\(\int \frac {\sqrt {e x} (a+b x^2)^p}{c+d x} \, dx\) [69]

Optimal result

Integrand size = 24, antiderivative size = 142 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )^p}{c+d x} \, dx=\frac {2 (e x)^{3/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},-p,1,\frac {7}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 c e}-\frac {2 d (e x)^{5/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{4},-p,1,\frac {9}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{5 c^2 e^2} \] Output:

2/3*(e*x)^(3/2)*(b*x^2+a)^p*AppellF1(3/4,1,-p,7/4,d^2*x^2/c^2,-b*x^2/a)/c/ 
e/((1+b*x^2/a)^p)-2/5*d*(e*x)^(5/2)*(b*x^2+a)^p*AppellF1(5/4,1,-p,9/4,d^2* 
x^2/c^2,-b*x^2/a)/c^2/e^2/((1+b*x^2/a)^p)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {616, 27, 1675, 2009}

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.

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

\(\Big \downarrow \) 616

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1675

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (-\frac {\sqrt {e x} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},1,-p,\frac {5}{4},\frac {d^2 x^2}{c^2},-\frac {b x^2}{a}\right )}{d}+\frac {(e x)^{3/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},1,-p,\frac {7}{4},\frac {d^2 x^2}{c^2},-\frac {b x^2}{a}\right )}{3 c e}+\frac {\sqrt {e x} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^2}{a}\right )}{d}\right )\)

Input:

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

Output:

2*(-((Sqrt[e*x]*(a + b*x^2)^p*AppellF1[1/4, 1, -p, 5/4, (d^2*x^2)/c^2, -(( 
b*x^2)/a)])/(d*(1 + (b*x^2)/a)^p)) + ((e*x)^(3/2)*(a + b*x^2)^p*AppellF1[3 
/4, 1, -p, 7/4, (d^2*x^2)/c^2, -((b*x^2)/a)])/(3*c*e*(1 + (b*x^2)/a)^p) + 
(Sqrt[e*x]*(a + b*x^2)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^2)/a)])/(d 
*(1 + (b*x^2)/a)^p))
 

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 616

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*(m + 1) - 
 1)*(c + d*(x^k/e))^n*(a + b*(x^(2*k)/e^2))^p, x], x, (e*x)^(1/k)], x]] /; 
FreeQ[{a, b, c, d, e, p}, x] && ILtQ[n, 0] && FractionQ[m]
 

rule 1675

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p 
_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, 
x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && (IGtQ[p, 0] || IGtQ[q, 0] | 
| IntegersQ[m, q])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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