\(\int x^3 (c+d \sqrt {a+b x^2})^{3/2} \, dx\) [256]

Optimal result

Integrand size = 23, antiderivative size = 141 \[ \int x^3 \left (c+d \sqrt {a+b x^2}\right )^{3/2} \, dx=-\frac {2 c \left (c^2-a d^2\right ) \left (c+d \sqrt {a+b x^2}\right )^{5/2}}{5 b^2 d^4}+\frac {2 \left (3 c^2-a d^2\right ) \left (c+d \sqrt {a+b x^2}\right )^{7/2}}{7 b^2 d^4}-\frac {2 c \left (c+d \sqrt {a+b x^2}\right )^{9/2}}{3 b^2 d^4}+\frac {2 \left (c+d \sqrt {a+b x^2}\right )^{11/2}}{11 b^2 d^4} \] Output:

-2/5*c*(-a*d^2+c^2)*(c+d*(b*x^2+a)^(1/2))^(5/2)/b^2/d^4+2/7*(-a*d^2+3*c^2) 
*(c+d*(b*x^2+a)^(1/2))^(7/2)/b^2/d^4-2/3*c*(c+d*(b*x^2+a)^(1/2))^(9/2)/b^2 
/d^4+2/11*(c+d*(b*x^2+a)^(1/2))^(11/2)/b^2/d^4
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.15 \[ \int x^3 \left (c+d \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \sqrt {c+d \sqrt {a+b x^2}} \left (-16 c^5+6 c^3 d^2 \left (10 a-b x^2\right )+8 c^4 d \sqrt {a+b x^2}+c^2 d^3 \sqrt {a+b x^2} \left (-28 a+5 b x^2\right )-4 c d^4 \left (31 a^2-4 a b x^2-35 b^2 x^4\right )+15 d^5 \sqrt {a+b x^2} \left (-4 a^2+3 a b x^2+7 b^2 x^4\right )\right )}{1155 b^2 d^4} \] Input:

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

Output:

(2*Sqrt[c + d*Sqrt[a + b*x^2]]*(-16*c^5 + 6*c^3*d^2*(10*a - b*x^2) + 8*c^4 
*d*Sqrt[a + b*x^2] + c^2*d^3*Sqrt[a + b*x^2]*(-28*a + 5*b*x^2) - 4*c*d^4*( 
31*a^2 - 4*a*b*x^2 - 35*b^2*x^4) + 15*d^5*Sqrt[a + b*x^2]*(-4*a^2 + 3*a*b* 
x^2 + 7*b^2*x^4)))/(1155*b^2*d^4)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 134, 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.

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

\(\Big \downarrow \) 7283

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1732

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

\(\Big \downarrow \) 522

\(\displaystyle -\frac {\int \left (-\frac {\left (c+d \sqrt {b x^2+a}\right )^{9/2}}{d^3}+\frac {3 c \left (c+d \sqrt {b x^2+a}\right )^{7/2}}{d^3}+\frac {\left (a d^2-3 c^2\right ) \left (c+d \sqrt {b x^2+a}\right )^{5/2}}{d^3}+\frac {\left (c^3-a c d^2\right ) \left (c+d \sqrt {b x^2+a}\right )^{3/2}}{d^3}\right )d\sqrt {b x^2+a}}{b^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {2 \left (3 c^2-a d^2\right ) \left (d \sqrt {a+b x^2}+c\right )^{7/2}}{7 d^4}+\frac {2 c \left (c^2-a d^2\right ) \left (d \sqrt {a+b x^2}+c\right )^{5/2}}{5 d^4}-\frac {2 \left (d \sqrt {a+b x^2}+c\right )^{11/2}}{11 d^4}+\frac {2 c \left (d \sqrt {a+b x^2}+c\right )^{9/2}}{3 d^4}}{b^2}\)

Input:

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

Output:

-(((2*c*(c^2 - a*d^2)*(c + d*Sqrt[a + b*x^2])^(5/2))/(5*d^4) - (2*(3*c^2 - 
 a*d^2)*(c + d*Sqrt[a + b*x^2])^(7/2))/(7*d^4) + (2*c*(c + d*Sqrt[a + b*x^ 
2])^(9/2))/(3*d^4) - (2*(c + d*Sqrt[a + b*x^2])^(11/2))/(11*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]

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

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 = 0.17 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.05 \[ \int x^3 \left (c+d \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \, {\left (140 \, b^{2} c d^{4} x^{4} - 124 \, a^{2} c d^{4} + 60 \, a c^{3} d^{2} - 16 \, c^{5} + 2 \, {\left (8 \, a b c d^{4} - 3 \, b c^{3} d^{2}\right )} x^{2} + {\left (105 \, b^{2} d^{5} x^{4} - 60 \, a^{2} d^{5} - 28 \, a c^{2} d^{3} + 8 \, c^{4} d + 5 \, {\left (9 \, a b d^{5} + b c^{2} d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a}\right )} \sqrt {\sqrt {b x^{2} + a} d + c}}{1155 \, b^{2} d^{4}} \] Input:

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

Output:

2/1155*(140*b^2*c*d^4*x^4 - 124*a^2*c*d^4 + 60*a*c^3*d^2 - 16*c^5 + 2*(8*a 
*b*c*d^4 - 3*b*c^3*d^2)*x^2 + (105*b^2*d^5*x^4 - 60*a^2*d^5 - 28*a*c^2*d^3 
 + 8*c^4*d + 5*(9*a*b*d^5 + b*c^2*d^3)*x^2)*sqrt(b*x^2 + a))*sqrt(sqrt(b*x 
^2 + a)*d + c)/(b^2*d^4)
 

Sympy [F]

\[ \int x^3 \left (c+d \sqrt {a+b x^2}\right )^{3/2} \, dx=\int x^{3} \left (c + d \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}\, dx \] 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 = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.72 \[ \int x^3 \left (c+d \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \, {\left (105 \, {\left (\sqrt {b x^{2} + a} d + c\right )}^{\frac {11}{2}} - 385 \, {\left (\sqrt {b x^{2} + a} d + c\right )}^{\frac {9}{2}} c - 165 \, {\left (a d^{2} - 3 \, c^{2}\right )} {\left (\sqrt {b x^{2} + a} d + c\right )}^{\frac {7}{2}} + 231 \, {\left (a c d^{2} - c^{3}\right )} {\left (\sqrt {b x^{2} + a} d + c\right )}^{\frac {5}{2}}\right )}}{1155 \, b^{2} d^{4}} \] Input:

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

Output:

2/1155*(105*(sqrt(b*x^2 + a)*d + c)^(11/2) - 385*(sqrt(b*x^2 + a)*d + c)^( 
9/2)*c - 165*(a*d^2 - 3*c^2)*(sqrt(b*x^2 + a)*d + c)^(7/2) + 231*(a*c*d^2 
- c^3)*(sqrt(b*x^2 + a)*d + c)^(5/2))/(b^2*d^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1488 vs. \(2 (117) = 234\).

Time = 0.20 (sec) , antiderivative size = 1488, normalized size of antiderivative = 10.55 \[ \int x^3 \left (c+d \sqrt {a+b x^2}\right )^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

