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

Optimal. Leaf size=44 \[ \frac {(c x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a c (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {364} \[ \frac {(c x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a c (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^m/(a + b*x^2),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 42, normalized size = 0.95 \[ \frac {x (c x)^m \, _2F_1\left (1,\frac {m+1}{2};\frac {m+1}{2}+1;-\frac {b x^2}{a}\right )}{a (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^m/(a + b*x^2),x]

[Out]

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

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (c x\right )^{m}}{b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x)^m/(b*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m/(b*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m/(b*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m/(a + b*x^2),x)

[Out]

int((c*x)^m/(a + b*x^2), x)

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sympy [C]  time = 1.25, size = 95, normalized size = 2.16 \[ \frac {c^{m} m x x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {c^{m} x x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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