\(\int x^2 (c+d x)^2 (a+b x^2)^{3/2} \, dx\) [1048]

Optimal result

Integrand size = 22, antiderivative size = 214 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {a^2 \left (8 b c^2-3 a d^2\right ) x \sqrt {a+b x^2}}{128 b^2}+\frac {a \left (8 b c^2-3 a d^2\right ) x^3 \sqrt {a+b x^2}}{64 b}+\frac {\left (8 b c^2-3 a d^2\right ) x^3 \left (a+b x^2\right )^{3/2}}{48 b}-\frac {2 a c d \left (a+b x^2\right )^{5/2}}{5 b^2}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {2 c d \left (a+b x^2\right )^{7/2}}{7 b^2}-\frac {a^3 \left (8 b c^2-3 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}} \] Output:

1/128*a^2*(-3*a*d^2+8*b*c^2)*x*(b*x^2+a)^(1/2)/b^2+1/64*a*(-3*a*d^2+8*b*c^ 
2)*x^3*(b*x^2+a)^(1/2)/b+1/48*(-3*a*d^2+8*b*c^2)*x^3*(b*x^2+a)^(3/2)/b-2/5 
*a*c*d*(b*x^2+a)^(5/2)/b^2+1/8*d^2*x^3*(b*x^2+a)^(5/2)/b+2/7*c*d*(b*x^2+a) 
^(7/2)/b^2-1/128*a^3*(-3*a*d^2+8*b*c^2)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2)) 
/b^(5/2)
 

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.79 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {\sqrt {a+b x^2} \left (-3 a^3 d (512 c+105 d x)+80 b^3 x^5 \left (28 c^2+48 c d x+21 d^2 x^2\right )+6 a^2 b x \left (140 c^2+128 c d x+35 d^2 x^2\right )+8 a b^2 x^3 \left (490 c^2+768 c d x+315 d^2 x^2\right )\right )}{13440 b^2}+\frac {a^3 \left (-8 b c^2+3 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{64 b^{5/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(-3*a^3*d*(512*c + 105*d*x) + 80*b^3*x^5*(28*c^2 + 48*c*d 
*x + 21*d^2*x^2) + 6*a^2*b*x*(140*c^2 + 128*c*d*x + 35*d^2*x^2) + 8*a*b^2* 
x^3*(490*c^2 + 768*c*d*x + 315*d^2*x^2)))/(13440*b^2) + (a^3*(-8*b*c^2 + 3 
*a*d^2)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(64*b^(5/2))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {541, 533, 27, 533, 25, 27, 455, 211, 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.

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

\(\Big \downarrow \) 541

\(\displaystyle \frac {\int x^2 \left (8 b c^2+16 b d x c-3 a d^2\right ) \left (b x^2+a\right )^{3/2}dx}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}-\frac {\int b x \left (32 a c d-7 \left (8 b c^2-3 a d^2\right ) x\right ) \left (b x^2+a\right )^{3/2}dx}{7 b}}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}-\frac {1}{7} \int x \left (32 a c d-7 \left (8 b c^2-3 a d^2\right ) x\right ) \left (b x^2+a\right )^{3/2}dx}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {1}{7} \left (\frac {\int -a \left (7 \left (8 b c^2-3 a d^2\right )+192 b c d x\right ) \left (b x^2+a\right )^{3/2}dx}{6 b}+\frac {7 x \left (a+b x^2\right )^{5/2} \left (8 b c^2-3 a d^2\right )}{6 b}\right )+\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x \left (a+b x^2\right )^{5/2} \left (8 b c^2-3 a d^2\right )}{6 b}-\frac {\int a \left (7 \left (8 b c^2-3 a d^2\right )+192 b c d x\right ) \left (b x^2+a\right )^{3/2}dx}{6 b}\right )+\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x \left (a+b x^2\right )^{5/2} \left (8 b c^2-3 a d^2\right )}{6 b}-\frac {a \int \left (7 \left (8 b c^2-3 a d^2\right )+192 b c d x\right ) \left (b x^2+a\right )^{3/2}dx}{6 b}\right )+\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x \left (a+b x^2\right )^{5/2} \left (8 b c^2-3 a d^2\right )}{6 b}-\frac {a \left (7 \left (8 b c^2-3 a d^2\right ) \int \left (b x^2+a\right )^{3/2}dx+\frac {192}{5} c d \left (a+b x^2\right )^{5/2}\right )}{6 b}\right )+\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x \left (a+b x^2\right )^{5/2} \left (8 b c^2-3 a d^2\right )}{6 b}-\frac {a \left (7 \left (8 b c^2-3 a d^2\right ) \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {192}{5} c d \left (a+b x^2\right )^{5/2}\right )}{6 b}\right )+\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x \left (a+b x^2\right )^{5/2} \left (8 b c^2-3 a d^2\right )}{6 b}-\frac {a \left (7 \left (8 b c^2-3 a d^2\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {192}{5} c d \left (a+b x^2\right )^{5/2}\right )}{6 b}\right )+\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x \left (a+b x^2\right )^{5/2} \left (8 b c^2-3 a d^2\right )}{6 b}-\frac {a \left (7 \left (8 b c^2-3 a d^2\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {192}{5} c d \left (a+b x^2\right )^{5/2}\right )}{6 b}\right )+\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{7} \left (\frac {7 x \left (a+b x^2\right )^{5/2} \left (8 b c^2-3 a d^2\right )}{6 b}-\frac {a \left (7 \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right ) \left (8 b c^2-3 a d^2\right )+\frac {192}{5} c d \left (a+b x^2\right )^{5/2}\right )}{6 b}\right )+\frac {16}{7} c d x^2 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {d^2 x^3 \left (a+b x^2\right )^{5/2}}{8 b}\)

