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

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Rubi [A]  time = 0.02, antiderivative size = 44, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.02, 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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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (c x\right )^{m} \left (\frac {x^{2} + b}{x^{2}}\right )^{p}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c x\right )^{m} {\left (\frac {b}{x^{2}} + 1\right )}^{p}\,{d x} \]

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)

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int \left (c x \right )^{m} \left (\frac {b}{x^{2}}+1\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c x\right )^{m} {\left (\frac {b}{x^{2}} + 1\right )}^{p}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (c\,x\right )}^m\,{\left (\frac {b}{x^2}+1\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m*(b/x^2 + 1)^p,x)

[Out]

int((c*x)^m*(b/x^2 + 1)^p, x)

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sympy [C]  time = 17.75, size = 54, normalized size = 1.23 \[ - \frac {c^{m} x x^{m} \Gamma \left (- \frac {m}{2} - \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - \frac {m}{2} - \frac {1}{2} \\ \frac {1}{2} - \frac {m}{2} \end {matrix}\middle | {\frac {b e^{i \pi }}{x^{2}}} \right )}}{2 \Gamma \left (\frac {1}{2} - \frac {m}{2}\right )} \]

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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