\(\int \frac {(e x)^{3/2} (a+b x^2)^p}{(c+d x)^2} \, dx\) [72]

Optimal result
Mathematica [F]
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 215 \[ \int \frac {(e x)^{3/2} \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=\frac {2 (e x)^{5/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{4},-p,2,\frac {9}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{5 c^2 e}-\frac {4 d (e x)^{7/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {7}{4},-p,2,\frac {11}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{7 c^3 e^2}+\frac {2 d^2 (e x)^{9/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {9}{4},-p,2,\frac {13}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{9 c^4 e^3} \] Output:

2/5*(e*x)^(5/2)*(b*x^2+a)^p*AppellF1(5/4,2,-p,9/4,d^2*x^2/c^2,-b*x^2/a)/c^ 
2/e/((1+b*x^2/a)^p)-4/7*d*(e*x)^(7/2)*(b*x^2+a)^p*AppellF1(7/4,2,-p,11/4,d 
^2*x^2/c^2,-b*x^2/a)/c^3/e^2/((1+b*x^2/a)^p)+2/9*d^2*(e*x)^(9/2)*(b*x^2+a) 
^p*AppellF1(9/4,2,-p,13/4,d^2*x^2/c^2,-b*x^2/a)/c^4/e^3/((1+b*x^2/a)^p)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.87, 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.

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

\(\Big \downarrow \) 616

\(\displaystyle \frac {2 \int \frac {e^4 x^2 \left (b x^2+a\right )^p}{(c e+d x e)^2}d\sqrt {e x}}{e}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1675

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 e \left (\frac {(e x)^{5/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{4},2,-p,\frac {9}{4},\frac {d^2 x^2}{c^2},-\frac {b x^2}{a}\right )}{5 c^2 e^2}-\frac {2 \sqrt {e x} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},1,-p,\frac {5}{4},\frac {d^2 x^2}{c^2},-\frac {b x^2}{a}\right )}{d^2}+\frac {\sqrt {e x} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},2,-p,\frac {5}{4},\frac {d^2 x^2}{c^2},-\frac {b x^2}{a}\right )}{d^2}+\frac {2 (e x)^{3/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},1,-p,\frac {7}{4},\frac {d^2 x^2}{c^2},-\frac {b x^2}{a}\right )}{3 c d e}-\frac {2 (e x)^{3/2} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},2,-p,\frac {7}{4},\frac {d^2 x^2}{c^2},-\frac {b x^2}{a}\right )}{3 c d e}+\frac {\sqrt {e x} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^2}{a}\right )}{d^2}\right )\)

Input:

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

Output:

2*e*((-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^2*(1 + (b*x^2)/a)^p) + (Sqrt[e*x]*(a + b*x^2)^p*AppellF1 
[1/4, 2, -p, 5/4, (d^2*x^2)/c^2, -((b*x^2)/a)])/(d^2*(1 + (b*x^2)/a)^p) + 
(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*d*e*(1 + (b*x^2)/a)^p) - (2*(e*x)^(3/2)*(a + b*x^2)^p*Appe 
llF1[3/4, 2, -p, 7/4, (d^2*x^2)/c^2, -((b*x^2)/a)])/(3*c*d*e*(1 + (b*x^2)/ 
a)^p) + ((e*x)^(5/2)*(a + b*x^2)^p*AppellF1[5/4, 2, -p, 9/4, (d^2*x^2)/c^2 
, -((b*x^2)/a)])/(5*c^2*e^2*(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^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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(e)*e*(8*sqrt(x)*(a + b*x**2)**p*a*d*p + 8*sqrt(x)*(a + b*x**2)**p*b* 
c*p*x + 2*sqrt(x)*(a + b*x**2)**p*b*c*x - 64*int((a + b*x**2)**p/(16*sqrt( 
x)*a*c**2*p**2 + 8*sqrt(x)*a*c**2*p + sqrt(x)*a*c**2 + 32*sqrt(x)*a*c*d*p* 
*2*x + 16*sqrt(x)*a*c*d*p*x + 2*sqrt(x)*a*c*d*x + 16*sqrt(x)*a*d**2*p**2*x 
**2 + 8*sqrt(x)*a*d**2*p*x**2 + sqrt(x)*a*d**2*x**2 + 16*sqrt(x)*b*c**2*p* 
*2*x**2 + 8*sqrt(x)*b*c**2*p*x**2 + sqrt(x)*b*c**2*x**2 + 32*sqrt(x)*b*c*d 
*p**2*x**3 + 16*sqrt(x)*b*c*d*p*x**3 + 2*sqrt(x)*b*c*d*x**3 + 16*sqrt(x)*b 
*d**2*p**2*x**4 + 8*sqrt(x)*b*d**2*p*x**4 + sqrt(x)*b*d**2*x**4),x)*a**2*c 
**2*d*p**3 - 32*int((a + b*x**2)**p/(16*sqrt(x)*a*c**2*p**2 + 8*sqrt(x)*a* 
c**2*p + sqrt(x)*a*c**2 + 32*sqrt(x)*a*c*d*p**2*x + 16*sqrt(x)*a*c*d*p*x + 
 2*sqrt(x)*a*c*d*x + 16*sqrt(x)*a*d**2*p**2*x**2 + 8*sqrt(x)*a*d**2*p*x**2 
 + sqrt(x)*a*d**2*x**2 + 16*sqrt(x)*b*c**2*p**2*x**2 + 8*sqrt(x)*b*c**2*p* 
x**2 + sqrt(x)*b*c**2*x**2 + 32*sqrt(x)*b*c*d*p**2*x**3 + 16*sqrt(x)*b*c*d 
*p*x**3 + 2*sqrt(x)*b*c*d*x**3 + 16*sqrt(x)*b*d**2*p**2*x**4 + 8*sqrt(x)*b 
*d**2*p*x**4 + sqrt(x)*b*d**2*x**4),x)*a**2*c**2*d*p**2 - 4*int((a + b*x** 
2)**p/(16*sqrt(x)*a*c**2*p**2 + 8*sqrt(x)*a*c**2*p + sqrt(x)*a*c**2 + 32*s 
qrt(x)*a*c*d*p**2*x + 16*sqrt(x)*a*c*d*p*x + 2*sqrt(x)*a*c*d*x + 16*sqrt(x 
)*a*d**2*p**2*x**2 + 8*sqrt(x)*a*d**2*p*x**2 + sqrt(x)*a*d**2*x**2 + 16*sq 
rt(x)*b*c**2*p**2*x**2 + 8*sqrt(x)*b*c**2*p*x**2 + sqrt(x)*b*c**2*x**2 + 3 
2*sqrt(x)*b*c*d*p**2*x**3 + 16*sqrt(x)*b*c*d*p*x**3 + 2*sqrt(x)*b*c*d*x...