\(\int \frac {x^2 \sqrt {c+d x}}{(a-b x^2)^3} \, dx\) [691]

Optimal result

Integrand size = 23, antiderivative size = 280 \[ \int \frac {x^2 \sqrt {c+d x}}{\left (a-b x^2\right )^3} \, dx=\frac {x \sqrt {c+d x}}{4 b \left (a-b x^2\right )^2}-\frac {\sqrt {c+d x} \left (2 b c^2 x+a d (c-3 d x)\right )}{16 a b \left (b c^2-a d^2\right ) \left (a-b x^2\right )}+\frac {\left (4 b c^2-6 \sqrt {a} \sqrt {b} c d+3 a d^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{32 a^{3/2} b^{7/4} \left (\sqrt {b} c-\sqrt {a} d\right )^{3/2}}-\frac {\left (4 b c^2+6 \sqrt {a} \sqrt {b} c d+3 a d^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c+\sqrt {a} d}}\right )}{32 a^{3/2} b^{7/4} \left (\sqrt {b} c+\sqrt {a} d\right )^{3/2}} \] Output:

1/4*x*(d*x+c)^(1/2)/b/(-b*x^2+a)^2-1/16*(d*x+c)^(1/2)*(2*b*c^2*x+a*d*(-3*d 
*x+c))/a/b/(-a*d^2+b*c^2)/(-b*x^2+a)+1/32*(4*b*c^2-6*a^(1/2)*b^(1/2)*c*d+3 
*a*d^2)*arctanh(b^(1/4)*(d*x+c)^(1/2)/(b^(1/2)*c-a^(1/2)*d)^(1/2))/a^(3/2) 
/b^(7/4)/(b^(1/2)*c-a^(1/2)*d)^(3/2)-1/32*(4*b*c^2+6*a^(1/2)*b^(1/2)*c*d+3 
*a*d^2)*arctanh(b^(1/4)*(d*x+c)^(1/2)/(b^(1/2)*c+a^(1/2)*d)^(1/2))/a^(3/2) 
/b^(7/4)/(b^(1/2)*c+a^(1/2)*d)^(3/2)
 

Mathematica [A] (verified)

Time = 1.63 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.21 \[ \int \frac {x^2 \sqrt {c+d x}}{\left (a-b x^2\right )^3} \, dx=\frac {-\frac {2 \sqrt {a} b \sqrt {c+d x} \left (2 b^2 c^2 x^3-a^2 d (c+d x)+a b x \left (2 c^2+c d x-3 d^2 x^2\right )\right )}{\left (-b c^2+a d^2\right ) \left (a-b x^2\right )^2}+\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \left (4 b c^2+6 \sqrt {a} \sqrt {b} c d+3 a d^2\right ) \arctan \left (\frac {\sqrt {-b c-\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c+\sqrt {a} d}\right )}{\left (\sqrt {b} c+\sqrt {a} d\right )^2}+\frac {\sqrt {b} \left (4 b c^2-6 \sqrt {a} \sqrt {b} c d+3 a d^2\right ) \arctan \left (\frac {\sqrt {-b c+\sqrt {a} \sqrt {b} d} \sqrt {c+d x}}{\sqrt {b} c-\sqrt {a} d}\right )}{\left (\sqrt {b} c-\sqrt {a} d\right ) \sqrt {-b c+\sqrt {a} \sqrt {b} d}}}{32 a^{3/2} b^2} \] Input:

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

Output:

((-2*Sqrt[a]*b*Sqrt[c + d*x]*(2*b^2*c^2*x^3 - a^2*d*(c + d*x) + a*b*x*(2*c 
^2 + c*d*x - 3*d^2*x^2)))/((-(b*c^2) + a*d^2)*(a - b*x^2)^2) + (Sqrt[-(b*c 
) - Sqrt[a]*Sqrt[b]*d]*(4*b*c^2 + 6*Sqrt[a]*Sqrt[b]*c*d + 3*a*d^2)*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 + (Sqrt[b]*(4*b*c^2 - 6*Sqrt[a]*Sqrt[b]*c*d + 3 
*a*d^2)*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)*Sqrt[-(b*c) + Sqrt[a]*Sqrt[b]*d]) 
)/(32*a^(3/2)*b^2)
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.61, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {561, 27, 1672, 27, 25, 27, 1492, 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.

