\(\int \frac {(e x)^{3/2} (a+b x^2)^p}{c+d x} \, dx\) [68]

Optimal result

Integrand size = 24, antiderivative size = 142 \[ \int \frac {(e x)^{3/2} \left (a+b x^2\right )^p}{c+d x} \, dx=\frac {2 (e x)^{5/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{4},-p,1,\frac {9}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{5 c e}-\frac {2 d (e x)^{7/2} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {7}{4},-p,1,\frac {11}{4},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{7 c^2 e^2} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.75, 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.

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

\(\Big \downarrow \) 616

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1675

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((e*x)^(3/2)*(b*x^2+a)^p/(d*x+c),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 - 256*int((sqrt(x)*(a + b*x**2)**p 
*x**2)/(16*a*c*p**2 + 16*a*c*p + 3*a*c + 16*a*d*p**2*x + 16*a*d*p*x + 3*a* 
d*x + 16*b*c*p**2*x**2 + 16*b*c*p*x**2 + 3*b*c*x**2 + 16*b*d*p**2*x**3 + 1 
6*b*d*p*x**3 + 3*b*d*x**3),x)*a*b*d**2*p**4 - 320*int((sqrt(x)*(a + b*x**2 
)**p*x**2)/(16*a*c*p**2 + 16*a*c*p + 3*a*c + 16*a*d*p**2*x + 16*a*d*p*x + 
3*a*d*x + 16*b*c*p**2*x**2 + 16*b*c*p*x**2 + 3*b*c*x**2 + 16*b*d*p**2*x**3 
 + 16*b*d*p*x**3 + 3*b*d*x**3),x)*a*b*d**2*p**3 - 112*int((sqrt(x)*(a + b* 
x**2)**p*x**2)/(16*a*c*p**2 + 16*a*c*p + 3*a*c + 16*a*d*p**2*x + 16*a*d*p* 
x + 3*a*d*x + 16*b*c*p**2*x**2 + 16*b*c*p*x**2 + 3*b*c*x**2 + 16*b*d*p**2* 
x**3 + 16*b*d*p*x**3 + 3*b*d*x**3),x)*a*b*d**2*p**2 - 12*int((sqrt(x)*(a + 
 b*x**2)**p*x**2)/(16*a*c*p**2 + 16*a*c*p + 3*a*c + 16*a*d*p**2*x + 16*a*d 
*p*x + 3*a*d*x + 16*b*c*p**2*x**2 + 16*b*c*p*x**2 + 3*b*c*x**2 + 16*b*d*p* 
*2*x**3 + 16*b*d*p*x**3 + 3*b*d*x**3),x)*a*b*d**2*p - 256*int((sqrt(x)*(a 
+ b*x**2)**p*x**2)/(16*a*c*p**2 + 16*a*c*p + 3*a*c + 16*a*d*p**2*x + 16*a* 
d*p*x + 3*a*d*x + 16*b*c*p**2*x**2 + 16*b*c*p*x**2 + 3*b*c*x**2 + 16*b*d*p 
**2*x**3 + 16*b*d*p*x**3 + 3*b*d*x**3),x)*b**2*c**2*p**4 - 512*int((sqrt(x 
)*(a + b*x**2)**p*x**2)/(16*a*c*p**2 + 16*a*c*p + 3*a*c + 16*a*d*p**2*x + 
16*a*d*p*x + 3*a*d*x + 16*b*c*p**2*x**2 + 16*b*c*p*x**2 + 3*b*c*x**2 + 16* 
b*d*p**2*x**3 + 16*b*d*p*x**3 + 3*b*d*x**3),x)*b**2*c**2*p**3 - 352*int...