3.1963 \(\int (1+\frac{b}{x^2})^p (c x)^m \, dx\)

Optimal. Leaf size=44 \[ \frac{(c x)^{m+1} \, _2F_1\left (\frac{1}{2} (-m-1),-p;\frac{1-m}{2};-\frac{b}{x^2}\right )}{c (m+1)} \]

[Out]

((c*x)^(1 + m)*Hypergeometric2F1[(-1 - m)/2, -p, (1 - m)/2, -(b/x^2)])/(c*(1 + m))

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Rubi [A]  time = 0.0169594, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {339, 364} \[ \frac{(c x)^{m+1} \, _2F_1\left (\frac{1}{2} (-m-1),-p;\frac{1-m}{2};-\frac{b}{x^2}\right )}{c (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(1 + b/x^2)^p*(c*x)^m,x]

[Out]

((c*x)^(1 + m)*Hypergeometric2F1[(-1 - m)/2, -p, (1 - m)/2, -(b/x^2)])/(c*(1 + m))

Rule 339

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Dist[((c*x)^(m + 1)*(1/x)^(m + 1))/c, Subst
[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]

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 \left (1+\frac{b}{x^2}\right )^p (c x)^m \, dx &=-\frac{\left (\left (\frac{1}{x}\right )^{1+m} (c x)^{1+m}\right ) \operatorname{Subst}\left (\int x^{-2-m} \left (1+b x^2\right )^p \, dx,x,\frac{1}{x}\right )}{c}\\ &=\frac{(c x)^{1+m} \, _2F_1\left (\frac{1}{2} (-1-m),-p;\frac{1-m}{2};-\frac{b}{x^2}\right )}{c (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0171649, size = 71, normalized size = 1.61 \[ \frac{x \left (\frac{b}{x^2}+1\right )^p \left (\frac{x^2}{b}+1\right )^{-p} (c x)^m \, _2F_1\left (\frac{1}{2} (m-2 p+1),-p;\frac{1}{2} (m-2 p+1)+1;-\frac{x^2}{b}\right )}{m-2 p+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + b/x^2)^p*(c*x)^m,x]

[Out]

((1 + b/x^2)^p*x*(c*x)^m*Hypergeometric2F1[(1 + m - 2*p)/2, -p, 1 + (1 + m - 2*p)/2, -(x^2/b)])/((1 + m - 2*p)
*(1 + x^2/b)^p)

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Maple [F]  time = 0.101, size = 0, normalized size = 0. \begin{align*} \int \left ( 1+{\frac{b}{{x}^{2}}} \right ) ^{p} \left ( cx \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+1/x^2*b)^p*(c*x)^m,x)

[Out]

int((1+1/x^2*b)^p*(c*x)^m,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m*(b/x^2 + 1)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (c x\right )^{m} \left (\frac{x^{2} + b}{x^{2}}\right )^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x)^m*((x^2 + b)/x^2)^p, x)

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Sympy [C]  time = 20.4743, size = 54, normalized size = 1.23 \begin{align*} - \frac{c^{m} x x^{m} \Gamma \left (- \frac{m}{2} - \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - \frac{m}{2} - \frac{1}{2} \\ \frac{1}{2} - \frac{m}{2} \end{matrix}\middle |{\frac{b e^{i \pi }}{x^{2}}} \right )}}{2 \Gamma \left (\frac{1}{2} - \frac{m}{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+b/x**2)**p*(c*x)**m,x)

[Out]

-c**m*x*x**m*gamma(-m/2 - 1/2)*hyper((-p, -m/2 - 1/2), (1/2 - m/2,), b*exp_polar(I*pi)/x**2)/(2*gamma(1/2 - m/
2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m*(b/x^2 + 1)^p, x)