\(\int \frac {x^3 \sqrt {c+d x}}{(a-b x^2)^2} \, dx\) [641]

Optimal result

Integrand size = 23, antiderivative size = 204 \[ \int \frac {x^3 \sqrt {c+d x}}{\left (a-b x^2\right )^2} \, dx=\frac {2 \sqrt {c+d x}}{b^2}+\frac {a \sqrt {c+d x}}{2 b^2 \left (a-b x^2\right )}-\frac {\left (4 \sqrt {b} c-5 \sqrt {a} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{4 b^{9/4} \sqrt {\sqrt {b} c-\sqrt {a} d}}-\frac {\left (4 \sqrt {b} c+5 \sqrt {a} d\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{4 b^{9/4} \sqrt {\sqrt {b} c+\sqrt {a} d}} \] Output:

2*(d*x+c)^(1/2)/b^2+1/2*a*(d*x+c)^(1/2)/b^2/(-b*x^2+a)-1/4*(4*b^(1/2)*c-5* 
a^(1/2)*d)*arctanh(b^(1/4)*(d*x+c)^(1/2)/(b^(1/2)*c-a^(1/2)*d)^(1/2))/b^(9 
/4)/(b^(1/2)*c-a^(1/2)*d)^(1/2)-1/4*(4*b^(1/2)*c+5*a^(1/2)*d)*arctanh(b^(1 
/4)*(d*x+c)^(1/2)/(b^(1/2)*c+a^(1/2)*d)^(1/2))/b^(9/4)/(b^(1/2)*c+a^(1/2)* 
d)^(1/2)
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.28 \[ \int \frac {x^3 \sqrt {c+d x}}{\left (a-b x^2\right )^2} \, dx=\frac {\frac {2 \sqrt {b} \sqrt {c+d x} \left (-5 a+4 b x^2\right )}{-a+b x^2}+\frac {\left (4 \sqrt {b} c+5 \sqrt {a} d\right ) \sqrt {-b c-\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )}{\sqrt {b} c+\sqrt {a} d}+\frac {\left (4 \sqrt {b} c-5 \sqrt {a} d\right ) \sqrt {-b c+\sqrt {a} \sqrt {b} d} \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{\sqrt {b} c-\sqrt {a} d}}{4 b^{5/2}} \] Input:

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

Output:

((2*Sqrt[b]*Sqrt[c + d*x]*(-5*a + 4*b*x^2))/(-a + b*x^2) + ((4*Sqrt[b]*c + 
 5*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) + ((4*Sqrt[b]*c - 5*Sqrt[a]*d)*Sqrt[-(b*c) + Sqrt[a]*Sqrt[b]*d]*ArcTa 
n[(Sqrt[-(b*c) + Sqrt[a]*Sqrt[b]*d]*Sqrt[c + d*x])/(Sqrt[b]*c - Sqrt[a]*d) 
])/(Sqrt[b]*c - Sqrt[a]*d))/(4*b^(5/2))
 

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.54, 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, 1672, 27, 2205, 2009}

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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1672

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2205

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (\frac {d^4 \left (\frac {a \left (4 \sqrt {b} c-5 \sqrt {a} d\right ) \sqrt {\sqrt {b} c-\sqrt {a} d} \left (\sqrt {a} d+\sqrt {b} c\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{2 b^{5/4}}+\frac {a \left (\sqrt {b} c-\sqrt {a} d\right ) \sqrt {\sqrt {a} d+\sqrt {b} c} \left (5 \sqrt {a} d+4 \sqrt {b} c\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {a} d+\sqrt {b} c}}\right )}{2 b^{5/4}}-\frac {4 a \sqrt {c+d x} \left (b c^2-a d^2\right )}{b}\right )}{4 a b \left (b c^2-a d^2\right )}-\frac {a d^4 \sqrt {c+d x}}{4 b^2 \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{d^4}\)

Input:

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

Output:

(-2*(-1/4*(a*d^4*Sqrt[c + d*x])/(b^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*a*(b*c^2 - a*d^2)*Sqrt[c + d*x])/b 
 + (a*(4*Sqrt[b]*c - 5*Sqrt[a]*d)*Sqrt[Sqrt[b]*c - Sqrt[a]*d]*(Sqrt[b]*c + 
 Sqrt[a]*d)*ArcTanh[(b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c - Sqrt[a]*d]])/ 
(2*b^(5/4)) + (a*(Sqrt[b]*c - Sqrt[a]*d)*Sqrt[Sqrt[b]*c + Sqrt[a]*d]*(4*Sq 
rt[b]*c + 5*Sqrt[a]*d)*ArcTanh[(b^(1/4)*Sqrt[c + d*x])/Sqrt[Sqrt[b]*c + Sq 
rt[a]*d]])/(2*b^(5/4))))/(4*a*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 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 1672

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2205

Int[(Px_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandInte 
grand[Px/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c}, x] && PolyQ[Px, x^ 
2] && Expon[Px, x^2] > 1
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {2 \sqrt {d x +c}}{b^{2}}+\frac {-\frac {\sqrt {d x +c}\, a \,d^{2}}{2 \left (b \left (d x +c \right )^{2}-2 b c \left (d x +c \right )-a \,d^{2}+b \,c^{2}\right )}+\frac {b \left (-\frac {\left (5 a \,d^{2}+4 \sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (-5 a \,d^{2}+4 \sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}}{b^{2}}\)	\(210\)
derivativedivides	\(\frac {2 \sqrt {d x +c}}{b^{2}}-\frac {2 \left (-\frac {a \,d^{2} \sqrt {d x +c}}{4 \left (-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}\right )}+\frac {b \left (-\frac {\left (-5 a \,d^{2}-4 \sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (5 a \,d^{2}-4 \sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}\right )}{b^{2}}\)	\(212\)
default	\(\frac {2 \sqrt {d x +c}}{b^{2}}-\frac {2 \left (-\frac {a \,d^{2} \sqrt {d x +c}}{4 \left (-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}\right )}+\frac {b \left (-\frac {\left (-5 a \,d^{2}-4 \sqrt {a b \,d^{2}}\, c \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (5 a \,d^{2}-4 \sqrt {a b \,d^{2}}\, c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}\right )}{b^{2}}\)	\(212\)
pseudoelliptic	\(\frac {-\frac {5 \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, b \left (a \,d^{2}-\frac {4 \sqrt {a b \,d^{2}}\, c}{5}\right ) \left (-b \,x^{2}+a \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{4}+\frac {5 \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (-\frac {b \left (a \,d^{2}+\frac {4 \sqrt {a b \,d^{2}}\, c}{5}\right ) \left (-b \,x^{2}+a \right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}+\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {d x +c}\, \sqrt {a b \,d^{2}}\, \left (-\frac {4 b \,x^{2}}{5}+a \right )\right )}{2}}{b^{2} \left (-b \,x^{2}+a \right ) \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\)	\(242\)

Input:

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

Output:

2*(d*x+c)^(1/2)/b^2+1/b^2*(-1/2*(d*x+c)^(1/2)*a*d^2/(b*(d*x+c)^2-2*b*c*(d* 
x+c)-a*d^2+b*c^2)+1/2*b*(-1/2*(5*a*d^2+4*(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*(-5*a*d^2+4*(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. 1712 vs. \(2 (151) = 302\).

