3.351 \(\int \frac {(c x)^{-3+m}}{a+b x^2} \, dx\)

Optimal. Leaf size=45 \[ -\frac {(c x)^{m-2} \, _2F_1\left (1,\frac {m-2}{2};\frac {m}{2};-\frac {b x^2}{a}\right )}{a c (2-m)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {364} \[ -\frac {(c x)^{m-2} \, _2F_1\left (1,\frac {m-2}{2};\frac {m}{2};-\frac {b x^2}{a}\right )}{a c (2-m)} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(-3 + m)/(a + b*x^2),x]

[Out]

-(((c*x)^(-2 + m)*Hypergeometric2F1[1, (-2 + m)/2, m/2, -((b*x^2)/a)])/(a*c*(2 - 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 \frac {(c x)^{-3+m}}{a+b x^2} \, dx &=-\frac {(c x)^{-2+m} \, _2F_1\left (1,\frac {1}{2} (-2+m);\frac {m}{2};-\frac {b x^2}{a}\right )}{a c (2-m)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 44, normalized size = 0.98 \[ \frac {x (c x)^{m-3} \, _2F_1\left (1,\frac {m-2}{2};\frac {m-2}{2}+1;-\frac {b x^2}{a}\right )}{a (m-2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(-3 + m)/(a + b*x^2),x]

[Out]

(x*(c*x)^(-3 + m)*Hypergeometric2F1[1, (-2 + m)/2, 1 + (-2 + m)/2, -((b*x^2)/a)])/(a*(-2 + m))

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (c x\right )^{m - 3}}{b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-3+m)/(b*x^2+a),x, algorithm="fricas")

[Out]

integral((c*x)^(m - 3)/(b*x^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{m - 3}}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-3+m)/(b*x^2+a),x, algorithm="giac")

[Out]

integrate((c*x)^(m - 3)/(b*x^2 + a), x)

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maple [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x \right )^{m -3}}{b \,x^{2}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(m-3)/(b*x^2+a),x)

[Out]

int((c*x)^(m-3)/(b*x^2+a),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{m - 3}}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(-3+m)/(b*x^2+a),x, algorithm="maxima")

[Out]

integrate((c*x)^(m - 3)/(b*x^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (c\,x\right )}^{m-3}}{b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(m - 3)/(a + b*x^2),x)

[Out]

int((c*x)^(m - 3)/(a + b*x^2), x)

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sympy [C]  time = 21.42, size = 92, normalized size = 2.04 \[ \frac {c^{m} m x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} - 1\right ) \Gamma \left (\frac {m}{2} - 1\right )}{4 a c^{3} x^{2} \Gamma \left (\frac {m}{2}\right )} - \frac {c^{m} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} - 1\right ) \Gamma \left (\frac {m}{2} - 1\right )}{2 a c^{3} x^{2} \Gamma \left (\frac {m}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(-3+m)/(b*x**2+a),x)

[Out]

c**m*m*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 - 1)*gamma(m/2 - 1)/(4*a*c**3*x**2*gamma(m/2)) - c**m*x*
*m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 - 1)*gamma(m/2 - 1)/(2*a*c**3*x**2*gamma(m/2))

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