\(\int x^3 \sqrt {a+b (c x^2)^{3/2}} \, dx\) [41]

Optimal result

Integrand size = 21, antiderivative size = 340 \[ \int x^3 \sqrt {a+b \left (c x^2\right )^{3/2}} \, dx=\frac {2}{11} x^4 \sqrt {a+b \left (c x^2\right )^{3/2}}+\frac {6 a \sqrt {c x^2} \sqrt {a+b \left (c x^2\right )^{3/2}}}{55 b c^2}-\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}+b^{2/3} c x^2-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}}\right ),-7-4 \sqrt {3}\right )}{55 b^{4/3} c^2 \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}} \] Output:

2/11*x^4*(a+b*(c*x^2)^(3/2))^(1/2)+6/55*a*(c*x^2)^(1/2)*(a+b*(c*x^2)^(3/2) 
)^(1/2)/b/c^2-4/55*3^(3/4)*(1/2*6^(1/2)+1/2*2^(1/2))*a^2*(a^(1/3)+b^(1/3)* 
(c*x^2)^(1/2))*((a^(2/3)+b^(2/3)*c*x^2-a^(1/3)*b^(1/3)*(c*x^2)^(1/2))/((1+ 
3^(1/2))*a^(1/3)+b^(1/3)*(c*x^2)^(1/2))^2)^(1/2)*EllipticF(((1-3^(1/2))*a^ 
(1/3)+b^(1/3)*(c*x^2)^(1/2))/((1+3^(1/2))*a^(1/3)+b^(1/3)*(c*x^2)^(1/2)),I 
*3^(1/2)+2*I)/b^(4/3)/c^2/(a^(1/3)*(a^(1/3)+b^(1/3)*(c*x^2)^(1/2))/((1+3^( 
1/2))*a^(1/3)+b^(1/3)*(c*x^2)^(1/2))^2)^(1/2)/(a+b*(c*x^2)^(3/2))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.95 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.32 \[ \int x^3 \sqrt {a+b \left (c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {c x^2} \sqrt {a+b \left (c x^2\right )^{3/2}} \left (a \left (\frac {a+b \left (c x^2\right )^{3/2}}{a}\right )^{3/2}-a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{3},\frac {4}{3},-\frac {b \left (c x^2\right )^{3/2}}{a}\right )\right )}{11 b c^2 \sqrt {\frac {a+b \left (c x^2\right )^{3/2}}{a}}} \] Input:

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

Output:

(2*Sqrt[c*x^2]*Sqrt[a + b*(c*x^2)^(3/2)]*(a*((a + b*(c*x^2)^(3/2))/a)^(3/2 
) - a*Hypergeometric2F1[-1/2, 1/3, 4/3, -((b*(c*x^2)^(3/2))/a)]))/(11*b*c^ 
2*Sqrt[(a + b*(c*x^2)^(3/2))/a])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {892, 811, 843, 759}

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.

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

\(\Big \downarrow \) 892

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

\(\Big \downarrow \) 811

\(\displaystyle \frac {\frac {3}{11} a \int \frac {\left (c x^2\right )^{3/2}}{\sqrt {b \left (c x^2\right )^{3/2}+a}}d\sqrt {c x^2}+\frac {2}{11} c^2 x^4 \sqrt {a+b \left (c x^2\right )^{3/2}}}{c^2}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {\frac {3}{11} a \left (\frac {2 \sqrt {c x^2} \sqrt {a+b \left (c x^2\right )^{3/2}}}{5 b}-\frac {2 a \int \frac {1}{\sqrt {b \left (c x^2\right )^{3/2}+a}}d\sqrt {c x^2}}{5 b}\right )+\frac {2}{11} c^2 x^4 \sqrt {a+b \left (c x^2\right )^{3/2}}}{c^2}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {\frac {3}{11} a \left (\frac {2 \sqrt {c x^2} \sqrt {a+b \left (c x^2\right )^{3/2}}}{5 b}-\frac {4 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}+b^{2/3} c x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} \sqrt {c x^2}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} \sqrt {c x^2}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} b^{4/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}}\right )+\frac {2}{11} c^2 x^4 \sqrt {a+b \left (c x^2\right )^{3/2}}}{c^2}\)

