\(\int \frac {x^3}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx\) [248]

Optimal result

Integrand size = 26, antiderivative size = 135 \[ \int \frac {x^3}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {a x^2+b x^3}}{b d^2 x}+\frac {c^2 \sqrt {a x^2+b x^3}}{d^2 (b c-a d) x (c+d x)}-\frac {c (3 b c-4 a d) \arctan \left (\frac {\sqrt {d} \sqrt {a x^2+b x^3}}{\sqrt {b c-a d} x}\right )}{d^{5/2} (b c-a d)^{3/2}} \] Output:

2*(b*x^3+a*x^2)^(1/2)/b/d^2/x+c^2*(b*x^3+a*x^2)^(1/2)/d^2/(-a*d+b*c)/x/(d* 
x+c)-c*(-4*a*d+3*b*c)*arctan(d^(1/2)*(b*x^3+a*x^2)^(1/2)/(-a*d+b*c)^(1/2)/ 
x)/d^(5/2)/(-a*d+b*c)^(3/2)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.99 \[ \int \frac {x^3}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\frac {x \left (\frac {\sqrt {d} (a+b x) (-2 a d (c+d x)+b c (3 c+2 d x))}{b (b c-a d) (c+d x)}-\frac {c (3 b c-4 a d) \sqrt {a+b x} \arctan \left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}\right )}{d^{5/2} \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

(x*((Sqrt[d]*(a + b*x)*(-2*a*d*(c + d*x) + b*c*(3*c + 2*d*x)))/(b*(b*c - a 
*d)*(c + d*x)) - (c*(3*b*c - 4*a*d)*Sqrt[a + b*x]*ArcTan[(Sqrt[d]*Sqrt[a + 
 b*x])/Sqrt[b*c - a*d]])/(b*c - a*d)^(3/2)))/(d^(5/2)*Sqrt[x^2*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1948, 100, 27, 90, 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.

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

\(\Big \downarrow \) 1948

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(x*Sqrt[a + b*x]*((c^2*Sqrt[a + b*x])/(d^2*(b*c - a*d)*(c + d*x)) - ((-4*( 
b*c - a*d)*Sqrt[a + b*x])/b + (2*c*(3*b*c - 4*a*d)*ArcTan[(Sqrt[d]*Sqrt[a 
+ b*x])/Sqrt[b*c - a*d]])/(Sqrt[d]*Sqrt[b*c - a*d]))/(2*d^2*(b*c - a*d)))) 
/Sqrt[a*x^2 + b*x^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 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 90

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

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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 = 0.66 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {-\frac {2 \sqrt {b x +a}\, \left (-b d x +2 a d +6 b c \right )}{3}+\frac {b^{2} c^{2} \left (\frac {c \sqrt {b x +a}}{d x +c}-\frac {\left (6 a d -5 b c \right ) \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right )}{\sqrt {d \left (a d -b c \right )}}\right )}{a d -b c}}{d^{3} b^{2}}\)	\(111\)
risch	\(\frac {2 \left (b x +a \right ) x}{d^{2} b \sqrt {x^{2} \left (b x +a \right )}}-\frac {c \left (\frac {b c \sqrt {b x +a}}{\left (a d -b c \right ) \left (d \left (b x +a \right )-a d +b c \right )}-\frac {\left (4 a d -3 b c \right ) \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right )}{\left (a d -b c \right ) \sqrt {d \left (a d -b c \right )}}\right ) \sqrt {b x +a}\, x}{d^{2} \sqrt {x^{2} \left (b x +a \right )}}\)	\(145\)
default	\(\frac {x \sqrt {b x +a}\, \left (4 \,\operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) a b c \,d^{2} x -3 \,\operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) b^{2} c^{2} d x +2 \sqrt {b x +a}\, \sqrt {d \left (a d -b c \right )}\, a \,d^{2} x -2 \sqrt {b x +a}\, \sqrt {d \left (a d -b c \right )}\, b c d x +4 \,\operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) a b \,c^{2} d -3 \,\operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) b^{2} c^{3}+2 \sqrt {b x +a}\, \sqrt {d \left (a d -b c \right )}\, a c d -3 \sqrt {b x +a}\, \sqrt {d \left (a d -b c \right )}\, b \,c^{2}\right )}{b \sqrt {b \,x^{3}+a \,x^{2}}\, d^{2} \left (a d -b c \right ) \left (d x +c \right ) \sqrt {d \left (a d -b c \right )}}\)	\(282\)

Input:

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

Output:

1/d^3*(-2/3*(b*x+a)^(1/2)*(-b*d*x+2*a*d+6*b*c)+b^2*c^2/(a*d-b*c)*(c*(b*x+a 
)^(1/2)/(d*x+c)-(6*a*d-5*b*c)/(d*(a*d-b*c))^(1/2)*arctanh(d*(b*x+a)^(1/2)/ 
(d*(a*d-b*c))^(1/2))))/b^2
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 503, normalized size of antiderivative = 3.73 \[ \int \frac {x^3}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\left [-\frac {\sqrt {-b c d + a d^{2}} {\left ({\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{2} + {\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d\right )} x\right )} \log \left (\frac {b d x^{2} - {\left (b c - 2 \, a d\right )} x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {-b c d + a d^{2}}}{d x^{2} + c x}\right ) - 2 \, {\left (3 \, b^{2} c^{3} d - 5 \, a b c^{2} d^{2} + 2 \, a^{2} c d^{3} + 2 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{2 \, {\left ({\left (b^{3} c^{2} d^{4} - 2 \, a b^{2} c d^{5} + a^{2} b d^{6}\right )} x^{2} + {\left (b^{3} c^{3} d^{3} - 2 \, a b^{2} c^{2} d^{4} + a^{2} b c d^{5}\right )} x\right )}}, \frac {\sqrt {b c d - a d^{2}} {\left ({\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{2} + {\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d\right )} x\right )} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {b c d - a d^{2}}}{b d x^{2} + a d x}\right ) + {\left (3 \, b^{2} c^{3} d - 5 \, a b c^{2} d^{2} + 2 \, a^{2} c d^{3} + 2 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{{\left (b^{3} c^{2} d^{4} - 2 \, a b^{2} c d^{5} + a^{2} b d^{6}\right )} x^{2} + {\left (b^{3} c^{3} d^{3} - 2 \, a b^{2} c^{2} d^{4} + a^{2} b c d^{5}\right )} x}\right ] \] Input:

integrate(x^3/(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)*((3*b^2*c^2*d - 4*a*b*c*d^2)*x^2 + (3*b^2*c^3 
- 4*a*b*c^2*d)*x)*log((b*d*x^2 - (b*c - 2*a*d)*x + 2*sqrt(b*x^3 + a*x^2)*s 
qrt(-b*c*d + a*d^2))/(d*x^2 + c*x)) - 2*(3*b^2*c^3*d - 5*a*b*c^2*d^2 + 2*a 
^2*c*d^3 + 2*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x)*sqrt(b*x^3 + a*x^2)) 
/((b^3*c^2*d^4 - 2*a*b^2*c*d^5 + a^2*b*d^6)*x^2 + (b^3*c^3*d^3 - 2*a*b^2*c 
^2*d^4 + a^2*b*c*d^5)*x), (sqrt(b*c*d - a*d^2)*((3*b^2*c^2*d - 4*a*b*c*d^2 
)*x^2 + (3*b^2*c^3 - 4*a*b*c^2*d)*x)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(b*c*d 
 - a*d^2)/(b*d*x^2 + a*d*x)) + (3*b^2*c^3*d - 5*a*b*c^2*d^2 + 2*a^2*c*d^3 
+ 2*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x)*sqrt(b*x^3 + a*x^2))/((b^3*c^ 
2*d^4 - 2*a*b^2*c*d^5 + a^2*b*d^6)*x^2 + (b^3*c^3*d^3 - 2*a*b^2*c^2*d^4 + 
a^2*b*c*d^5)*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 329, normalized size of antiderivative = 2.44 \[ \int \frac {x^3}{(c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\frac {4 \sqrt {d}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) a b \,c^{2} d +4 \sqrt {d}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) a b c \,d^{2} x -3 \sqrt {d}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) b^{2} c^{3}-3 \sqrt {d}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) b^{2} c^{2} d x +2 \sqrt {b x +a}\, a^{2} c \,d^{3}+2 \sqrt {b x +a}\, a^{2} d^{4} x -5 \sqrt {b x +a}\, a b \,c^{2} d^{2}-4 \sqrt {b x +a}\, a b c \,d^{3} x +3 \sqrt {b x +a}\, b^{2} c^{3} d +2 \sqrt {b x +a}\, b^{2} c^{2} d^{2} x}{b \,d^{3} \left (a^{2} d^{3} x -2 a b c \,d^{2} x +b^{2} c^{2} d x +a^{2} c \,d^{2}-2 a b \,c^{2} d +b^{2} c^{3}\right )} \] Input:

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

Output:

(4*sqrt(d)*sqrt( - a*d + b*c)*atan((sqrt(a + b*x)*d)/(sqrt(d)*sqrt( - a*d 
+ b*c)))*a*b*c**2*d + 4*sqrt(d)*sqrt( - a*d + b*c)*atan((sqrt(a + b*x)*d)/ 
(sqrt(d)*sqrt( - a*d + b*c)))*a*b*c*d**2*x - 3*sqrt(d)*sqrt( - a*d + b*c)* 
atan((sqrt(a + b*x)*d)/(sqrt(d)*sqrt( - a*d + b*c)))*b**2*c**3 - 3*sqrt(d) 
*sqrt( - a*d + b*c)*atan((sqrt(a + b*x)*d)/(sqrt(d)*sqrt( - a*d + b*c)))*b 
**2*c**2*d*x + 2*sqrt(a + b*x)*a**2*c*d**3 + 2*sqrt(a + b*x)*a**2*d**4*x - 
 5*sqrt(a + b*x)*a*b*c**2*d**2 - 4*sqrt(a + b*x)*a*b*c*d**3*x + 3*sqrt(a + 
 b*x)*b**2*c**3*d + 2*sqrt(a + b*x)*b**2*c**2*d**2*x)/(b*d**3*(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 
))