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

\(\Big \downarrow \) 561

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1672

\(\displaystyle \frac {2 \left (\frac {d^4 \int -\frac {2 a \left (c \left (a-\frac {b c^2}{d^2}\right ) d^2+3 \left (b c^2-a d^2\right ) (c+d x)\right )}{d^2 \left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )^2}d\sqrt {c+d x}}{16 a b \left (b c^2-a d^2\right )}-\frac {d^5 x \sqrt {c+d x} \left (a-\frac {b c^2}{d^2}\right )}{8 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (-\frac {d^2 \int -\frac {c \left (b c^2-a d^2\right )-3 \left (b c^2-a d^2\right ) (c+d x)}{\left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )^2}d\sqrt {c+d x}}{8 b \left (b c^2-a d^2\right )}-\frac {d^5 x \sqrt {c+d x} \left (a-\frac {b c^2}{d^2}\right )}{8 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {d^2 \int \frac {\left (b c^2-a d^2\right ) (c-3 (c+d x))}{\left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )^2}d\sqrt {c+d x}}{8 b \left (b c^2-a d^2\right )}-\frac {d^5 x \sqrt {c+d x} \left (a-\frac {b c^2}{d^2}\right )}{8 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {d^2 \int \frac {c-3 (c+d x)}{\left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )^2}d\sqrt {c+d x}}{8 b}-\frac {d^5 x \sqrt {c+d x} \left (a-\frac {b c^2}{d^2}\right )}{8 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^3}\)

\(\Big \downarrow \) 1492

\(\displaystyle \frac {2 \left (\frac {d^2 \left (\frac {d^4 \int -\frac {2 b \left (2 b c^3+\left (2 b c^2-3 a d^2\right ) (c+d x)\right )}{d^4 \left (-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a\right )}d\sqrt {c+d x}}{8 a b \left (b c^2-a d^2\right )}+\frac {\sqrt {c+d x} \left (2 c \left (b c^2-2 a d^2\right )-(c+d x) \left (2 b c^2-3 a d^2\right )\right )}{4 a \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}\right )}{8 b}-\frac {d^5 x \sqrt {c+d x} \left (a-\frac {b c^2}{d^2}\right )}{8 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {d^2 \left (\frac {\sqrt {c+d x} \left (2 c \left (b c^2-2 a d^2\right )-(c+d x) \left (2 b c^2-3 a d^2\right )\right )}{4 a \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}-\frac {\int \frac {2 b c^3+\left (2 b c^2-3 a d^2\right ) (c+d x)}{-\frac {b c^2}{d^2}+\frac {2 b (c+d x) c}{d^2}-\frac {b (c+d x)^2}{d^2}+a}d\sqrt {c+d x}}{4 a \left (b c^2-a d^2\right )}\right )}{8 b}-\frac {d^5 x \sqrt {c+d x} \left (a-\frac {b c^2}{d^2}\right )}{8 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^3}\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {2 \left (\frac {d^2 \left (\frac {\sqrt {c+d x} \left (2 c \left (b c^2-2 a d^2\right )-(c+d x) \left (2 b c^2-3 a d^2\right )\right )}{4 a \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}-\frac {\frac {1}{2} \left (-\frac {4 b^{3/2} c^3}{\sqrt {a} d}+3 \sqrt {a} \sqrt {b} c d-3 a d^2+2 b c^2\right ) \int \frac {1}{\frac {\sqrt {b} \left (\sqrt {b} c-\sqrt {a} d\right )}{d^2}-\frac {b (c+d x)}{d^2}}d\sqrt {c+d x}+\frac {\left (\sqrt {b} c-\sqrt {a} d\right ) \left (6 \sqrt {a} \sqrt {b} c d+3 a d^2+4 b c^2\right ) \int \frac {1}{\frac {\sqrt {b} \left (\sqrt {b} c+\sqrt {a} d\right )}{d^2}-\frac {b (c+d x)}{d^2}}d\sqrt {c+d x}}{2 \sqrt {a} d}}{4 a \left (b c^2-a d^2\right )}\right )}{8 b}-\frac {d^5 x \sqrt {c+d x} \left (a-\frac {b c^2}{d^2}\right )}{8 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \left (\frac {d^2 \left (\frac {\sqrt {c+d x} \left (2 c \left (b c^2-2 a d^2\right )-(c+d x) \left (2 b c^2-3 a d^2\right )\right )}{4 a \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )}-\frac {\frac {d \left (\sqrt {b} c-\sqrt {a} d\right ) \left (6 \sqrt {a} \sqrt {b} c d+3 a d^2+4 b c^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {a} d+\sqrt {b} c}}\right )}{2 \sqrt {a} b^{3/4} \sqrt {\sqrt {a} d+\sqrt {b} c}}+\frac {d^2 \left (-\frac {4 b^{3/2} c^3}{\sqrt {a} d}+3 \sqrt {a} \sqrt {b} c d-3 a d^2+2 b c^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c+d x}}{\sqrt {\sqrt {b} c-\sqrt {a} d}}\right )}{2 b^{3/4} \sqrt {\sqrt {b} c-\sqrt {a} d}}}{4 a \left (b c^2-a d^2\right )}\right )}{8 b}-\frac {d^5 x \sqrt {c+d x} \left (a-\frac {b c^2}{d^2}\right )}{8 b \left (b c^2-a d^2\right ) \left (a-\frac {b c^2}{d^2}+\frac {2 b c (c+d x)}{d^2}-\frac {b (c+d x)^2}{d^2}\right )^2}\right )}{d^3}\)

