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

Optimal. Leaf size=65 \[ \frac {2 b \tan ^{n+1}(e+f x) \sqrt {b \tan ^n(e+f x)} \, _2F_1\left (1,\frac {1}{4} (3 n+2);\frac {3 (n+2)}{4};-\tan ^2(e+f x)\right )}{f (3 n+2)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 65, 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} \[ \frac {2 b \tan ^{n+1}(e+f x) \sqrt {b \tan ^n(e+f x)} \, _2F_1\left (1,\frac {1}{4} (3 n+2);\frac {3 (n+2)}{4};-\tan ^2(e+f x)\right )}{f (3 n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^n)^(3/2),x]

[Out]

(2*b*Hypergeometric2F1[1, (2 + 3*n)/4, (3*(2 + n))/4, -Tan[e + f*x]^2]*Tan[e + f*x]^(1 + n)*Sqrt[b*Tan[e + f*x
]^n])/(f*(2 + 3*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 \left (b \tan ^n(e+f x)\right )^{3/2} \, dx &=\left (b \tan ^{-\frac {n}{2}}(e+f x) \sqrt {b \tan ^n(e+f x)}\right ) \int \tan ^{\frac {3 n}{2}}(e+f x) \, dx\\ &=\frac {\left (b \tan ^{-\frac {n}{2}}(e+f x) \sqrt {b \tan ^n(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {x^{3 n/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 b \, _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) \sqrt {b \tan ^n(e+f x)}}{f (2+3 n)}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 60, normalized size = 0.92 \[ \frac {2 \tan (e+f x) \left (b \tan ^n(e+f x)\right )^{3/2} \, _2F_1\left (1,\frac {1}{4} (3 n+2);\frac {3 (n+2)}{4};-\tan ^2(e+f x)\right )}{f (3 n+2)} \]

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

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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((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 \[ \int \left (b \tan \left (f x + e\right )^{n}\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 1.22, size = 0, normalized size = 0.00 \[ \int \left (b \left (\tan ^{n}\left (f x +e \right )\right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = -0.02 \[ \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^n\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x)^n)^(3/2),x)

[Out]

int((b*tan(e + f*x)^n)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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