\(\int \frac {x (d+e x^2)}{\sqrt {a x+b x^3+c x^5}} \, dx\) [74]

Optimal result

Integrand size = 27, antiderivative size = 287 \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {2 d x^2 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 \sqrt {a x+b x^3+c x^5}}+\frac {2 e x^4 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},\frac {1}{2},\frac {11}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{7 \sqrt {a x+b x^3+c x^5}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 11.17 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.83 \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \left (7 d x^2 \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+3 e x^4 \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},\frac {1}{2},\frac {11}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{21 \sqrt {x \left (a+b x^2+c x^4\right )}} \] Input:

Integrate[(x*(d + e*x^2))/Sqrt[a*x + b*x^3 + c*x^5],x]
 

Output:

(2*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b 
 + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*(7*d*x^2*AppellF1 
[3/4, 1/2, 1/2, 7/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + S 
qrt[b^2 - 4*a*c])] + 3*e*x^4*AppellF1[7/4, 1/2, 1/2, 11/4, (-2*c*x^2)/(b + 
 Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(21*Sqrt[x*(a + 
 b*x^2 + c*x^4)])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2001, 1674, 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.

\(\displaystyle \int \frac {x \left (d+e x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx\)

\(\Big \downarrow \) 2001

\(\displaystyle \frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \int \frac {\sqrt {x} \left (e x^2+d\right )}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {a x+b x^3+c x^5}}\)

\(\Big \downarrow \) 1674

\(\displaystyle \frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \int \left (\frac {e x^{5/2}}{\sqrt {c x^4+b x^2+a}}+\frac {d \sqrt {x}}{\sqrt {c x^4+b x^2+a}}\right )dx}{\sqrt {a x+b x^3+c x^5}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (\frac {2 d x^{3/2} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 \sqrt {a+b x^2+c x^4}}+\frac {2 e x^{7/2} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},\frac {1}{2},\frac {11}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{7 \sqrt {a+b x^2+c x^4}}\right )}{\sqrt {a x+b x^3+c x^5}}\)

Input:

Int[(x*(d + e*x^2))/Sqrt[a*x + b*x^3 + c*x^5],x]
 

Output:

(Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*((2*d*x^(3/2)*Sqrt[1 + (2*c*x^2)/(b - Sqr 
t[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[3/4, 
 1/2, 1/2, 7/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b 
^2 - 4*a*c])])/(3*Sqrt[a + b*x^2 + c*x^4]) + (2*e*x^(7/2)*Sqrt[1 + (2*c*x^ 
2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Ap 
pellF1[7/4, 1/2, 1/2, 11/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2) 
/(b + Sqrt[b^2 - 4*a*c])])/(7*Sqrt[a + b*x^2 + c*x^4])))/Sqrt[a*x + b*x^3 
+ c*x^5]
 

Defintions of rubi rules used

rule 1674

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && N 
eQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])
 

rule 2001

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(k_.) + (c_.)*(x_)^(n_.))^(p_ 
)*((A_) + (B_.)*(x_)^(q_)), x_Symbol] :> Simp[(a*x^j + b*x^k + c*x^n)^p/(x^ 
(j*p)*(a + b*x^(k - j) + c*x^(2*(k - j)))^p)   Int[x^(m + j*p)*(A + B*x^(k 
- j))*(a + b*x^(k - j) + c*x^(2*(k - j)))^p, x], x] /; FreeQ[{a, b, c, A, B 
, j, k, m, p}, x] && EqQ[q, k - j] && EqQ[n, 2*k - j] &&  !IntegerQ[p] && P 
osQ[k - j]
 

rule 2009

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

Maple [F]

\[\int \frac {x \left (e \,x^{2}+d \right )}{\sqrt {c \,x^{5}+b \,x^{3}+x a}}d x\]

Input:

int(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x)
 

Output:

int(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x)
 

Fricas [F]

\[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {{\left (e x^{2} + d\right )} x}{\sqrt {c x^{5} + b x^{3} + a x}} \,d x } \] Input:

integrate(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(c*x^5 + b*x^3 + a*x)*(e*x^2 + d)/(c*x^4 + b*x^2 + a), x)
 

Sympy [F]

\[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx=\int \frac {x \left (d + e x^{2}\right )}{\sqrt {x \left (a + b x^{2} + c x^{4}\right )}}\, dx \] Input:

integrate(x*(e*x**2+d)/(c*x**5+b*x**3+a*x)**(1/2),x)
 

Output:

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

Maxima [F]

\[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {{\left (e x^{2} + d\right )} x}{\sqrt {c x^{5} + b x^{3} + a x}} \,d x } \] Input:

integrate(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="maxima")
 

Output:

integrate((e*x^2 + d)*x/sqrt(c*x^5 + b*x^3 + a*x), x)
 

Giac [F]

\[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {{\left (e x^{2} + d\right )} x}{\sqrt {c x^{5} + b x^{3} + a x}} \,d x } \] Input:

integrate(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="giac")
 

Output:

integrate((e*x^2 + d)*x/sqrt(c*x^5 + b*x^3 + a*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx=\int \frac {x\,\left (e\,x^2+d\right )}{\sqrt {c\,x^5+b\,x^3+a\,x}} \,d x \] Input:

int((x*(d + e*x^2))/(a*x + b*x^3 + c*x^5)^(1/2),x)
 

Output:

int((x*(d + e*x^2))/(a*x + b*x^3 + c*x^5)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x \left (d+e x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx=\left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) e +\left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{4}+b \,x^{2}+a}d x \right ) d \] Input:

int(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x)
 

Output:

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