\(\int \frac {x^3}{(c+d x) \sqrt {a x^2+b x^3}} \, dx\) [241]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \frac {x^3}{(c+d x) \sqrt {a x^2+b x^3}} \, dx=-\frac {2 (b c+a d) \sqrt {a x^2+b x^3}}{b^2 d^2 x}+\frac {2 \left (a x^2+b x^3\right )^{3/2}}{3 b^2 d x^3}+\frac {2 c^2 \arctan \left (\frac {\sqrt {d} \sqrt {a x^2+b x^3}}{\sqrt {b c-a d} x}\right )}{d^{5/2} \sqrt {b c-a d}} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.87 \[ \int \frac {x^3}{(c+d x) \sqrt {a x^2+b x^3}} \, dx=\frac {2 x \left (\frac {\sqrt {d} (a+b x) (-3 b c-2 a d+b d x)}{b^2}+\frac {3 c^2 \sqrt {a+b x} \arctan \left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d}}\right )}{3 d^{5/2} \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1948, 99, 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.

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

\(\Big \downarrow \) 1948

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

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])
 

rule 2009

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

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {2 \left (-b d x +2 a d +3 b c \right ) \left (b x +a \right ) x}{3 b^{2} d^{2} \sqrt {x^{2} \left (b x +a \right )}}-\frac {2 c^{2} \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) \sqrt {b x +a}\, x}{d^{2} \sqrt {d \left (a d -b c \right )}\, \sqrt {x^{2} \left (b x +a \right )}}\)	\(102\)
pseudoelliptic	\(\frac {2 b^{3} c^{3} \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right )+\frac {16 \sqrt {b x +a}\, \sqrt {d \left (a d -b c \right )}\, \left (\frac {\left (3 d^{2} x^{2}-5 c d x +15 c^{2}\right ) b^{2}}{8}+\frac {5 d \left (-\frac {2 d x}{5}+c \right ) a b}{4}+a^{2} d^{2}\right )}{15}}{d^{3} \sqrt {d \left (a d -b c \right )}\, b^{3}}\)	\(115\)
default	\(\frac {2 x \sqrt {b x +a}\, \left (\sqrt {d \left (a d -b c \right )}\, \left (b x +a \right )^{\frac {3}{2}} d -3 b^{2} c^{2} \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right )-3 \sqrt {d \left (a d -b c \right )}\, \sqrt {b x +a}\, a d -3 \sqrt {d \left (a d -b c \right )}\, \sqrt {b x +a}\, b c \right )}{3 \sqrt {b \,x^{3}+a \,x^{2}}\, b^{2} d^{2} \sqrt {d \left (a d -b c \right )}}\)	\(140\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.61 \[ \int \frac {x^3}{(c+d x) \sqrt {a x^2+b x^3}} \, dx=\left [-\frac {3 \, \sqrt {-b c d + a d^{2}} b^{2} c^{2} x \log \left (\frac {b d x^{2} - {\left (b c - 2 \, a d\right )} x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {-b c d + a d^{2}}}{d x^{2} + c x}\right ) + 2 \, {\left (3 \, b^{2} c^{2} d - a b c d^{2} - 2 \, a^{2} d^{3} - {\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{3 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x}, -\frac {2 \, {\left (3 \, \sqrt {b c d - a d^{2}} b^{2} c^{2} x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {b c d - a d^{2}}}{b d x^{2} + a d x}\right ) + {\left (3 \, b^{2} c^{2} d - a b c d^{2} - 2 \, a^{2} d^{3} - {\left (b^{2} c d^{2} - a b d^{3}\right )} x\right )} \sqrt {b x^{3} + a x^{2}}\right )}}{3 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(x^3/(d*x+c)/(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)), x)
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.11 \[ \int \frac {x^3}{(c+d x) \sqrt {a x^2+b x^3}} \, dx=\frac {-2 \sqrt {d}\, \sqrt {-a d +b c}\, \mathit {atan} \left (\frac {\sqrt {b x +a}\, d}{\sqrt {d}\, \sqrt {-a d +b c}}\right ) b^{2} c^{2}-\frac {4 \sqrt {b x +a}\, a^{2} d^{3}}{3}-\frac {2 \sqrt {b x +a}\, a b c \,d^{2}}{3}+\frac {2 \sqrt {b x +a}\, a b \,d^{3} x}{3}+2 \sqrt {b x +a}\, b^{2} c^{2} d -\frac {2 \sqrt {b x +a}\, b^{2} c \,d^{2} x}{3}}{b^{2} d^{3} \left (a d -b c \right )} \] Input:

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

Output:

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