\(\int x^3 (a+\frac {b}{c+d x^2})^{3/2} \, dx\) [177]

Optimal result

Integrand size = 21, antiderivative size = 140 \[ \int x^3 \left (a+\frac {b}{c+d x^2}\right )^{3/2} \, dx=\frac {b c \sqrt {a+\frac {b}{c+d x^2}}}{d^2}+\frac {(5 b-4 a c) \left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}{8 d^2}+\frac {a \left (c+d x^2\right )^2 \sqrt {a+\frac {b}{c+d x^2}}}{4 d^2}+\frac {3 b (b-4 a c) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{8 \sqrt {a} d^2} \] Output:

b*c*(a+b/(d*x^2+c))^(1/2)/d^2+1/8*(-4*a*c+5*b)*(d*x^2+c)*(a+b/(d*x^2+c))^( 
1/2)/d^2+1/4*a*(d*x^2+c)^2*(a+b/(d*x^2+c))^(1/2)/d^2+3/8*b*(-4*a*c+b)*arct 
anh((a+b/(d*x^2+c))^(1/2)/a^(1/2))/a^(1/2)/d^2
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.81 \[ \int x^3 \left (a+\frac {b}{c+d x^2}\right )^{3/2} \, dx=\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (13 b c-2 a c^2+5 b d x^2+2 a d^2 x^4\right )}{8 d^2}-\frac {3 b (-b+4 a c) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{8 \sqrt {a} d^2} \] Input:

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

Output:

(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(13*b*c - 2*a*c^2 + 5*b*d*x^2 + 2*a 
*d^2*x^4))/(8*d^2) - (3*b*(-b + 4*a*c)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c 
 + d*x^2)]/Sqrt[a]])/(8*Sqrt[a]*d^2)
 

Rubi [A] (warning: unable to verify)

Time = 0.64 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.25, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {2057, 2053, 2052, 25, 27, 360, 25, 1471, 27, 299, 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.

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

\(\Big \downarrow \) 2057

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

\(\Big \downarrow \) 2053

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

\(\Big \downarrow \) 2052

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1471

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(b*((a*b*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(4*(a - x^4)^2) + (-1/2*(( 
5*b - 4*a*c)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(a - x^4) + (8*c*Sqrt[ 
(b + a*c + a*d*x^2)/(c + d*x^2)] + (3*(b - 4*a*c)*ArcTanh[Sqrt[(b + a*c + 
a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/Sqrt[a])/2)/4))/d^2
 

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 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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 360

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & 
& (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

rule 2052

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S 
ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d)   Subst[Int[x^(q* 
(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* 
x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] 
&& IntegerQ[m]
 

rule 2053

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) 
))^(p_), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( 
(a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, 
 x] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2057

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* 
((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.73

method	result	size
risch	\(-\frac {\left (-2 a d \,x^{2}+2 a c -5 b \right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{8 d^{2}}-\frac {b \left (\frac {\left (12 a c -3 b \right ) \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {a \,c^{2}+b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right )}{2 \sqrt {a \,d^{2}}}-\frac {8 c \left (a d \,x^{2}+a c +b \right )}{d \sqrt {a \,d^{2} x^{4}+2 a d \,x^{2} c +b d \,x^{2}+a \,c^{2}+b c}}\right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}}{8 d \left (a d \,x^{2}+a c +b \right )}\)	\(242\)
default	\(\frac {\left (4 \sqrt {a \,d^{2} x^{4}+2 a d \,x^{2} c +b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a \,d^{2} x^{4}-12 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a d \,x^{2} c +b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b c \,d^{2} x^{2}+3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a d \,x^{2} c +b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{2} d^{2} x^{2}+10 \sqrt {a \,d^{2} x^{4}+2 a d \,x^{2} c +b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b d \,x^{2}-12 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a d \,x^{2} c +b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b \,c^{2} d -4 \sqrt {a \,d^{2} x^{4}+2 a d \,x^{2} c +b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a \,c^{2}+3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a d \,x^{2} c +b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{2} c d +16 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {a \,d^{2}}\, b c +10 \sqrt {a \,d^{2} x^{4}+2 a d \,x^{2} c +b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{16 d^{2} \sqrt {a \,d^{2}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}}\)	\(593\)

Input:

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

Output:

