3.1.19 \(\int (b \tan ^n(e+f x))^{5/2} \, dx\) [19]

Optimal. Leaf size=71 \[ \frac {2 b^2 \, _2F_1\left (1,\frac {1}{4} (2+5 n);\frac {1}{4} (6+5 n);-\tan ^2(e+f x)\right ) \tan ^{1+2 n}(e+f x) \sqrt {b \tan ^n(e+f x)}}{f (2+5 n)} \]

[Out]

2*b^2*hypergeom([1, 1/2+5/4*n],[3/2+5/4*n],-tan(f*x+e)^2)*(b*tan(f*x+e)^n)^(1/2)*tan(f*x+e)^(1+2*n)/f/(2+5*n)

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Rubi [A]
time = 0.04, 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 = {3740, 3557, 371} \begin {gather*} \frac {2 b^2 \tan ^{2 n+1}(e+f x) \sqrt {b \tan ^n(e+f x)} \, _2F_1\left (1,\frac {1}{4} (5 n+2);\frac {1}{4} (5 n+6);-\tan ^2(e+f x)\right )}{f (5 n+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^n)^(5/2),x]

[Out]

(2*b^2*Hypergeometric2F1[1, (2 + 5*n)/4, (6 + 5*n)/4, -Tan[e + f*x]^2]*Tan[e + f*x]^(1 + 2*n)*Sqrt[b*Tan[e + f
*x]^n])/(f*(2 + 5*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 \left (b \tan ^n(e+f x)\right )^{5/2} \, dx &=\left (b^2 \tan ^{-\frac {n}{2}}(e+f x) \sqrt {b \tan ^n(e+f x)}\right ) \int \tan ^{\frac {5 n}{2}}(e+f x) \, dx\\ &=\frac {\left (b^2 \tan ^{-\frac {n}{2}}(e+f x) \sqrt {b \tan ^n(e+f x)}\right ) \text {Subst}\left (\int \frac {x^{5 n/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 b^2 \, _2F_1\left (1,\frac {1}{4} (2+5 n);\frac {1}{4} (6+5 n);-\tan ^2(e+f x)\right ) \tan ^{1+2 n}(e+f x) \sqrt {b \tan ^n(e+f x)}}{f (2+5 n)}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 62, normalized size = 0.87 \begin {gather*} \frac {2 \, _2F_1\left (1,\frac {1}{4} (2+5 n);\frac {1}{4} (6+5 n);-\tan ^2(e+f x)\right ) \tan (e+f x) \left (b \tan ^n(e+f x)\right )^{5/2}}{f (2+5 n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^n)^(5/2),x]

[Out]

(2*Hypergeometric2F1[1, (2 + 5*n)/4, (6 + 5*n)/4, -Tan[e + f*x]^2]*Tan[e + f*x]*(b*Tan[e + f*x]^n)^(5/2))/(f*(
2 + 5*n))

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \left (b \left (\tan ^{n}\left (f x +e \right )\right )\right )^{\frac {5}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^n)^(5/2),x)

[Out]

int((b*tan(f*x+e)^n)^(5/2),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((b*tan(f*x+e)^n)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^n)^(5/2), 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((b*tan(f*x+e)^n)^(5/2),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 \left (b \tan ^{n}{\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**n)**(5/2),x)

[Out]

Integral((b*tan(e + f*x)**n)**(5/2), 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((b*tan(f*x+e)^n)^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^n)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^n\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x)^n)^(5/2),x)

[Out]

int((b*tan(e + f*x)^n)^(5/2), x)

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