\(\int (d x)^m (a+b x^2+c x^4)^{3/2} \, dx\) [1088]

Optimal result

Integrand size = 22, antiderivative size = 158 \[ \int (d x)^m \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {a (d x)^{1+m} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {3}{2},-\frac {3}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m) \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \] Output:

a*(d*x)^(1+m)*(c*x^4+b*x^2+a)^(1/2)*AppellF1(1/2+1/2*m,-3/2,-3/2,3/2+1/2*m 
,-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+m)/ 
(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. \(357\) vs. \(2(158)=316\).

Time = 1.74 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.26 \[ \int (d x)^m \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {x (d x)^m \sqrt {a+b x^2+c x^4} \left (a \left (15+8 m+m^2\right ) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},-\frac {1}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+(1+m) x^2 \left (b (5+m) \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {1}{2},-\frac {1}{2},\frac {5+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+c (3+m) x^2 \operatorname {AppellF1}\left (\frac {5+m}{2},-\frac {1}{2},-\frac {1}{2},\frac {7+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{(1+m) (3+m) (5+m) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}}} \] Input:

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

Output:

(x*(d*x)^m*Sqrt[a + b*x^2 + c*x^4]*(a*(15 + 8*m + m^2)*AppellF1[(1 + m)/2, 
 -1/2, -1/2, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b 
+ Sqrt[b^2 - 4*a*c])] + (1 + m)*x^2*(b*(5 + m)*AppellF1[(3 + m)/2, -1/2, - 
1/2, (5 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b 
^2 - 4*a*c])] + c*(3 + m)*x^2*AppellF1[(5 + m)/2, -1/2, -1/2, (7 + m)/2, ( 
-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])])))/ 
((1 + m)*(3 + m)*(5 + m)*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])])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.

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

\(\Big \downarrow \) 1461

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (d x)^m \left (a+b x^2+c x^4\right )^{3/2} \, dx=\text {too large to display} \] Input:

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

Output:

(d**m*(x**m*sqrt(a + b*x**2 + c*x**4)*a*c*m**2*x + 14*x**m*sqrt(a + b*x**2 
 + c*x**4)*a*c*m*x + 45*x**m*sqrt(a + b*x**2 + c*x**4)*a*c*x + 3*x**m*sqrt 
(a + b*x**2 + c*x**4)*b**2*x + x**m*sqrt(a + b*x**2 + c*x**4)*b*c*m**2*x** 
3 + 11*x**m*sqrt(a + b*x**2 + c*x**4)*b*c*m*x**3 + 24*x**m*sqrt(a + b*x**2 
 + c*x**4)*b*c*x**3 + x**m*sqrt(a + b*x**2 + c*x**4)*c**2*m**2*x**5 + 8*x* 
*m*sqrt(a + b*x**2 + c*x**4)*c**2*m*x**5 + 15*x**m*sqrt(a + b*x**2 + c*x** 
4)*c**2*x**5 + 12*int((x**m*sqrt(a + b*x**2 + c*x**4)*x**2)/(a*m**3 + 15*a 
*m**2 + 71*a*m + 105*a + b*m**3*x**2 + 15*b*m**2*x**2 + 71*b*m*x**2 + 105* 
b*x**2 + c*m**3*x**4 + 15*c*m**2*x**4 + 71*c*m*x**4 + 105*c*x**4),x)*a*b*c 
*m**4 + 228*int((x**m*sqrt(a + b*x**2 + c*x**4)*x**2)/(a*m**3 + 15*a*m**2 
+ 71*a*m + 105*a + b*m**3*x**2 + 15*b*m**2*x**2 + 71*b*m*x**2 + 105*b*x**2 
 + c*m**3*x**4 + 15*c*m**2*x**4 + 71*c*m*x**4 + 105*c*x**4),x)*a*b*c*m**3 
+ 1572*int((x**m*sqrt(a + b*x**2 + c*x**4)*x**2)/(a*m**3 + 15*a*m**2 + 71* 
a*m + 105*a + b*m**3*x**2 + 15*b*m**2*x**2 + 71*b*m*x**2 + 105*b*x**2 + c* 
m**3*x**4 + 15*c*m**2*x**4 + 71*c*m*x**4 + 105*c*x**4),x)*a*b*c*m**2 + 466 
8*int((x**m*sqrt(a + b*x**2 + c*x**4)*x**2)/(a*m**3 + 15*a*m**2 + 71*a*m + 
 105*a + b*m**3*x**2 + 15*b*m**2*x**2 + 71*b*m*x**2 + 105*b*x**2 + c*m**3* 
x**4 + 15*c*m**2*x**4 + 71*c*m*x**4 + 105*c*x**4),x)*a*b*c*m + 5040*int((x 
**m*sqrt(a + b*x**2 + c*x**4)*x**2)/(a*m**3 + 15*a*m**2 + 71*a*m + 105*a + 
 b*m**3*x**2 + 15*b*m**2*x**2 + 71*b*m*x**2 + 105*b*x**2 + c*m**3*x**4 ...