\(\int \frac {x^3}{\sqrt {c+d x} (a-b x^2)^2} \, dx\) [664]

Optimal result

Integrand size = 23, antiderivative size = 210 \[ \int \frac {x^3}{\sqrt {c+d x} \left (a-b x^2\right )^2} \, dx=\frac {a (c-d x) \sqrt {c+d x}}{2 b \left (b c^2-a d^2\right ) \left (a-b x^2\right )}-\frac {\left (4 \sqrt {b} c-3 \sqrt {a} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{4 b^{7/4} \left (\sqrt {b} c-\sqrt {a} d\right )^{3/2}}-\frac {\left (4 \sqrt {b} c+3 \sqrt {a} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{4 b^{7/4} \left (\sqrt {b} c+\sqrt {a} d\right )^{3/2}} \] Output:

1/2*a*(-d*x+c)*(d*x+c)^(1/2)/b/(-a*d^2+b*c^2)/(-b*x^2+a)-1/4*(4*b^(1/2)*c- 
3*a^(1/2)*d)*arctanh(b^(1/4)*(d*x+c)^(1/2)/(b^(1/2)*c-a^(1/2)*d)^(1/2))/b^ 
(7/4)/(b^(1/2)*c-a^(1/2)*d)^(3/2)-1/4*(4*b^(1/2)*c+3*a^(1/2)*d)*arctanh(b^ 
(1/4)*(d*x+c)^(1/2)/(b^(1/2)*c+a^(1/2)*d)^(1/2))/b^(7/4)/(b^(1/2)*c+a^(1/2 
)*d)^(3/2)
 

Mathematica [A] (verified)

Time = 1.41 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.26 \[ \int \frac {x^3}{\sqrt {c+d x} \left (a-b x^2\right )^2} \, dx=\frac {-\frac {2 a b (c-d x) \sqrt {c+d x}}{\left (-b c^2+a d^2\right ) \left (a-b x^2\right )}+\frac {\left (4 \sqrt {b} c+3 \sqrt {a} d\right ) \sqrt {-b c-\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )}{\left (\sqrt {b} c+\sqrt {a} d\right )^2}+\frac {\left (4 \sqrt {b} c-3 \sqrt {a} d\right ) \sqrt {-b c+\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{\left (\sqrt {b} c-\sqrt {a} d\right )^2}}{4 b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.38, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {561, 25, 27, 1517, 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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1517

\(\displaystyle -\frac {2 \left (\frac {d^4 \int \frac {2 a \left (2 c \left (a-\frac {2 b c^2}{d^2}\right )-\left (3 a-\frac {4 b c^2}{d^2}\right ) (c+d x)\right )}{-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a}d\sqrt {c+d x}}{8 a b \left (b c^2-a d^2\right )}-\frac {a d^4 (c-d x) \sqrt {c+d x}}{4 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{d^4}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 1517

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x 
_Symbol] :> With[{f = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + 
c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + 
c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a*b*g - f*(b^2 - 2* 
a*c) - c*(b*f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a* 
(p + 1)*(b^2 - 4*a*c))   Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p 
 + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] 
 + b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)* 
x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ 
[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.33

method	result	size
pseudoelliptic	\(-\frac {\frac {\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (-b \,x^{2}+a \right ) \left (\left (-3 a \,d^{2}+4 b \,c^{2}\right ) \sqrt {a b \,d^{2}}+a b c \,d^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}+\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (\frac {\left (\left (3 a \,d^{2}-4 b \,c^{2}\right ) \sqrt {a b \,d^{2}}+a b c \,d^{2}\right ) \left (-b \,x^{2}+a \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}+\sqrt {d x +c}\, \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, a \left (-d x +c \right )\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, b \left (a \,d^{2}-b \,c^{2}\right ) \left (-b \,x^{2}+a \right )}\)	\(279\)
derivativedivides	\(-\frac {2 \left (-\frac {a \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{4 b \left (a \,d^{2}-b \,c^{2}\right )}+\frac {a c \,d^{2} \sqrt {d x +c}}{2 b \left (a \,d^{2}-b \,c^{2}\right )}\right )}{-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}}-\frac {-\frac {\left (-a b c \,d^{2}-3 \sqrt {a b \,d^{2}}\, a \,d^{2}+4 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (a b c \,d^{2}-3 \sqrt {a b \,d^{2}}\, a \,d^{2}+4 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}}{2 \left (a \,d^{2}-b \,c^{2}\right )}\)	\(301\)
default	\(-\frac {2 \left (-\frac {a \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{4 b \left (a \,d^{2}-b \,c^{2}\right )}+\frac {a c \,d^{2} \sqrt {d x +c}}{2 b \left (a \,d^{2}-b \,c^{2}\right )}\right )}{-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}}-\frac {-\frac {\left (-a b c \,d^{2}-3 \sqrt {a b \,d^{2}}\, a \,d^{2}+4 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (a b c \,d^{2}-3 \sqrt {a b \,d^{2}}\, a \,d^{2}+4 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}}{2 \left (a \,d^{2}-b \,c^{2}\right )}\)	\(301\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3697 vs. \(2 (160) = 320\).

Time = 0.65 (sec) , antiderivative size = 3697, normalized size of antiderivative = 17.60 \[ \int \frac {x^3}{\sqrt {c+d x} \left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 860 vs. \(2 (160) = 320\).

