\(\int \frac {(a+b x^2)^p}{(e x)^{3/2} (c+d x)} \, dx\) [1867]

Optimal result

Integrand size = 24, antiderivative size = 138 \[ \int \frac {\left (a+b x^2\right )^p}{(e x)^{3/2} (c+d x)} \, dx=-\frac {2 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{4},-p,1,\frac {3}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c e \sqrt {e x}}-\frac {2 d \sqrt {e x} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},-p,1,\frac {5}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 e^2} \] Output:

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

Mathematica [F]

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

Integrate[(a + b*x^2)^p/((e*x)^(3/2)*(c + d*x)),x]
 

Output:

Integrate[(a + b*x^2)^p/((e*x)^(3/2)*(c + d*x)), x]
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.46, 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.

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

\(\Big \downarrow \) 616

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1675

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

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*x^2)^p/((e*x)^(3/2)*(c + d*x)),x]
 

Output:

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

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

Input:

int((b*x^2+a)^p/(e*x)^(3/2)/(d*x+c),x)
 

Output:

int((b*x^2+a)^p/(e*x)^(3/2)/(d*x+c),x)
 

Fricas [F]

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

integrate((b*x^2+a)^p/(e*x)^(3/2)/(d*x+c),x, algorithm="fricas")
 

Output:

integral(sqrt(e*x)*(b*x^2 + a)^p/(d*e^2*x^3 + c*e^2*x^2), x)
 

Sympy [F(-1)]

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

integrate((b*x**2+a)**p/(e*x)**(3/2)/(d*x+c),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((b*x^2+a)^p/(e*x)^(3/2)/(d*x+c),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^p/((d*x + c)*(e*x)^(3/2)), x)
 

Giac [F]

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

integrate((b*x^2+a)^p/(e*x)^(3/2)/(d*x+c),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^p/((d*x + c)*(e*x)^(3/2)), x)
 

Mupad [F(-1)]

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

int((a + b*x^2)^p/((e*x)^(3/2)*(c + d*x)),x)
 

Output:

int((a + b*x^2)^p/((e*x)^(3/2)*(c + d*x)), x)
 

Reduce [F]

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

int((b*x^2+a)^p/(e*x)^(3/2)/(d*x+c),x)
 

Output:

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