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

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

[Out]

2*b^2*hypergeom([1, 1/2+5/4*p],[3/2+5/4*p],-tan(d*x+c)^2)*(b*tan(d*x+c)^p)^(1/2)*tan(d*x+c)^(1+2*p)/d/(2+5*p)

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

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^p)^(5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.11, size = 62, normalized size = 0.87 \[ \frac {2 \tan (c+d x) \left (b \tan ^p(c+d x)\right )^{5/2} \, _2F_1\left (1,\frac {1}{4} (5 p+2);\frac {1}{4} (5 p+6);-\tan ^2(c+d x)\right )}{d (5 p+2)} \]

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

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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(d*x+c)^p)^(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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 3.41, size = 0, normalized size = 0.00 \[ \int \left (b \left (\tan ^{p}\left (d x +c \right )\right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan \left (d x + c\right )^{p}\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^p\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(c + d*x)^p)^(5/2),x)

[Out]

int((b*tan(c + d*x)^p)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan ^{p}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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