\(\int \frac {x^4 (c+d x)^2}{(a+b x^2)^{3/2}} \, dx\) [1218]

Optimal result

Integrand size = 22, antiderivative size = 162 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {a x (c+d x)^2}{b^2 \sqrt {a+b x^2}}-\frac {6 a c d \sqrt {a+b x^2}}{b^3}+\frac {\left (4 b c^2-15 a d^2\right ) x \sqrt {a+b x^2}}{8 b^3}+\frac {d^2 x^3 \sqrt {a+b x^2}}{4 b^2}+\frac {2 c d \left (a+b x^2\right )^{3/2}}{3 b^3}-\frac {3 a \left (4 b c^2-5 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \] Output:

a*x*(d*x+c)^2/b^2/(b*x^2+a)^(1/2)-6*a*c*d*(b*x^2+a)^(1/2)/b^3+1/8*(-15*a*d 
^2+4*b*c^2)*x*(b*x^2+a)^(1/2)/b^3+1/4*d^2*x^3*(b*x^2+a)^(1/2)/b^2+2/3*c*d* 
(b*x^2+a)^(3/2)/b^3-3/8*a*(-5*a*d^2+4*b*c^2)*arctanh(b^(1/2)*x/(b*x^2+a)^( 
1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.84 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-a^2 d (128 c+45 d x)+a b x \left (36 c^2-64 c d x-15 d^2 x^2\right )+2 b^2 x^3 \left (6 c^2+8 c d x+3 d^2 x^2\right )}{24 b^3 \sqrt {a+b x^2}}+\frac {3 a \left (-4 b c^2+5 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{4 b^{7/2}} \] Input:

Integrate[(x^4*(c + d*x)^2)/(a + b*x^2)^(3/2),x]
 

Output:

(-(a^2*d*(128*c + 45*d*x)) + a*b*x*(36*c^2 - 64*c*d*x - 15*d^2*x^2) + 2*b^ 
2*x^3*(6*c^2 + 8*c*d*x + 3*d^2*x^2))/(24*b^3*Sqrt[a + b*x^2]) + (3*a*(-4*b 
*c^2 + 5*a*d^2)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(4*b^(7 
/2))
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.33, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {531, 2185, 2185, 27, 676, 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.

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

\(\Big \downarrow \) 531

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

\(\Big \downarrow \) 2185

\(\displaystyle \frac {a x (c+d x)^2}{b^2 \sqrt {a+b x^2}}-\frac {\frac {\int \frac {(c+d x) \left (a b c x^2 d^3+7 a^2 c d^3+a \left (b c^2+15 a d^2\right ) x d^2\right )}{\sqrt {b x^2+a}}dx}{4 b d^3}-\frac {a \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {a x (c+d x)^2}{b^2 \sqrt {a+b x^2}}-\frac {\frac {\frac {\int \frac {a b d^4 (c+d x) \left (19 a c d+\left (2 b c^2+45 a d^2\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b d^2}+\frac {1}{3} a c d^2 \sqrt {a+b x^2} (c+d x)^2}{4 b d^3}-\frac {a \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a x (c+d x)^2}{b^2 \sqrt {a+b x^2}}-\frac {\frac {\frac {1}{3} a d^2 \int \frac {(c+d x) \left (19 a c d+\left (2 b c^2+45 a d^2\right ) x\right )}{\sqrt {b x^2+a}}dx+\frac {1}{3} a c d^2 \sqrt {a+b x^2} (c+d x)^2}{4 b d^3}-\frac {a \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {a x (c+d x)^2}{b^2 \sqrt {a+b x^2}}-\frac {\frac {\frac {1}{3} a d^2 \left (\frac {9 a d \left (4 b c^2-5 a d^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {2 c \sqrt {a+b x^2} \left (32 a d^2+b c^2\right )}{b}+\frac {d x \sqrt {a+b x^2} \left (45 a d^2+2 b c^2\right )}{2 b}\right )+\frac {1}{3} a c d^2 \sqrt {a+b x^2} (c+d x)^2}{4 b d^3}-\frac {a \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {a x (c+d x)^2}{b^2 \sqrt {a+b x^2}}-\frac {\frac {\frac {1}{3} a d^2 \left (\frac {9 a d \left (4 b c^2-5 a d^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {2 c \sqrt {a+b x^2} \left (32 a d^2+b c^2\right )}{b}+\frac {d x \sqrt {a+b x^2} \left (45 a d^2+2 b c^2\right )}{2 b}\right )+\frac {1}{3} a c d^2 \sqrt {a+b x^2} (c+d x)^2}{4 b d^3}-\frac {a \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {a x (c+d x)^2}{b^2 \sqrt {a+b x^2}}-\frac {\frac {\frac {1}{3} a d^2 \left (\frac {9 a d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (4 b c^2-5 a d^2\right )}{2 b^{3/2}}+\frac {2 c \sqrt {a+b x^2} \left (32 a d^2+b c^2\right )}{b}+\frac {d x \sqrt {a+b x^2} \left (45 a d^2+2 b c^2\right )}{2 b}\right )+\frac {1}{3} a c d^2 \sqrt {a+b x^2} (c+d x)^2}{4 b d^3}-\frac {a \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}\)