Input:

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

Output:

(2*(-1/8*((a - (b*c^2)/d^2)*d^5*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)^2) + (d^2*((Sqrt 
[c + d*x]*(2*c*(b*c^2 - 2*a*d^2) - (2*b*c^2 - 3*a*d^2)*(c + d*x)))/(4*a*(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^2*(2*b*c^2 - (4*b^(3/2)*c^3)/(Sqrt[a]*d) + 3*Sqrt[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]) + (d*(Sqrt[b]*c - Sqrt[a]*d)*(4*b 
*c^2 + 6*Sqrt[a]*Sqrt[b]*c*d + 3*a*d^2)*ArcTanh[(b^(1/4)*Sqrt[c + d*x])/Sq 
rt[Sqrt[b]*c + Sqrt[a]*d]])/(2*Sqrt[a]*b^(3/4)*Sqrt[Sqrt[b]*c + Sqrt[a]*d] 
))/(4*a*(b*c^2 - a*d^2))))/(8*b)))/d^3
 

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 1492

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb 
ol] :> Simp[x*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)*((a + b*x^2 + 
 c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 
 - 4*a*c))   Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 
7)*(d*b - 2*a*e)*c*x^2, x]*(a + b*x^2 + c*x^4)^(p + 1), 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] && 
 LtQ[p, -1] && IntegerQ[2*p]
 

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]
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.20

