\(\int \frac {\sqrt {a+b x^2+c x^4}}{(d x)^{3/2}} \, dx\) [1070]

Optimal result

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

-2*(c*x^4+b*x^2+a)^(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)/(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. \(345\) vs. \(2(145)=290\).

Time = 11.29 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.38 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{(d x)^{3/2}} \, dx=\frac {x \left (-42 \left (a+b x^2+c x^4\right )+28 b x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+24 c x^4 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},\frac {1}{2},\frac {11}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{21 (d x)^{3/2} \sqrt {a+b x^2+c x^4}} \] Input:

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

Output:

(x*(-42*(a + b*x^2 + c*x^4) + 28*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])] + 24*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*(d* 
x)^(3/2)*Sqrt[a + b*x^2 + c*x^4])
 

Rubi [A] (verified)

Time = 0.48 (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.

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

\(\Big \downarrow \) 1461

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

(-2*Sqrt[a + b*x^2 + c*x^4]*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]*Sq 
rt[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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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