-1/8/d^2*(-2*a*d*x^2+2*a*c-5*b)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2 
)-1/8*b/d*(1/2*(12*a*c-3*b)*ln((a*c*d+1/2*b*d+a*d^2*x^2)/(a*d^2)^(1/2)+(a* 
c^2+b*c+(2*a*c*d+b*d)*x^2+a*d^2*x^4)^(1/2))/(a*d^2)^(1/2)-8*c*(a*d*x^2+a*c 
+b)/d/(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2))*((a*d*x^2+a*c+b)/(d 
*x^2+c))^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)/(a*d*x^2+a*c+b)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.39 \[ \int x^3 \left (a+\frac {b}{c+d x^2}\right )^{3/2} \, dx=\left [\frac {3 \, {\left (4 \, a b c - b^{2}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (2 \, a^{2} d^{2} x^{4} + 5 \, a b d x^{2} - 2 \, a^{2} c^{2} + 13 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, a d^{2}}, \frac {3 \, {\left (4 \, a b c - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \, {\left (2 \, a^{2} d^{2} x^{4} + 5 \, a b d x^{2} - 2 \, a^{2} c^{2} + 13 \, a b c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, a d^{2}}\right ] \] Input:

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

Output:

[1/32*(3*(4*a*b*c - b^2)*sqrt(a)*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a^2* 
c + a*b)*d*x^2 + 8*a*b*c + b^2 - 4*(2*a*d^2*x^4 + (4*a*c + b)*d*x^2 + 2*a* 
c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))) + 4*(2*a^2*d^2*x 
^4 + 5*a*b*d*x^2 - 2*a^2*c^2 + 13*a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + 
 c)))/(a*d^2), 1/16*(3*(4*a*b*c - b^2)*sqrt(-a)*arctan(1/2*(2*a*d*x^2 + 2* 
a*c + b)*sqrt(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c 
 + a*b)) + 2*(2*a^2*d^2*x^4 + 5*a*b*d*x^2 - 2*a^2*c^2 + 13*a*b*c)*sqrt((a* 
d*x^2 + a*c + b)/(d*x^2 + c)))/(a*d^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.76 \[ \int x^3 \left (a+\frac {b}{c+d x^2}\right )^{3/2} \, dx=\frac {b c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{d^{2}} + \frac {3 \, {\left (4 \, a c - b\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{16 \, \sqrt {a} d^{2}} - \frac {{\left (4 \, a b c - 5 \, b^{2}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} - {\left (4 \, a^{2} b c - 3 \, a b^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, {\left (a^{2} d^{2} - \frac {2 \, {\left (a d x^{2} + a c + b\right )} a d^{2}}{d x^{2} + c} + \frac {{\left (a d x^{2} + a c + b\right )}^{2} d^{2}}{{\left (d x^{2} + c\right )}^{2}}\right )}} \] Input:

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

Output:

b*c*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/d^2 + 3/16*(4*a*c - b)*b*log(-(s 
qrt(a) - sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(sqrt(a) + sqrt((a*d*x^2 + 
 a*c + b)/(d*x^2 + c))))/(sqrt(a)*d^2) - 1/8*((4*a*b*c - 5*b^2)*((a*d*x^2 
+ a*c + b)/(d*x^2 + c))^(3/2) - (4*a^2*b*c - 3*a*b^2)*sqrt((a*d*x^2 + a*c 
+ b)/(d*x^2 + c)))/(a^2*d^2 - 2*(a*d*x^2 + a*c + b)*a*d^2/(d*x^2 + c) + (a 
*d*x^2 + a*c + b)^2*d^2/(d*x^2 + c)^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (124) = 248\).

Time = 0.38 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.39 \[ \int x^3 \left (a+\frac {b}{c+d x^2}\right )^{3/2} \, dx=\frac {1}{8} \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} {\left (\frac {2 \, a x^{2} \mathrm {sgn}\left (d x^{2} + c\right )}{d} - \frac {2 \, a^{2} c d^{2} \mathrm {sgn}\left (d x^{2} + c\right ) - 5 \, a b d^{2} \mathrm {sgn}\left (d x^{2} + c\right )}{a d^{4}}\right )} + \frac {{\left (4 \, a b c \mathrm {sgn}\left (d x^{2} + c\right ) - b^{2} \mathrm {sgn}\left (d x^{2} + c\right )\right )} \log \left ({\left | 2 \, a^{2} c^{3} d + 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a^{\frac {3}{2}} c^{2} {\left | d \right |} + 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} a c d + a b c^{2} d + 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{3} \sqrt {a} {\left | d \right |} + 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} \sqrt {a} b c {\left | d \right |} + {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} b d \right |}\right )}{16 \, \sqrt {a} d {\left | d \right |}} + \frac {{\left (4 \, a b c d^{2} {\left | d \right |} \mathrm {sgn}\left (d x^{2} + c\right ) - b^{2} d^{2} {\left | d \right |} \mathrm {sgn}\left (d x^{2} + c\right )\right )} \log \left (96\right )}{8 \, \sqrt {a} d^{5}} \] Input:

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

Output:

1/8*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)*(2*a*x^2*sgn(d*x 
^2 + c)/d - (2*a^2*c*d^2*sgn(d*x^2 + c) - 5*a*b*d^2*sgn(d*x^2 + c))/(a*d^4 
)) + 1/16*(4*a*b*c*sgn(d*x^2 + c) - b^2*sgn(d*x^2 + c))*log(abs(2*a^2*c^3* 
d + 6*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + 
b*c))*a^(3/2)*c^2*abs(d) + 6*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x 
^2 + b*d*x^2 + a*c^2 + b*c))^2*a*c*d + a*b*c^2*d + 2*(sqrt(a*d^2)*x^2 - sq 
rt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))^3*sqrt(a)*abs(d) + 2* 
(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))* 
sqrt(a)*b*c*abs(d) + (sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d 
*x^2 + a*c^2 + b*c))^2*b*d))/(sqrt(a)*d*abs(d)) + 1/8*(4*a*b*c*d^2*abs(d)* 
sgn(d*x^2 + c) - b^2*d^2*abs(d)*sgn(d*x^2 + c))*log(96)/(sqrt(a)*d^5)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.01 \[ \int x^3 \left (a+\frac {b}{c+d x^2}\right )^{3/2} \, dx=\frac {-2 \sqrt {d \,x^{2}+c}\, \sqrt {a d \,x^{2}+a c +b}\, a^{2} c^{2}+2 \sqrt {d \,x^{2}+c}\, \sqrt {a d \,x^{2}+a c +b}\, a^{2} d^{2} x^{4}+13 \sqrt {d \,x^{2}+c}\, \sqrt {a d \,x^{2}+a c +b}\, a b c +5 \sqrt {d \,x^{2}+c}\, \sqrt {a d \,x^{2}+a c +b}\, a b d \,x^{2}+12 \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}\, \sqrt {a d \,x^{2}+a c +b}+\sqrt {d \,x^{2}+c}\, a \right ) a b \,c^{2}+12 \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}\, \sqrt {a d \,x^{2}+a c +b}+\sqrt {d \,x^{2}+c}\, a \right ) a b c d \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}\, \sqrt {a d \,x^{2}+a c +b}+\sqrt {d \,x^{2}+c}\, a \right ) b^{2} c -3 \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}\, \sqrt {a d \,x^{2}+a c +b}+\sqrt {d \,x^{2}+c}\, a \right ) b^{2} d \,x^{2}}{8 a \,d^{2} \left (d \,x^{2}+c \right )} \] Input:

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

Output:

( - 2*sqrt(c + d*x**2)*sqrt(a*c + a*d*x**2 + b)*a**2*c**2 + 2*sqrt(c + d*x 
**2)*sqrt(a*c + a*d*x**2 + b)*a**2*d**2*x**4 + 13*sqrt(c + d*x**2)*sqrt(a* 
c + a*d*x**2 + b)*a*b*c + 5*sqrt(c + d*x**2)*sqrt(a*c + a*d*x**2 + b)*a*b* 
d*x**2 + 12*sqrt(a)*log( - sqrt(a)*sqrt(a*c + a*d*x**2 + b) + sqrt(c + d*x 
**2)*a)*a*b*c**2 + 12*sqrt(a)*log( - sqrt(a)*sqrt(a*c + a*d*x**2 + b) + sq 
rt(c + d*x**2)*a)*a*b*c*d*x**2 - 3*sqrt(a)*log( - sqrt(a)*sqrt(a*c + a*d*x 
**2 + b) + sqrt(c + d*x**2)*a)*b**2*c - 3*sqrt(a)*log( - sqrt(a)*sqrt(a*c 
+ a*d*x**2 + b) + sqrt(c + d*x**2)*a)*b**2*d*x**2)/(8*a*d**2*(c + d*x**2))