\(\int x^3 (c+d x)^2 \sqrt {a+b x^2} \, dx\) [984]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int x^3 (c+d x)^2 \sqrt {a+b x^2} \, dx=-\frac {a^2 c d x \sqrt {a+b x^2}}{8 b^2}+\frac {a c d x^3 \sqrt {a+b x^2}}{12 b}+\frac {1}{3} c d x^5 \sqrt {a+b x^2}-\frac {a \left (7 b c^2-4 a d^2\right ) \left (a+b x^2\right )^{3/2}}{21 b^3}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {\left (7 b c^2-4 a d^2\right ) \left (a+b x^2\right )^{5/2}}{35 b^3}+\frac {a^3 c d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Output:

-1/8*a^2*c*d*x*(b*x^2+a)^(1/2)/b^2+1/12*a*c*d*x^3*(b*x^2+a)^(1/2)/b+1/3*c* 
d*x^5*(b*x^2+a)^(1/2)-1/21*a*(-4*a*d^2+7*b*c^2)*(b*x^2+a)^(3/2)/b^3+1/7*d^ 
2*x^4*(b*x^2+a)^(3/2)/b+1/35*(-4*a*d^2+7*b*c^2)*(b*x^2+a)^(5/2)/b^3+1/8*a^ 
3*c*d*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.75 \[ \int x^3 (c+d x)^2 \sqrt {a+b x^2} \, dx=\frac {\sqrt {a+b x^2} \left (64 a^3 d^2+2 a b^2 x^2 \left (28 c^2+35 c d x+12 d^2 x^2\right )+8 b^3 x^4 \left (21 c^2+35 c d x+15 d^2 x^2\right )-a^2 b \left (112 c^2+105 c d x+32 d^2 x^2\right )\right )-105 a^3 \sqrt {b} c d \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{840 b^3} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(64*a^3*d^2 + 2*a*b^2*x^2*(28*c^2 + 35*c*d*x + 12*d^2*x^2 
) + 8*b^3*x^4*(21*c^2 + 35*c*d*x + 15*d^2*x^2) - a^2*b*(112*c^2 + 105*c*d* 
x + 32*d^2*x^2)) - 105*a^3*Sqrt[b]*c*d*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]] 
)/(840*b^3)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {541, 533, 27, 533, 25, 27, 533, 27, 455, 211, 224, 219}

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.

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

\(\Big \downarrow \) 541

\(\displaystyle \frac {\int x^3 \left (7 b c^2+14 b d x c-4 a d^2\right ) \sqrt {b x^2+a}dx}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}-\frac {\int 6 b x^2 \left (7 a c d-\left (7 b c^2-4 a d^2\right ) x\right ) \sqrt {b x^2+a}dx}{6 b}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {\int -a x \left (2 \left (7 b c^2-4 a d^2\right )+35 b c d x\right ) \sqrt {b x^2+a}dx}{5 b}+\frac {x^2 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{5 b}+\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int a x \left (2 \left (7 b c^2-4 a d^2\right )+35 b c d x\right ) \sqrt {b x^2+a}dx}{5 b}+\frac {x^2 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{5 b}+\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {a \int x \left (2 \left (7 b c^2-4 a d^2\right )+35 b c d x\right ) \sqrt {b x^2+a}dx}{5 b}+\frac {x^2 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{5 b}+\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {-\frac {a \left (\frac {35}{4} c d x \left (a+b x^2\right )^{3/2}-\frac {\int b \left (35 a c d-8 \left (7 b c^2-4 a d^2\right ) x\right ) \sqrt {b x^2+a}dx}{4 b}\right )}{5 b}+\frac {x^2 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{5 b}+\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {a \left (\frac {35}{4} c d x \left (a+b x^2\right )^{3/2}-\frac {1}{4} \int \left (35 a c d-8 \left (7 b c^2-4 a d^2\right ) x\right ) \sqrt {b x^2+a}dx\right )}{5 b}+\frac {x^2 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{5 b}+\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {-\frac {a \left (\frac {1}{4} \left (\frac {8 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{3 b}-35 a c d \int \sqrt {b x^2+a}dx\right )+\frac {35}{4} c d x \left (a+b x^2\right )^{3/2}\right )}{5 b}+\frac {x^2 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{5 b}+\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {-\frac {a \left (\frac {1}{4} \left (\frac {8 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{3 b}-35 a c d \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )\right )+\frac {35}{4} c d x \left (a+b x^2\right )^{3/2}\right )}{5 b}+\frac {x^2 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{5 b}+\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {-\frac {a \left (\frac {1}{4} \left (\frac {8 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{3 b}-35 a c d \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )\right )+\frac {35}{4} c d x \left (a+b x^2\right )^{3/2}\right )}{5 b}+\frac {x^2 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{5 b}+\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {a \left (\frac {1}{4} \left (\frac {8 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{3 b}-35 a c d \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )\right )+\frac {35}{4} c d x \left (a+b x^2\right )^{3/2}\right )}{5 b}+\frac {x^2 \left (a+b x^2\right )^{3/2} \left (7 b c^2-4 a d^2\right )}{5 b}+\frac {7}{3} c d x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {d^2 x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

