3.1.47 \(\int (b \tan ^p(c+d x))^{3/2} \, dx\) [47]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {3740, 3557, 371} \begin {gather*} \frac {2 b \tan ^{p+1}(c+d x) \sqrt {b \tan ^p(c+d x)} \, _2F_1\left (1,\frac {1}{4} (3 p+2);\frac {3 (p+2)}{4};-\tan ^2(c+d x)\right )}{d (3 p+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^p)^(3/2),x]

[Out]

(2*b*Hypergeometric2F1[1, (2 + 3*p)/4, (3*(2 + p))/4, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + p)*Sqrt[b*Tan[c + d*x
]^p])/(d*(2 + 3*p))

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 ^p(c+d x)\right )^{3/2} \, dx &=\left (b \tan ^{-\frac {p}{2}}(c+d x) \sqrt {b \tan ^p(c+d x)}\right ) \int \tan ^{\frac {3 p}{2}}(c+d x) \, dx\\ &=\frac {\left (b \tan ^{-\frac {p}{2}}(c+d x) \sqrt {b \tan ^p(c+d x)}\right ) \text {Subst}\left (\int \frac {x^{3 p/2}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {2 b \, _2F_1\left (1,\frac {1}{4} (2+3 p);\frac {3 (2+p)}{4};-\tan ^2(c+d x)\right ) \tan ^{1+p}(c+d x) \sqrt {b \tan ^p(c+d x)}}{d (2+3 p)}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 60, normalized size = 0.92 \begin {gather*} \frac {2 \, _2F_1\left (1,\frac {1}{4} (2+3 p);\frac {3 (2+p)}{4};-\tan ^2(c+d x)\right ) \tan (c+d x) \left (b \tan ^p(c+d x)\right )^{3/2}}{d (2+3 p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x]^p)^(3/2),x]

[Out]

(2*Hypergeometric2F1[1, (2 + 3*p)/4, (3*(2 + p))/4, -Tan[c + d*x]^2]*Tan[c + d*x]*(b*Tan[c + d*x]^p)^(3/2))/(d
*(2 + 3*p))

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Maple [F]
time = 0.22, size = 0, normalized size = 0.00 \[\int \left (b \left (\tan ^{p}\left (d x +c \right )\right )\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(d*x+c)^p)^(3/2),x)

[Out]

int((b*tan(d*x+c)^p)^(3/2),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(d*x+c)^p)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c)^p)^(3/2), 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(d*x+c)^p)^(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \tan ^{p}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(d*x+c)**p)**(3/2),x)

[Out]

Integral((b*tan(c + d*x)**p)**(3/2), 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(d*x+c)^p)^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c)^p)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^p\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(c + d*x)^p)^(3/2),x)

[Out]

int((b*tan(c + d*x)^p)^(3/2), x)

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