\(\int \frac {x}{(c+d x)^{3/2} (a-b x^2)^2} \, dx\) [674]

Optimal result

Integrand size = 21, antiderivative size = 213 \[ \int \frac {x}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=\frac {3 c d^2}{\left (b c^2-a d^2\right )^2 \sqrt {c+d x}}+\frac {c-d x}{2 \left (b c^2-a d^2\right ) \sqrt {c+d x} \left (a-b x^2\right )}-\frac {3 d \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{4 \sqrt {a} \sqrt [4]{b} \left (\sqrt {b} c-\sqrt {a} d\right )^{5/2}}+\frac {3 d \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{4 \sqrt {a} \sqrt [4]{b} \left (\sqrt {b} c+\sqrt {a} d\right )^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.93 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.31 \[ \int \frac {x}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=-\frac {a d^2 (5 c+d x)+b c \left (c^2-c d x-6 d^2 x^2\right )}{2 \left (b c^2-a d^2\right )^2 \sqrt {c+d x} \left (-a+b x^2\right )}-\frac {3 d \sqrt {-b c-\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )}{4 \sqrt {a} \sqrt {b} \left (\sqrt {b} c+\sqrt {a} d\right )^3}-\frac {3 d \sqrt {-b c+\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{4 \sqrt {a} \sqrt {b} \left (-\sqrt {b} c+\sqrt {a} d\right )^3} \] Input:

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

Output:

-1/2*(a*d^2*(5*c + d*x) + b*c*(c^2 - c*d*x - 6*d^2*x^2))/((b*c^2 - a*d^2)^ 
2*Sqrt[c + d*x]*(-a + b*x^2)) - (3*d*Sqrt[-(b*c) - Sqrt[a]*Sqrt[b]*d]*ArcT 
an[(Sqrt[-(b*c) - Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/(Sqrt[b]*c + Sqrt[a]*d 
)])/(4*Sqrt[a]*Sqrt[b]*(Sqrt[b]*c + Sqrt[a]*d)^3) - (3*d*Sqrt[-(b*c) + Sqr 
t[a]*Sqrt[b]*d]*ArcTan[(Sqrt[-(b*c) + Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/(S 
qrt[b]*c - Sqrt[a]*d)])/(4*Sqrt[a]*Sqrt[b]*(-(Sqrt[b]*c) + Sqrt[a]*d)^3)
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.51, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {561, 25, 27, 1600, 27, 1604, 27, 1480, 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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1600

\(\displaystyle -\frac {2 \left (\frac {d^4 \int -\frac {6 a b (c-d x)}{d^2 (c+d x) \left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )}d\sqrt {c+d x}}{8 a b \left (b c^2-a d^2\right )}-\frac {d^2 (c-d x)}{4 \sqrt {c+d x} \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{d^2}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1604

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-2*(-1/4*(d^2*(c - d*x))/((b*c^2 - a*d^2)*Sqrt[c + d*x]*(a - (b*c^2)/d^2 
+ (2*b*c*(c + d*x))/d^2 - (b*(c + d*x)^2)/d^2)) - (3*d^2*((-2*c)/((a - (b* 
c^2)/d^2)*Sqrt[c + d*x]) - (-1/2*(d*(Sqrt[b]*c + Sqrt[a]*d)^2*ArcTanh[(b^( 
1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[a]*d]])/(Sqrt[a]*b^(1/4)*Sqrt[Sq 
rt[b]*c - Sqrt[a]*d]) + (d*(Sqrt[b]*c - Sqrt[a]*d)^2*ArcTanh[(b^(1/4)*Sqrt 
[c + d*x])/Sqrt[Sqrt[b]*c + Sqrt[a]*d]])/(2*Sqrt[a]*b^(1/4)*Sqrt[Sqrt[b]*c 
 + Sqrt[a]*d]))/((a - (b*c^2)/d^2)*d^2)))/(4*(b*c^2 - a*d^2))))/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 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 561