method	result	size
pseudoelliptic	\(\frac {-\frac {3 d \left (\left (-a \,d^{2}+\frac {2 b \,c^{2}}{3}\right ) \sqrt {a b \,d^{2}}+b c \left (a \,d^{2}-\frac {4 b \,c^{2}}{3}\right )\right ) \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (-b \,x^{2}+a \right )^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}+\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, \left (-\frac {3 d \left (-b \,x^{2}+a \right )^{2} \left (\left (a \,d^{2}-\frac {2 b \,c^{2}}{3}\right ) \sqrt {a b \,d^{2}}+b c \left (a \,d^{2}-\frac {4 b \,c^{2}}{3}\right )\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2}+\sqrt {d x +c}\, \left (-2 b^{2} c^{2} x^{3}-2 \left (\frac {3 d x}{2}+c \right ) x a \left (-d x +c \right ) b +a^{2} d \left (d x +c \right )\right ) \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\right )}{16 \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}\, a \left (a \,d^{2}-b \,c^{2}\right ) b \left (-b \,x^{2}+a \right )^{2}}\)	\(335\)
default	\(2 d^{3} \left (\frac {\frac {\left (3 a \,d^{2}-2 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{32 a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right )}-\frac {\left (5 a \,d^{2}-3 b \,c^{2}\right ) c \left (d x +c \right )^{\frac {5}{2}}}{16 a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right )}+\frac {\left (a^{2} d^{4}+9 b \,c^{2} d^{2} a -6 b^{2} c^{4}\right ) \left (d x +c \right )^{\frac {3}{2}}}{32 a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right ) b}-\frac {c^{3} \sqrt {d x +c}}{16 a \,d^{2}}}{\left (-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}\right )^{2}}+\frac {-\frac {\left (3 a b c \,d^{2}-4 c^{3} b^{2}+3 \sqrt {a b \,d^{2}}\, a \,d^{2}-2 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (-3 a b c \,d^{2}+4 c^{3} b^{2}+3 \sqrt {a b \,d^{2}}\, a \,d^{2}-2 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}}{32 a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right )}\right )\)	\(428\)
derivativedivides	\(-2 d^{3} \left (-\frac {\frac {\left (3 a \,d^{2}-2 b \,c^{2}\right ) \left (d x +c \right )^{\frac {7}{2}}}{32 a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right )}-\frac {\left (5 a \,d^{2}-3 b \,c^{2}\right ) c \left (d x +c \right )^{\frac {5}{2}}}{16 a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right )}+\frac {\left (a^{2} d^{4}+9 b \,c^{2} d^{2} a -6 b^{2} c^{4}\right ) \left (d x +c \right )^{\frac {3}{2}}}{32 a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right ) b}-\frac {c^{3} \sqrt {d x +c}}{16 a \,d^{2}}}{\left (-b \left (d x +c \right )^{2}+2 b c \left (d x +c \right )+a \,d^{2}-b \,c^{2}\right )^{2}}-\frac {-\frac {\left (3 a b c \,d^{2}-4 c^{3} b^{2}+3 \sqrt {a b \,d^{2}}\, a \,d^{2}-2 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \operatorname {arctanh}\left (\frac {b \sqrt {d x +c}}{\sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (b c +\sqrt {a b \,d^{2}}\right ) b}}+\frac {\left (-3 a b c \,d^{2}+4 c^{3} b^{2}+3 \sqrt {a b \,d^{2}}\, a \,d^{2}-2 \sqrt {a b \,d^{2}}\, b \,c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}\right )}{2 b \sqrt {a b \,d^{2}}\, \sqrt {\left (-b c +\sqrt {a b \,d^{2}}\right ) b}}}{32 a \,d^{2} \left (a \,d^{2}-b \,c^{2}\right )}\right )\)	\(429\)

Input:

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

Output:

1/16/(a*b*d^2)^(1/2)*(-3/2*d*((-a*d^2+2/3*b*c^2)*(a*b*d^2)^(1/2)+b*c*(a*d^ 
2-4/3*b*c^2))*((b*c+(a*b*d^2)^(1/2))*b)^(1/2)*(-b*x^2+a)^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) 
*(-3/2*d*(-b*x^2+a)^2*((a*d^2-2/3*b*c^2)*(a*b*d^2)^(1/2)+b*c*(a*d^2-4/3*b* 
c^2))*arctanh(b*(d*x+c)^(1/2)/((b*c+(a*b*d^2)^(1/2))*b)^(1/2))+(d*x+c)^(1/ 
2)*(-2*b^2*c^2*x^3-2*(3/2*d*x+c)*x*a*(-d*x+c)*b+a^2*d*(d*x+c))*(a*b*d^2)^( 
1/2)*((b*c+(a*b*d^2)^(1/2))*b)^(1/2)))/((b*c+(a*b*d^2)^(1/2))*b)^(1/2)/((- 
b*c+(a*b*d^2)^(1/2))*b)^(1/2)/a/(a*d^2-b*c^2)/b/(-b*x^2+a)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4237 vs. \(2 (225) = 450\).

Time = 0.63 (sec) , antiderivative size = 4237, normalized size of antiderivative = 15.13 \[ \int \frac {x^2 \sqrt {c+d x}}{\left (a-b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1029 vs. \(2 (225) = 450\).

