\(\int \frac {(h x)^m (d+e x+f x^2)}{(a+b x^2)^{3/2}} \, dx\) [15]

Optimal result

Integrand size = 27, antiderivative size = 191 \[ \int \frac {(h x)^m \left (d+e x+f x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {f (h x)^{1+m}}{b h m \sqrt {a+b x^2}}+\frac {(b d m-a f (1+m)) (h x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a b h m (1+m) \sqrt {a+b x^2}}+\frac {e (h x)^{2+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a h^2 (2+m) \sqrt {a+b x^2}} \] Output:

f*(h*x)^(1+m)/b/h/m/(b*x^2+a)^(1/2)+(b*d*m-a*f*(1+m))*(h*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)/a/b/h/m/(1+m)/( 
b*x^2+a)^(1/2)+e*(h*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)/a/h^2/(2+m)/(b*x^2+a)^(1/2)
 

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.83 \[ \int \frac {(h x)^m \left (d+e x+f x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x (h x)^m \sqrt {1+\frac {b x^2}{a}} \left (d \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+(1+m) x \left (e (3+m) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )+f (2+m) x \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )\right )\right )}{a (1+m) (2+m) (3+m) \sqrt {a+b x^2}} \] Input:

Integrate[((h*x)^m*(d + e*x + f*x^2))/(a + b*x^2)^(3/2),x]
 

Output:

(x*(h*x)^m*Sqrt[1 + (b*x^2)/a]*(d*(6 + 5*m + m^2)*Hypergeometric2F1[3/2, ( 
1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + (1 + m)*x*(e*(3 + m)*Hypergeometric2F 
1[3/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)] + f*(2 + m)*x*Hypergeometric2F1 
[3/2, (3 + m)/2, (5 + m)/2, -((b*x^2)/a)])))/(a*(1 + m)*(2 + m)*(3 + m)*Sq 
rt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2337, 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.

\(\displaystyle \int \frac {(h x)^m \left (d+e x+f x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2337

\(\displaystyle \frac {(h x)^{m+1} \left (-\frac {a f}{b}+d+e x\right )}{a h \sqrt {a+b x^2}}-\frac {\int \frac {(h x)^m \left (d m-\frac {a f (m+1)}{b}+e (m+1) x\right )}{\sqrt {b x^2+a}}dx}{a}\)

\(\Big \downarrow \) 557

\(\displaystyle \frac {(h x)^{m+1} \left (-\frac {a f}{b}+d+e x\right )}{a h \sqrt {a+b x^2}}-\frac {\left (d m-\frac {a f (m+1)}{b}\right ) \int \frac {(h x)^m}{\sqrt {b x^2+a}}dx+\frac {e (m+1) \int \frac {(h x)^{m+1}}{\sqrt {b x^2+a}}dx}{h}}{a}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {(h x)^{m+1} \left (-\frac {a f}{b}+d+e x\right )}{a h \sqrt {a+b x^2}}-\frac {\frac {\sqrt {\frac {b x^2}{a}+1} \left (d m-\frac {a f (m+1)}{b}\right ) \int \frac {(h x)^m}{\sqrt {\frac {b x^2}{a}+1}}dx}{\sqrt {a+b x^2}}+\frac {e (m+1) \sqrt {\frac {b x^2}{a}+1} \int \frac {(h x)^{m+1}}{\sqrt {\frac {b x^2}{a}+1}}dx}{h \sqrt {a+b x^2}}}{a}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {(h x)^{m+1} \left (-\frac {a f}{b}+d+e x\right )}{a h \sqrt {a+b x^2}}-\frac {\frac {\sqrt {\frac {b x^2}{a}+1} (h x)^{m+1} \left (d m-\frac {a f (m+1)}{b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{h (m+1) \sqrt {a+b x^2}}+\frac {e (m+1) \sqrt {\frac {b x^2}{a}+1} (h x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-\frac {b x^2}{a}\right )}{h^2 (m+2) \sqrt {a+b x^2}}}{a}\)

Input:

Int[((h*x)^m*(d + e*x + f*x^2))/(a + b*x^2)^(3/2),x]
 

Output:

((h*x)^(1 + m)*(d - (a*f)/b + e*x))/(a*h*Sqrt[a + b*x^2]) - (((d*m - (a*f* 
(1 + m))/b)*(h*x)^(1 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, (1 + 
m)/2, (3 + m)/2, -((b*x^2)/a)])/(h*(1 + m)*Sqrt[a + b*x^2]) + (e*(1 + m)*( 
h*x)^(2 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m) 
/2, -((b*x^2)/a)])/(h^2*(2 + m)*Sqrt[a + b*x^2]))/a
 

Defintions of rubi rules used

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]

\[\int \frac {\left (h x \right )^{m} \left (f \,x^{2}+e x +d \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}d x\]

Input:

int((h*x)^m*(f*x^2+e*x+d)/(b*x^2+a)^(3/2),x)
 

Output:

int((h*x)^m*(f*x^2+e*x+d)/(b*x^2+a)^(3/2),x)
 

Fricas [F]

\[ \int \frac {(h x)^m \left (d+e x+f x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{2} + e x + d\right )} \left (h x\right )^{m}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((h*x)^m*(f*x^2+e*x+d)/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x^2 + a)*(f*x^2 + e*x + d)*(h*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 = 6.70 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.89 \[ \int \frac {(h x)^m \left (d+e x+f x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {d h^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {e h^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + 2\right )} + \frac {f h^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \] Input:

integrate((h*x)**m*(f*x**2+e*x+d)/(b*x**2+a)**(3/2),x)
 

Output:

d*h**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)) + e*h**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)) + f*h**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))
 

Maxima [F]

\[ \int \frac {(h x)^m \left (d+e x+f x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{2} + e x + d\right )} \left (h x\right )^{m}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((h*x)^m*(f*x^2+e*x+d)/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((f*x^2 + e*x + d)*(h*x)^m/(b*x^2 + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {(h x)^m \left (d+e x+f x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{2} + e x + d\right )} \left (h x\right )^{m}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((h*x)^m*(f*x^2+e*x+d)/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((f*x^2 + e*x + d)*(h*x)^m/(b*x^2 + a)^(3/2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(h x)^m \left (d+e x+f x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {{\left (h\,x\right )}^m\,\left (f\,x^2+e\,x+d\right )}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int(((h*x)^m*(d + e*x + f*x^2))/(a + b*x^2)^(3/2),x)
 

Output:

int(((h*x)^m*(d + e*x + f*x^2))/(a + b*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(h x)^m \left (d+e x+f x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=h^{m} \left (\left (\int \frac {x^{m}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) d +\left (\int \frac {x^{m} x^{2}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) f +\left (\int \frac {x^{m} x}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) e \right ) \] Input:

int((h*x)^m*(f*x^2+e*x+d)/(b*x^2+a)^(3/2),x)
 

Output:

h**m*(int(x**m/(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)*f + int((x**m 
*x)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*e)