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{k = Denominator[n]}, Simp[k/d   Subst[Int[x^(k*(n + 1) - 1)*(-c 
/d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), 
 x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac 
tionQ[n] && IntegerQ[p] && IntegerQ[m]
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

rule 1600

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(a + b*x^2 + c*x^4)^(p + 1) 
*((d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^2)/(2*a*f*(p + 1)*(b^2 - 4*a 
*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c))   Int[(f*x)^m*(a + b*x^2 + c 
*x^4)^(p + 1)*Simp[d*(b^2*(m + 2*(p + 1) + 1) - 2*a*c*(m + 4*(p + 1) + 1)) 
- a*b*e*(m + 1) + c*(m + 2*(2*p + 3) + 1)*(b*d - 2*a*e)*x^2, x], x], x] /; 
FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && Int 
egerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1604

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1) 
/(a*f*(m + 1))), x] + Simp[1/(a*f^2*(m + 1))   Int[(f*x)^(m + 2)*(a + b*x^2 
 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x 
], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[ 
m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(2 d^{2} \left (\frac {c}{\left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}+\frac {\frac {-\frac {b c \left (d x +c \right )^{\frac {3}{2}}}{2}+\left (\frac {a \,d^{2}}{4}+\frac {3 b \,c^{2}}{4}\right ) \sqrt {d x +c}}{-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}}+\frac {3 b \left (-\frac {\left (-a \,d^{2}-b \,c^{2}+2 \sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (a \,d^{2}+b \,c^{2}+2 \sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}}{\left (a \,d^{2}-b \,c^{2}\right )^{2}}\right )\)	\(269\)
default	\(2 d^{2} \left (\frac {c}{\left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {d x +c}}+\frac {\frac {-\frac {b c \left (d x +c \right )^{\frac {3}{2}}}{2}+\left (\frac {a \,d^{2}}{4}+\frac {3 b \,c^{2}}{4}\right ) \sqrt {d x +c}}{-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}}+\frac {3 b \left (-\frac {\left (-a \,d^{2}-b \,c^{2}+2 \sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (a \,d^{2}+b \,c^{2}+2 \sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}}{\left (a \,d^{2}-b \,c^{2}\right )^{2}}\right )\)	\(269\)
pseudoelliptic	\(\frac {\frac {3 d^{2} \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, b \left (a \,d^{2}+b \,c^{2}+2 \sqrt {a b \,d^{2}}\, c \right ) \sqrt {d x +c}\, \left (-b \,x^{2}+a \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}+\frac {3 \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (\sqrt {d x +c}\, b \,d^{2} \left (-b \,x^{2}+a \right ) \left (a \,d^{2}+b \,c^{2}-2 \sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )+\frac {10 \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {a b \,d^{2}}\, \left (\frac {c \left (2 d x +c \right ) \left (-3 d x +c \right ) b}{5}+a \,d^{2} \left (\frac {d x}{5}+c \right )\right )}{3}\right )}{4}}{\left (-b \,x^{2}+a \right ) \left (a \,d^{2}-b \,c^{2}\right )^{2} \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {d x +c}}\)	\(302\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5303 vs. \(2 (165) = 330\).

Time = 0.31 (sec) , antiderivative size = 5303, normalized size of antiderivative = 24.90 \[ \int \frac {x}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1323 vs. \(2 (165) = 330\).

