3.51 \(\int \frac{1}{(b \tan ^p(c+d x))^{5/2}} \, dx\)

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

[Out]

(2*Hypergeometric2F1[1, (2 - 5*p)/4, (6 - 5*p)/4, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 - 2*p))/(b^2*d*(2 - 5*p)*Sq
rt[b*Tan[c + d*x]^p])

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Rubi [A]  time = 0.0466852, antiderivative size = 71, normalized size of antiderivative = 1., 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} \[ \frac{2 \tan ^{1-2 p}(c+d x) \, _2F_1\left (1,\frac{1}{4} (2-5 p);\frac{1}{4} (6-5 p);-\tan ^2(c+d x)\right )}{b^2 d (2-5 p) \sqrt{b \tan ^p(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^p)^(-5/2),x]

[Out]

(2*Hypergeometric2F1[1, (2 - 5*p)/4, (6 - 5*p)/4, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 - 2*p))/(b^2*d*(2 - 5*p)*Sq
rt[b*Tan[c + d*x]^p])

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]])

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 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])

Rubi steps

\begin{align*} \int \frac{1}{\left (b \tan ^p(c+d x)\right )^{5/2}} \, dx &=\frac{\tan ^{\frac{p}{2}}(c+d x) \int \tan ^{-\frac{5 p}{2}}(c+d x) \, dx}{b^2 \sqrt{b \tan ^p(c+d x)}}\\ &=\frac{\tan ^{\frac{p}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{x^{-5 p/2}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{b^2 d \sqrt{b \tan ^p(c+d x)}}\\ &=\frac{2 \, _2F_1\left (1,\frac{1}{4} (2-5 p);\frac{1}{4} (6-5 p);-\tan ^2(c+d x)\right ) \tan ^{1-2 p}(c+d x)}{b^2 d (2-5 p) \sqrt{b \tan ^p(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0675633, size = 62, normalized size = 0.87 \[ -\frac{2 \tan (c+d x) \, _2F_1\left (1,\frac{1}{4} (2-5 p);\frac{1}{4} (6-5 p);-\tan ^2(c+d x)\right )}{d (5 p-2) \left (b \tan ^p(c+d x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x]^p)^(-5/2),x]

[Out]

(-2*Hypergeometric2F1[1, (2 - 5*p)/4, (6 - 5*p)/4, -Tan[c + d*x]^2]*Tan[c + d*x])/(d*(-2 + 5*p)*(b*Tan[c + d*x
]^p)^(5/2))

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Maple [F]  time = 0.117, size = 0, normalized size = 0. \begin{align*} \int \left ( b \left ( \tan \left ( dx+c \right ) \right ) ^{p} \right ) ^{-{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(d*x+c)^p)^(5/2),x)

[Out]

int(1/(b*tan(d*x+c)^p)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tan \left (d x + c\right )^{p}\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c)^p)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c)^p)^(-5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c)^p)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tan ^{p}{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c)**p)**(5/2),x)

[Out]

Integral((b*tan(c + d*x)**p)**(-5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tan \left (d x + c\right )^{p}\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c)^p)^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c)^p)^(-5/2), x)