\(\int \frac {(c x)^m (A+C x^2)}{a+b x^2} \, dx\) [60]

Optimal result

Integrand size = 22, antiderivative size = 76 \[ \int \frac {(c x)^m \left (A+C x^2\right )}{a+b x^2} \, dx=\frac {C (c x)^{1+m}}{b c (1+m)}+\frac {(A b-a C) (c x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a b c (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {470, 371} \[ \int \frac {(c x)^m \left (A+C x^2\right )}{a+b x^2} \, dx=\frac {(c x)^{m+1} (A b-a C) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a b c (m+1)}+\frac {C (c x)^{m+1}}{b c (m+1)} \]

[In]

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

[Out]

(C*(c*x)^(1 + m))/(b*c*(1 + m)) + ((A*b - a*C)*(c*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x
^2)/a)])/(a*b*c*(1 + 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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {C (c x)^{1+m}}{b c (1+m)}-\frac {(-A b (1+m)+a C (1+m)) \int \frac {(c x)^m}{a+b x^2} \, dx}{b (1+m)} \\ & = \frac {C (c x)^{1+m}}{b c (1+m)}+\frac {(A b-a C) (c x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a b c (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int \frac {(c x)^m \left (A+C x^2\right )}{a+b x^2} \, dx=\frac {x (c x)^m \left (a C+(A b-a C) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )}{a b (1+m)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.82 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.64 \[ \int \frac {(c x)^m \left (A+C x^2\right )}{a+b x^2} \, dx=\frac {A c^{m} m x^{m + 1} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A c^{m} x^{m + 1} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {C c^{m} m x^{m + 3} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 C c^{m} x^{m + 3} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \]

[In]

integrate((c*x)**m*(C*x**2+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)) +
C*c**m*m*x**(m + 3)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*a*gamma(m/2 + 5/2)) +
3*C*c**m*x**(m + 3)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*a*gamma(m/2 + 5/2))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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