\(\int \frac {(c x)^m (A+B x+C x^2+D x^3)}{(a+b x^2)^{3/2}} \, dx\) [145]

Optimal result

Integrand size = 32, antiderivative size = 228 \[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {C (c x)^{1+m}}{b c m \sqrt {a+b x^2}}+\frac {D (c x)^{2+m}}{b c^2 (1+m) \sqrt {a+b x^2}}-\frac {\left (\frac {C}{b m}-\frac {A}{a+a m}\right ) (c x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{c \sqrt {a+b x^2}}+\frac {\left (\frac {B}{a (2+m)}-\frac {D}{b+b m}\right ) (c x)^{2+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{c^2 \sqrt {a+b x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.79 \[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x (c x)^m \sqrt {1+\frac {b x^2}{a}} \left (\frac {A \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+x \left (\frac {B \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{2+m}+x \left (\frac {C \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )}{3+m}+\frac {D x \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {4+m}{2},\frac {6+m}{2},-\frac {b x^2}{a}\right )}{4+m}\right )\right )\right )}{a \sqrt {a+b x^2}} \] Input:

Integrate[((c*x)^m*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^(3/2),x]
 

Output:

(x*(c*x)^m*Sqrt[1 + (b*x^2)/a]*((A*Hypergeometric2F1[3/2, (1 + m)/2, (3 + 
m)/2, -((b*x^2)/a)])/(1 + m) + x*((B*Hypergeometric2F1[3/2, (2 + m)/2, (4 
+ m)/2, -((b*x^2)/a)])/(2 + m) + x*((C*Hypergeometric2F1[3/2, (3 + m)/2, ( 
5 + m)/2, -((b*x^2)/a)])/(3 + m) + (D*x*Hypergeometric2F1[3/2, (4 + m)/2, 
(6 + m)/2, -((b*x^2)/a)])/(4 + m)))))/(a*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2337, 27, 557, 279, 278}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2337

\(\displaystyle \frac {(c x)^{m+1} \left (x \left (B-\frac {a D}{b}\right )-\frac {a C}{b}+A\right )}{a c \sqrt {a+b x^2}}-\frac {\int \frac {(c x)^m \left (b \left (A m-\frac {a C (m+1)}{b}\right )+(b B (m+1)-a D (m+2)) x\right )}{b \sqrt {b x^2+a}}dx}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(c x)^{m+1} \left (x \left (B-\frac {a D}{b}\right )-\frac {a C}{b}+A\right )}{a c \sqrt {a+b x^2}}-\frac {\int \frac {(c x)^m (A b m-a C (m+1)+(b B (m+1)-a D (m+2)) x)}{\sqrt {b x^2+a}}dx}{a b}\)

\(\Big \downarrow \) 557

\(\displaystyle \frac {(c x)^{m+1} \left (x \left (B-\frac {a D}{b}\right )-\frac {a C}{b}+A\right )}{a c \sqrt {a+b x^2}}-\frac {(A b m-a C (m+1)) \int \frac {(c x)^m}{\sqrt {b x^2+a}}dx+\frac {(b B (m+1)-a D (m+2)) \int \frac {(c x)^{m+1}}{\sqrt {b x^2+a}}dx}{c}}{a b}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {(c x)^{m+1} \left (x \left (B-\frac {a D}{b}\right )-\frac {a C}{b}+A\right )}{a c \sqrt {a+b x^2}}-\frac {\frac {\sqrt {\frac {b x^2}{a}+1} (A b m-a C (m+1)) \int \frac {(c x)^m}{\sqrt {\frac {b x^2}{a}+1}}dx}{\sqrt {a+b x^2}}+\frac {\sqrt {\frac {b x^2}{a}+1} (b B (m+1)-a D (m+2)) \int \frac {(c x)^{m+1}}{\sqrt {\frac {b x^2}{a}+1}}dx}{c \sqrt {a+b x^2}}}{a b}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {(c x)^{m+1} \left (x \left (B-\frac {a D}{b}\right )-\frac {a C}{b}+A\right )}{a c \sqrt {a+b x^2}}-\frac {\frac {\sqrt {\frac {b x^2}{a}+1} (c x)^{m+1} (A b m-a C (m+1)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{c (m+1) \sqrt {a+b x^2}}+\frac {\sqrt {\frac {b x^2}{a}+1} (c x)^{m+2} (b B (m+1)-a D (m+2)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-\frac {b x^2}{a}\right )}{c^2 (m+2) \sqrt {a+b x^2}}}{a b}\)

