\(\int \frac {(a+b x^2)^p}{\sqrt {e x} (c+d x)} \, dx\) [70]

Optimal result

Integrand size = 24, antiderivative size = 140 \[ \int \frac {\left (a+b x^2\right )^p}{\sqrt {e x} (c+d x)} \, dx=\frac {2 \sqrt {e x} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},-p,1,\frac {5}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c e}-\frac {2 d (e x)^{3/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},-p,1,\frac {7}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 c^2 e^2} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {616, 27, 1569, 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.

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

\(\Big \downarrow \) 616

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1569

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

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*x^2)^p/(Sqrt[e*x]*(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)])/(c*e*(1 + (b*x^2)/a)^p) - (d*(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^2*e^2*(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 1569

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int 
[ExpandIntegrand[(a + c*x^4)^p, (d/(d^2 - e^2*x^4) - e*(x^2/(d^2 - e^2*x^4) 
))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  ! 
IntegerQ[p] && ILtQ[q, 0]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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