\(\int \sqrt {d x} (a+b x^2+c x^4)^{3/2} \, dx\) [1072]

Optimal result

Integrand size = 24, antiderivative size = 148 \[ \int \sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {2 a (d x)^{3/2} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {3}{2},-\frac {3}{2},\frac {7}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 d \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(417\) vs. \(2(148)=296\).

Time = 11.53 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.82 \[ \int \sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {2 x \sqrt {d x} \left (7 \left (12 a b^2+209 a^2 c+12 b^3 x^2+328 a b c x^2+131 b^2 c x^4+286 a c^2 x^4+196 b c^2 x^6+77 c^3 x^8\right )-28 a \left (3 b^2-44 a c\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+12 b \left (-5 b^2+36 a c\right ) x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},\frac {1}{2},\frac {11}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{8085 c \sqrt {a+b x^2+c x^4}} \] Input:

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

Output:

(2*x*Sqrt[d*x]*(7*(12*a*b^2 + 209*a^2*c + 12*b^3*x^2 + 328*a*b*c*x^2 + 131 
*b^2*c*x^4 + 286*a*c^2*x^4 + 196*b*c^2*x^6 + 77*c^3*x^8) - 28*a*(3*b^2 - 4 
4*a*c)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqr 
t[(b + Sqrt[b^2 - 4*a*c] + 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])] + 12*b*(-5*b^2 + 36*a*c)*x^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])]*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])]))/(8085*c*Sqrt[a + b*x^ 
2 + c*x^4])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1461, 394}

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.

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

\(\Big \downarrow \) 1461

\(\displaystyle \frac {a \sqrt {a+b x^2+c x^4} \int \sqrt {d x} \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}+1\right )^{3/2}dx}{\sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}\)

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1461

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[a^IntPart[p]*((a + b*x^2 + c*x^4)^FracPart[p]/((1 + 2*c*(x^2/(b + R 
t[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^2/(b - Rt[b^2 - 4*a*c, 2])))^F 
racPart[p]))   Int[(d*x)^m*(1 + 2*c*(x^2/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2 
*c*(x^2/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {d x} \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {2 \sqrt {d}\, \left (209 \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, a c x +12 \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} x +119 \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, b c \,x^{3}+77 \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2} x^{5}+216 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a b c -30 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) b^{3}+264 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a^{2} c -18 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a \,b^{2}\right )}{1155 c} \] Input:

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

Output:

(2*sqrt(d)*(209*sqrt(x)*sqrt(a + b*x**2 + c*x**4)*a*c*x + 12*sqrt(x)*sqrt( 
a + b*x**2 + c*x**4)*b**2*x + 119*sqrt(x)*sqrt(a + b*x**2 + c*x**4)*b*c*x* 
*3 + 77*sqrt(x)*sqrt(a + b*x**2 + c*x**4)*c**2*x**5 + 216*int((sqrt(x)*sqr 
t(a + b*x**2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*a*b*c - 30*int((sqrt 
(x)*sqrt(a + b*x**2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*b**3 + 264*in 
t((sqrt(x)*sqrt(a + b*x**2 + c*x**4))/(a + b*x**2 + c*x**4),x)*a**2*c - 18 
*int((sqrt(x)*sqrt(a + b*x**2 + c*x**4))/(a + b*x**2 + c*x**4),x)*a*b**2)) 
/(1155*c)