Time = 0.30 (sec) , antiderivative size = 1029, normalized size of antiderivative = 3.68 \[ \int \frac {x^2 \sqrt {c+d x}}{\left (a-b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

1/32*((a*b^2*c^2*d - a^2*b*d^3)^2*(2*b*c^2*d - 3*a*d^3)*abs(b) + 2*(sqrt(a 
*b)*b^3*c^5*d - sqrt(a*b)*a*b^2*c^3*d^3)*abs(a*b^2*c^2*d - a^2*b*d^3)*abs( 
b) - (4*a*b^6*c^8*d - 11*a^2*b^5*c^6*d^3 + 10*a^3*b^4*c^4*d^5 - 3*a^4*b^3* 
c^2*d^7)*abs(b))*arctan(sqrt(d*x + c)/sqrt(-(a*b^3*c^3 - a^2*b^2*c*d^2 + s 
qrt((a*b^3*c^3 - a^2*b^2*c*d^2)^2 - (a*b^3*c^4 - 2*a^2*b^2*c^2*d^2 + a^3*b 
*d^4)*(a*b^3*c^2 - a^2*b^2*d^2)))/(a*b^3*c^2 - a^2*b^2*d^2)))/((a^2*b^5*c^ 
4*d - 2*a^3*b^4*c^2*d^3 + a^4*b^3*d^5 - sqrt(a*b)*a*b^5*c^5 + 2*sqrt(a*b)* 
a^2*b^4*c^3*d^2 - sqrt(a*b)*a^3*b^3*c*d^4)*sqrt(-b^2*c - sqrt(a*b)*b*d)*ab 
s(a*b^2*c^2*d - a^2*b*d^3)) + 1/32*((a*b^2*c^2*d - a^2*b*d^3)^2*(2*b*c^2*d 
 - 3*a*d^3)*abs(b) - 2*(sqrt(a*b)*b^3*c^5*d - sqrt(a*b)*a*b^2*c^3*d^3)*abs 
(a*b^2*c^2*d - a^2*b*d^3)*abs(b) - (4*a*b^6*c^8*d - 11*a^2*b^5*c^6*d^3 + 1 
0*a^3*b^4*c^4*d^5 - 3*a^4*b^3*c^2*d^7)*abs(b))*arctan(sqrt(d*x + c)/sqrt(- 
(a*b^3*c^3 - a^2*b^2*c*d^2 - sqrt((a*b^3*c^3 - a^2*b^2*c*d^2)^2 - (a*b^3*c 
^4 - 2*a^2*b^2*c^2*d^2 + a^3*b*d^4)*(a*b^3*c^2 - a^2*b^2*d^2)))/(a*b^3*c^2 
 - a^2*b^2*d^2)))/((a^2*b^5*c^4*d - 2*a^3*b^4*c^2*d^3 + a^4*b^3*d^5 + sqrt 
(a*b)*a*b^5*c^5 - 2*sqrt(a*b)*a^2*b^4*c^3*d^2 + sqrt(a*b)*a^3*b^3*c*d^4)*s 
qrt(-b^2*c + sqrt(a*b)*b*d)*abs(a*b^2*c^2*d - a^2*b*d^3)) + 1/16*(2*(d*x + 
 c)^(7/2)*b^2*c^2*d - 6*(d*x + c)^(5/2)*b^2*c^3*d + 6*(d*x + c)^(3/2)*b^2* 
c^4*d - 2*sqrt(d*x + c)*b^2*c^5*d - 3*(d*x + c)^(7/2)*a*b*d^3 + 10*(d*x + 
c)^(5/2)*a*b*c*d^3 - 9*(d*x + c)^(3/2)*a*b*c^2*d^3 + 2*sqrt(d*x + c)*a*...
 

Mupad [B] (verification not implemented)