-2/3465*(231*(15*sqrt(sqrt(b*x^2 + a)*d + c)*a - (3*(sqrt(b*x^2 + a)*d + c 
)^(5/2) - 10*(sqrt(b*x^2 + a)*d + c)^(3/2)*c + 15*sqrt(sqrt(b*x^2 + a)*d + 
 c)*c^2)/d^2)*a*c*d/b + 33*(35*(sqrt(b*x^2 + a)*d + c)^(3/2)*a*d^2*sgn((sq 
rt(b*x^2 + a)*d + c)*d - c*d) - 105*sqrt(sqrt(b*x^2 + a)*d + c)*a*c*d^2*sg 
n((sqrt(b*x^2 + a)*d + c)*d - c*d) - 15*(sqrt(b*x^2 + a)*d + c)^(7/2)*sgn( 
(sqrt(b*x^2 + a)*d + c)*d - c*d) + 63*(sqrt(b*x^2 + a)*d + c)^(5/2)*c*sgn( 
(sqrt(b*x^2 + a)*d + c)*d - c*d) - 105*(sqrt(b*x^2 + a)*d + c)^(3/2)*c^2*s 
gn((sqrt(b*x^2 + a)*d + c)*d - c*d) + 105*sqrt(sqrt(b*x^2 + a)*d + c)*c^3* 
sgn((sqrt(b*x^2 + a)*d + c)*d - c*d))*a/(b*d) + 33*(35*(sqrt(b*x^2 + a)*d 
+ c)^(3/2)*a*d^2*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) - 105*sqrt(sqrt(b*x^ 
2 + a)*d + c)*a*c*d^2*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) - 15*(sqrt(b*x^ 
2 + a)*d + c)^(7/2)*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) + 63*(sqrt(b*x^2 
+ a)*d + c)^(5/2)*c*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) - 105*(sqrt(b*x^2 
 + a)*d + c)^(3/2)*c^2*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d) + 105*sqrt(sqr 
t(b*x^2 + a)*d + c)*c^3*sgn((sqrt(b*x^2 + a)*d + c)*d - c*d))*c^2/(b*d^3) 
- 11*(315*sqrt(sqrt(b*x^2 + a)*d + c)*a^2*d^4 - 126*(sqrt(b*x^2 + a)*d + c 
)^(5/2)*a*d^2 + 420*(sqrt(b*x^2 + a)*d + c)^(3/2)*a*c*d^2 - 630*sqrt(sqrt( 
b*x^2 + a)*d + c)*a*c^2*d^2 + 35*(sqrt(b*x^2 + a)*d + c)^(9/2) - 180*(sqrt 
(b*x^2 + a)*d + c)^(7/2)*c + 378*(sqrt(b*x^2 + a)*d + c)^(5/2)*c^2 - 420*( 
sqrt(b*x^2 + a)*d + c)^(3/2)*c^3 + 315*sqrt(sqrt(b*x^2 + a)*d + c)*c^4)...
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (c+d \sqrt {a+b x^2}\right )^{3/2} \, dx=\int x^3\,{\left (c+d\,\sqrt {b\,x^2+a}\right )}^{3/2} \,d x \] 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 = 0.22 (sec) , antiderivative size = 461, normalized size of antiderivative = 3.27 \[ \int x^3 \left (c+d \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \sqrt {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d x +\sqrt {b \,x^{2}+a}\, c +\sqrt {b}\, c x +a d +b d \,x^{2}}\, \sqrt {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}\, \left (-124 \sqrt {b \,x^{2}+a}\, a^{2} c \,d^{4}+60 \sqrt {b \,x^{2}+a}\, a \,c^{3} d^{2}-15 a^{2} b \,d^{5} x^{2}+150 a \,b^{2} d^{5} x^{4}+5 b^{2} c^{2} d^{3} x^{4}+8 b \,c^{4} d \,x^{2}+16 \sqrt {b}\, c^{5} x -28 a^{2} c^{2} d^{3}+8 a \,c^{4} d +105 b^{3} d^{5} x^{6}+60 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{2} d^{5} x -105 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b^{2} d^{5} x^{5}-8 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, c^{4} d x +140 \sqrt {b \,x^{2}+a}\, b^{2} c \,d^{4} x^{4}-6 \sqrt {b \,x^{2}+a}\, b \,c^{3} d^{2} x^{2}+124 \sqrt {b}\, a^{2} c \,d^{4} x -60 \sqrt {b}\, a \,c^{3} d^{2} x -140 \sqrt {b}\, b^{2} c \,d^{4} x^{5}+6 \sqrt {b}\, b \,c^{3} d^{2} x^{3}-23 a b \,c^{2} d^{3} x^{2}-16 \sqrt {b \,x^{2}+a}\, c^{5}-60 a^{3} d^{5}-45 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a b \,d^{5} x^{3}+28 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a \,c^{2} d^{3} x -5 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b \,c^{2} d^{3} x^{3}+16 \sqrt {b \,x^{2}+a}\, a b c \,d^{4} x^{2}-16 \sqrt {b}\, a b c \,d^{4} x^{3}\right )}{1155 a \,b^{2} d^{4}} \] 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)*(60*sqrt(b)*sqrt(a + b* 
x**2)*a**2*d**5*x - 45*sqrt(b)*sqrt(a + b*x**2)*a*b*d**5*x**3 + 28*sqrt(b) 
*sqrt(a + b*x**2)*a*c**2*d**3*x - 105*sqrt(b)*sqrt(a + b*x**2)*b**2*d**5*x 
**5 - 5*sqrt(b)*sqrt(a + b*x**2)*b*c**2*d**3*x**3 - 8*sqrt(b)*sqrt(a + b*x 
**2)*c**4*d*x - 124*sqrt(a + b*x**2)*a**2*c*d**4 + 16*sqrt(a + b*x**2)*a*b 
*c*d**4*x**2 + 60*sqrt(a + b*x**2)*a*c**3*d**2 + 140*sqrt(a + b*x**2)*b**2 
*c*d**4*x**4 - 6*sqrt(a + b*x**2)*b*c**3*d**2*x**2 - 16*sqrt(a + b*x**2)*c 
**5 + 124*sqrt(b)*a**2*c*d**4*x - 16*sqrt(b)*a*b*c*d**4*x**3 - 60*sqrt(b)* 
a*c**3*d**2*x - 140*sqrt(b)*b**2*c*d**4*x**5 + 6*sqrt(b)*b*c**3*d**2*x**3 
+ 16*sqrt(b)*c**5*x - 60*a**3*d**5 - 15*a**2*b*d**5*x**2 - 28*a**2*c**2*d* 
*3 + 150*a*b**2*d**5*x**4 - 23*a*b*c**2*d**3*x**2 + 8*a*c**4*d + 105*b**3* 
d**5*x**6 + 5*b**2*c**2*d**3*x**4 + 8*b*c**4*d*x**2))/(1155*a*b**2*d**4)