Input:

Int[((c*x)^m*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^(3/2),x]
 

Output:

((c*x)^(1 + m)*(A - (a*C)/b + (B - (a*D)/b)*x))/(a*c*Sqrt[a + b*x^2]) - (( 
(A*b*m - a*C*(1 + m))*(c*x)^(1 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[ 
1/2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(c*(1 + m)*Sqrt[a + b*x^2]) + (( 
b*B*(1 + m) - a*D*(2 + m))*(c*x)^(2 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometri 
c2F1[1/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)])/(c^2*(2 + m)*Sqrt[a + b*x^2 
]))/(a*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

rule 2337

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq 
, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 1]}, Simp[(-(c*x)^(m + 1))*(f + g*x)*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))) 
, x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2 
*a*(p + 1)*Q + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a 
, b, c, m}, x] && PolyQ[Pq, x] && LtQ[p, -1] &&  !GtQ[m, 0]
 

Maple [F]

\[\int \frac {\left (c x \right )^{m} \left (D x^{3}+C \,x^{2}+B x +A \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}d x\]

Input:

int((c*x)^m*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x)
 

Output:

int((c*x)^m*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x)
 

Fricas [F]

\[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} \left (c x\right )^{m}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c*x)^m*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x, algorithm="fricas 
")
 

Output:

integral((D*x^3 + C*x^2 + B*x + A)*sqrt(b*x^2 + a)*(c*x)^m/(b^2*x^4 + 2*a* 
b*x^2 + a^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.76 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.98 \[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {A c^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B c^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + 2\right )} + \frac {C c^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {D c^{m} x^{m + 4} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + 3\right )} \] Input:

integrate((c*x)**m*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**(3/2),x)
 

Output:

A*c**m*x**(m + 1)*gamma(m/2 + 1/2)*hyper((3/2, m/2 + 1/2), (m/2 + 3/2,), b 
*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(m/2 + 3/2)) + B*c**m*x**(m + 2) 
*gamma(m/2 + 1)*hyper((3/2, m/2 + 1), (m/2 + 2,), b*x**2*exp_polar(I*pi)/a 
)/(2*a**(3/2)*gamma(m/2 + 2)) + C*c**m*x**(m + 3)*gamma(m/2 + 3/2)*hyper(( 
3/2, m/2 + 3/2), (m/2 + 5/2,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma 
(m/2 + 5/2)) + D*c**m*x**(m + 4)*gamma(m/2 + 2)*hyper((3/2, m/2 + 2), (m/2 
 + 3,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(m/2 + 3))
 

Maxima [F]

\[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} \left (c x\right )^{m}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c*x)^m*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x, algorithm="maxima 
")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(c*x)^m/(b*x^2 + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} \left (c x\right )^{m}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c*x)^m*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(c*x)^m/(b*x^2 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {{\left (c\,x\right )}^m\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int(((c*x)^m*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(3/2),x)
 

Output:

int(((c*x)^m*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(c x)^m \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=c^{m} \left (\left (\int \frac {x^{m}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) a +\left (\int \frac {x^{m} x^{3}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) d +\left (\int \frac {x^{m} x^{2}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) c +\left (\int \frac {x^{m} x}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) b \right ) \] Input:

int((c*x)^m*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^(3/2),x)
 

Output:

c**m*(int(x**m/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*a + int(( 
x**m*x**3)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*d + int((x**m 
*x**2)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*c + int((x**m*x)/ 
(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*b)