\(\int \frac {(e x)^m (c+d x)^2}{\sqrt {a+b x^2}} \, dx\) [168]

Optimal result

Integrand size = 24, antiderivative size = 192 \[ \int \frac {(e x)^m (c+d x)^2}{\sqrt {a+b x^2}} \, dx=\frac {d^2 (e x)^{1+m} \sqrt {a+b x^2}}{b e (2+m)}+\frac {\left (\frac {c^2}{1+m}-\frac {a d^2}{b (2+m)}\right ) (e x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{e \sqrt {a+b x^2}}+\frac {2 c d (e x)^{2+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{e^2 (2+m) \sqrt {a+b x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.83 \[ \int \frac {(e x)^m (c+d x)^2}{\sqrt {a+b x^2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {b x^2}{a}} \left (c^2 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+d (1+m) x \left (2 c (3+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )+d (2+m) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )\right )\right )}{(1+m) (2+m) (3+m) \sqrt {a+b x^2}} \] Input:

Integrate[((e*x)^m*(c + d*x)^2)/Sqrt[a + b*x^2],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {559, 25, 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.

\(\displaystyle \int \frac {(c+d x)^2 (e x)^m}{\sqrt {a+b x^2}} \, dx\)

\(\Big \downarrow \) 559

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

Int[((e*x)^m*(c + d*x)^2)/Sqrt[a + b*x^2],x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 559

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[d^n*(e*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*e^(n - 1)*( 
m + n + 2*p + 1))), x] + Simp[1/(b*(m + n + 2*p + 1))   Int[(e*x)^m*(a + b* 
x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1 
)*x^n - a*d^n*(m + n - 1)*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, e, m, 
 p}, x] && IGtQ[n, 1] &&  !IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
 

Maple [F]

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

Input:

int((e*x)^m*(d*x+c)^2/(b*x^2+a)^(1/2),x)
 

Output:

int((e*x)^m*(d*x+c)^2/(b*x^2+a)^(1/2),x)
 

Fricas [F]

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

integrate((e*x)^m*(d*x+c)^2/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

integral((d^2*x^2 + 2*c*d*x + c^2)*(e*x)^m/sqrt(b*x^2 + a), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.92 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.90 \[ \int \frac {(e x)^m (c+d x)^2}{\sqrt {a+b x^2}} \, dx=\frac {c^{2} e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {c d e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\sqrt {a} \Gamma \left (\frac {m}{2} + 2\right )} + \frac {d^{2} e^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \] Input:

integrate((e*x)**m*(d*x+c)**2/(b*x**2+a)**(1/2),x)
 

Output:

c**2*e**m*x**(m + 1)*gamma(m/2 + 1/2)*hyper((1/2, m/2 + 1/2), (m/2 + 3/2,) 
, b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(m/2 + 3/2)) + c*d*e**m*x**(m 
+ 2)*gamma(m/2 + 1)*hyper((1/2, m/2 + 1), (m/2 + 2,), b*x**2*exp_polar(I*p 
i)/a)/(sqrt(a)*gamma(m/2 + 2)) + d**2*e**m*x**(m + 3)*gamma(m/2 + 3/2)*hyp 
er((1/2, m/2 + 3/2), (m/2 + 5/2,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*ga 
mma(m/2 + 5/2))
 

Maxima [F]

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

integrate((e*x)^m*(d*x+c)^2/(b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate((d*x + c)^2*(e*x)^m/sqrt(b*x^2 + a), x)
 

Giac [F]

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

integrate((e*x)^m*(d*x+c)^2/(b*x^2+a)^(1/2),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x + c)^2*(e*x)^m/sqrt(b*x^2 + a), x)
 

Mupad [F(-1)]

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

int(((e*x)^m*(c + d*x)^2)/(a + b*x^2)^(1/2),x)
 

Output:

int(((e*x)^m*(c + d*x)^2)/(a + b*x^2)^(1/2), x)
 

Reduce [F]

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

int((e*x)^m*(d*x+c)^2/(b*x^2+a)^(1/2),x)
 

Output:

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