\(\int \frac {x^2 (A+B x)}{\sqrt {c+d x} \sqrt {a x+b x^2}} \, dx\) [14]

Optimal result

Integrand size = 31, antiderivative size = 388 \[ \int \frac {x^2 (A+B x)}{\sqrt {c+d x} \sqrt {a x+b x^2}} \, dx=-\frac {2 (9 a b B c d-2 (b c+a d) (4 b B c-5 A b d+4 a B d)) x \sqrt {c+d x}}{15 b^2 d^3 \sqrt {a x+b x^2}}+\frac {2 (5 A b d-4 B (b c+a d)) \sqrt {c+d x} \sqrt {a x+b x^2}}{15 b^2 d^2}+\frac {2 B x \sqrt {c+d x} \sqrt {a x+b x^2}}{5 b d}+\frac {2 \sqrt {a} (9 a b B c d-2 (b c+a d) (4 b B c-5 A b d+4 a B d)) \sqrt {x} \sqrt {c+d x} E\left (\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )|1-\frac {a d}{b c}\right )}{15 b^{5/2} d^3 \sqrt {\frac {a (c+d x)}{c (a+b x)}} \sqrt {a x+b x^2}}+\frac {2 a^{3/2} (4 b B c-5 A b d+4 a B d) \sqrt {x} \sqrt {c+d x} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ),1-\frac {a d}{b c}\right )}{15 b^{5/2} d^2 \sqrt {\frac {a (c+d x)}{c (a+b x)}} \sqrt {a x+b x^2}} \] Output:

-2/15*(9*a*b*B*c*d-2*(a*d+b*c)*(-5*A*b*d+4*B*a*d+4*B*b*c))*x*(d*x+c)^(1/2) 
/b^2/d^3/(b*x^2+a*x)^(1/2)+2/15*(5*A*b*d-4*B*(a*d+b*c))*(d*x+c)^(1/2)*(b*x 
^2+a*x)^(1/2)/b^2/d^2+2/5*B*x*(d*x+c)^(1/2)*(b*x^2+a*x)^(1/2)/b/d+2/15*a^( 
1/2)*(9*a*b*B*c*d-2*(a*d+b*c)*(-5*A*b*d+4*B*a*d+4*B*b*c))*x^(1/2)*(d*x+c)^ 
(1/2)*EllipticE(b^(1/2)*x^(1/2)/a^(1/2)/(1+b*x/a)^(1/2),(1-a*d/b/c)^(1/2)) 
/b^(5/2)/d^3/(a*(d*x+c)/c/(b*x+a))^(1/2)/(b*x^2+a*x)^(1/2)+2/15*a^(3/2)*(- 
5*A*b*d+4*B*a*d+4*B*b*c)*x^(1/2)*(d*x+c)^(1/2)*InverseJacobiAM(arctan(b^(1 
/2)*x^(1/2)/a^(1/2)),(1-a*d/b/c)^(1/2))/b^(5/2)/d^2/(a*(d*x+c)/c/(b*x+a))^ 
(1/2)/(b*x^2+a*x)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 23.40 (sec) , antiderivative size = 353, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 (A+B x)}{\sqrt {c+d x} \sqrt {a x+b x^2}} \, dx=\frac {2 \sqrt {x} \left (\frac {\left (8 a^2 B d^2+a b d (7 B c-10 A d)+2 b^2 c (4 B c-5 A d)\right ) (a+b x) (c+d x)}{b \sqrt {x}}+d \sqrt {x} (a+b x) (c+d x) (-4 a B d+b (-4 B c+5 A d+3 B d x))+i \sqrt {\frac {a}{b}} d \left (8 a^2 B d^2+a b d (7 B c-10 A d)+2 b^2 c (4 B c-5 A d)\right ) \sqrt {1+\frac {a}{b x}} \sqrt {1+\frac {c}{d x}} x E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {a}{b}}}{\sqrt {x}}\right )|\frac {b c}{a d}\right )-i \sqrt {\frac {a}{b}} d \left (8 a^2 B d^2+a b d (3 B c-10 A d)+b^2 c (4 B c-5 A d)\right ) \sqrt {1+\frac {a}{b x}} \sqrt {1+\frac {c}{d x}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {a}{b}}}{\sqrt {x}}\right ),\frac {b c}{a d}\right )\right )}{15 b^2 d^3 \sqrt {x (a+b x)} \sqrt {c+d x}} \] Input:

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

Output:

(2*Sqrt[x]*(((8*a^2*B*d^2 + a*b*d*(7*B*c - 10*A*d) + 2*b^2*c*(4*B*c - 5*A* 
d))*(a + b*x)*(c + d*x))/(b*Sqrt[x]) + d*Sqrt[x]*(a + b*x)*(c + d*x)*(-4*a 
*B*d + b*(-4*B*c + 5*A*d + 3*B*d*x)) + I*Sqrt[a/b]*d*(8*a^2*B*d^2 + a*b*d* 
(7*B*c - 10*A*d) + 2*b^2*c*(4*B*c - 5*A*d))*Sqrt[1 + a/(b*x)]*Sqrt[1 + c/( 
d*x)]*x*EllipticE[I*ArcSinh[Sqrt[a/b]/Sqrt[x]], (b*c)/(a*d)] - I*Sqrt[a/b] 
*d*(8*a^2*B*d^2 + a*b*d*(3*B*c - 10*A*d) + b^2*c*(4*B*c - 5*A*d))*Sqrt[1 + 
 a/(b*x)]*Sqrt[1 + c/(d*x)]*x*EllipticF[I*ArcSinh[Sqrt[a/b]/Sqrt[x]], (b*c 
)/(a*d)]))/(15*b^2*d^3*Sqrt[x*(a + b*x)]*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 1.65 (sec) , antiderivative size = 371, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2184, 27, 2184, 27, 1269, 1169, 122, 120, 127, 126}

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.

\(\displaystyle \int \frac {x^2 (A+B x)}{\sqrt {a x+b x^2} \sqrt {c+d x}} \, dx\)

\(\Big \downarrow \) 2184

\(\displaystyle \frac {2 \int -\frac {a B d c^2+B d (2 b c+5 a d) x c+d^2 (7 b B c-5 A b d+4 a B d) x^2}{2 \sqrt {c+d x} \sqrt {b x^2+a x}}dx}{5 b d^3}+\frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}-\frac {\int \frac {a B d c^2+B d (2 b c+5 a d) x c+d^2 (7 b B c-5 A b d+4 a B d) x^2}{\sqrt {c+d x} \sqrt {b x^2+a x}}dx}{5 b d^3}\)

\(\Big \downarrow \) 2184

\(\displaystyle \frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}-\frac {\frac {2 \int -\frac {d^3 \left (a c (4 b B c-5 A b d+4 a B d)+\left (2 c (4 B c-5 A d) b^2+a d (7 B c-10 A d) b+8 a^2 B d^2\right ) x\right )}{2 \sqrt {c+d x} \sqrt {b x^2+a x}}dx}{3 b d^2}+\frac {2 d \sqrt {a x+b x^2} \sqrt {c+d x} (4 a B d-5 A b d+7 b B c)}{3 b}}{5 b d^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}-\frac {\frac {2 d \sqrt {a x+b x^2} \sqrt {c+d x} (4 a B d-5 A b d+7 b B c)}{3 b}-\frac {d \int \frac {a c (4 b B c-5 A b d+4 a B d)+\left (2 c (4 B c-5 A d) b^2+a d (7 B c-10 A d) b+8 a^2 B d^2\right ) x}{\sqrt {c+d x} \sqrt {b x^2+a x}}dx}{3 b}}{5 b d^3}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}-\frac {\frac {2 d \sqrt {a x+b x^2} \sqrt {c+d x} (4 a B d-5 A b d+7 b B c)}{3 b}-\frac {d \left (\frac {\left (8 a^2 B d^2+a b d (7 B c-10 A d)+2 b^2 c (4 B c-5 A d)\right ) \int \frac {\sqrt {c+d x}}{\sqrt {b x^2+a x}}dx}{d}-\frac {c \left (4 a^2 B d^2+a b d (3 B c-5 A d)+2 b^2 c (4 B c-5 A d)\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {b x^2+a x}}dx}{d}\right )}{3 b}}{5 b d^3}\)

