\(\int \frac {1}{(d x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx\) [1078]

Optimal result

Integrand size = 24, antiderivative size = 145 \[ \int \frac {1}{(d x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=-\frac {2 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (-\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {3}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d \sqrt {d x} \sqrt {a+b x^2+c x^4}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(348\) vs. \(2(145)=290\).

Time = 11.37 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.40 \[ \int \frac {1}{(d x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\frac {x \left (-42 \left (a+b x^2+c x^4\right )+14 b x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+18 c x^4 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},\frac {1}{2},\frac {11}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{21 a (d x)^{3/2} \sqrt {a+b x^2+c x^4}} \] Input:

Integrate[1/((d*x)^(3/2)*Sqrt[a + b*x^2 + c*x^4]),x]
 

Output:

(x*(-42*(a + b*x^2 + c*x^4) + 14*b*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 + S 
qrt[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])] + 18*c*x^4*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])]))/(21*a*( 
d*x)^(3/2)*Sqrt[a + b*x^2 + c*x^4])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 145, 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.

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

\(\Big \downarrow \) 1461

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

\(\Big \downarrow \) 394

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

Input:

Int[1/((d*x)^(3/2)*Sqrt[a + b*x^2 + c*x^4]),x]
 

Output:

(-2*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sq 
rt[b^2 - 4*a*c])]*AppellF1[-1/4, 1/2, 1/2, 3/4, (-2*c*x^2)/(b - Sqrt[b^2 - 
 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(d*Sqrt[d*x]*Sqrt[a + b*x^2 
 + c*x^4])
 

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]

\[\int \frac {1}{\left (d x \right )^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {1}{(d x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{2} + a} \left (d x\right )^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

integral(sqrt(c*x^4 + b*x^2 + a)*sqrt(d*x)/(c*d^2*x^6 + b*d^2*x^4 + a*d^2* 
x^2), x)
 

Sympy [F]

\[ \int \frac {1}{(d x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {1}{\left (d x\right )^{\frac {3}{2}} \sqrt {a + b x^{2} + c x^{4}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{(d x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{2} + a} \left (d x\right )^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{(d x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{2} + a} \left (d x\right )^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {1}{{\left (d\,x\right )}^{3/2}\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(d x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{6}+b \,x^{4}+a \,x^{2}}d x \right )}{d^{2}} \] Input:

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

Output:

(sqrt(d)*int((sqrt(x)*sqrt(a + b*x**2 + c*x**4))/(a*x**2 + b*x**4 + c*x**6 
),x))/d**2