\(\int \frac {x^3}{(c+d \sqrt {a+b x^2})^{3/2}} \, dx\) [281]

Optimal result

Integrand size = 23, antiderivative size = 135 \[ \int \frac {x^3}{\left (c+d \sqrt {a+b x^2}\right )^{3/2}} \, dx=\frac {2 c \left (c^2-a d^2\right )}{b^2 d^4 \sqrt {c+d \sqrt {a+b x^2}}}+\frac {2 \left (3 c^2-a d^2\right ) \sqrt {c+d \sqrt {a+b x^2}}}{b^2 d^4}-\frac {2 c \left (c+d \sqrt {a+b x^2}\right )^{3/2}}{b^2 d^4}+\frac {2 \left (c+d \sqrt {a+b x^2}\right )^{5/2}}{5 b^2 d^4} \] Output:

2*c*(-a*d^2+c^2)/b^2/d^4/(c+d*(b*x^2+a)^(1/2))^(1/2)+2*(-a*d^2+3*c^2)*(c+d 
*(b*x^2+a)^(1/2))^(1/2)/b^2/d^4-2*c*(c+d*(b*x^2+a)^(1/2))^(3/2)/b^2/d^4+2/ 
5*(c+d*(b*x^2+a)^(1/2))^(5/2)/b^2/d^4
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.63 \[ \int \frac {x^3}{\left (c+d \sqrt {a+b x^2}\right )^{3/2}} \, dx=\frac {32 c^3-24 a c d^2-4 b c d^2 x^2+2 \sqrt {a+b x^2} \left (8 c^2 d+d^3 \left (-4 a+b x^2\right )\right )}{5 b^2 d^4 \sqrt {c+d \sqrt {a+b x^2}}} \] Input:

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

Output:

(32*c^3 - 24*a*c*d^2 - 4*b*c*d^2*x^2 + 2*Sqrt[a + b*x^2]*(8*c^2*d + d^3*(- 
4*a + b*x^2)))/(5*b^2*d^4*Sqrt[c + d*Sqrt[a + b*x^2]])
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {7283, 896, 25, 1732, 522, 2009}

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.

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

\(\Big \downarrow \) 7283

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1732

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

\(\Big \downarrow \) 522

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-(((-2*c*(c^2 - a*d^2))/(d^4*Sqrt[c + d*Sqrt[a + b*x^2]]) - (2*(3*c^2 - a* 
d^2)*Sqrt[c + d*Sqrt[a + b*x^2]])/d^4 + (2*c*(c + d*Sqrt[a + b*x^2])^(3/2) 
)/d^4 - (2*(c + d*Sqrt[a + b*x^2])^(5/2))/(5*d^4))/b^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

rule 1732

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb 
ol] :> With[{g = Denominator[n]}, Simp[g   Subst[Int[x^(g - 1)*(d + e*x^(g* 
n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} 
, x] && EqQ[n2, 2*n] && FractionQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7283

Int[(u_)*(x_)^(m_.), x_Symbol] :> With[{lst = PowerVariableExpn[u, m + 1, x 
]}, Simp[1/lst[[2]]   Subst[Int[NormalizeIntegrand[Simplify[lst[[1]]/x], x] 
, x], x, (lst[[3]]*x)^lst[[2]]], x] /;  !FalseQ[lst] && NeQ[lst[[2]], m + 1 
]] /; IntegerQ[m] && NeQ[m, -1] && NonsumQ[u] && (GtQ[m, 0] ||  !AlgebraicF 
unctionQ[u, x])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.03 \[ \int \frac {x^3}{\left (c+d \sqrt {a+b x^2}\right )^{3/2}} \, dx=\frac {2 \, {\left (b^{2} d^{4} x^{4} - 4 \, a^{2} d^{4} + 20 \, a c^{2} d^{2} - 16 \, c^{4} - {\left (3 \, a b d^{4} - 10 \, b c^{2} d^{2}\right )} x^{2} - {\left (3 \, b c d^{3} x^{2} + 8 \, a c d^{3} - 8 \, c^{3} d\right )} \sqrt {b x^{2} + a}\right )} \sqrt {\sqrt {b x^{2} + a} d + c}}{5 \, {\left (b^{3} d^{6} x^{2} + a b^{2} d^{6} - b^{2} c^{2} d^{4}\right )}} \] Input:

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

Output:

2/5*(b^2*d^4*x^4 - 4*a^2*d^4 + 20*a*c^2*d^2 - 16*c^4 - (3*a*b*d^4 - 10*b*c 
^2*d^2)*x^2 - (3*b*c*d^3*x^2 + 8*a*c*d^3 - 8*c^3*d)*sqrt(b*x^2 + a))*sqrt( 
sqrt(b*x^2 + a)*d + c)/(b^3*d^6*x^2 + a*b^2*d^6 - b^2*c^2*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.79 \[ \int \frac {x^3}{\left (c+d \sqrt {a+b x^2}\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {{\left (\sqrt {b x^{2} + a} d + c\right )}^{\frac {5}{2}} - 5 \, {\left (\sqrt {b x^{2} + a} d + c\right )}^{\frac {3}{2}} c - 5 \, {\left (a d^{2} - 3 \, c^{2}\right )} \sqrt {\sqrt {b x^{2} + a} d + c}}{d^{2}} - \frac {5 \, {\left (a c d^{2} - c^{3}\right )}}{\sqrt {\sqrt {b x^{2} + a} d + c} d^{2}}\right )}}{5 \, b^{2} d^{2}} \] Input:

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

Output:

2/5*(((sqrt(b*x^2 + a)*d + c)^(5/2) - 5*(sqrt(b*x^2 + a)*d + c)^(3/2)*c - 
5*(a*d^2 - 3*c^2)*sqrt(sqrt(b*x^2 + a)*d + c))/d^2 - 5*(a*c*d^2 - c^3)/(sq 
rt(sqrt(b*x^2 + a)*d + c)*d^2))/(b^2*d^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (117) = 234\).

Time = 0.18 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.87 \[ \int \frac {x^3}{\left (c+d \sqrt {a+b x^2}\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {15 \, {\left (a c d^{2} - c^{3}\right )} \arctan \left (\frac {\sqrt {\sqrt {b x^{2} + a} d + c} \mathrm {sgn}\left ({\left (\sqrt {b x^{2} + a} d + c\right )} d - c d\right )}{\sqrt {c \mathrm {sgn}\left ({\left (\sqrt {b x^{2} + a} d + c\right )} d - c d\right ) - c}}\right )}{\sqrt {c \mathrm {sgn}\left ({\left (\sqrt {b x^{2} + a} d + c\right )} d - c d\right ) - c} b d^{2}} - \frac {15 \, \sqrt {\sqrt {b x^{2} + a} d + c} a b^{4} d^{10} \mathrm {sgn}\left ({\left (\sqrt {b x^{2} + a} d + c\right )} d - c d\right ) - 3 \, {\left (\sqrt {b x^{2} + a} d + c\right )}^{\frac {5}{2}} b^{4} d^{8} \mathrm {sgn}\left ({\left (\sqrt {b x^{2} + a} d + c\right )} d - c d\right ) + 10 \, {\left (\sqrt {b x^{2} + a} d + c\right )}^{\frac {3}{2}} b^{4} c d^{8} \mathrm {sgn}\left ({\left (\sqrt {b x^{2} + a} d + c\right )} d - c d\right ) - 30 \, \sqrt {\sqrt {b x^{2} + a} d + c} b^{4} c^{2} d^{8} \mathrm {sgn}\left ({\left (\sqrt {b x^{2} + a} d + c\right )} d - c d\right ) + 5 \, {\left (\sqrt {b x^{2} + a} d + c\right )}^{\frac {3}{2}} b^{4} c d^{8} - 15 \, \sqrt {\sqrt {b x^{2} + a} d + c} b^{4} c^{2} d^{8}}{b^{5} d^{10} \mathrm {sgn}\left ({\left (\sqrt {b x^{2} + a} d + c\right )} d - c d\right )}\right )}}{15 \, b d^{2}} \] Input:

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

Output:

2/15*(15*(a*c*d^2 - c^3)*arctan(sqrt(sqrt(b*x^2 + a)*d + c)*sgn((sqrt(b*x^ 
2 + a)*d + c)*d - c*d)/sqrt(c*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) - c))/( 
sqrt(c*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) - c)*b*d^2) - (15*sqrt(sqrt(b* 
x^2 + a)*d + c)*a*b^4*d^10*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) - 3*(sqrt( 
b*x^2 + a)*d + c)^(5/2)*b^4*d^8*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) + 10* 
(sqrt(b*x^2 + a)*d + c)^(3/2)*b^4*c*d^8*sgn((sqrt(b*x^2 + a)*d + c)*d - c* 
d) - 30*sqrt(sqrt(b*x^2 + a)*d + c)*b^4*c^2*d^8*sgn((sqrt(b*x^2 + a)*d + c 
)*d - c*d) + 5*(sqrt(b*x^2 + a)*d + c)^(3/2)*b^4*c*d^8 - 15*sqrt(sqrt(b*x^ 
2 + a)*d + c)*b^4*c^2*d^8)/(b^5*d^10*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d)) 
)/(b*d^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.68 \[ \int \frac {x^3}{\left (c+d \sqrt {a+b x^2}\right )^{3/2}} \, dx=\frac {2 \sqrt {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d x +\sqrt {b \,x^{2}+a}\, c +\sqrt {b}\, c x +a d +b d \,x^{2}}\, \sqrt {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}\, \left (8 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a c \,d^{3} x +3 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b c \,d^{3} x^{3}-8 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, c^{3} d x -4 \sqrt {b \,x^{2}+a}\, a^{2} d^{4}-3 \sqrt {b \,x^{2}+a}\, a b \,d^{4} x^{2}+20 \sqrt {b \,x^{2}+a}\, a \,c^{2} d^{2}+\sqrt {b \,x^{2}+a}\, b^{2} d^{4} x^{4}+10 \sqrt {b \,x^{2}+a}\, b \,c^{2} d^{2} x^{2}-16 \sqrt {b \,x^{2}+a}\, c^{4}+4 \sqrt {b}\, a^{2} d^{4} x +3 \sqrt {b}\, a b \,d^{4} x^{3}-20 \sqrt {b}\, a \,c^{2} d^{2} x -\sqrt {b}\, b^{2} d^{4} x^{5}-10 \sqrt {b}\, b \,c^{2} d^{2} x^{3}+16 \sqrt {b}\, c^{4} x -8 a^{2} c \,d^{3}-11 a b c \,d^{3} x^{2}+8 a \,c^{3} d -3 b^{2} c \,d^{3} x^{4}+8 b \,c^{3} d \,x^{2}\right )}{5 a \,b^{2} d^{4} \left (b \,d^{2} x^{2}+a \,d^{2}-c^{2}\right )} \] Input:

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

Output:

(2*sqrt(sqrt(b)*sqrt(a + b*x**2)*d*x + sqrt(a + b*x**2)*c + sqrt(b)*c*x + 
a*d + b*d*x**2)*sqrt(sqrt(a + b*x**2) + sqrt(b)*x)*(8*sqrt(b)*sqrt(a + b*x 
**2)*a*c*d**3*x + 3*sqrt(b)*sqrt(a + b*x**2)*b*c*d**3*x**3 - 8*sqrt(b)*sqr 
t(a + b*x**2)*c**3*d*x - 4*sqrt(a + b*x**2)*a**2*d**4 - 3*sqrt(a + b*x**2) 
*a*b*d**4*x**2 + 20*sqrt(a + b*x**2)*a*c**2*d**2 + sqrt(a + b*x**2)*b**2*d 
**4*x**4 + 10*sqrt(a + b*x**2)*b*c**2*d**2*x**2 - 16*sqrt(a + b*x**2)*c**4 
 + 4*sqrt(b)*a**2*d**4*x + 3*sqrt(b)*a*b*d**4*x**3 - 20*sqrt(b)*a*c**2*d** 
2*x - sqrt(b)*b**2*d**4*x**5 - 10*sqrt(b)*b*c**2*d**2*x**3 + 16*sqrt(b)*c* 
*4*x - 8*a**2*c*d**3 - 11*a*b*c*d**3*x**2 + 8*a*c**3*d - 3*b**2*c*d**3*x** 
4 + 8*b*c**3*d*x**2))/(5*a*b**2*d**4*(a*d**2 + b*d**2*x**2 - c**2))