\(\int \frac {x^2}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx\) [249]

Optimal result

Integrand size = 26, antiderivative size = 105 \[ \int \frac {x^2}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=-\frac {c \sqrt {a x^2+b x^3}}{d (b c-a d) x (c+d x)}+\frac {(b c-2 a d) \arctan \left (\frac {\sqrt {d} \sqrt {a x^2+b x^3}}{\sqrt {b c-a d} x}\right )}{d^{3/2} (b c-a d)^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11 \[ \int \frac {x^2}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\frac {-c \sqrt {d} \sqrt {b c-a d} x (a+b x)+(b c-2 a d) x \sqrt {a+b x} (c+d x) \arctan \left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{d^{3/2} (b c-a d)^{3/2} \sqrt {x^2 (a+b x)} (c+d x)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1948, 87, 73, 218}

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.

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

\(\Big \downarrow \) 1948

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1948

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Simp[e^IntPart[m]*(e*x)^FracPart[m]*( 
(a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x 
^n)^FracPart[p]))   Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; 
FreeQ[{a, b, c, d, e, j, m, n, p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] 
 && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])
 

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(\frac {2 \sqrt {b x +a}-\frac {b c \left (\frac {c \sqrt {b x +a}}{d x +c}-\frac {\left (4 a d -3 b c \right ) \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right )}{\sqrt {d \left (a d -b c \right )}}\right )}{a d -b c}}{d^{2} b}\)	\(94\)
default	\(\frac {x \sqrt {b x +a}\, \left (-2 \,\operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) a \,d^{2} x +\operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) b c d x -2 \,\operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) a c d +\operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) b \,c^{2}+c \sqrt {b x +a}\, \sqrt {d \left (a d -b c \right )}\right )}{\sqrt {b \,x^{3}+a \,x^{2}}\, d \left (a d -b c \right ) \left (d x +c \right ) \sqrt {d \left (a d -b c \right )}}\)	\(187\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 378, normalized size of antiderivative = 3.60 \[ \int \frac {x^2}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\left [-\frac {\sqrt {-b c d + a d^{2}} {\left ({\left (b c d - 2 \, a d^{2}\right )} x^{2} + {\left (b c^{2} - 2 \, a c d\right )} x\right )} \log \left (\frac {b d x^{2} - {\left (b c - 2 \, a d\right )} x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {-b c d + a d^{2}}}{d x^{2} + c x}\right ) + 2 \, {\left (b c^{2} d - a c d^{2}\right )} \sqrt {b x^{3} + a x^{2}}}{2 \, {\left ({\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} x^{2} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x\right )}}, -\frac {\sqrt {b c d - a d^{2}} {\left ({\left (b c d - 2 \, a d^{2}\right )} x^{2} + {\left (b c^{2} - 2 \, a c d\right )} x\right )} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {b c d - a d^{2}}}{b d x^{2} + a d x}\right ) + {\left (b c^{2} d - a c d^{2}\right )} \sqrt {b x^{3} + a x^{2}}}{{\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} x^{2} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x}\right ] \] Input:

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

Output:

[-1/2*(sqrt(-b*c*d + a*d^2)*((b*c*d - 2*a*d^2)*x^2 + (b*c^2 - 2*a*c*d)*x)* 
log((b*d*x^2 - (b*c - 2*a*d)*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(-b*c*d + a*d^2 
))/(d*x^2 + c*x)) + 2*(b*c^2*d - a*c*d^2)*sqrt(b*x^3 + a*x^2))/((b^2*c^2*d 
^3 - 2*a*b*c*d^4 + a^2*d^5)*x^2 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4 
)*x), -(sqrt(b*c*d - a*d^2)*((b*c*d - 2*a*d^2)*x^2 + (b*c^2 - 2*a*c*d)*x)* 
arctan(sqrt(b*x^3 + a*x^2)*sqrt(b*c*d - a*d^2)/(b*d*x^2 + a*d*x)) + (b*c^2 
*d - a*c*d^2)*sqrt(b*x^3 + a*x^2))/((b^2*c^2*d^3 - 2*a*b*c*d^4 + a^2*d^5)* 
x^2 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.32 \[ \int \frac {x^2}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\frac {-2 \sqrt {d}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) a c d -2 \sqrt {d}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) a \,d^{2} x +\sqrt {d}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) b \,c^{2}+\sqrt {d}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) b c d x +\sqrt {b x +a}\, a c \,d^{2}-\sqrt {b x +a}\, b \,c^{2} d}{d^{2} \left (a^{2} d^{3} x -2 a b c \,d^{2} x +b^{2} c^{2} d x +a^{2} c \,d^{2}-2 a b \,c^{2} d +b^{2} c^{3}\right )} \] Input:

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

Output:

( - 2*sqrt(d)*sqrt( - a*d + b*c)*atan((sqrt(a + b*x)*d)/(sqrt(d)*sqrt( - a 
*d + b*c)))*a*c*d - 2*sqrt(d)*sqrt( - a*d + b*c)*atan((sqrt(a + b*x)*d)/(s 
qrt(d)*sqrt( - a*d + b*c)))*a*d**2*x + sqrt(d)*sqrt( - a*d + b*c)*atan((sq 
rt(a + b*x)*d)/(sqrt(d)*sqrt( - a*d + b*c)))*b*c**2 + sqrt(d)*sqrt( - a*d 
+ b*c)*atan((sqrt(a + b*x)*d)/(sqrt(d)*sqrt( - a*d + b*c)))*b*c*d*x + sqrt 
(a + b*x)*a*c*d**2 - sqrt(a + b*x)*b*c**2*d)/(d**2*(a**2*c*d**2 + a**2*d** 
3*x - 2*a*b*c**2*d - 2*a*b*c*d**2*x + b**2*c**3 + b**2*c**2*d*x))