Time = 0.15 (sec) , antiderivative size = 1712, normalized size of antiderivative = 8.39 \[ \int \frac {x^3 \sqrt {c+d x}}{\left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/8*((b^3*x^2 - a*b^2)*sqrt((16*b*c^3 - 15*a*c*d^2 + (b^5*c^2 - a*b^4*d^2 
)*sqrt((576*a*b^2*c^4*d^2 - 1200*a^2*b*c^2*d^4 + 625*a^3*d^6)/(b^11*c^4 - 
2*a*b^10*c^2*d^2 + a^2*b^9*d^4)))/(b^5*c^2 - a*b^4*d^2))*log((384*b^2*c^4 
- 1000*a*b*c^2*d^2 + 625*a^2*d^4)*sqrt(d*x + c) + (96*b^4*c^4 - 220*a*b^3* 
c^2*d^2 + 125*a^2*b^2*d^4 - (b^8*c^3 - a*b^7*c*d^2)*sqrt((576*a*b^2*c^4*d^ 
2 - 1200*a^2*b*c^2*d^4 + 625*a^3*d^6)/(b^11*c^4 - 2*a*b^10*c^2*d^2 + a^2*b 
^9*d^4)))*sqrt((16*b*c^3 - 15*a*c*d^2 + (b^5*c^2 - a*b^4*d^2)*sqrt((576*a* 
b^2*c^4*d^2 - 1200*a^2*b*c^2*d^4 + 625*a^3*d^6)/(b^11*c^4 - 2*a*b^10*c^2*d 
^2 + a^2*b^9*d^4)))/(b^5*c^2 - a*b^4*d^2))) - (b^3*x^2 - a*b^2)*sqrt((16*b 
*c^3 - 15*a*c*d^2 + (b^5*c^2 - a*b^4*d^2)*sqrt((576*a*b^2*c^4*d^2 - 1200*a 
^2*b*c^2*d^4 + 625*a^3*d^6)/(b^11*c^4 - 2*a*b^10*c^2*d^2 + a^2*b^9*d^4)))/ 
(b^5*c^2 - a*b^4*d^2))*log((384*b^2*c^4 - 1000*a*b*c^2*d^2 + 625*a^2*d^4)* 
sqrt(d*x + c) - (96*b^4*c^4 - 220*a*b^3*c^2*d^2 + 125*a^2*b^2*d^4 - (b^8*c 
^3 - a*b^7*c*d^2)*sqrt((576*a*b^2*c^4*d^2 - 1200*a^2*b*c^2*d^4 + 625*a^3*d 
^6)/(b^11*c^4 - 2*a*b^10*c^2*d^2 + a^2*b^9*d^4)))*sqrt((16*b*c^3 - 15*a*c* 
d^2 + (b^5*c^2 - a*b^4*d^2)*sqrt((576*a*b^2*c^4*d^2 - 1200*a^2*b*c^2*d^4 + 
 625*a^3*d^6)/(b^11*c^4 - 2*a*b^10*c^2*d^2 + a^2*b^9*d^4)))/(b^5*c^2 - a*b 
^4*d^2))) + (b^3*x^2 - a*b^2)*sqrt((16*b*c^3 - 15*a*c*d^2 - (b^5*c^2 - a*b 
^4*d^2)*sqrt((576*a*b^2*c^4*d^2 - 1200*a^2*b*c^2*d^4 + 625*a^3*d^6)/(b^11* 
c^4 - 2*a*b^10*c^2*d^2 + a^2*b^9*d^4)))/(b^5*c^2 - a*b^4*d^2))*log((384...
 

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]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (151) = 302\).

