\(\int \frac {(h x)^m (d+e x+f x^2)^2}{(a+b x^2)^{3/2}} \, dx\) [14]

Optimal result

Integrand size = 29, antiderivative size = 338 \[ \int \frac {(h x)^m \left (d+e x+f x^2\right )^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\left (b \left (e^2+2 d f\right ) (2+m)-a f^2 (3+m)\right ) (h x)^{1+m}}{b^2 h m (2+m) \sqrt {a+b x^2}}+\frac {2 e f (h x)^{2+m}}{b h^2 (1+m) \sqrt {a+b x^2}}+\frac {f^2 (h x)^{3+m}}{b h^3 (2+m) \sqrt {a+b x^2}}+\frac {\left (b^2 d^2 m (2+m)-a b \left (e^2+2 d f\right ) \left (2+3 m+m^2\right )+a^2 f^2 \left (3+4 m+m^2\right )\right ) (h x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a b^2 h m (1+m) (2+m) \sqrt {a+b x^2}}+\frac {2 e \left (\frac {d}{a (2+m)}-\frac {f}{b+b m}\right ) (h x)^{2+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{h^2 \sqrt {a+b x^2}} \] Output:

(b*(2*d*f+e^2)*(2+m)-a*f^2*(3+m))*(h*x)^(1+m)/b^2/h/m/(2+m)/(b*x^2+a)^(1/2 
)+2*e*f*(h*x)^(2+m)/b/h^2/(1+m)/(b*x^2+a)^(1/2)+f^2*(h*x)^(3+m)/b/h^3/(2+m 
)/(b*x^2+a)^(1/2)+(b^2*d^2*m*(2+m)-a*b*(2*d*f+e^2)*(m^2+3*m+2)+a^2*f^2*(m^ 
2+4*m+3))*(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^2/h/m/(1+m)/(2+m)/(b*x^2+a)^(1/2)+2*e*(d/a/(2+m)-f/(b*m 
+b))*(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)/h^2/(b*x^2+a)^(1/2)
 

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.68 \[ \int \frac {(h x)^m \left (d+e x+f x^2\right )^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x (h x)^m \sqrt {1+\frac {b x^2}{a}} \left (\frac {d^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+x \left (\frac {2 d e \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{2+m}+x \left (\frac {\left (e^2+2 d f\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )}{3+m}+f x \left (\frac {2 e \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {4+m}{2},\frac {6+m}{2},-\frac {b x^2}{a}\right )}{4+m}+\frac {f x \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5+m}{2},\frac {7+m}{2},-\frac {b x^2}{a}\right )}{5+m}\right )\right )\right )\right )}{a \sqrt {a+b x^2}} \] Input:

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

Output:

(x*(h*x)^m*Sqrt[1 + (b*x^2)/a]*((d^2*Hypergeometric2F1[3/2, (1 + m)/2, (3 
+ m)/2, -((b*x^2)/a)])/(1 + m) + x*((2*d*e*Hypergeometric2F1[3/2, (2 + m)/ 
2, (4 + m)/2, -((b*x^2)/a)])/(2 + m) + x*(((e^2 + 2*d*f)*Hypergeometric2F1 
[3/2, (3 + m)/2, (5 + m)/2, -((b*x^2)/a)])/(3 + m) + f*x*((2*e*Hypergeomet 
ric2F1[3/2, (4 + m)/2, (6 + m)/2, -((b*x^2)/a)])/(4 + m) + (f*x*Hypergeome 
tric2F1[3/2, (5 + m)/2, (7 + m)/2, -((b*x^2)/a)])/(5 + m))))))/(a*Sqrt[a + 
 b*x^2])
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 312, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2337, 2340, 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.

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

\(\Big \downarrow \) 2337

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

((h*x)^(1 + m)*(d^2 + (a^2*f^2)/b^2 - (a*(e^2 + 2*d*f))/b + 2*e*(d - (a*f) 
/b)*x))/(a*h*Sqrt[a + b*x^2]) - (-((a*f^2*(h*x)^(1 + m)*Sqrt[a + b*x^2])/( 
b^2*h*(2 + m))) + (((b*d^2*m*(2 + m) - a*(e^2 + 2*d*f)*(2 + 3*m + m^2) + ( 
a^2*f^2*(3 + 4*m + m^2))/b)*(h*x)^(1 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometr 
ic2F1[1/2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(h*(1 + m)*Sqrt[a + b*x^2] 
) + (2*e*(b*d*(1 + m) - a*f*(2 + m))*(h*x)^(2 + m)*Sqrt[1 + (b*x^2)/a]*Hyp 
ergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)])/(h^2*Sqrt[a + b*x 
^2]))/(b*(2 + m)))/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]
 

rule 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((f^2*x^4 + 2*e*f*x^3 + 2*d*e*x + (e^2 + 2*d*f)*x^2 + d^2)*sqrt(b* 
x^2 + a)*(h*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(h x)^m \left (d+e x+f x^2\right )^2}{\left (a+b x^2\right )^{3/2}} \, dx=h^{m} \left (\left (\int \frac {x^{m}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) d^{2}+\left (\int \frac {x^{m} x^{4}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) f^{2}+2 \left (\int \frac {x^{m} x^{3}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) e f +2 \left (\int \frac {x^{m} x^{2}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) d f +\left (\int \frac {x^{m} x^{2}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) e^{2}+2 \left (\int \frac {x^{m} x}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \right ) d e \right ) \] Input:

int((h*x)^m*(f*x^2+e*x+d)^2/(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**2 + in 
t((x**m*x**4)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*f**2 + 2*i 
nt((x**m*x**3)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*e*f + 2*i 
nt((x**m*x**2)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*d*f + int 
((x**m*x**2)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*e**2 + 2*in 
t((x**m*x)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)*d*e)