Input:

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

Output:

(d^2*x^4*(a + b*x^2)^(3/2))/(7*b) + (((7*b*c^2 - 4*a*d^2)*x^2*(a + b*x^2)^ 
(3/2))/(5*b) + (7*c*d*x^3*(a + b*x^2)^(3/2))/3 - (a*((35*c*d*x*(a + b*x^2) 
^(3/2))/4 + ((8*(7*b*c^2 - 4*a*d^2)*(a + b*x^2)^(3/2))/(3*b) - 35*a*c*d*(( 
x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b] 
)))/4))/(5*b))/(7*b)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 533

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[2*p]
 

rule 541

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x 
] + Simp[1/(b*(m + n + 2*p + 1))   Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + 
 n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) 
*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt 
Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {\left (120 b^{3} d^{2} x^{6}+280 b^{3} c d \,x^{5}+24 a \,b^{2} d^{2} x^{4}+168 b^{3} c^{2} x^{4}+70 a \,b^{2} c d \,x^{3}-32 x^{2} a^{2} b \,d^{2}+56 a \,b^{2} c^{2} x^{2}-105 a^{2} x b c d +64 a^{3} d^{2}-112 a^{2} b \,c^{2}\right ) \sqrt {b \,x^{2}+a}}{840 b^{3}}+\frac {a^{3} c d \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\)	\(148\)
default	\(c^{2} \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )+d^{2} \left (\frac {x^{4} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{7 b}-\frac {4 a \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )}{7 b}\right )+2 c d \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )\)	\(185\)

Input:

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

Output:

1/840*(120*b^3*d^2*x^6+280*b^3*c*d*x^5+24*a*b^2*d^2*x^4+168*b^3*c^2*x^4+70 
*a*b^2*c*d*x^3-32*a^2*b*d^2*x^2+56*a*b^2*c^2*x^2-105*a^2*b*c*d*x+64*a^3*d^ 
2-112*a^2*b*c^2)*(b*x^2+a)^(1/2)/b^3+1/8*a^3/b^(5/2)*c*d*ln(b^(1/2)*x+(b*x 
^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.64 \[ \int x^3 (c+d x)^2 \sqrt {a+b x^2} \, dx=\left [\frac {105 \, a^{3} \sqrt {b} c d \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (120 \, b^{3} d^{2} x^{6} + 280 \, b^{3} c d x^{5} + 70 \, a b^{2} c d x^{3} - 105 \, a^{2} b c d x - 112 \, a^{2} b c^{2} + 64 \, a^{3} d^{2} + 24 \, {\left (7 \, b^{3} c^{2} + a b^{2} d^{2}\right )} x^{4} + 8 \, {\left (7 \, a b^{2} c^{2} - 4 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{1680 \, b^{3}}, -\frac {105 \, a^{3} \sqrt {-b} c d \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (120 \, b^{3} d^{2} x^{6} + 280 \, b^{3} c d x^{5} + 70 \, a b^{2} c d x^{3} - 105 \, a^{2} b c d x - 112 \, a^{2} b c^{2} + 64 \, a^{3} d^{2} + 24 \, {\left (7 \, b^{3} c^{2} + a b^{2} d^{2}\right )} x^{4} + 8 \, {\left (7 \, a b^{2} c^{2} - 4 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{840 \, b^{3}}\right ] \] Input:

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

Output:

[1/1680*(105*a^3*sqrt(b)*c*d*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - 
a) + 2*(120*b^3*d^2*x^6 + 280*b^3*c*d*x^5 + 70*a*b^2*c*d*x^3 - 105*a^2*b*c 
*d*x - 112*a^2*b*c^2 + 64*a^3*d^2 + 24*(7*b^3*c^2 + a*b^2*d^2)*x^4 + 8*(7* 
a*b^2*c^2 - 4*a^2*b*d^2)*x^2)*sqrt(b*x^2 + a))/b^3, -1/840*(105*a^3*sqrt(- 
b)*c*d*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (120*b^3*d^2*x^6 + 280*b^3*c*d 
*x^5 + 70*a*b^2*c*d*x^3 - 105*a^2*b*c*d*x - 112*a^2*b*c^2 + 64*a^3*d^2 + 2 
4*(7*b^3*c^2 + a*b^2*d^2)*x^4 + 8*(7*a*b^2*c^2 - 4*a^2*b*d^2)*x^2)*sqrt(b* 
x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.16 \[ \int x^3 (c+d x)^2 \sqrt {a+b x^2} \, dx=\begin {cases} \frac {a^{3} c d \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (- \frac {a^{2} c d x}{8 b^{2}} + \frac {a c d x^{3}}{12 b} - \frac {2 a \left (a c^{2} - \frac {4 a \left (\frac {a d^{2}}{7} + b c^{2}\right )}{5 b}\right )}{3 b^{2}} + \frac {c d x^{5}}{3} + \frac {d^{2} x^{6}}{7} + \frac {x^{4} \left (\frac {a d^{2}}{7} + b c^{2}\right )}{5 b} + \frac {x^{2} \left (a c^{2} - \frac {4 a \left (\frac {a d^{2}}{7} + b c^{2}\right )}{5 b}\right )}{3 b}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {c^{2} x^{4}}{4} + \frac {2 c d x^{5}}{5} + \frac {d^{2} x^{6}}{6}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a**3*c*d*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqr 
t(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(8*b**2) + sqrt(a + b*x**2 
)*(-a**2*c*d*x/(8*b**2) + a*c*d*x**3/(12*b) - 2*a*(a*c**2 - 4*a*(a*d**2/7 
+ b*c**2)/(5*b))/(3*b**2) + c*d*x**5/3 + d**2*x**6/7 + x**4*(a*d**2/7 + b* 
c**2)/(5*b) + x**2*(a*c**2 - 4*a*(a*d**2/7 + b*c**2)/(5*b))/(3*b)), Ne(b, 
0)), (sqrt(a)*(c**2*x**4/4 + 2*c*d*x**5/5 + d**2*x**6/6), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93 \[ \int x^3 (c+d x)^2 \sqrt {a+b x^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} d^{2} x^{4}}{7 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} c d x^{3}}{3 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} c^{2} x^{2}}{5 \, b} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a d^{2} x^{2}}{35 \, b^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a c d x}{4 \, b^{2}} + \frac {\sqrt {b x^{2} + a} a^{2} c d x}{8 \, b^{2}} + \frac {a^{3} c d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a c^{2}}{15 \, b^{2}} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} d^{2}}{105 \, b^{3}} \] Input:

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

Output:

