\(\int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx\) [275]

Optimal result

Integrand size = 28, antiderivative size = 71 \[ \int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx=\frac {2 (e x)^{5/2} \sqrt {c+d x^4} \operatorname {AppellF1}\left (\frac {5}{8},1,-\frac {1}{2},\frac {13}{8},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{5 a e \sqrt {1+\frac {d x^4}{c}}} \] Output:

2/5*(e*x)^(5/2)*(d*x^4+c)^(1/2)*AppellF1(5/8,1,-1/2,13/8,-b*x^4/a,-d*x^4/c 
)/a/e/(1+d*x^4/c)^(1/2)
 

Mathematica [A] (verified)

Time = 11.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99 \[ \int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx=\frac {2 x (e x)^{3/2} \sqrt {c+d x^4} \operatorname {AppellF1}\left (\frac {5}{8},-\frac {1}{2},1,\frac {13}{8},-\frac {d x^4}{c},-\frac {b x^4}{a}\right )}{5 a \sqrt {\frac {c+d x^4}{c}}} \] Input:

Integrate[((e*x)^(3/2)*Sqrt[c + d*x^4])/(a + b*x^4),x]
 

Output:

(2*x*(e*x)^(3/2)*Sqrt[c + d*x^4]*AppellF1[5/8, -1/2, 1, 13/8, -((d*x^4)/c) 
, -((b*x^4)/a)])/(5*a*Sqrt[(c + d*x^4)/c])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {966, 27, 1013, 1012}

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.

\(\displaystyle \int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx\)

\(\Big \downarrow \) 966

\(\displaystyle \frac {2 \int \frac {e^6 x^2 \sqrt {d x^4+c}}{b x^4 e^4+a e^4}d\sqrt {e x}}{e}\)

\(\Big \downarrow \) 27

\(\displaystyle 2 e^3 \int \frac {e^2 x^2 \sqrt {d x^4+c}}{b x^4 e^4+a e^4}d\sqrt {e x}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {2 e^3 \sqrt {c+d x^4} \int \frac {e^2 x^2 \sqrt {\frac {d x^4}{c}+1}}{b x^4 e^4+a e^4}d\sqrt {e x}}{\sqrt {\frac {d x^4}{c}+1}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {2 (e x)^{5/2} \sqrt {c+d x^4} \operatorname {AppellF1}\left (\frac {5}{8},1,-\frac {1}{2},\frac {13}{8},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{5 a e \sqrt {\frac {d x^4}{c}+1}}\)

Input:

Int[((e*x)^(3/2)*Sqrt[c + d*x^4])/(a + b*x^4),x]
 

Output:

(2*(e*x)^(5/2)*Sqrt[c + d*x^4]*AppellF1[5/8, 1, -1/2, 13/8, -((b*x^4)/a), 
-((d*x^4)/c)])/(5*a*e*Sqrt[1 + (d*x^4)/c])
 

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 966

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

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ 
n/a))^FracPart[p])   Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, n - 1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {3}{2}} \sqrt {d \,x^{4}+c}}{b \,x^{4}+a}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx=\int { \frac {\sqrt {d x^{4} + c} \left (e x\right )^{\frac {3}{2}}}{b x^{4} + a} \,d x } \] Input:

integrate((e*x)^(3/2)*(d*x^4+c)^(1/2)/(b*x^4+a),x, algorithm="fricas")
 

Output:

integral(sqrt(d*x^4 + c)*sqrt(e*x)*e*x/(b*x^4 + a), x)
 

Sympy [F]

\[ \int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx=\int \frac {\left (e x\right )^{\frac {3}{2}} \sqrt {c + d x^{4}}}{a + b x^{4}}\, dx \] Input:

integrate((e*x)**(3/2)*(d*x**4+c)**(1/2)/(b*x**4+a),x)
 

Output:

Integral((e*x)**(3/2)*sqrt(c + d*x**4)/(a + b*x**4), x)
 

Maxima [F]

\[ \int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx=\int { \frac {\sqrt {d x^{4} + c} \left (e x\right )^{\frac {3}{2}}}{b x^{4} + a} \,d x } \] Input:

integrate((e*x)^(3/2)*(d*x^4+c)^(1/2)/(b*x^4+a),x, algorithm="maxima")
 

Output:

integrate(sqrt(d*x^4 + c)*(e*x)^(3/2)/(b*x^4 + a), x)
 

Giac [F]

\[ \int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx=\int { \frac {\sqrt {d x^{4} + c} \left (e x\right )^{\frac {3}{2}}}{b x^{4} + a} \,d x } \] Input:

integrate((e*x)^(3/2)*(d*x^4+c)^(1/2)/(b*x^4+a),x, algorithm="giac")
 

Output:

integrate(sqrt(d*x^4 + c)*(e*x)^(3/2)/(b*x^4 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx=\int \frac {{\left (e\,x\right )}^{3/2}\,\sqrt {d\,x^4+c}}{b\,x^4+a} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^{3/2} \sqrt {c+d x^4}}{a+b x^4} \, dx=\sqrt {e}\, \left (\int \frac {\sqrt {x}\, \sqrt {d \,x^{4}+c}\, x}{b \,x^{4}+a}d x \right ) e \] Input:

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

Output:

sqrt(e)*int((sqrt(x)*sqrt(c + d*x**4)*x)/(a + b*x**4),x)*e