\(\int \frac {x^2}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\) [983]

Optimal result

Integrand size = 24, antiderivative size = 115 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {x}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {(2 b c+a d) x}{3 c (b c-a d)^2 \sqrt {c+d x^2}}-\frac {\sqrt {a} b \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.06 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {a d^2 x^3+b c x \left (3 c+2 d x^2\right )}{3 c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {\sqrt {a} b \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \] Input:

Integrate[x^2/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {373, 402, 27, 291, 218}

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.

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

Input:

Int[x^2/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

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

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 218

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

rule 291

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

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 402

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

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(\frac {-3 a b c \,\operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right ) \left (x^{2} d +c \right )^{\frac {3}{2}}+\sqrt {\left (a d -b c \right ) a}\, x \left (a \,d^{2} x^{2}+2 b c d \,x^{2}+3 b \,c^{2}\right )}{3 \sqrt {\left (a d -b c \right ) a}\, \left (x^{2} d +c \right )^{\frac {3}{2}} \left (a d -b c \right )^{2} c}\)	\(116\)
default	\(\text {Expression too large to display}\)	\(1439\)

Input:

int(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (97) = 194\).

Time = 0.25 (sec) , antiderivative size = 550, normalized size of antiderivative = 4.78 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b c d^{2} x^{4} + 2 \, b c^{2} d x^{2} + b c^{3}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (3 \, b c^{2} x + {\left (2 \, b c d + a d^{2}\right )} x^{3}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} x^{2}\right )}}, \frac {3 \, {\left (b c d^{2} x^{4} + 2 \, b c^{2} d x^{2} + b c^{3}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{3} + a c x\right )}}\right ) + 2 \, {\left (3 \, b c^{2} x + {\left (2 \, b c d + a d^{2}\right )} x^{3}\right )} \sqrt {d x^{2} + c}}{6 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} x^{2}\right )}}\right ] \] Input:

integrate(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/12*(3*(b*c*d^2*x^4 + 2*b*c^2*d*x^2 + b*c^3)*sqrt(-a/(b*c - a*d))*log((( 
b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d) 
*x^2 - 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^3 - (a*b*c^2 - a^2*c*d)*x)*s 
qrt(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(3*b 
*c^2*x + (2*b*c*d + a*d^2)*x^3)*sqrt(d*x^2 + c))/(b^2*c^5 - 2*a*b*c^4*d + 
a^2*c^3*d^2 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^4 + 2*(b^2*c^4*d 
 - 2*a*b*c^3*d^2 + a^2*c^2*d^3)*x^2), 1/6*(3*(b*c*d^2*x^4 + 2*b*c^2*d*x^2 
+ b*c^3)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d* 
x^2 + c)*sqrt(a/(b*c - a*d))/(a*d*x^3 + a*c*x)) + 2*(3*b*c^2*x + (2*b*c*d 
+ a*d^2)*x^3)*sqrt(d*x^2 + c))/(b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2 + (b^2 
*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^4 + 2*(b^2*c^4*d - 2*a*b*c^3*d^2 + 
 a^2*c^2*d^3)*x^2)]
 

Sympy [F]

\[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^{2}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**2/(b*x**2+a)/(d*x**2+c)**(5/2),x)
 

Output:

Integral(x**2/((a + b*x**2)*(c + d*x**2)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {x^{2}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(x^2/((b*x^2 + a)*(d*x^2 + c)^(5/2)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (97) = 194\).

Time = 0.14 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.53 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {a b \sqrt {d} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {{\left (\frac {{\left (2 \, b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + a^{3} d^{5}\right )} x^{2}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}} + \frac {3 \, {\left (b^{3} c^{4} d - 2 \, a b^{2} c^{3} d^{2} + a^{2} b c^{2} d^{3}\right )}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \] Input:

integrate(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

a*b*sqrt(d)*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/s 
qrt(a*b*c*d - a^2*d^2))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*b*c*d - a^ 
2*d^2)) + 1/3*((2*b^3*c^3*d^2 - 3*a*b^2*c^2*d^3 + a^3*d^5)*x^2/(b^4*c^5*d 
- 4*a*b^3*c^4*d^2 + 6*a^2*b^2*c^3*d^3 - 4*a^3*b*c^2*d^4 + a^4*c*d^5) + 3*( 
b^3*c^4*d - 2*a*b^2*c^3*d^2 + a^2*b*c^2*d^3)/(b^4*c^5*d - 4*a*b^3*c^4*d^2 
+ 6*a^2*b^2*c^3*d^3 - 4*a^3*b*c^2*d^4 + a^4*c*d^5))*x/(d*x^2 + c)^(3/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^2}{\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:

int(x^2/((a + b*x^2)*(c + d*x^2)^(5/2)),x)
 

Output:

int(x^2/((a + b*x^2)*(c + d*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 949, normalized size of antiderivative = 8.25 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x)
 

Output:

( - 3*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c 
) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b*c**3*d 
- 6*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) 
- 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b*c**2*d**2 
*x**2 - 3*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - 
 b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b*c*d 
**3*x**4 - 3*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - 
 b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b*c** 
3*d - 6*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) 
 - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b*c**2*d** 
2*x**2 - 3*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b 
*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*b*c*d** 
3*x**4 + 3*sqrt(a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 
 2*sqrt(d)*sqrt(c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*b*c**3*d + 6*sqrt(a) 
*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*sqrt(d)*sqrt(c 
+ d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*b*c**2*d**2*x**2 + 3*sqrt(a)*sqrt(a*d 
- b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*sqrt(d)*sqrt(c + d*x**2)* 
b*x + 2*a*d + 2*b*d*x**2)*b*c*d**3*x**4 + 2*sqrt(c + d*x**2)*a**2*d**4*x** 
3 + 6*sqrt(c + d*x**2)*a*b*c**2*d**2*x + 2*sqrt(c + d*x**2)*a*b*c*d**3*x** 
3 - 6*sqrt(c + d*x**2)*b**2*c**3*d*x - 4*sqrt(c + d*x**2)*b**2*c**2*d**...