\(\int \frac {x^3 (a+b x^2)^{3/2}}{\sqrt {c+d x^2}} \, dx\) [1179]

Optimal result

Integrand size = 26, antiderivative size = 187 \[ \int \frac {x^3 \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{16 b d^3}-\frac {(5 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{24 b d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 b d}-\frac {(b c-a d)^2 (5 b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{16 b^{3/2} d^{7/2}} \] Output:

1/16*(-a*d+b*c)*(a*d+5*b*c)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b/d^3-1/24*(a* 
d+5*b*c)*(b*x^2+a)^(3/2)*(d*x^2+c)^(1/2)/b/d^2+1/6*(b*x^2+a)^(5/2)*(d*x^2+ 
c)^(1/2)/b/d-1/16*(-a*d+b*c)^2*(a*d+5*b*c)*arctanh(d^(1/2)*(b*x^2+a)^(1/2) 
/b^(1/2)/(d*x^2+c)^(1/2))/b^(3/2)/d^(7/2)
 

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (3 a^2 d^2+2 a b d \left (-11 c+7 d x^2\right )+b^2 \left (15 c^2-10 c d x^2+8 d^2 x^4\right )\right )}{48 b d^3}-\frac {(b c-a d)^2 (5 b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )}{16 b^{3/2} d^{7/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(3*a^2*d^2 + 2*a*b*d*(-11*c + 7*d*x^2) + 
b^2*(15*c^2 - 10*c*d*x^2 + 8*d^2*x^4)))/(48*b*d^3) - ((b*c - a*d)^2*(5*b*c 
 + a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/(Sqrt[d]*Sqrt[a + b*x^2])])/(16* 
b^(3/2)*d^(7/2))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {354, 90, 60, 60, 66, 221}

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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(((a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(3*b*d) - ((5*b*c + a*d)*(((a + b*x^2 
)^(3/2)*Sqrt[c + d*x^2])/(2*d) - (3*(b*c - a*d)*((Sqrt[a + b*x^2]*Sqrt[c + 
 d*x^2])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[ 
c + d*x^2])])/(Sqrt[b]*d^(3/2))))/(4*d)))/(6*b*d))/2
 

Defintions of rubi rules used

rule 60

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

rule 66

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

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 221

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

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.13

method	result	size
risch	\(\frac {\left (8 b^{2} d^{2} x^{4}+14 x^{2} a b \,d^{2}-10 x^{2} b^{2} c d +3 a^{2} d^{2}-22 a b c d +15 b^{2} c^{2}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}{48 b \,d^{3}}-\frac {\left (a^{3} d^{3}+3 a^{2} b c \,d^{2}-9 a \,b^{2} c^{2} d +5 b^{3} c^{3}\right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}}{32 b \,d^{3} \sqrt {b d}\, \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(212\)
default	\(-\frac {\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}\, \left (-16 b^{2} d^{2} x^{4} \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}-28 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, a b \,d^{2} x^{2}+20 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, b^{2} c d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}+9 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}-27 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d +15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}-6 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, a^{2} d^{2}+44 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, a b c d -30 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}\, b^{2} c^{2}\right )}{96 b \,d^{3} \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {b d}}\)	\(455\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \left (\frac {a^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{16 b d}-\frac {a^{3} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{32 b \sqrt {b d}}-\frac {3 a^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c}{32 d \sqrt {b d}}+\frac {b \,x^{4} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{6 d}+\frac {7 x^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a}{24 d}-\frac {5 b \,x^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, c}{24 d^{2}}-\frac {11 \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a c}{24 d^{2}}+\frac {5 b \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, c^{2}}{16 d^{3}}+\frac {9 b \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a \,c^{2}}{32 d^{2} \sqrt {b d}}-\frac {5 b^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c^{3}}{32 d^{3} \sqrt {b d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(473\)

Input:

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

Output:

1/48/b*(8*b^2*d^2*x^4+14*a*b*d^2*x^2-10*b^2*c*d*x^2+3*a^2*d^2-22*a*b*c*d+1 
5*b^2*c^2)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^3-1/32/b*(a^3*d^3+3*a^2*b*c*d 
^2-9*a*b^2*c^2*d+5*b^3*c^3)/d^3*ln((1/2*a*d+1/2*b*c+b*d*x^2)/(b*d)^(1/2)+( 
b*d*x^4+(a*d+b*c)*x^2+a*c)^(1/2))/(b*d)^(1/2)*((b*x^2+a)*(d*x^2+c))^(1/2)/ 
(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.35 \[ \int \frac {x^3 \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\left [\frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{192 \, b^{2} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{96 \, b^{2} d^{4}}\right ] \] Input:

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

Output:

[1/192*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*sqrt(b*d)* 
log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)* 
x^2 - 4*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d)) 
 + 4*(8*b^3*d^3*x^4 + 15*b^3*c^2*d - 22*a*b^2*c*d^2 + 3*a^2*b*d^3 - 2*(5*b 
^3*c*d^2 - 7*a*b^2*d^3)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^4), 1 
/96*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*sqrt(-b*d)*ar 
ctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d 
)/(b^2*d^2*x^4 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^2)) + 2*(8*b^3*d^3*x^4 + 
15*b^3*c^2*d - 22*a*b^2*c*d^2 + 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 7*a*b^2*d^3 
)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^4)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3 \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.20 \[ \int \frac {x^3 \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (2 \, {\left (b x^{2} + a\right )} {\left (\frac {4 \, {\left (b x^{2} + a\right )}}{b^{2} d} - \frac {5 \, b^{3} c d^{3} + a b^{2} d^{4}}{b^{4} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{4} c^{2} d^{2} - 4 \, a b^{3} c d^{3} - a^{2} b^{2} d^{4}\right )}}{b^{4} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{3}}\right )} b}{48 \, {\left | b \right |}} \] Input:

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

Output:

1/48*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a) 
*(4*(b*x^2 + a)/(b^2*d) - (5*b^3*c*d^3 + a*b^2*d^4)/(b^4*d^5)) + 3*(5*b^4* 
c^2*d^2 - 4*a*b^3*c*d^3 - a^2*b^2*d^4)/(b^4*d^5)) + 3*(5*b^3*c^3 - 9*a*b^2 
*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*log(abs(-sqrt(b*x^2 + a)*sqrt(b*d) + sqr 
t(b^2*c + (b*x^2 + a)*b*d - a*b*d)))/(sqrt(b*d)*b*d^3))*b/abs(b)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.75 \[ \int \frac {x^3 \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\frac {3 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{3}-22 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,d^{2}+14 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{3} x^{2}+15 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{3} c^{2} d -10 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{3} c \,d^{2} x^{2}+8 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{3} d^{3} x^{4}+3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a^{3} d^{3}+9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a^{2} b c \,d^{2}-27 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a \,b^{2} c^{2} d +15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d +\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) b^{3} c^{3}}{48 b^{2} d^{4}} \] Input:

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

Output:

(3*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a**2*b*d**3 - 22*sqrt(c + d*x**2)*sqr 
t(a + b*x**2)*a*b**2*c*d**2 + 14*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*b**2* 
d**3*x**2 + 15*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**3*c**2*d - 10*sqrt(c + 
 d*x**2)*sqrt(a + b*x**2)*b**3*c*d**2*x**2 + 8*sqrt(c + d*x**2)*sqrt(a + b 
*x**2)*b**3*d**3*x**4 + 3*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)* 
d + sqrt(d)*sqrt(c + d*x**2)*b)*a**3*d**3 + 9*sqrt(d)*sqrt(b)*log( - sqrt( 
b)*sqrt(a + b*x**2)*d + sqrt(d)*sqrt(c + d*x**2)*b)*a**2*b*c*d**2 - 27*sqr 
t(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d + sqrt(d)*sqrt(c + d*x**2)* 
b)*a*b**2*c**2*d + 15*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d + 
sqrt(d)*sqrt(c + d*x**2)*b)*b**3*c**3)/(48*b**2*d**4)