\(\int \frac {(a+b x^2+c x^4)^{3/2}}{(d x)^{3/2}} \, dx\) [1074]

Optimal result

Integrand size = 24, antiderivative size = 146 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=-\frac {2 a \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (-\frac {1}{4},-\frac {3}{2},-\frac {3}{2},\frac {3}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d \sqrt {d x} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \] Output:

-2*a*(c*x^4+b*x^2+a)^(1/2)*AppellF1(-1/4,-3/2,-3/2,3/4,-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. \(384\) vs. \(2(146)=292\).

Time = 11.52 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.63 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=\frac {x \left (14 \left (-77 a^2-64 a b x^2+13 b^2 x^4-70 a c x^4+20 b c x^6+7 c^2 x^8\right )+896 a b x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+24 \left (b^2+28 a c\right ) x^4 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},\frac {1}{2},\frac {11}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{539 (d x)^{3/2} \sqrt {a+b x^2+c x^4}} \] Input:

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

Output:

(x*(14*(-77*a^2 - 64*a*b*x^2 + 13*b^2*x^4 - 70*a*c*x^4 + 20*b*c*x^6 + 7*c^ 
2*x^8) + 896*a*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 + 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])] + 24*(b^2 + 28*a*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])]))/(539*(d*x)^( 
3/2)*Sqrt[a + b*x^2 + c*x^4])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 146, 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.

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

\(\Big \downarrow \) 1461

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{(d x)^{3/2}} \, dx=\frac {2 \sqrt {d}\, \left (115 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a +13 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b \,x^{2}+7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c \,x^{4}-120 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a c +6 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) b^{2}+96 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{6}+b \,x^{4}+a \,x^{2}}d x \right ) a^{2}\right )}{77 \sqrt {x}\, d^{2}} \] Input:

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

Output:

(2*sqrt(d)*(115*sqrt(a + b*x**2 + c*x**4)*a + 13*sqrt(a + b*x**2 + c*x**4) 
*b*x**2 + 7*sqrt(a + b*x**2 + c*x**4)*c*x**4 - 120*sqrt(x)*int((sqrt(x)*sq 
rt(a + b*x**2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*a*c + 6*sqrt(x)*int 
((sqrt(x)*sqrt(a + b*x**2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*b**2 + 
96*sqrt(x)*int((sqrt(x)*sqrt(a + b*x**2 + c*x**4))/(a*x**2 + b*x**4 + c*x* 
*6),x)*a**2))/(77*sqrt(x)*d**2)