\(\int \frac {x^4}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\) [982]

Optimal result

Integrand size = 24, antiderivative size = 117 \[ \int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {c x}{3 d (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {(b c-4 a d) x}{3 d (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a^{3/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{5/2}} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(351\) vs. \(2(117)=234\).

Time = 1.79 (sec) , antiderivative size = 351, normalized size of antiderivative = 3.00 \[ \int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {-3 a c x+b c x^3-4 a d x^3}{3 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {\sqrt {a} \left (-b c+a d+\sqrt {b} \sqrt {c} \sqrt {b c-a d}\right ) \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{d (-b c+a d)^3}+\frac {\sqrt {a} \sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \left (b c-a d+\sqrt {b} \sqrt {c} \sqrt {b c-a d}\right ) \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (-\sqrt {c}+\sqrt {c+d x^2}\right )}\right )}{d (-b c+a d)^3} \] Input:

Integrate[x^4/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

(-3*a*c*x + b*c*x^3 - 4*a*d*x^3)/(3*(b*c - a*d)^2*(c + d*x^2)^(3/2)) + (Sq 
rt[a]*(-(b*c) + a*d + Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d])*Sqrt[2*b*c - a*d + 
2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*ArcTan[(Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sq 
rt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(Sqrt[c] - Sqrt[c + d*x^2]))])/(d*(-(b* 
c) + a*d)^3) + (Sqrt[a]*Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a* 
d]]*(b*c - a*d + Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d])*ArcTan[(Sqrt[2*b*c - a*d 
 - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(-Sqrt[c] + Sqrt[c + d*x 
^2]))])/(d*(-(b*c) + a*d)^3)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {372, 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.

\(\displaystyle \int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

Input:

Int[x^4/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

-1/3*(c*x)/(d*(b*c - a*d)*(c + d*x^2)^(3/2)) + (((b*c - 4*a*d)*x)/((b*c - 
a*d)*Sqrt[c + d*x^2]) + (3*a^(3/2)*d*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*S 
qrt[c + d*x^2])])/(b*c - a*d)^(3/2))/(3*d*(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 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[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 = 0.71 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91

method	result	size
pseudoelliptic	\(\frac {a^{2} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right ) \left (x^{2} d +c \right )^{\frac {3}{2}}-\left (\left (\frac {4 x^{2} d}{3}+c \right ) a -\frac {x^{2} b c}{3}\right ) \sqrt {\left (a d -b c \right ) a}\, x}{\sqrt {\left (a d -b c \right ) a}\, \left (x^{2} d +c \right )^{\frac {3}{2}} \left (a d -b c \right )^{2}}\)	\(107\)
default	\(\text {Expression too large to display}\)	\(1502\)

Input:

int(x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/((a*d-b*c)*a)^(1/2)/(d*x^2+c)^(3/2)*(a^2*arctanh((d*x^2+c)^(1/2)/x*a/((a 
*d-b*c)*a)^(1/2))*(d*x^2+c)^(3/2)-((4/3*x^2*d+c)*a-1/3*x^2*b*c)*((a*d-b*c) 
*a)^(1/2)*x)/(a*d-b*c)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (99) = 198\).

Time = 0.28 (sec) , antiderivative size = 524, normalized size of antiderivative = 4.48 \[ \int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a d^{2} x^{4} + 2 \, a c d x^{2} + a c^{2}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left ({\left (b c - 4 \, a d\right )} x^{3} - 3 \, a c x\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{2}\right )}}, -\frac {3 \, {\left (a d^{2} x^{4} + 2 \, a c d x^{2} + a c^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{3} + a c x\right )}}\right ) - 2 \, {\left ({\left (b c - 4 \, a d\right )} x^{3} - 3 \, a c x\right )} \sqrt {d x^{2} + c}}{6 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} + {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{2}\right )}}\right ] \] Input:

integrate(x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/12*(3*(a*d^2*x^4 + 2*a*c*d*x^2 + a*c^2)*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)*sqrt( 
d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*((b*c - 
4*a*d)*x^3 - 3*a*c*x)*sqrt(d*x^2 + c))/(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^ 
2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d 
^2 + a^2*c*d^3)*x^2), -1/6*(3*(a*d^2*x^4 + 2*a*c*d*x^2 + a*c^2)*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*((b*c - 4*a*d)*x^3 - 3*a*c*x)*sqrt(d*x^2 
+ c))/(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + 
a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*x^2)]
 

Sympy [F]

\[ \int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^{4}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**4/(b*x**2+a)/(d*x**2+c)**(5/2),x)
 

Output:

Integral(x**4/((a + b*x**2)*(c + d*x**2)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {x^{4}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(x^4/((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. 304 vs. \(2 (99) = 198\).

Time = 0.14 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.60 \[ \int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {a^{2} \sqrt {d} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {{\left (\frac {{\left (b^{3} c^{4} d - 6 \, a b^{2} c^{3} d^{2} + 9 \, a^{2} b c^{2} d^{3} - 4 \, a^{3} c d^{4}\right )} x^{2}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}} - \frac {3 \, {\left (a b^{2} c^{4} d - 2 \, a^{2} b c^{3} d^{2} + a^{3} c^{2} d^{3}\right )}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \] Input:

integrate(x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

-a^2*sqrt(d)*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/ 
sqrt(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*((b^3*c^4*d - 6*a*b^2*c^3*d^2 + 9*a^2*b*c^2*d^3 - 4*a^3*c*d 
^4)*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*(a*b^2*c^4*d - 2*a^2*b*c^3*d^2 + a^3*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. \[ \int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^4}{\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:

int(x^4/((a + b*x^2)*(c + d*x^2)^(5/2)),x)
 

Output:

int(x^4/((a + b*x^2)*(c + d*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 903, normalized size of antiderivative = 7.72 \[ \int \frac {x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int(x^4/(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)*a*c**2*d**2 
+ 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)*a*c*d**3*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)*a*d**4*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)*a*c**2*d** 
2 + 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)*a*c*d**3*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)*a*d**4*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)*a*c**2*d**2 - 6*sqrt(a)*sqr 
t(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)*a*c*d**3*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)*a*d**4*x**4 - 6*sqrt(c + d*x**2)*a**2*c*d**3*x - 8*sqrt 
(c + d*x**2)*a**2*d**4*x**3 + 6*sqrt(c + d*x**2)*a*b*c**2*d**2*x + 10*sqrt 
(c + d*x**2)*a*b*c*d**3*x**3 - 2*sqrt(c + d*x**2)*b**2*c**2*d**2*x**3 +...