Time = 0.22 (sec) , antiderivative size = 860, normalized size of antiderivative = 4.10 \[ \int \frac {x^3}{\sqrt {c+d x} \left (a-b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.93 (sec) , antiderivative size = 5461, normalized size of antiderivative = 26.00 \[ \int \frac {x^3}{\sqrt {c+d x} \left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((((256*a*b^6*c^5*d^2 + 128*a^3*b^4*c*d^6 - 384*a^2*b^5*c^3*d^4)/(8*( 
b^4*c^4 + a^2*b^2*d^4 - 2*a*b^3*c^2*d^2)) + ((c + d*x)^(1/2)*(64*a*b^6*c^5 
*d^2 + 64*a^3*b^4*c*d^6 - 128*a^2*b^5*c^3*d^4)*((16*b^6*c^5 - 9*a^2*d^5*(a 
*b^7)^(1/2) - 15*a*b^5*c^3*d^2 + 3*a^2*b^4*c*d^4 - 24*b^2*c^4*d*(a*b^7)^(1 
/2) + 29*a*b*c^2*d^3*(a*b^7)^(1/2))/(64*(b^10*c^6 - a^3*b^7*d^6 - 3*a*b^9* 
c^4*d^2 + 3*a^2*b^8*c^2*d^4)))^(1/2))/(a^2*d^4 + b^2*c^4 - 2*a*b*c^2*d^2)) 
*((16*b^6*c^5 - 9*a^2*d^5*(a*b^7)^(1/2) - 15*a*b^5*c^3*d^2 + 3*a^2*b^4*c*d 
^4 - 24*b^2*c^4*d*(a*b^7)^(1/2) + 29*a*b*c^2*d^3*(a*b^7)^(1/2))/(64*(b^10* 
c^6 - a^3*b^7*d^6 - 3*a*b^9*c^4*d^2 + 3*a^2*b^8*c^2*d^4)))^(1/2) - ((c + d 
*x)^(1/2)*(9*a^3*d^6 + 16*a*b^2*c^4*d^2 - 23*a^2*b*c^2*d^4))/(a^2*d^4 + b^ 
2*c^4 - 2*a*b*c^2*d^2))*((16*b^6*c^5 - 9*a^2*d^5*(a*b^7)^(1/2) - 15*a*b^5* 
c^3*d^2 + 3*a^2*b^4*c*d^4 - 24*b^2*c^4*d*(a*b^7)^(1/2) + 29*a*b*c^2*d^3*(a 
*b^7)^(1/2))/(64*(b^10*c^6 - a^3*b^7*d^6 - 3*a*b^9*c^4*d^2 + 3*a^2*b^8*c^2 
*d^4)))^(1/2)*1i - (((256*a*b^6*c^5*d^2 + 128*a^3*b^4*c*d^6 - 384*a^2*b^5* 
c^3*d^4)/(8*(b^4*c^4 + a^2*b^2*d^4 - 2*a*b^3*c^2*d^2)) - ((c + d*x)^(1/2)* 
(64*a*b^6*c^5*d^2 + 64*a^3*b^4*c*d^6 - 128*a^2*b^5*c^3*d^4)*((16*b^6*c^5 - 
 9*a^2*d^5*(a*b^7)^(1/2) - 15*a*b^5*c^3*d^2 + 3*a^2*b^4*c*d^4 - 24*b^2*c^4 
*d*(a*b^7)^(1/2) + 29*a*b*c^2*d^3*(a*b^7)^(1/2))/(64*(b^10*c^6 - a^3*b^7*d 
^6 - 3*a*b^9*c^4*d^2 + 3*a^2*b^8*c^2*d^4)))^(1/2))/(a^2*d^4 + b^2*c^4 - 2* 
a*b*c^2*d^2))*((16*b^6*c^5 - 9*a^2*d^5*(a*b^7)^(1/2) - 15*a*b^5*c^3*d^2...
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 1270, normalized size of antiderivative = 6.05 \[ \int \frac {x^3}{\sqrt {c+d x} \left (a-b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(6*sqrt(a)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*s 
qrt(sqrt(b)*sqrt(a)*d - b*c)))*a**2*d**3 - 10*sqrt(a)*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*c**2*d - 6*sqrt(a)*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*d**3*x**2 + 10*sqrt(a)*sqrt 
(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sqrt(b)*sqr 
t(a)*d - b*c)))*b**2*c**2*d*x**2 + 4*sqrt(b)*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*c*d* 
*2 - 8*sqrt(b)*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*c**3 - 4*sqrt(b)*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*c*d**2*x**2 + 8*sqrt(b)*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**2*c**3*x**2 + 3*sqrt( 
a)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*log( - sqrt(sqrt(b)*sqrt(a)*d + b*c) + sq 
rt(b)*sqrt(c + d*x))*a**2*d**3 - 5*sqrt(a)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*l 
og( - sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sqrt(c + d*x))*a*b*c**2*d - 
3*sqrt(a)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*log( - sqrt(sqrt(b)*sqrt(a)*d + b* 
c) + sqrt(b)*sqrt(c + d*x))*a*b*d**3*x**2 + 5*sqrt(a)*sqrt(sqrt(b)*sqrt(a) 
*d + b*c)*log( - sqrt(sqrt(b)*sqrt(a)*d + b*c) + sqrt(b)*sqrt(c + d*x))*b* 
*2*c**2*d*x**2 - 3*sqrt(a)*sqrt(sqrt(b)*sqrt(a)*d + b*c)*log(sqrt(sqrt(...