1/7*(b*x^2 + a)^(3/2)*d^2*x^4/b + 1/3*(b*x^2 + a)^(3/2)*c*d*x^3/b + 1/5*(b 
*x^2 + a)^(3/2)*c^2*x^2/b - 4/35*(b*x^2 + a)^(3/2)*a*d^2*x^2/b^2 - 1/4*(b* 
x^2 + a)^(3/2)*a*c*d*x/b^2 + 1/8*sqrt(b*x^2 + a)*a^2*c*d*x/b^2 + 1/8*a^3*c 
*d*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 2/15*(b*x^2 + a)^(3/2)*a*c^2/b^2 + 8/1 
05*(b*x^2 + a)^(3/2)*a^2*d^2/b^3
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.86 \[ \int x^3 (c+d x)^2 \sqrt {a+b x^2} \, dx=-\frac {a^{3} c d \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} - \frac {1}{840} \, \sqrt {b x^{2} + a} {\left ({\left (\frac {105 \, a^{2} c d}{b^{2}} - 2 \, {\left ({\left (\frac {35 \, a c d}{b} + 4 \, {\left (5 \, {\left (3 \, d^{2} x + 7 \, c d\right )} x + \frac {3 \, {\left (7 \, b^{5} c^{2} + a b^{4} d^{2}\right )}}{b^{5}}\right )} x\right )} x + \frac {4 \, {\left (7 \, a b^{4} c^{2} - 4 \, a^{2} b^{3} d^{2}\right )}}{b^{5}}\right )} x\right )} x + \frac {16 \, {\left (7 \, a^{2} b^{3} c^{2} - 4 \, a^{3} b^{2} d^{2}\right )}}{b^{5}}\right )} \] Input:

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

Output:

-1/8*a^3*c*d*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) - 1/840*sqrt(b 
*x^2 + a)*((105*a^2*c*d/b^2 - 2*((35*a*c*d/b + 4*(5*(3*d^2*x + 7*c*d)*x + 
3*(7*b^5*c^2 + a*b^4*d^2)/b^5)*x)*x + 4*(7*a*b^4*c^2 - 4*a^2*b^3*d^2)/b^5) 
*x)*x + 16*(7*a^2*b^3*c^2 - 4*a^3*b^2*d^2)/b^5)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.16 \[ \int x^3 (c+d x)^2 \sqrt {a+b x^2} \, dx=\frac {64 \sqrt {b \,x^{2}+a}\, a^{3} d^{2}-112 \sqrt {b \,x^{2}+a}\, a^{2} b \,c^{2}-105 \sqrt {b \,x^{2}+a}\, a^{2} b c d x -32 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{2} x^{2}+56 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{2} x^{2}+70 \sqrt {b \,x^{2}+a}\, a \,b^{2} c d \,x^{3}+24 \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{2} x^{4}+168 \sqrt {b \,x^{2}+a}\, b^{3} c^{2} x^{4}+280 \sqrt {b \,x^{2}+a}\, b^{3} c d \,x^{5}+120 \sqrt {b \,x^{2}+a}\, b^{3} d^{2} x^{6}+105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c d}{840 b^{3}} \] Input:

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

Output:

(64*sqrt(a + b*x**2)*a**3*d**2 - 112*sqrt(a + b*x**2)*a**2*b*c**2 - 105*sq 
rt(a + b*x**2)*a**2*b*c*d*x - 32*sqrt(a + b*x**2)*a**2*b*d**2*x**2 + 56*sq 
rt(a + b*x**2)*a*b**2*c**2*x**2 + 70*sqrt(a + b*x**2)*a*b**2*c*d*x**3 + 24 
*sqrt(a + b*x**2)*a*b**2*d**2*x**4 + 168*sqrt(a + b*x**2)*b**3*c**2*x**4 + 
 280*sqrt(a + b*x**2)*b**3*c*d*x**5 + 120*sqrt(a + b*x**2)*b**3*d**2*x**6 
+ 105*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*c*d)/(840*b 
**3)