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

Optimal. Leaf size=71 \[ \frac {2 \tan ^{1-n}(e+f x) \, _2F_1\left (1,\frac {1}{4} (2-3 n);\frac {3 (2-n)}{4};-\tan ^2(e+f x)\right )}{b f (2-3 n) \sqrt {b \tan ^n(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 71, 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 = {3659, 3476, 364} \[ \frac {2 \tan ^{1-n}(e+f x) \, _2F_1\left (1,\frac {1}{4} (2-3 n);\frac {3 (2-n)}{4};-\tan ^2(e+f x)\right )}{b f (2-3 n) \sqrt {b \tan ^n(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^n)^(-3/2),x]

[Out]

(2*Hypergeometric2F1[1, (2 - 3*n)/4, (3*(2 - n))/4, -Tan[e + f*x]^2]*Tan[e + f*x]^(1 - n))/(b*f*(2 - 3*n)*Sqrt
[b*Tan[e + f*x]^n])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 3476

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 3659

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}{\left (b \tan ^n(e+f x)\right )^{3/2}} \, dx &=\frac {\tan ^{\frac {n}{2}}(e+f x) \int \tan ^{-\frac {3 n}{2}}(e+f x) \, dx}{b \sqrt {b \tan ^n(e+f x)}}\\ &=\frac {\tan ^{\frac {n}{2}}(e+f x) \operatorname {Subst}\left (\int \frac {x^{-3 n/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{b f \sqrt {b \tan ^n(e+f x)}}\\ &=\frac {2 \, _2F_1\left (1,\frac {1}{4} (2-3 n);\frac {3 (2-n)}{4};-\tan ^2(e+f x)\right ) \tan ^{1-n}(e+f x)}{b f (2-3 n) \sqrt {b \tan ^n(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 60, normalized size = 0.85 \[ -\frac {2 \tan (e+f x) \, _2F_1\left (1,\frac {1}{4} (2-3 n);-\frac {3}{4} (n-2);-\tan ^2(e+f x)\right )}{f (3 n-2) \left (b \tan ^n(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Hypergeometric2F1[1, (2 - 3*n)/4, (-3*(-2 + n))/4, -Tan[e + f*x]^2]*Tan[e + f*x])/(f*(-2 + 3*n)*(b*Tan[e +
f*x]^n)^(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e)^n)^(-3/2), x)

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maple [F]  time = 1.23, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \left (\tan ^{n}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e)^n)^(-3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^n\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(e + f*x)^n)^(3/2),x)

[Out]

int(1/(b*tan(e + f*x)^n)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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