3.489 \(\int \frac {(e x)^m (A+B x)}{a+c x^2} \, dx\)

Optimal. Leaf size=91 \[ \frac {A (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {c x^2}{a}\right )}{a e (m+1)}+\frac {B (e x)^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\frac {c x^2}{a}\right )}{a e^2 (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {808, 364} \[ \frac {A (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {c x^2}{a}\right )}{a e (m+1)}+\frac {B (e x)^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\frac {c x^2}{a}\right )}{a e^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^m*(A + B*x))/(a + c*x^2),x]

[Out]

(A*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((c*x^2)/a)])/(a*e*(1 + m)) + (B*(e*x)^(2 + m)*Hy
pergeometric2F1[1, (2 + m)/2, (4 + m)/2, -((c*x^2)/a)])/(a*e^2*(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])

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {(e x)^m (A+B x)}{a+c x^2} \, dx &=A \int \frac {(e x)^m}{a+c x^2} \, dx+\frac {B \int \frac {(e x)^{1+m}}{a+c x^2} \, dx}{e}\\ &=\frac {A (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {c x^2}{a}\right )}{a e (1+m)}+\frac {B (e x)^{2+m} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-\frac {c x^2}{a}\right )}{a e^2 (2+m)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 82, normalized size = 0.90 \[ \frac {x (e x)^m \left (A (m+2) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {c x^2}{a}\right )+B (m+1) x \, _2F_1\left (1,\frac {m}{2}+1;\frac {m}{2}+2;-\frac {c x^2}{a}\right )\right )}{a (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^m*(A + B*x))/(a + c*x^2),x]

[Out]

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

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fricas [F]  time = 1.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x + A\right )} \left (e x\right )^{m}}{c x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x + A)*(e*x)^m/(c*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(e*x)^m/(c*x^2 + a), x)

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maple [F]  time = 0.71, size = 0, normalized size = 0.00 \[ \int \frac {\left (B x +A \right ) \left (e x \right )^{m}}{c \,x^{2}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(B*x+A)/(c*x^2+a),x)

[Out]

int((e*x)^m*(B*x+A)/(c*x^2+a),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(e*x)^m/(c*x^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x\right )}{c\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^m*(A + B*x))/(a + c*x^2),x)

[Out]

int(((e*x)^m*(A + B*x))/(a + c*x^2), x)

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sympy [C]  time = 4.21, size = 192, normalized size = 2.11 \[ \frac {A e^{m} m x x^{m} \Phi \left (\frac {c x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A e^{m} x x^{m} \Phi \left (\frac {c x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B e^{m} m x^{2} x^{m} \Phi \left (\frac {c x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{4 a \Gamma \left (\frac {m}{2} + 2\right )} + \frac {B e^{m} x^{2} x^{m} \Phi \left (\frac {c x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{2 a \Gamma \left (\frac {m}{2} + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(B*x+A)/(c*x**2+a),x)

[Out]

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

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