\(\Big \downarrow \) 1169

\(\displaystyle \frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}-\frac {\frac {2 d \sqrt {a x+b x^2} \sqrt {c+d x} (4 a B d-5 A b d+7 b B c)}{3 b}-\frac {d \left (\frac {\sqrt {x} \sqrt {a+b x} \left (8 a^2 B d^2+a b d (7 B c-10 A d)+2 b^2 c (4 B c-5 A d)\right ) \int \frac {\sqrt {c+d x}}{\sqrt {x} \sqrt {a+b x}}dx}{d \sqrt {a x+b x^2}}-\frac {c \sqrt {x} \sqrt {a+b x} \left (4 a^2 B d^2+a b d (3 B c-5 A d)+2 b^2 c (4 B c-5 A d)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x} \sqrt {c+d x}}dx}{d \sqrt {a x+b x^2}}\right )}{3 b}}{5 b d^3}\)

\(\Big \downarrow \) 122

\(\displaystyle \frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}-\frac {\frac {2 d \sqrt {a x+b x^2} \sqrt {c+d x} (4 a B d-5 A b d+7 b B c)}{3 b}-\frac {d \left (\frac {\sqrt {x} \sqrt {\frac {b x}{a}+1} \sqrt {c+d x} \left (8 a^2 B d^2+a b d (7 B c-10 A d)+2 b^2 c (4 B c-5 A d)\right ) \int \frac {\sqrt {\frac {d x}{c}+1}}{\sqrt {x} \sqrt {\frac {b x}{a}+1}}dx}{d \sqrt {a x+b x^2} \sqrt {\frac {d x}{c}+1}}-\frac {c \sqrt {x} \sqrt {a+b x} \left (4 a^2 B d^2+a b d (3 B c-5 A d)+2 b^2 c (4 B c-5 A d)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x} \sqrt {c+d x}}dx}{d \sqrt {a x+b x^2}}\right )}{3 b}}{5 b d^3}\)

\(\Big \downarrow \) 120

\(\displaystyle \frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}-\frac {\frac {2 d \sqrt {a x+b x^2} \sqrt {c+d x} (4 a B d-5 A b d+7 b B c)}{3 b}-\frac {d \left (\frac {2 \sqrt {-a} \sqrt {x} \sqrt {\frac {b x}{a}+1} \sqrt {c+d x} \left (8 a^2 B d^2+a b d (7 B c-10 A d)+2 b^2 c (4 B c-5 A d)\right ) E\left (\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-a}}\right )|\frac {a d}{b c}\right )}{\sqrt {b} d \sqrt {a x+b x^2} \sqrt {\frac {d x}{c}+1}}-\frac {c \sqrt {x} \sqrt {a+b x} \left (4 a^2 B d^2+a b d (3 B c-5 A d)+2 b^2 c (4 B c-5 A d)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x} \sqrt {c+d x}}dx}{d \sqrt {a x+b x^2}}\right )}{3 b}}{5 b d^3}\)