Input:

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

Output:

(d^2*x^3*(a + b*x^2)^(5/2))/(8*b) + ((16*c*d*x^2*(a + b*x^2)^(5/2))/7 + (( 
7*(8*b*c^2 - 3*a*d^2)*x*(a + b*x^2)^(5/2))/(6*b) - (a*((192*c*d*(a + b*x^2 
)^(5/2))/5 + 7*(8*b*c^2 - 3*a*d^2)*((x*(a + b*x^2)^(3/2))/4 + (3*a*((x*Sqr 
t[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4 
)))/(6*b))/7)/(8*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 = 0.33 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {\left (-1680 b^{3} d^{2} x^{7}-3840 b^{3} c d \,x^{6}-2520 a \,b^{2} d^{2} x^{5}-2240 b^{3} c^{2} x^{5}-6144 c d a \,b^{2} x^{4}-210 d^{2} x^{3} a^{2} b -3920 a \,b^{2} c^{2} x^{3}-768 x^{2} a^{2} b c d +315 a^{3} d^{2} x -840 a^{2} b \,c^{2} x +1536 a^{3} d c \right ) \sqrt {b \,x^{2}+a}}{13440 b^{2}}+\frac {a^{3} \left (3 a \,d^{2}-8 b \,c^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {5}{2}}}\)	\(170\)
default	\(c^{2} \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )+d^{2} \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )+2 c d \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 b^{2}}\right )\)	\(217\)

Input:

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

Output:

-1/13440/b^2*(-1680*b^3*d^2*x^7-3840*b^3*c*d*x^6-2520*a*b^2*d^2*x^5-2240*b 
^3*c^2*x^5-6144*a*b^2*c*d*x^4-210*a^2*b*d^2*x^3-3920*a*b^2*c^2*x^3-768*a^2 
*b*c*d*x^2+315*a^3*d^2*x-840*a^2*b*c^2*x+1536*a^3*c*d)*(b*x^2+a)^(1/2)+1/1 
28*a^3*(3*a*d^2-8*b*c^2)/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.77 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=\left [-\frac {105 \, {\left (8 \, a^{3} b c^{2} - 3 \, a^{4} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (1680 \, b^{4} d^{2} x^{7} + 3840 \, b^{4} c d x^{6} + 6144 \, a b^{3} c d x^{4} + 768 \, a^{2} b^{2} c d x^{2} - 1536 \, a^{3} b c d + 280 \, {\left (8 \, b^{4} c^{2} + 9 \, a b^{3} d^{2}\right )} x^{5} + 70 \, {\left (56 \, a b^{3} c^{2} + 3 \, a^{2} b^{2} d^{2}\right )} x^{3} + 105 \, {\left (8 \, a^{2} b^{2} c^{2} - 3 \, a^{3} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{26880 \, b^{3}}, \frac {105 \, {\left (8 \, a^{3} b c^{2} - 3 \, a^{4} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (1680 \, b^{4} d^{2} x^{7} + 3840 \, b^{4} c d x^{6} + 6144 \, a b^{3} c d x^{4} + 768 \, a^{2} b^{2} c d x^{2} - 1536 \, a^{3} b c d + 280 \, {\left (8 \, b^{4} c^{2} + 9 \, a b^{3} d^{2}\right )} x^{5} + 70 \, {\left (56 \, a b^{3} c^{2} + 3 \, a^{2} b^{2} d^{2}\right )} x^{3} + 105 \, {\left (8 \, a^{2} b^{2} c^{2} - 3 \, a^{3} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{13440 \, b^{3}}\right ] \] Input:

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

Output:

[-1/26880*(105*(8*a^3*b*c^2 - 3*a^4*d^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x 
^2 + a)*sqrt(b)*x - a) - 2*(1680*b^4*d^2*x^7 + 3840*b^4*c*d*x^6 + 6144*a*b 
^3*c*d*x^4 + 768*a^2*b^2*c*d*x^2 - 1536*a^3*b*c*d + 280*(8*b^4*c^2 + 9*a*b 
^3*d^2)*x^5 + 70*(56*a*b^3*c^2 + 3*a^2*b^2*d^2)*x^3 + 105*(8*a^2*b^2*c^2 - 
 3*a^3*b*d^2)*x)*sqrt(b*x^2 + a))/b^3, 1/13440*(105*(8*a^3*b*c^2 - 3*a^4*d 
^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (1680*b^4*d^2*x^7 + 3840 
*b^4*c*d*x^6 + 6144*a*b^3*c*d*x^4 + 768*a^2*b^2*c*d*x^2 - 1536*a^3*b*c*d + 
 280*(8*b^4*c^2 + 9*a*b^3*d^2)*x^5 + 70*(56*a*b^3*c^2 + 3*a^2*b^2*d^2)*x^3 
 + 105*(8*a^2*b^2*c^2 - 3*a^3*b*d^2)*x)*sqrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.54 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=\begin {cases} - \frac {a \left (a^{2} c^{2} - \frac {3 a \left (a^{2} d^{2} + 2 a b c^{2} - \frac {5 a \left (\frac {9 a b d^{2}}{8} + b^{2} c^{2}\right )}{6 b}\right )}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (- \frac {4 a^{3} c d}{35 b^{2}} + \frac {2 a^{2} c d x^{2}}{35 b} + \frac {16 a c d x^{4}}{35} + \frac {2 b c d x^{6}}{7} + \frac {b d^{2} x^{7}}{8} + \frac {x^{5} \cdot \left (\frac {9 a b d^{2}}{8} + b^{2} c^{2}\right )}{6 b} + \frac {x^{3} \left (a^{2} d^{2} + 2 a b c^{2} - \frac {5 a \left (\frac {9 a b d^{2}}{8} + b^{2} c^{2}\right )}{6 b}\right )}{4 b} + \frac {x \left (a^{2} c^{2} - \frac {3 a \left (a^{2} d^{2} + 2 a b c^{2} - \frac {5 a \left (\frac {9 a b d^{2}}{8} + b^{2} c^{2}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {c^{2} x^{3}}{3} + \frac {c d x^{4}}{2} + \frac {d^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a*(a**2*c**2 - 3*a*(a**2*d**2 + 2*a*b*c**2 - 5*a*(9*a*b*d**2/8 
 + b**2*c**2)/(6*b))/(4*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2* 
b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(2*b) + sqrt(a + b 
*x**2)*(-4*a**3*c*d/(35*b**2) + 2*a**2*c*d*x**2/(35*b) + 16*a*c*d*x**4/35 
+ 2*b*c*d*x**6/7 + b*d**2*x**7/8 + x**5*(9*a*b*d**2/8 + b**2*c**2)/(6*b) + 
 x**3*(a**2*d**2 + 2*a*b*c**2 - 5*a*(9*a*b*d**2/8 + b**2*c**2)/(6*b))/(4*b 
) + x*(a**2*c**2 - 3*a*(a**2*d**2 + 2*a*b*c**2 - 5*a*(9*a*b*d**2/8 + b**2* 
c**2)/(6*b))/(4*b))/(2*b)), Ne(b, 0)), (a**(3/2)*(c**2*x**3/3 + c*d*x**4/2 
 + d**2*x**5/5), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.01 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} d^{2} x^{3}}{8 \, b} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} c d x^{2}}{7 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} c^{2} x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a c^{2} x}{24 \, b} - \frac {\sqrt {b x^{2} + a} a^{2} c^{2} x}{16 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a d^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} d^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} a^{3} d^{2} x}{128 \, b^{2}} - \frac {a^{3} c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3 \, a^{4} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a c d}{35 \, b^{2}} \] Input:

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

Output:

1/8*(b*x^2 + a)^(5/2)*d^2*x^3/b + 2/7*(b*x^2 + a)^(5/2)*c*d*x^2/b + 1/6*(b 
*x^2 + a)^(5/2)*c^2*x/b - 1/24*(b*x^2 + a)^(3/2)*a*c^2*x/b - 1/16*sqrt(b*x 
^2 + a)*a^2*c^2*x/b - 1/16*(b*x^2 + a)^(5/2)*a*d^2*x/b^2 + 1/64*(b*x^2 + a 
)^(3/2)*a^2*d^2*x/b^2 + 3/128*sqrt(b*x^2 + a)*a^3*d^2*x/b^2 - 1/16*a^3*c^2 
*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 3/128*a^4*d^2*arcsinh(b*x/sqrt(a*b))/b^( 
5/2) - 4/35*(b*x^2 + a)^(5/2)*a*c*d/b^2
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.89 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=-\frac {1}{13440} \, {\left (\frac {1536 \, a^{3} c d}{b^{2}} - {\left (2 \, {\left (\frac {384 \, a^{2} c d}{b} + {\left (4 \, {\left (768 \, a c d + 5 \, {\left (6 \, {\left (7 \, b d^{2} x + 16 \, b c d\right )} x + \frac {7 \, {\left (8 \, b^{7} c^{2} + 9 \, a b^{6} d^{2}\right )}}{b^{6}}\right )} x\right )} x + \frac {35 \, {\left (56 \, a b^{6} c^{2} + 3 \, a^{2} b^{5} d^{2}\right )}}{b^{6}}\right )} x\right )} x + \frac {105 \, {\left (8 \, a^{2} b^{5} c^{2} - 3 \, a^{3} b^{4} d^{2}\right )}}{b^{6}}\right )} x\right )} \sqrt {b x^{2} + a} + \frac {{\left (8 \, a^{3} b c^{2} - 3 \, a^{4} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {5}{2}}} \] Input:

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

Output:

-1/13440*(1536*a^3*c*d/b^2 - (2*(384*a^2*c*d/b + (4*(768*a*c*d + 5*(6*(7*b 
*d^2*x + 16*b*c*d)*x + 7*(8*b^7*c^2 + 9*a*b^6*d^2)/b^6)*x)*x + 35*(56*a*b^ 
6*c^2 + 3*a^2*b^5*d^2)/b^6)*x)*x + 105*(8*a^2*b^5*c^2 - 3*a^3*b^4*d^2)/b^6 
)*x)*sqrt(b*x^2 + a) + 1/128*(8*a^3*b*c^2 - 3*a^4*d^2)*log(abs(-sqrt(b)*x 
+ sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.29 \[ \int x^2 (c+d x)^2 \left (a+b x^2\right )^{3/2} \, dx=\frac {-1536 \sqrt {b \,x^{2}+a}\, a^{3} b c d -315 \sqrt {b \,x^{2}+a}\, a^{3} b \,d^{2} x +840 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c^{2} x +768 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c d \,x^{2}+210 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d^{2} x^{3}+3920 \sqrt {b \,x^{2}+a}\, a \,b^{3} c^{2} x^{3}+6144 \sqrt {b \,x^{2}+a}\, a \,b^{3} c d \,x^{4}+2520 \sqrt {b \,x^{2}+a}\, a \,b^{3} d^{2} x^{5}+2240 \sqrt {b \,x^{2}+a}\, b^{4} c^{2} x^{5}+3840 \sqrt {b \,x^{2}+a}\, b^{4} c d \,x^{6}+1680 \sqrt {b \,x^{2}+a}\, b^{4} d^{2} x^{7}+315 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} d^{2}-840 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b \,c^{2}}{13440 b^{3}} \] Input:

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

Output:

( - 1536*sqrt(a + b*x**2)*a**3*b*c*d - 315*sqrt(a + b*x**2)*a**3*b*d**2*x 
+ 840*sqrt(a + b*x**2)*a**2*b**2*c**2*x + 768*sqrt(a + b*x**2)*a**2*b**2*c 
*d*x**2 + 210*sqrt(a + b*x**2)*a**2*b**2*d**2*x**3 + 3920*sqrt(a + b*x**2) 
*a*b**3*c**2*x**3 + 6144*sqrt(a + b*x**2)*a*b**3*c*d*x**4 + 2520*sqrt(a + 
b*x**2)*a*b**3*d**2*x**5 + 2240*sqrt(a + b*x**2)*b**4*c**2*x**5 + 3840*sqr 
t(a + b*x**2)*b**4*c*d*x**6 + 1680*sqrt(a + b*x**2)*b**4*d**2*x**7 + 315*s 
qrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4*d**2 - 840*sqrt(b) 
*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b*c**2)/(13440*b**3)