\(\int \frac {(c x)^m (A+B x)}{a+b x^2} \, dx\) [59]

Optimal result

Integrand size = 20, antiderivative size = 91 \[ \int \frac {(c x)^m (A+B x)}{a+b x^2} \, dx=\frac {A (c x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a c (1+m)}+\frac {B (c x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a c^2 (2+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {822, 371} \[ \int \frac {(c x)^m (A+B x)}{a+b x^2} \, dx=\frac {A (c x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a c (m+1)}+\frac {B (c x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {b x^2}{a}\right )}{a c^2 (m+2)} \]

[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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 822

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*} \text {integral}& = 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] (verified)

Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int \frac {(c x)^m (A+B x)}{a+b x^2} \, dx=\frac {x (c x)^m \left (B (1+m) x \operatorname {Hypergeometric2F1}\left (1,1+\frac {m}{2},2+\frac {m}{2},-\frac {b x^2}{a}\right )+A (2+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )}{a (1+m) (2+m)} \]

[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))

Maple [F]

\[\int \frac {\left (c x \right )^{m} \left (B x +A \right )}{b \,x^{2}+a}d x\]

[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)

Fricas [F]

\[ \int \frac {(c x)^m (A+B x)}{a+b x^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (c x\right )^{m}}{b x^{2} + a} \,d x } \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.67 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.08 \[ \int \frac {(c x)^m (A+B x)}{a+b x^2} \, dx=\frac {A c^{m} m x^{m + 1} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A c^{m} x^{m + 1} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B c^{m} m x^{m + 2} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{4 a \Gamma \left (\frac {m}{2} + 2\right )} + \frac {B c^{m} x^{m + 2} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{2 a \Gamma \left (\frac {m}{2} + 2\right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(c x)^m (A+B x)}{a+b x^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (c x\right )^{m}}{b x^{2} + a} \,d x } \]

[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)

Giac [F]

\[ \int \frac {(c x)^m (A+B x)}{a+b x^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (c x\right )^{m}}{b x^{2} + a} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^m (A+B x)}{a+b x^2} \, dx=\int \frac {{\left (c\,x\right )}^m\,\left (A+B\,x\right )}{b\,x^2+a} \,d x \]

[In]

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

[Out]

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