\(\Big \downarrow \) 127

\(\displaystyle \frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}-\frac {\frac {2 d \sqrt {a x+b x^2} \sqrt {c+d x} (4 a B d-5 A b d+7 b B c)}{3 b}-\frac {d \left (\frac {2 \sqrt {-a} \sqrt {x} \sqrt {\frac {b x}{a}+1} \sqrt {c+d x} \left (8 a^2 B d^2+a b d (7 B c-10 A d)+2 b^2 c (4 B c-5 A d)\right ) E\left (\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-a}}\right )|\frac {a d}{b c}\right )}{\sqrt {b} d \sqrt {a x+b x^2} \sqrt {\frac {d x}{c}+1}}-\frac {c \sqrt {x} \sqrt {\frac {b x}{a}+1} \sqrt {\frac {d x}{c}+1} \left (4 a^2 B d^2+a b d (3 B c-5 A d)+2 b^2 c (4 B c-5 A d)\right ) \int \frac {1}{\sqrt {x} \sqrt {\frac {b x}{a}+1} \sqrt {\frac {d x}{c}+1}}dx}{d \sqrt {a x+b x^2} \sqrt {c+d x}}\right )}{3 b}}{5 b d^3}\)

\(\Big \downarrow \) 126

\(\displaystyle \frac {2 B \sqrt {a x+b x^2} (c+d x)^{3/2}}{5 b d^2}-\frac {\frac {2 d \sqrt {a x+b x^2} \sqrt {c+d x} (4 a B d-5 A b d+7 b B c)}{3 b}-\frac {d \left (\frac {2 \sqrt {-a} \sqrt {x} \sqrt {\frac {b x}{a}+1} \sqrt {c+d x} \left (8 a^2 B d^2+a b d (7 B c-10 A d)+2 b^2 c (4 B c-5 A d)\right ) E\left (\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-a}}\right )|\frac {a d}{b c}\right )}{\sqrt {b} d \sqrt {a x+b x^2} \sqrt {\frac {d x}{c}+1}}-\frac {2 \sqrt {-a} c \sqrt {x} \sqrt {\frac {b x}{a}+1} \sqrt {\frac {d x}{c}+1} \left (4 a^2 B d^2+a b d (3 B c-5 A d)+2 b^2 c (4 B c-5 A d)\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-a}}\right ),\frac {a d}{b c}\right )}{\sqrt {b} d \sqrt {a x+b x^2} \sqrt {c+d x}}\right )}{3 b}}{5 b d^3}\)

Input:

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

Output:

(2*B*(c + d*x)^(3/2)*Sqrt[a*x + b*x^2])/(5*b*d^2) - ((2*d*(7*b*B*c - 5*A*b 
*d + 4*a*B*d)*Sqrt[c + d*x]*Sqrt[a*x + b*x^2])/(3*b) - (d*((2*Sqrt[-a]*(8* 
a^2*B*d^2 + a*b*d*(7*B*c - 10*A*d) + 2*b^2*c*(4*B*c - 5*A*d))*Sqrt[x]*Sqrt 
[1 + (b*x)/a]*Sqrt[c + d*x]*EllipticE[ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[-a]], 
(a*d)/(b*c)])/(Sqrt[b]*d*Sqrt[1 + (d*x)/c]*Sqrt[a*x + b*x^2]) - (2*Sqrt[-a 
]*c*(4*a^2*B*d^2 + a*b*d*(3*B*c - 5*A*d) + 2*b^2*c*(4*B*c - 5*A*d))*Sqrt[x 
]*Sqrt[1 + (b*x)/a]*Sqrt[1 + (d*x)/c]*EllipticF[ArcSin[(Sqrt[b]*Sqrt[x])/S 
qrt[-a]], (a*d)/(b*c)])/(Sqrt[b]*d*Sqrt[c + d*x]*Sqrt[a*x + b*x^2])))/(3*b 
))/(5*b*d^3)
 

