tan(f*x+e)^2)*tan(f*x+e)/f/(2-n)/(b*tan(f*x+e)^n)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 62, 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 = {3740, 3557, 371} \begin {gather*} \frac {2 \tan (e+f x) \, _2F_1\left (1,\frac {2-n}{4};\frac {6-n}{4};-\tan ^2(e+f x)\right )}{f (2-n) \sqrt {b \tan ^n(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Tan[e + f*x]^n],x]

[Out]

(2*Hypergeometric2F1[1, (2 - n)/4, (6 - n)/4, -Tan[e + f*x]^2]*Tan[e + f*x])/(f*(2 - n)*Sqrt[b*Tan[e + f*x]^n]
)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3740

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Tan[e + f*x
])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*FracPart[p])), Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, 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 ^n(e+f x)}} \, dx &=\frac {\tan ^{\frac {n}{2}}(e+f x) \int \tan ^{-\frac {n}{2}}(e+f x) \, dx}{\sqrt {b \tan ^n(e+f x)}}\\ &=\frac {\tan ^{\frac {n}{2}}(e+f x) \text {Subst}\left (\int \frac {x^{-n/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f \sqrt {b \tan ^n(e+f x)}}\\ &=\frac {2 \, _2F_1\left (1,\frac {2-n}{4};\frac {6-n}{4};-\tan ^2(e+f x)\right ) \tan (e+f x)}{f (2-n) \sqrt {b \tan ^n(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 60, normalized size = 0.97 \begin {gather*} -\frac {2 \, _2F_1\left (1,\frac {2-n}{4};\frac {6-n}{4};-\tan ^2(e+f x)\right ) \tan (e+f x)}{f (-2+n) \sqrt {b \tan ^n(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Tan[e + f*x]^n],x]

[Out]

(-2*Hypergeometric2F1[1, (2 - n)/4, (6 - n)/4, -Tan[e + f*x]^2]*Tan[e + f*x])/(f*(-2 + n)*Sqrt[b*Tan[e + f*x]^
n])

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {1}{\sqrt {b \left (\tan ^{n}\left (f x +e \right )\right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*tan(f*x + e)^n), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b \tan ^{n}{\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)**n)**(1/2),x)

[Out]

Integral(1/sqrt(b*tan(e + f*x)**n), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*tan(f*x + e)^n), x)

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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 )}^n}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(e + f*x)^n)^(1/2),x)

[Out]

int(1/(b*tan(e + f*x)^n)^(1/2), x)

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