\(\int \frac {(e x)^m \sqrt {a+b x^2}}{c+d x} \, dx\) [159]

Optimal result

Integrand size = 24, antiderivative size = 167 \[ \int \frac {(e x)^m \sqrt {a+b x^2}}{c+d x} \, dx=\frac {(e x)^{1+m} \sqrt {a+b x^2} \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1,\frac {3+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c e (1+m) \sqrt {1+\frac {b x^2}{a}}}-\frac {d (e x)^{2+m} \sqrt {a+b x^2} \operatorname {AppellF1}\left (\frac {2+m}{2},-\frac {1}{2},1,\frac {4+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 e^2 (2+m) \sqrt {1+\frac {b x^2}{a}}} \] Output:

(e*x)^(1+m)*(b*x^2+a)^(1/2)*AppellF1(1/2+1/2*m,1,-1/2,3/2+1/2*m,d^2*x^2/c^ 
2,-b*x^2/a)/c/e/(1+m)/(1+b*x^2/a)^(1/2)-d*(e*x)^(2+m)*(b*x^2+a)^(1/2)*Appe 
llF1(1+1/2*m,1,-1/2,2+1/2*m,d^2*x^2/c^2,-b*x^2/a)/c^2/e^2/(2+m)/(1+b*x^2/a 
)^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {623, 621, 395, 394}

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.

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

\(\Big \downarrow \) 623

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

\(\Big \downarrow \) 621

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

\(\displaystyle x^{-m} (e x)^m \left (\frac {x^{m+1} \sqrt {a+b x^2} \operatorname {AppellF1}\left (\frac {m+1}{2},-\frac {1}{2},1,\frac {m+3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c (m+1) \sqrt {\frac {b x^2}{a}+1}}-\frac {d x^{m+2} \sqrt {a+b x^2} \operatorname {AppellF1}\left (\frac {m+2}{2},-\frac {1}{2},1,\frac {m+4}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 (m+2) \sqrt {\frac {b x^2}{a}+1}}\right )\)

Input:

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

Output:

((e*x)^m*((x^(1 + m)*Sqrt[a + b*x^2]*AppellF1[(1 + m)/2, -1/2, 1, (3 + m)/ 
2, -((b*x^2)/a), (d^2*x^2)/c^2])/(c*(1 + m)*Sqrt[1 + (b*x^2)/a]) - (d*x^(2 
 + m)*Sqrt[a + b*x^2]*AppellF1[(2 + m)/2, -1/2, 1, (4 + m)/2, -((b*x^2)/a) 
, (d^2*x^2)/c^2])/(c^2*(2 + m)*Sqrt[1 + (b*x^2)/a])))/x^m
 

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

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

rule 621

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

rule 623

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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