Time = 0.31 (sec) , antiderivative size = 1323, normalized size of antiderivative = 6.21 \[ \int \frac {x}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(3*(2*(b^2*c^4*d - 2*a*b*c^2*d^3 + a^2*d^5)^2*sqrt(a*b)*a*c*d^3*abs(b 
) - (3*a*b^3*c^6*d^3 - 5*a^2*b^2*c^4*d^5 + a^3*b*c^2*d^7 + a^4*d^9)*abs(b^ 
2*c^4*d - 2*a*b*c^2*d^3 + a^2*d^5)*abs(b) + (sqrt(a*b)*b^5*c^11*d^3 - 3*sq 
rt(a*b)*a*b^4*c^9*d^5 + 2*sqrt(a*b)*a^2*b^3*c^7*d^7 + 2*sqrt(a*b)*a^3*b^2* 
c^5*d^9 - 3*sqrt(a*b)*a^4*b*c^3*d^11 + sqrt(a*b)*a^5*c*d^13)*abs(b))*arcta 
n(sqrt(d*x + c)/sqrt(-(b^3*c^5 - 2*a*b^2*c^3*d^2 + a^2*b*c*d^4 + sqrt((b^3 
*c^5 - 2*a*b^2*c^3*d^2 + a^2*b*c*d^4)^2 - (b^3*c^6 - 3*a*b^2*c^4*d^2 + 3*a 
^2*b*c^2*d^4 - a^3*d^6)*(b^3*c^4 - 2*a*b^2*c^2*d^2 + a^2*b*d^4)))/(b^3*c^4 
 - 2*a*b^2*c^2*d^2 + a^2*b*d^4)))/((a*b^5*c^9 - 4*a^2*b^4*c^7*d^2 + 6*a^3* 
b^3*c^5*d^4 - 4*a^4*b^2*c^3*d^6 + a^5*b*c*d^8 - sqrt(a*b)*a*b^4*c^8*d + 4* 
sqrt(a*b)*a^2*b^3*c^6*d^3 - 6*sqrt(a*b)*a^3*b^2*c^4*d^5 + 4*sqrt(a*b)*a^4* 
b*c^2*d^7 - sqrt(a*b)*a^5*d^9)*sqrt(-b^2*c - sqrt(a*b)*b*d)*abs(b^2*c^4*d 
- 2*a*b*c^2*d^3 + a^2*d^5)) - 3*(2*(b^2*c^4*d - 2*a*b*c^2*d^3 + a^2*d^5)^2 
*sqrt(a*b)*a*c*d^3*abs(b) + (3*a*b^3*c^6*d^3 - 5*a^2*b^2*c^4*d^5 + a^3*b*c 
^2*d^7 + a^4*d^9)*abs(b^2*c^4*d - 2*a*b*c^2*d^3 + a^2*d^5)*abs(b) + (sqrt( 
a*b)*b^5*c^11*d^3 - 3*sqrt(a*b)*a*b^4*c^9*d^5 + 2*sqrt(a*b)*a^2*b^3*c^7*d^ 
7 + 2*sqrt(a*b)*a^3*b^2*c^5*d^9 - 3*sqrt(a*b)*a^4*b*c^3*d^11 + sqrt(a*b)*a 
^5*c*d^13)*abs(b))*arctan(sqrt(d*x + c)/sqrt(-(b^3*c^5 - 2*a*b^2*c^3*d^2 + 
 a^2*b*c*d^4 - sqrt((b^3*c^5 - 2*a*b^2*c^3*d^2 + a^2*b*c*d^4)^2 - (b^3*c^6 
 - 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d^4 - a^3*d^6)*(b^3*c^4 - 2*a*b^2*c^2*...
 

Mupad [B] (verification not implemented)