Time = 0.20 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.56 \[ \int \frac {x^3 \sqrt {c+d x}}{\left (a-b x^2\right )^2} \, dx=-\frac {\sqrt {d x + c} a d^{2}}{2 \, {\left ({\left (d x + c\right )}^{2} b - 2 \, {\left (d x + c\right )} b c + b c^{2} - a d^{2}\right )} b^{2}} + \frac {{\left (\sqrt {a b} c d^{2} {\left | b \right |} + {\left (4 \, b c^{2} - 5 \, a d^{2}\right )} {\left | b \right |} {\left | d \right |}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {b^{3} c + \sqrt {b^{6} c^{2} - {\left (b^{3} c^{2} - a b^{2} d^{2}\right )} b^{3}}}{b^{3}}}}\right )}{4 \, {\left (b^{3} c - \sqrt {a b} b^{2} d\right )} \sqrt {-b^{2} c - \sqrt {a b} b d} {\left | d \right |}} - \frac {{\left (\sqrt {a b} c d^{2} {\left | b \right |} - {\left (4 \, b c^{2} - 5 \, a d^{2}\right )} {\left | b \right |} {\left | d \right |}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-\frac {b^{3} c - \sqrt {b^{6} c^{2} - {\left (b^{3} c^{2} - a b^{2} d^{2}\right )} b^{3}}}{b^{3}}}}\right )}{4 \, {\left (b^{3} c + \sqrt {a b} b^{2} d\right )} \sqrt {-b^{2} c + \sqrt {a b} b d} {\left | d \right |}} + \frac {2 \, \sqrt {d x + c}}{b^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 2583, normalized size of antiderivative = 12.66 \[ \int \frac {x^3 \sqrt {c+d x}}{\left (a-b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*(c + d*x)^(1/2))/b^2 + atan(((((80*a^2*b^4*d^4 - 64*a*b^5*c^2*d^2)/(2*b 
^3) - 64*a*b^4*c*d^2*(c + d*x)^(1/2)*((16*b^6*c^3 - 25*a*d^3*(a*b^9)^(1/2) 
 - 15*a*b^5*c*d^2 + 24*b*c^2*d*(a*b^9)^(1/2))/(64*(b^10*c^2 - a*b^9*d^2))) 
^(1/2))*((16*b^6*c^3 - 25*a*d^3*(a*b^9)^(1/2) - 15*a*b^5*c*d^2 + 24*b*c^2* 
d*(a*b^9)^(1/2))/(64*(b^10*c^2 - a*b^9*d^2)))^(1/2) + ((25*a^2*d^4 + 16*a* 
b*c^2*d^2)*(c + d*x)^(1/2))/b)*((16*b^6*c^3 - 25*a*d^3*(a*b^9)^(1/2) - 15* 
a*b^5*c*d^2 + 24*b*c^2*d*(a*b^9)^(1/2))/(64*(b^10*c^2 - a*b^9*d^2)))^(1/2) 
*1i - (((80*a^2*b^4*d^4 - 64*a*b^5*c^2*d^2)/(2*b^3) + 64*a*b^4*c*d^2*(c + 
d*x)^(1/2)*((16*b^6*c^3 - 25*a*d^3*(a*b^9)^(1/2) - 15*a*b^5*c*d^2 + 24*b*c 
^2*d*(a*b^9)^(1/2))/(64*(b^10*c^2 - a*b^9*d^2)))^(1/2))*((16*b^6*c^3 - 25* 
a*d^3*(a*b^9)^(1/2) - 15*a*b^5*c*d^2 + 24*b*c^2*d*(a*b^9)^(1/2))/(64*(b^10 
*c^2 - a*b^9*d^2)))^(1/2) - ((25*a^2*d^4 + 16*a*b*c^2*d^2)*(c + d*x)^(1/2) 
)/b)*((16*b^6*c^3 - 25*a*d^3*(a*b^9)^(1/2) - 15*a*b^5*c*d^2 + 24*b*c^2*d*( 
a*b^9)^(1/2))/(64*(b^10*c^2 - a*b^9*d^2)))^(1/2)*1i)/((((80*a^2*b^4*d^4 - 
64*a*b^5*c^2*d^2)/(2*b^3) - 64*a*b^4*c*d^2*(c + d*x)^(1/2)*((16*b^6*c^3 - 
25*a*d^3*(a*b^9)^(1/2) - 15*a*b^5*c*d^2 + 24*b*c^2*d*(a*b^9)^(1/2))/(64*(b 
^10*c^2 - a*b^9*d^2)))^(1/2))*((16*b^6*c^3 - 25*a*d^3*(a*b^9)^(1/2) - 15*a 
*b^5*c*d^2 + 24*b*c^2*d*(a*b^9)^(1/2))/(64*(b^10*c^2 - a*b^9*d^2)))^(1/2) 
+ ((25*a^2*d^4 + 16*a*b*c^2*d^2)*(c + d*x)^(1/2))/b)*((16*b^6*c^3 - 25*a*d 
^3*(a*b^9)^(1/2) - 15*a*b^5*c*d^2 + 24*b*c^2*d*(a*b^9)^(1/2))/(64*(b^10...
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 940, normalized size of antiderivative = 4.61 \[ \int \frac {x^3 \sqrt {c+d x}}{\left (a-b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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