\(\int \frac {x^3}{(a+b x^2)^2 (c+d x^2)^{3/2}} \, dx\) [1024]

Optimal result

Integrand size = 24, antiderivative size = 118 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {c}{(b c-a d)^2 \sqrt {c+d x^2}}+\frac {a \sqrt {c+d x^2}}{2 (b c-a d)^2 \left (a+b x^2\right )}-\frac {(2 b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 \sqrt {b} (b c-a d)^{5/2}} \] Output:

c/(-a*d+b*c)^2/(d*x^2+c)^(1/2)+1/2*a*(d*x^2+c)^(1/2)/(-a*d+b*c)^2/(b*x^2+a 
)-1/2*(a*d+2*b*c)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(1/2 
)/(-a*d+b*c)^(5/2)
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.93 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {1}{2} \left (\frac {3 a c+2 b c x^2+a d x^2}{(b c-a d)^2 \left (a+b x^2\right ) \sqrt {c+d x^2}}+\frac {(2 b c+a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{\sqrt {b} (-b c+a d)^{5/2}}\right ) \] Input:

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

Output:

((3*a*c + 2*b*c*x^2 + a*d*x^2)/((b*c - a*d)^2*(a + b*x^2)*Sqrt[c + d*x^2]) 
 + ((2*b*c + a*d)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/(S 
qrt[b]*(-(b*c) + a*d)^(5/2)))/2
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {354, 87, 61, 73, 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.

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 61

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

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

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

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 = 0.75 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.06

method	result	size
pseudoelliptic	\(\frac {\frac {\sqrt {x^{2} d +c}\, \left (b \,x^{2}+a \right ) \left (a d +2 b c \right ) \arctan \left (\frac {\sqrt {x^{2} d +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{2}+\frac {3 \sqrt {\left (a d -b c \right ) b}\, \left (\left (\frac {x^{2} d}{3}+c \right ) a +\frac {2 x^{2} b c}{3}\right )}{2}}{\left (b \,x^{2}+a \right ) \left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}\, \sqrt {x^{2} d +c}}\)	\(125\)
default	\(\text {Expression too large to display}\)	\(1922\)

Input:

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

Output:

3/2*(1/3*(d*x^2+c)^(1/2)*(b*x^2+a)*(a*d+2*b*c)*arctan((d*x^2+c)^(1/2)*b/(( 
a*d-b*c)*b)^(1/2))+((a*d-b*c)*b)^(1/2)*((1/3*x^2*d+c)*a+2/3*x^2*b*c))/((a* 
d-b*c)*b)^(1/2)/(d*x^2+c)^(1/2)/(b*x^2+a)/(a*d-b*c)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (100) = 200\).

Time = 0.14 (sec) , antiderivative size = 732, normalized size of antiderivative = 6.20 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (2 \, b^{2} c d + a b d^{2}\right )} x^{4} + 2 \, a b c^{2} + a^{2} c d + {\left (2 \, b^{2} c^{2} + 3 \, a b c d + a^{2} d^{2}\right )} x^{2}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (3 \, a b^{2} c^{2} - 3 \, a^{2} b c d + {\left (2 \, b^{3} c^{2} - a b^{2} c d - a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{4} c^{4} - 3 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} - a^{4} b c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{4} + {\left (b^{5} c^{4} - 2 \, a b^{4} c^{3} d + 2 \, a^{3} b^{2} c d^{3} - a^{4} b d^{4}\right )} x^{2}\right )}}, -\frac {{\left ({\left (2 \, b^{2} c d + a b d^{2}\right )} x^{4} + 2 \, a b c^{2} + a^{2} c d + {\left (2 \, b^{2} c^{2} + 3 \, a b c d + a^{2} d^{2}\right )} x^{2}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (3 \, a b^{2} c^{2} - 3 \, a^{2} b c d + {\left (2 \, b^{3} c^{2} - a b^{2} c d - a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{4} c^{4} - 3 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} - a^{4} b c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{4} + {\left (b^{5} c^{4} - 2 \, a b^{4} c^{3} d + 2 \, a^{3} b^{2} c d^{3} - a^{4} b d^{4}\right )} x^{2}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^3/(b*x^2+a)^2/(d*x^2+c)^(3/2),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 = 0.12 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.53 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {{\left (2 \, b c d + a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (d x^{2} + c\right )} b c d - 2 \, b c^{2} d + {\left (d x^{2} + c\right )} a d^{2} + 2 \, a c d^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left ({\left (d x^{2} + c\right )}^{\frac {3}{2}} b - \sqrt {d x^{2} + c} b c + \sqrt {d x^{2} + c} a d\right )}}}{2 \, d} \] Input:

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

Output:

1/2*((2*b*c*d + a*d^2)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((b^ 
2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-b^2*c + a*b*d)) + (2*(d*x^2 + c)*b*c*d 
- 2*b*c^2*d + (d*x^2 + c)*a*d^2 + 2*a*c*d^2)/((b^2*c^2 - 2*a*b*c*d + a^2*d 
^2)*((d*x^2 + c)^(3/2)*b - sqrt(d*x^2 + c)*b*c + sqrt(d*x^2 + c)*a*d)))/d
 

Mupad [B] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.20 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {c}{a\,d-b\,c}+\frac {\left (d\,x^2+c\right )\,\left (a\,d+2\,b\,c\right )}{2\,{\left (a\,d-b\,c\right )}^2}}{b\,{\left (d\,x^2+c\right )}^{3/2}+\sqrt {d\,x^2+c}\,\left (a\,d-b\,c\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^{5/2}}\right )\,\left (a\,d+2\,b\,c\right )}{2\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{5/2}} \] Input:

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

Output:

(c/(a*d - b*c) + ((c + d*x^2)*(a*d + 2*b*c))/(2*(a*d - b*c)^2))/(b*(c + d* 
x^2)^(3/2) + (c + d*x^2)^(1/2)*(a*d - b*c)) + (atan((b^(1/2)*(c + d*x^2)^( 
1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(a*d - b*c)^(5/2))*(a*d + 2*b*c))/(2 
*b^(1/2)*(a*d - b*c)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 819, normalized size of antiderivative = 6.94 \[ \int \frac {x^3}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(d)*sqrt(c + d*x**2)*b*x + b*c + b*d*x* 
*2)/(sqrt(b)*sqrt(c + d*x**2)*sqrt(a*d - b*c) + sqrt(d)*sqrt(b)*sqrt(a*d - 
 b*c)*x))*a**2*c*d + sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(d)*sqrt(c + d*x**2 
)*b*x + b*c + b*d*x**2)/(sqrt(b)*sqrt(c + d*x**2)*sqrt(a*d - b*c) + sqrt(d 
)*sqrt(b)*sqrt(a*d - b*c)*x))*a**2*d**2*x**2 + 2*sqrt(b)*sqrt(a*d - b*c)*a 
tan((sqrt(d)*sqrt(c + d*x**2)*b*x + b*c + b*d*x**2)/(sqrt(b)*sqrt(c + d*x* 
*2)*sqrt(a*d - b*c) + sqrt(d)*sqrt(b)*sqrt(a*d - b*c)*x))*a*b*c**2 + 3*sqr 
t(b)*sqrt(a*d - b*c)*atan((sqrt(d)*sqrt(c + d*x**2)*b*x + b*c + b*d*x**2)/ 
(sqrt(b)*sqrt(c + d*x**2)*sqrt(a*d - b*c) + sqrt(d)*sqrt(b)*sqrt(a*d - b*c 
)*x))*a*b*c*d*x**2 + sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(d)*sqrt(c + d*x**2 
)*b*x + b*c + b*d*x**2)/(sqrt(b)*sqrt(c + d*x**2)*sqrt(a*d - b*c) + sqrt(d 
)*sqrt(b)*sqrt(a*d - b*c)*x))*a*b*d**2*x**4 + 2*sqrt(b)*sqrt(a*d - b*c)*at 
an((sqrt(d)*sqrt(c + d*x**2)*b*x + b*c + b*d*x**2)/(sqrt(b)*sqrt(c + d*x** 
2)*sqrt(a*d - b*c) + sqrt(d)*sqrt(b)*sqrt(a*d - b*c)*x))*b**2*c**2*x**2 + 
2*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(d)*sqrt(c + d*x**2)*b*x + b*c + b*d*x 
**2)/(sqrt(b)*sqrt(c + d*x**2)*sqrt(a*d - b*c) + sqrt(d)*sqrt(b)*sqrt(a*d 
- b*c)*x))*b**2*c*d*x**4 + 3*sqrt(c + d*x**2)*a**2*b*c*d + sqrt(c + d*x**2 
)*a**2*b*d**2*x**2 - 3*sqrt(c + d*x**2)*a*b**2*c**2 + sqrt(c + d*x**2)*a*b 
**2*c*d*x**2 - 2*sqrt(c + d*x**2)*b**3*c**2*x**2)/(2*b*(a**4*c*d**3 + a**4 
*d**4*x**2 - 3*a**3*b*c**2*d**2 - 2*a**3*b*c*d**3*x**2 + a**3*b*d**4*x*...