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 120

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_] 
 :> Simp[2*(Sqrt[e]/b)*Rt[-b/d, 2]*EllipticE[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[- 
b/d, 2])], c*(f/(d*e))], x] /; FreeQ[{b, c, d, e, f}, x] && GtQ[c, 0] && Gt 
Q[e, 0] &&  !LtQ[-b/d, 0]
 

rule 122

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_] 
 :> Simp[Sqrt[e + f*x]*(Sqrt[1 + d*(x/c)]/(Sqrt[c + d*x]*Sqrt[1 + f*(x/e)]) 
)   Int[Sqrt[1 + f*(x/e)]/(Sqrt[b*x]*Sqrt[1 + d*(x/c)]), x], x] /; FreeQ[{b 
, c, d, e, f}, x] &&  !(GtQ[c, 0] && GtQ[e, 0])
 

rule 126

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x 
_] :> Simp[(2/(b*Sqrt[e]))*Rt[-b/d, 2]*EllipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]* 
Rt[-b/d, 2])], c*(f/(d*e))], x] /; FreeQ[{b, c, d, e, f}, x] && GtQ[c, 0] & 
& GtQ[e, 0] && (PosQ[-b/d] || NegQ[-b/f])
 

rule 127

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x 
_] :> Simp[Sqrt[1 + d*(x/c)]*(Sqrt[1 + f*(x/e)]/(Sqrt[c + d*x]*Sqrt[e + f*x 
]))   Int[1/(Sqrt[b*x]*Sqrt[1 + d*(x/c)]*Sqrt[1 + f*(x/e)]), x], x] /; Free 
Q[{b, c, d, e, f}, x] &&  !(GtQ[c, 0] && GtQ[e, 0])
 

rule 1169

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> 
 Simp[Sqrt[x]*(Sqrt[b + c*x]/Sqrt[b*x + c*x^2])   Int[(d + e*x)^m/(Sqrt[x]* 
Sqrt[b + c*x]), x], x] /; FreeQ[{b, c, d, e}, x] && NeQ[c*d - b*e, 0] && Eq 
Q[m^2, 1/4]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 2184

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S 
imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q 
+ 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + 
b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 
1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q - 1) - c 
*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[ 
q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && Pol 
yQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !(IGt 
Q[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

Maple [A] (verified)

Time = 1.23 (sec) , antiderivative size = 481, normalized size of antiderivative = 1.24

method	result	size
elliptic	\(\frac {\sqrt {x \left (b x +a \right ) \left (d x +c \right )}\, \left (\frac {2 B x \sqrt {b d \,x^{3}+a d \,x^{2}+b c \,x^{2}+a c x}}{5 b d}+\frac {2 \left (A -\frac {2 B \left (2 a d +2 b c \right )}{5 b d}\right ) \sqrt {b d \,x^{3}+a d \,x^{2}+b c \,x^{2}+a c x}}{3 b d}-\frac {2 \left (A -\frac {2 B \left (2 a d +2 b c \right )}{5 b d}\right ) a \,c^{2} \sqrt {\frac {\left (x +\frac {c}{d}\right ) d}{c}}\, \sqrt {\frac {x +\frac {a}{b}}{-\frac {c}{d}+\frac {a}{b}}}\, \sqrt {-\frac {x d}{c}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {c}{d}\right ) d}{c}}, \sqrt {-\frac {c}{d \left (-\frac {c}{d}+\frac {a}{b}\right )}}\right )}{3 b \,d^{2} \sqrt {b d \,x^{3}+a d \,x^{2}+b c \,x^{2}+a c x}}+\frac {2 \left (-\frac {3 a B c}{5 b d}-\frac {2 \left (A -\frac {2 B \left (2 a d +2 b c \right )}{5 b d}\right ) \left (a d +b c \right )}{3 b d}\right ) c \sqrt {\frac {\left (x +\frac {c}{d}\right ) d}{c}}\, \sqrt {\frac {x +\frac {a}{b}}{-\frac {c}{d}+\frac {a}{b}}}\, \sqrt {-\frac {x d}{c}}\, \left (\left (-\frac {c}{d}+\frac {a}{b}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {c}{d}\right ) d}{c}}, \sqrt {-\frac {c}{d \left (-\frac {c}{d}+\frac {a}{b}\right )}}\right )-\frac {a \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {c}{d}\right ) d}{c}}, \sqrt {-\frac {c}{d \left (-\frac {c}{d}+\frac {a}{b}\right )}}\right )}{b}\right )}{d \sqrt {b d \,x^{3}+a d \,x^{2}+b c \,x^{2}+a c x}}\right )}{\sqrt {x \left (b x +a \right )}\, \sqrt {d x +c}}\)	\(481\)
default	\(\text {Expression too large to display}\)	\(1063\)