Time = 11.11 (sec) , antiderivative size = 7948, normalized size of antiderivative = 37.31 \[ \int \frac {x}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*c*d^2)/(a*d^2 - b*c^2) + ((a*d^4 + 11*b*c^2*d^2)*(c + d*x))/(2*(a*d^2 
- b*c^2)^2) - (3*b*c*d^2*(c + d*x)^2)/(a*d^2 - b*c^2)^2)/((a*d^2 - b*c^2)* 
(c + d*x)^(1/2) - b*(c + d*x)^(5/2) + 2*b*c*(c + d*x)^(3/2)) - atan(-((((9 
*(a^2*d^7*(a^3*b)^(1/2) - a*b^3*c^5*d^2 - 10*a^2*b^2*c^3*d^4 + 5*b^2*c^4*d 
^3*(a^3*b)^(1/2) - 5*a^3*b*c*d^6 + 10*a*b*c^2*d^5*(a^3*b)^(1/2)))/(64*(a^7 
*b*d^10 - a^2*b^6*c^10 + 5*a^3*b^5*c^8*d^2 - 10*a^4*b^4*c^6*d^4 + 10*a^5*b 
^3*c^4*d^6 - 5*a^6*b^2*c^2*d^8)))^(1/2)*((c + d*x)^(1/2)*((9*(a^2*d^7*(a^3 
*b)^(1/2) - a*b^3*c^5*d^2 - 10*a^2*b^2*c^3*d^4 + 5*b^2*c^4*d^3*(a^3*b)^(1/ 
2) - 5*a^3*b*c*d^6 + 10*a*b*c^2*d^5*(a^3*b)^(1/2)))/(64*(a^7*b*d^10 - a^2* 
b^6*c^10 + 5*a^3*b^5*c^8*d^2 - 10*a^4*b^4*c^6*d^4 + 10*a^5*b^3*c^4*d^6 - 5 
*a^6*b^2*c^2*d^8)))^(1/2)*(2048*a*b^14*c^21*d^2 + 2048*a^11*b^4*c*d^22 - 2 
0480*a^2*b^13*c^19*d^4 + 92160*a^3*b^12*c^17*d^6 - 245760*a^4*b^11*c^15*d^ 
8 + 430080*a^5*b^10*c^13*d^10 - 516096*a^6*b^9*c^11*d^12 + 430080*a^7*b^8* 
c^9*d^14 - 245760*a^8*b^7*c^7*d^16 + 92160*a^9*b^6*c^5*d^18 - 20480*a^10*b 
^5*c^3*d^20) + 768*a^10*b^3*d^22 + 2304*a*b^12*c^18*d^4 - 17664*a^2*b^11*c 
^16*d^6 + 58368*a^3*b^10*c^14*d^8 - 107520*a^4*b^9*c^12*d^10 + 118272*a^5* 
b^8*c^10*d^12 - 75264*a^6*b^7*c^8*d^14 + 21504*a^7*b^6*c^6*d^16 + 3072*a^8 
*b^5*c^4*d^18 - 3840*a^9*b^4*c^2*d^20) - (c + d*x)^(1/2)*(288*a^8*b^3*d^20 
 + 288*b^11*c^16*d^4 - 5760*a^2*b^9*c^12*d^8 + 18432*a^3*b^8*c^10*d^10 - 2 
5920*a^4*b^7*c^8*d^12 + 18432*a^5*b^6*c^6*d^14 - 5760*a^6*b^5*c^4*d^16)...
 

Reduce [B] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 1539, normalized size of antiderivative = 7.23 \[ \int \frac {x}{(c+d x)^{3/2} \left (a-b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(18*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x 
)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a**2*b*c*d**3 + 6*sqrt(a)*sq 
rt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)* 
sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a*b**2*c**3*d - 18*sqrt(a)*sqrt(c + d*x)*s 
qrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)* 
sqrt(a)*d - b*c)))*a*b**2*c*d**3*x**2 - 6*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt( 
b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d 
 - b*c)))*b**3*c**3*d*x**2 + 6*sqrt(b)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)* 
d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a 
**3*d**4 + 18*sqrt(b)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sq 
rt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a**2*b*c**2*d**2 - 
 6*sqrt(b)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x) 
*b)/(sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a**2*b*d**4*x**2 - 18*sqrt(b) 
*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt( 
b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a*b**2*c**2*d**2*x**2 + 9*sqrt(a)*sqrt( 
c + d*x)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*log( - sqrt(sqrt(b)*sqrt(a)*d + b*c 
) + sqrt(b)*sqrt(c + d*x))*a**2*b*c*d**3 + 3*sqrt(a)*sqrt(c + d*x)*sqrt(sq 
rt(b)*sqrt(a)*d + b*c)*log( - sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sqrt 
(c + d*x))*a*b**2*c**3*d - 9*sqrt(a)*sqrt(c + d*x)*sqrt(sqrt(b)*sqrt(a)*d 
+ b*c)*log( - sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sqrt(c + d*x))*a*...