Time = 10.95 (sec) , antiderivative size = 6670, normalized size of antiderivative = 23.82 \[ \int \frac {x^2 \sqrt {c+d x}}{\left (a-b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(((3*a*d^3 - 2*b*c^2*d)*(c + d*x)^(7/2))/(16*a*(a*d^2 - b*c^2)) - (c^3*d*( 
c + d*x)^(1/2))/(8*a) + ((c + d*x)^(3/2)*(a^2*d^5 - 6*b^2*c^4*d + 9*a*b*c^ 
2*d^3))/(16*a*b*(a*d^2 - b*c^2)) - (c*(5*a*d^3 - 3*b*c^2*d)*(c + d*x)^(5/2 
))/(8*a*(a*d^2 - b*c^2)))/(b^2*(c + d*x)^4 + a^2*d^4 + b^2*c^4 + (6*b^2*c^ 
2 - 2*a*b*d^2)*(c + d*x)^2 - (4*b^2*c^3 - 4*a*b*c*d^2)*(c + d*x) - 4*b^2*c 
*(c + d*x)^3 - 2*a*b*c^2*d^2) + atan(((((8192*a^3*b^6*c^5*d^3 - 8192*a^4*b 
^5*c^3*d^5)/(4096*(a^3*b^4*c^4 + a^5*b^2*d^4 - 2*a^4*b^3*c^2*d^2)) - ((c + 
 d*x)^(1/2)*(-(9*a^2*d^7*(a^9*b^7)^(1/2) - 16*a^3*b^7*c^7 + 9*a^6*b^4*c*d^ 
6 + 36*a^4*b^6*c^5*d^2 - 33*a^5*b^5*c^3*d^4 + 16*b^2*c^4*d^3*(a^9*b^7)^(1/ 
2) - 21*a*b*c^2*d^5*(a^9*b^7)^(1/2))/(4096*(a^6*b^10*c^6 - a^9*b^7*d^6 - 3 
*a^7*b^9*c^4*d^2 + 3*a^8*b^8*c^2*d^4)))^(1/2)*(4096*a^5*b^4*c*d^6 + 4096*a 
^3*b^6*c^5*d^2 - 8192*a^4*b^5*c^3*d^4))/(64*(a^4*d^4 + a^2*b^2*c^4 - 2*a^3 
*b*c^2*d^2)))*(-(9*a^2*d^7*(a^9*b^7)^(1/2) - 16*a^3*b^7*c^7 + 9*a^6*b^4*c* 
d^6 + 36*a^4*b^6*c^5*d^2 - 33*a^5*b^5*c^3*d^4 + 16*b^2*c^4*d^3*(a^9*b^7)^( 
1/2) - 21*a*b*c^2*d^5*(a^9*b^7)^(1/2))/(4096*(a^6*b^10*c^6 - a^9*b^7*d^6 - 
 3*a^7*b^9*c^4*d^2 + 3*a^8*b^8*c^2*d^4)))^(1/2) + ((c + d*x)^(1/2)*(9*a^3* 
d^8 + 16*b^3*c^6*d^2 - 20*a*b^2*c^4*d^4 - 3*a^2*b*c^2*d^6))/(64*(a^4*d^4 + 
 a^2*b^2*c^4 - 2*a^3*b*c^2*d^2)))*(-(9*a^2*d^7*(a^9*b^7)^(1/2) - 16*a^3*b^ 
7*c^7 + 9*a^6*b^4*c*d^6 + 36*a^4*b^6*c^5*d^2 - 33*a^5*b^5*c^3*d^4 + 16*b^2 
*c^4*d^3*(a^9*b^7)^(1/2) - 21*a*b*c^2*d^5*(a^9*b^7)^(1/2))/(4096*(a^6*b...
 

Reduce [B] (verification not implemented)

Time = 3.43 (sec) , antiderivative size = 2118, normalized size of antiderivative = 7.56 \[ \int \frac {x^2 \sqrt {c+d x}}{\left (a-b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^2*(d*x+c)^(1/2)/(-b*x^2+a)^3,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**4*d**4 - 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**3*b*c**2*d**2 - 12*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**3*b*d**4*x**2 + 8*sqr 
t(a)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(sq 
rt(b)*sqrt(a)*d - b*c)))*a**2*b**2*c**4 + 20*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 
**2*b**2*c**2*d**2*x**2 + 6*sqrt(a)*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**2*d**4*x** 
4 - 16*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**3*c**4*x**2 - 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**3*c**2*d**2*x**4 + 8*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**4*c**4 
*x**4 + 4*sqrt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)*atan((sqrt(c + d*x)*b)/(sq 
rt(b)*sqrt(sqrt(b)*sqrt(a)*d - b*c)))*a**3*b*c**3*d - 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**2*b**2*c**3*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*b**3*...