Input:

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

Output:

(x*(b*x+a)*(d*x+c))^(1/2)/(x*(b*x+a))^(1/2)/(d*x+c)^(1/2)*(2/5*B/b/d*x*(b* 
d*x^3+a*d*x^2+b*c*x^2+a*c*x)^(1/2)+2/3*(A-2/5*B/b/d*(2*a*d+2*b*c))/b/d*(b* 
d*x^3+a*d*x^2+b*c*x^2+a*c*x)^(1/2)-2/3*(A-2/5*B/b/d*(2*a*d+2*b*c))/b/d^2*a 
*c^2*((x+c/d)/c*d)^(1/2)*((x+a/b)/(-c/d+a/b))^(1/2)*(-1/c*x*d)^(1/2)/(b*d* 
x^3+a*d*x^2+b*c*x^2+a*c*x)^(1/2)*EllipticF(((x+c/d)/c*d)^(1/2),(-c/d/(-c/d 
+a/b))^(1/2))+2*(-3/5*a*B*c/b/d-2/3*(A-2/5*B/b/d*(2*a*d+2*b*c))/b/d*(a*d+b 
*c))*c/d*((x+c/d)/c*d)^(1/2)*((x+a/b)/(-c/d+a/b))^(1/2)*(-1/c*x*d)^(1/2)/( 
b*d*x^3+a*d*x^2+b*c*x^2+a*c*x)^(1/2)*((-c/d+a/b)*EllipticE(((x+c/d)/c*d)^( 
1/2),(-c/d/(-c/d+a/b))^(1/2))-a/b*EllipticF(((x+c/d)/c*d)^(1/2),(-c/d/(-c/ 
d+a/b))^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.20 \[ \int \frac {x^2 (A+B x)}{\sqrt {c+d x} \sqrt {a x+b x^2}} \, dx=-\frac {2 \, {\left ({\left (8 \, B b^{3} c^{3} + {\left (3 \, B a b^{2} - 10 \, A b^{3}\right )} c^{2} d + {\left (3 \, B a^{2} b - 5 \, A a b^{2}\right )} c d^{2} + 2 \, {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} d^{3}\right )} \sqrt {b d} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (b^{2} c^{2} - a b c d + a^{2} d^{2}\right )}}{3 \, b^{2} d^{2}}, -\frac {4 \, {\left (2 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )}}{27 \, b^{3} d^{3}}, \frac {3 \, b d x + b c + a d}{3 \, b d}\right ) + 3 \, {\left (8 \, B b^{3} c^{2} d + {\left (7 \, B a b^{2} - 10 \, A b^{3}\right )} c d^{2} + 2 \, {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} d^{3}\right )} \sqrt {b d} {\rm weierstrassZeta}\left (\frac {4 \, {\left (b^{2} c^{2} - a b c d + a^{2} d^{2}\right )}}{3 \, b^{2} d^{2}}, -\frac {4 \, {\left (2 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )}}{27 \, b^{3} d^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (b^{2} c^{2} - a b c d + a^{2} d^{2}\right )}}{3 \, b^{2} d^{2}}, -\frac {4 \, {\left (2 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )}}{27 \, b^{3} d^{3}}, \frac {3 \, b d x + b c + a d}{3 \, b d}\right )\right ) - 3 \, {\left (3 \, B b^{3} d^{3} x - 4 \, B b^{3} c d^{2} - {\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} d^{3}\right )} \sqrt {b x^{2} + a x} \sqrt {d x + c}\right )}}{45 \, b^{4} d^{4}} \] Input:

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

Output:

-2/45*((8*B*b^3*c^3 + (3*B*a*b^2 - 10*A*b^3)*c^2*d + (3*B*a^2*b - 5*A*a*b^ 
2)*c*d^2 + 2*(4*B*a^3 - 5*A*a^2*b)*d^3)*sqrt(b*d)*weierstrassPInverse(4/3* 
(b^2*c^2 - a*b*c*d + a^2*d^2)/(b^2*d^2), -4/27*(2*b^3*c^3 - 3*a*b^2*c^2*d 
- 3*a^2*b*c*d^2 + 2*a^3*d^3)/(b^3*d^3), 1/3*(3*b*d*x + b*c + a*d)/(b*d)) + 
 3*(8*B*b^3*c^2*d + (7*B*a*b^2 - 10*A*b^3)*c*d^2 + 2*(4*B*a^2*b - 5*A*a*b^ 
2)*d^3)*sqrt(b*d)*weierstrassZeta(4/3*(b^2*c^2 - a*b*c*d + a^2*d^2)/(b^2*d 
^2), -4/27*(2*b^3*c^3 - 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + 2*a^3*d^3)/(b^3*d^ 
3), weierstrassPInverse(4/3*(b^2*c^2 - a*b*c*d + a^2*d^2)/(b^2*d^2), -4/27 
*(2*b^3*c^3 - 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + 2*a^3*d^3)/(b^3*d^3), 1/3*(3 
*b*d*x + b*c + a*d)/(b*d))) - 3*(3*B*b^3*d^3*x - 4*B*b^3*c*d^2 - (4*B*a*b^ 
2 - 5*A*b^3)*d^3)*sqrt(b*x^2 + a*x)*sqrt(d*x + c))/(b^4*d^4)
 

Sympy [F]

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

integrate(x**2*(B*x+A)/(d*x+c)**(1/2)/(b*x**2+a*x)**(1/2),x)
 

Output:

Integral(x**2*(A + B*x)/(sqrt(x*(a + b*x))*sqrt(c + d*x)), x)
 

Maxima [F]

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

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

Output:

integrate((B*x + A)*x^2/(sqrt(b*x^2 + a*x)*sqrt(d*x + c)), x)
 

Giac [F]

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

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

Output:

integrate((B*x + A)*x^2/(sqrt(b*x^2 + a*x)*sqrt(d*x + c)), x)
 

Mupad [F(-1)]

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

int((x^2*(A + B*x))/((a*x + b*x^2)^(1/2)*(c + d*x)^(1/2)),x)
                                                                                    
                                                                                    
 

Output:

int((x^2*(A + B*x))/((a*x + b*x^2)^(1/2)*(c + d*x)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {x^2 (A+B x)}{\sqrt {c+d x} \sqrt {a x+b x^2}} \, dx=\frac {-6 \sqrt {x}\, \sqrt {d x +c}\, \sqrt {b x +a}\, a c +4 \sqrt {x}\, \sqrt {d x +c}\, \sqrt {b x +a}\, a d x +4 \sqrt {x}\, \sqrt {d x +c}\, \sqrt {b x +a}\, b c x +2 \left (\int \frac {\sqrt {x}\, \sqrt {d x +c}\, \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+b^{2} c d \,x^{2}+a^{2} d^{2} x +2 a b c d x +b^{2} c^{2} x +a^{2} c d +a b \,c^{2}}d x \right ) a^{3} d^{3}+5 \left (\int \frac {\sqrt {x}\, \sqrt {d x +c}\, \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+b^{2} c d \,x^{2}+a^{2} d^{2} x +2 a b c d x +b^{2} c^{2} x +a^{2} c d +a b \,c^{2}}d x \right ) a^{2} b c \,d^{2}-5 \left (\int \frac {\sqrt {x}\, \sqrt {d x +c}\, \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+b^{2} c d \,x^{2}+a^{2} d^{2} x +2 a b c d x +b^{2} c^{2} x +a^{2} c d +a b \,c^{2}}d x \right ) a \,b^{2} c^{2} d -8 \left (\int \frac {\sqrt {x}\, \sqrt {d x +c}\, \sqrt {b x +a}\, x}{a b \,d^{2} x^{2}+b^{2} c d \,x^{2}+a^{2} d^{2} x +2 a b c d x +b^{2} c^{2} x +a^{2} c d +a b \,c^{2}}d x \right ) b^{3} c^{3}+3 \left (\int \frac {\sqrt {x}\, \sqrt {d x +c}\, \sqrt {b x +a}}{a b \,d^{2} x^{3}+b^{2} c d \,x^{3}+a^{2} d^{2} x^{2}+2 a b c d \,x^{2}+b^{2} c^{2} x^{2}+a^{2} c d x +a b \,c^{2} x}d x \right ) a^{3} c^{2} d +3 \left (\int \frac {\sqrt {x}\, \sqrt {d x +c}\, \sqrt {b x +a}}{a b \,d^{2} x^{3}+b^{2} c d \,x^{3}+a^{2} d^{2} x^{2}+2 a b c d \,x^{2}+b^{2} c^{2} x^{2}+a^{2} c d x +a b \,c^{2} x}d x \right ) a^{2} b \,c^{3}}{10 d \left (a d +b c \right )} \] Input:

int(x^2*(B*x+A)/(d*x+c)^(1/2)/(b*x^2+a*x)^(1/2),x)
 

Output:

( - 6*sqrt(x)*sqrt(c + d*x)*sqrt(a + b*x)*a*c + 4*sqrt(x)*sqrt(c + d*x)*sq 
rt(a + b*x)*a*d*x + 4*sqrt(x)*sqrt(c + d*x)*sqrt(a + b*x)*b*c*x + 2*int((s 
qrt(x)*sqrt(c + d*x)*sqrt(a + b*x)*x)/(a**2*c*d + a**2*d**2*x + a*b*c**2 + 
 2*a*b*c*d*x + a*b*d**2*x**2 + b**2*c**2*x + b**2*c*d*x**2),x)*a**3*d**3 + 
 5*int((sqrt(x)*sqrt(c + d*x)*sqrt(a + b*x)*x)/(a**2*c*d + a**2*d**2*x + a 
*b*c**2 + 2*a*b*c*d*x + a*b*d**2*x**2 + b**2*c**2*x + b**2*c*d*x**2),x)*a* 
*2*b*c*d**2 - 5*int((sqrt(x)*sqrt(c + d*x)*sqrt(a + b*x)*x)/(a**2*c*d + a* 
*2*d**2*x + a*b*c**2 + 2*a*b*c*d*x + a*b*d**2*x**2 + b**2*c**2*x + b**2*c* 
d*x**2),x)*a*b**2*c**2*d - 8*int((sqrt(x)*sqrt(c + d*x)*sqrt(a + b*x)*x)/( 
a**2*c*d + a**2*d**2*x + a*b*c**2 + 2*a*b*c*d*x + a*b*d**2*x**2 + b**2*c** 
2*x + b**2*c*d*x**2),x)*b**3*c**3 + 3*int((sqrt(x)*sqrt(c + d*x)*sqrt(a + 
b*x))/(a**2*c*d*x + a**2*d**2*x**2 + a*b*c**2*x + 2*a*b*c*d*x**2 + a*b*d** 
2*x**3 + b**2*c**2*x**2 + b**2*c*d*x**3),x)*a**3*c**2*d + 3*int((sqrt(x)*s 
qrt(c + d*x)*sqrt(a + b*x))/(a**2*c*d*x + a**2*d**2*x**2 + a*b*c**2*x + 2* 
a*b*c*d*x**2 + a*b*d**2*x**3 + b**2*c**2*x**2 + b**2*c*d*x**3),x)*a**2*b*c 
**3)/(10*d*(a*d + b*c))