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

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

[Out]

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

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Rubi [A]  time = 0.0427198, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {808, 364} \[ \frac{A (c x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a c (m+1)}+\frac{B (c x)^{m+2} \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\frac{b x^2}{a}\right )}{a c^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

Mathematica [A]  time = 0.0290382, size = 82, normalized size = 0.9 \[ \frac{x (c x)^m \left (A (m+2) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )+B (m+1) x \, _2F_1\left (1,\frac{m}{2}+1;\frac{m}{2}+2;-\frac{b x^2}{a}\right )\right )}{a (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.031, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( Bx+A \right ) \left ( cx \right ) ^{m}}{b{x}^{2}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 4.73584, size = 192, normalized size = 2.11 \begin{align*} \frac{A 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{A 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 )} + \frac{B c^{m} m x^{2} 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 \Gamma \left (\frac{m}{2} + 2\right )} + \frac{B c^{m} x^{2} 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 \Gamma \left (\frac{m}{2} + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*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)) + A*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)) + B*c**m*m
*x**2*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1)*gamma(m/2 + 1)/(4*a*gamma(m/2 + 2)) + B*c**m*x**2*x*
*m*lerchphi(b*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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )} \left (c x\right )^{m}}{b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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