Input:

Int[(x^4*(c + d*x)^2)/(a + b*x^2)^(3/2),x]
 

Output:

(a*x*(c + d*x)^2)/(b^2*Sqrt[a + b*x^2]) - (-1/4*(a*(c + d*x)^3*Sqrt[a + b* 
x^2])/(b*d) + ((a*c*d^2*(c + d*x)^2*Sqrt[a + b*x^2])/3 + (a*d^2*((2*c*(b*c 
^2 + 32*a*d^2)*Sqrt[a + b*x^2])/b + (d*(2*b*c^2 + 45*a*d^2)*x*Sqrt[a + b*x 
^2])/(2*b) + (9*a*d*(4*b*c^2 - 5*a*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2 
]])/(2*b^(3/2))))/3)/(4*b*d^3))/(a*b)
 

Defintions of rubi rules used

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 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 531

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi 
alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
 + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 
2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b 
*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p 
 + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt 
Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

rule 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {\left (-6 b \,d^{2} x^{3}-16 b c d \,x^{2}+21 a \,d^{2} x -12 c^{2} b x +80 a c d \right ) \sqrt {b \,x^{2}+a}}{24 b^{3}}+\frac {a \left (3 b \left (5 a \,d^{2}-4 b \,c^{2}\right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )+\frac {7 a \,d^{2} x}{\sqrt {b \,x^{2}+a}}-\frac {4 b \,c^{2} x}{\sqrt {b \,x^{2}+a}}-\frac {16 a c d}{\sqrt {b \,x^{2}+a}}\right )}{8 b^{3}}\)	\(158\)
default	\(c^{2} \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+d^{2} \left (\frac {x^{5}}{4 b \sqrt {b \,x^{2}+a}}-\frac {5 a \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )+2 c d \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )\)	\(214\)

Input:

int(x^4*(d*x+c)^2/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(-6*b*d^2*x^3-16*b*c*d*x^2+21*a*d^2*x-12*b*c^2*x+80*a*c*d)*(b*x^2+a) 
^(1/2)/b^3+1/8*a/b^3*(3*b*(5*a*d^2-4*b*c^2)*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2 
)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))+7*a*d^2*x/(b*x^2+a)^(1/2)-4*b*c^2*x/(b*x^ 
2+a)^(1/2)-16*a*c*d/(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.27 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {9 \, {\left (4 \, a^{2} b c^{2} - 5 \, a^{3} d^{2} + {\left (4 \, a b^{2} c^{2} - 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (6 \, b^{3} d^{2} x^{5} + 16 \, b^{3} c d x^{4} - 64 \, a b^{2} c d x^{2} - 128 \, a^{2} b c d + 3 \, {\left (4 \, b^{3} c^{2} - 5 \, a b^{2} d^{2}\right )} x^{3} + 9 \, {\left (4 \, a b^{2} c^{2} - 5 \, a^{2} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, {\left (b^{5} x^{2} + a b^{4}\right )}}, \frac {9 \, {\left (4 \, a^{2} b c^{2} - 5 \, a^{3} d^{2} + {\left (4 \, a b^{2} c^{2} - 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, b^{3} d^{2} x^{5} + 16 \, b^{3} c d x^{4} - 64 \, a b^{2} c d x^{2} - 128 \, a^{2} b c d + 3 \, {\left (4 \, b^{3} c^{2} - 5 \, a b^{2} d^{2}\right )} x^{3} + 9 \, {\left (4 \, a b^{2} c^{2} - 5 \, a^{2} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{24 \, {\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \] Input:

integrate(x^4*(d*x+c)^2/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

[-1/48*(9*(4*a^2*b*c^2 - 5*a^3*d^2 + (4*a*b^2*c^2 - 5*a^2*b*d^2)*x^2)*sqrt 
(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(6*b^3*d^2*x^5 + 1 
6*b^3*c*d*x^4 - 64*a*b^2*c*d*x^2 - 128*a^2*b*c*d + 3*(4*b^3*c^2 - 5*a*b^2* 
d^2)*x^3 + 9*(4*a*b^2*c^2 - 5*a^2*b*d^2)*x)*sqrt(b*x^2 + a))/(b^5*x^2 + a* 
b^4), 1/24*(9*(4*a^2*b*c^2 - 5*a^3*d^2 + (4*a*b^2*c^2 - 5*a^2*b*d^2)*x^2)* 
sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (6*b^3*d^2*x^5 + 16*b^3*c*d* 
x^4 - 64*a*b^2*c*d*x^2 - 128*a^2*b*c*d + 3*(4*b^3*c^2 - 5*a*b^2*d^2)*x^3 + 
 9*(4*a*b^2*c^2 - 5*a^2*b*d^2)*x)*sqrt(b*x^2 + a))/(b^5*x^2 + a*b^4)]
 

Sympy [F]

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

integrate(x**4*(d*x+c)**2/(b*x**2+a)**(3/2),x)
 

Output:

Integral(x**4*(c + d*x)**2/(a + b*x**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.22 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {d^{2} x^{5}}{4 \, \sqrt {b x^{2} + a} b} + \frac {2 \, c d x^{4}}{3 \, \sqrt {b x^{2} + a} b} + \frac {c^{2} x^{3}}{2 \, \sqrt {b x^{2} + a} b} - \frac {5 \, a d^{2} x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} - \frac {8 \, a c d x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {3 \, a c^{2} x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {15 \, a^{2} d^{2} x}{8 \, \sqrt {b x^{2} + a} b^{3}} - \frac {3 \, a c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {15 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {7}{2}}} - \frac {16 \, a^{2} c d}{3 \, \sqrt {b x^{2} + a} b^{3}} \] Input:

integrate(x^4*(d*x+c)^2/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

1/4*d^2*x^5/(sqrt(b*x^2 + a)*b) + 2/3*c*d*x^4/(sqrt(b*x^2 + a)*b) + 1/2*c^ 
2*x^3/(sqrt(b*x^2 + a)*b) - 5/8*a*d^2*x^3/(sqrt(b*x^2 + a)*b^2) - 8/3*a*c* 
d*x^2/(sqrt(b*x^2 + a)*b^2) + 3/2*a*c^2*x/(sqrt(b*x^2 + a)*b^2) - 15/8*a^2 
*d^2*x/(sqrt(b*x^2 + a)*b^3) - 3/2*a*c^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) + 
15/8*a^2*d^2*arcsinh(b*x/sqrt(a*b))/b^(7/2) - 16/3*a^2*c*d/(sqrt(b*x^2 + a 
)*b^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.91 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left ({\left (2 \, {\left (\frac {3 \, d^{2} x}{b} + \frac {8 \, c d}{b}\right )} x + \frac {3 \, {\left (4 \, b^{5} c^{2} - 5 \, a b^{4} d^{2}\right )}}{b^{6}}\right )} x - \frac {64 \, a c d}{b^{2}}\right )} x + \frac {9 \, {\left (4 \, a b^{4} c^{2} - 5 \, a^{2} b^{3} d^{2}\right )}}{b^{6}}\right )} x - \frac {128 \, a^{2} c d}{b^{3}}}{24 \, \sqrt {b x^{2} + a}} + \frac {3 \, {\left (4 \, a b c^{2} - 5 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {7}{2}}} \] Input:

integrate(x^4*(d*x+c)^2/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

1/24*((((2*(3*d^2*x/b + 8*c*d/b)*x + 3*(4*b^5*c^2 - 5*a*b^4*d^2)/b^6)*x - 
64*a*c*d/b^2)*x + 9*(4*a*b^4*c^2 - 5*a^2*b^3*d^2)/b^6)*x - 128*a^2*c*d/b^3 
)/sqrt(b*x^2 + a) + 3/8*(4*a*b*c^2 - 5*a^2*d^2)*log(abs(-sqrt(b)*x + sqrt( 
b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x^4\,{\left (c+d\,x\right )}^2}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int((x^4*(c + d*x)^2)/(a + b*x^2)^(3/2),x)
 

Output:

int((x^4*(c + d*x)^2)/(a + b*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.07 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-128 \sqrt {b \,x^{2}+a}\, a^{2} b c d -45 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{2} x +36 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{2} x -64 \sqrt {b \,x^{2}+a}\, a \,b^{2} c d \,x^{2}-15 \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{2} x^{3}+12 \sqrt {b \,x^{2}+a}\, b^{3} c^{2} x^{3}+16 \sqrt {b \,x^{2}+a}\, b^{3} c d \,x^{4}+6 \sqrt {b \,x^{2}+a}\, b^{3} d^{2} x^{5}+45 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d^{2}-36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,c^{2}+45 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,d^{2} x^{2}-36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c^{2} x^{2}-30 \sqrt {b}\, a^{3} d^{2}+27 \sqrt {b}\, a^{2} b \,c^{2}-30 \sqrt {b}\, a^{2} b \,d^{2} x^{2}+27 \sqrt {b}\, a \,b^{2} c^{2} x^{2}}{24 b^{4} \left (b \,x^{2}+a \right )} \] Input:

int(x^4*(d*x+c)^2/(b*x^2+a)^(3/2),x)
 

Output:

( - 128*sqrt(a + b*x**2)*a**2*b*c*d - 45*sqrt(a + b*x**2)*a**2*b*d**2*x + 
36*sqrt(a + b*x**2)*a*b**2*c**2*x - 64*sqrt(a + b*x**2)*a*b**2*c*d*x**2 - 
15*sqrt(a + b*x**2)*a*b**2*d**2*x**3 + 12*sqrt(a + b*x**2)*b**3*c**2*x**3 
+ 16*sqrt(a + b*x**2)*b**3*c*d*x**4 + 6*sqrt(a + b*x**2)*b**3*d**2*x**5 + 
45*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*d**2 - 36*sqrt 
(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*c**2 + 45*sqrt(b)*l 
og((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*d**2*x**2 - 36*sqrt(b)*l 
og((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*c**2*x**2 - 30*sqrt(b)*a 
**3*d**2 + 27*sqrt(b)*a**2*b*c**2 - 30*sqrt(b)*a**2*b*d**2*x**2 + 27*sqrt( 
b)*a*b**2*c**2*x**2)/(24*b**4*(a + b*x**2))