Input:

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

Output:

((2*c^2*x^4*Sqrt[a + b*(c*x^2)^(3/2)])/11 + (3*a*((2*Sqrt[c*x^2]*Sqrt[a + 
b*(c*x^2)^(3/2)])/(5*b) - (4*Sqrt[2 + Sqrt[3]]*a*(a^(1/3) + b^(1/3)*Sqrt[c 
*x^2])*Sqrt[(a^(2/3) + b^(2/3)*c*x^2 - a^(1/3)*b^(1/3)*Sqrt[c*x^2])/((1 + 
Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])^2]*EllipticF[ArcSin[((1 - Sqrt[3]) 
*a^(1/3) + b^(1/3)*Sqrt[c*x^2])/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^ 
2])], -7 - 4*Sqrt[3]])/(5*3^(1/4)*b^(4/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3) 
*Sqrt[c*x^2]))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])^2]*Sqrt[a + b 
*(c*x^2)^(3/2)])))/11)/c^2
 

Defintions of rubi rules used

rule 759

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* 
((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s 
+ r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & 
& PosQ[a]
 

rule 811

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 
))   Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m 
, p, x]
 

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 892

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo 
l] :> Simp[(d*x)^(m + 1)/(d*((c*x^q)^(1/q))^(m + 1))   Subst[Int[x^m*(a + b 
*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x 
] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.28 \[ \int x^3 \sqrt {a+b \left (c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (6 \, \sqrt {\frac {\sqrt {c x^{2}} b c}{x}} a^{2} {\rm weierstrassPInverse}\left (0, -\frac {4 \, \sqrt {c x^{2}} a}{b c^{2} x}, x\right ) - {\left (5 \, b^{2} c^{3} x^{4} + 3 \, \sqrt {c x^{2}} a b c\right )} \sqrt {\sqrt {c x^{2}} b c x^{2} + a}\right )}}{55 \, b^{2} c^{3}} \] Input:

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

Output:

-2/55*(6*sqrt(sqrt(c*x^2)*b*c/x)*a^2*weierstrassPInverse(0, -4*sqrt(c*x^2) 
*a/(b*c^2*x), x) - (5*b^2*c^3*x^4 + 3*sqrt(c*x^2)*a*b*c)*sqrt(sqrt(c*x^2)* 
b*c*x^2 + a))/(b^2*c^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^3 \sqrt {a+b \left (c x^2\right )^{3/2}} \, dx=\frac {\frac {6 \sqrt {c}\, \sqrt {\sqrt {c}\, b c \,x^{3}+a}\, a x}{55}+\frac {2 \sqrt {\sqrt {c}\, b c \,x^{3}+a}\, b \,c^{2} x^{4}}{11}-\frac {6 \sqrt {c}\, \left (\int \frac {\sqrt {\sqrt {c}\, b c \,x^{3}+a}}{-b^{2} c^{3} x^{6}+a^{2}}d x \right ) a^{3}}{55}+\frac {6 \left (\int \frac {\sqrt {\sqrt {c}\, b c \,x^{3}+a}\, x^{3}}{-b^{2} c^{3} x^{6}+a^{2}}d x \right ) a^{2} b \,c^{2}}{55}}{b \,c^{2}} \] Input:

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

Output:

(2*(3*sqrt(c)*sqrt(sqrt(c)*b*c*x**3 + a)*a*x + 5*sqrt(sqrt(c)*b*c*x**3 + a 
)*b*c**2*x**4 - 3*sqrt(c)*int(sqrt(sqrt(c)*b*c*x**3 + a)/(a**2 - b**2*c**3 
*x**6),x)*a**3 + 3*int((sqrt(sqrt(c)*b*c*x**3 + a)*x**3)/(a**2 - b**2*c**3 
*x**6),x)*a**2*b*c**2))/(55*b*c**2)