\(\int \frac {A+B x+C x^2+D x^3}{(a+b x) \sqrt {c+d x}} \, dx\) [83]

Optimal result

Integrand size = 32, antiderivative size = 188 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) \sqrt {c+d x}} \, dx=\frac {2 \left (a^2 d^2 D-a b d (C d-c D)-b^2 \left (c C d-B d^2-c^2 D\right )\right ) \sqrt {c+d x}}{b^3 d^3}+\frac {2 (b C d-2 b c D-a d D) (c+d x)^{3/2}}{3 b^2 d^3}+\frac {2 D (c+d x)^{5/2}}{5 b d^3}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{7/2} \sqrt {b c-a d}} \] Output:

2*(a^2*d^2*D-a*b*d*(C*d-D*c)-b^2*(-B*d^2+C*c*d-D*c^2))*(d*x+c)^(1/2)/b^3/d 
^3+2/3*(C*b*d-D*a*d-2*D*b*c)*(d*x+c)^(3/2)/b^2/d^3+2/5*D*(d*x+c)^(5/2)/b/d 
^3-2*(A*b^3-a*(B*b^2-C*a*b+D*a^2))*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c 
)^(1/2))/b^(7/2)/(-a*d+b*c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) \sqrt {c+d x}} \, dx=\frac {2 \sqrt {c+d x} \left (15 a^2 d^2 D-5 a b d (3 C d-2 c D+d D x)+b^2 \left (8 c^2 D-2 c d (5 C+2 D x)+d^2 \left (15 B+5 C x+3 D x^2\right )\right )\right )}{15 b^3 d^3}+\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{7/2} \sqrt {-b c+a d}} \] Input:

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

Output:

(2*Sqrt[c + d*x]*(15*a^2*d^2*D - 5*a*b*d*(3*C*d - 2*c*D + d*D*x) + b^2*(8* 
c^2*D - 2*c*d*(5*C + 2*D*x) + d^2*(15*B + 5*C*x + 3*D*x^2))))/(15*b^3*d^3) 
 + (2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/S 
qrt[-(b*c) + a*d]])/(b^(7/2)*Sqrt[-(b*c) + a*d])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2123, 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.

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

\(\Big \downarrow \) 2123

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{7/2} \sqrt {b c-a d}}+\frac {2 \sqrt {c+d x} \left (a^2 d^2 D-a b d (C d-c D)-\left (b^2 \left (-B d^2+c^2 (-D)+c C d\right )\right )\right )}{b^3 d^3}+\frac {2 (c+d x)^{3/2} (-a d D-2 b c D+b C d)}{3 b^2 d^3}+\frac {2 D (c+d x)^{5/2}}{5 b d^3}\)

Input:

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

Output:

(2*(a^2*d^2*D - a*b*d*(C*d - c*D) - b^2*(c*C*d - B*d^2 - c^2*D))*Sqrt[c + 
d*x])/(b^3*d^3) + (2*(b*C*d - 2*b*c*D - a*d*D)*(c + d*x)^(3/2))/(3*b^2*d^3 
) + (2*D*(c + d*x)^(5/2))/(5*b*d^3) - (2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D 
))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(7/2)*Sqrt[b*c - a 
*d])
 

Defintions of rubi rules used

rule 2009

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

rule 2123

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+2 \sqrt {\left (a d -b c \right ) b}\, \left (\left (\left (\frac {1}{5} D x^{2}+\frac {1}{3} C x +B \right ) b^{2}-\left (\frac {D x}{3}+C \right ) a b +D a^{2}\right ) d^{2}-\frac {2 \left (\left (\frac {2 D x}{5}+C \right ) b -D a \right ) c b d}{3}+\frac {8 D b^{2} c^{2}}{15}\right ) \sqrt {x d +c}}{d^{3} b^{3} \sqrt {\left (a d -b c \right ) b}}\)	\(157\)
derivativedivides	\(\frac {\frac {2 \left (\frac {D \left (x d +c \right )^{\frac {5}{2}} b^{2}}{5}+\frac {C \,b^{2} d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {D a b d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {2 D b^{2} c \left (x d +c \right )^{\frac {3}{2}}}{3}+B \,d^{2} b^{2} \sqrt {x d +c}-C a \,d^{2} b \sqrt {x d +c}-C \,b^{2} c d \sqrt {x d +c}+D a^{2} d^{2} \sqrt {x d +c}+D a b c d \sqrt {x d +c}+D b^{2} c^{2} \sqrt {x d +c}\right )}{b^{3}}+\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{3} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\)	\(220\)
default	\(\frac {\frac {2 \left (\frac {D \left (x d +c \right )^{\frac {5}{2}} b^{2}}{5}+\frac {C \,b^{2} d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {D a b d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {2 D b^{2} c \left (x d +c \right )^{\frac {3}{2}}}{3}+B \,d^{2} b^{2} \sqrt {x d +c}-C a \,d^{2} b \sqrt {x d +c}-C \,b^{2} c d \sqrt {x d +c}+D a^{2} d^{2} \sqrt {x d +c}+D a b c d \sqrt {x d +c}+D b^{2} c^{2} \sqrt {x d +c}\right )}{b^{3}}+\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{3} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\)	\(220\)

Input:

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

Output:

2/((a*d-b*c)*b)^(1/2)*(d^3*(A*b^3-B*a*b^2+C*a^2*b-D*a^3)*arctan(b*(d*x+c)^ 
(1/2)/((a*d-b*c)*b)^(1/2))+((a*d-b*c)*b)^(1/2)*(((1/5*D*x^2+1/3*C*x+B)*b^2 
-(1/3*D*x+C)*a*b+D*a^2)*d^2-2/3*((2/5*D*x+C)*b-D*a)*c*b*d+8/15*D*b^2*c^2)* 
(d*x+c)^(1/2))/d^3/b^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 551, normalized size of antiderivative = 2.93 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) \sqrt {c+d x}} \, dx=\left [-\frac {15 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \sqrt {b^{2} c - a b d} d^{3} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) - 2 \, {\left (8 \, D b^{4} c^{3} + 2 \, {\left (D a b^{3} - 5 \, C b^{4}\right )} c^{2} d + 5 \, {\left (D a^{2} b^{2} - C a b^{3} + 3 \, B b^{4}\right )} c d^{2} - 15 \, {\left (D a^{3} b - C a^{2} b^{2} + B a b^{3}\right )} d^{3} + 3 \, {\left (D b^{4} c d^{2} - D a b^{3} d^{3}\right )} x^{2} - {\left (4 \, D b^{4} c^{2} d + {\left (D a b^{3} - 5 \, C b^{4}\right )} c d^{2} - 5 \, {\left (D a^{2} b^{2} - C a b^{3}\right )} d^{3}\right )} x\right )} \sqrt {d x + c}}{15 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )}}, -\frac {2 \, {\left (15 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \sqrt {-b^{2} c + a b d} d^{3} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - {\left (8 \, D b^{4} c^{3} + 2 \, {\left (D a b^{3} - 5 \, C b^{4}\right )} c^{2} d + 5 \, {\left (D a^{2} b^{2} - C a b^{3} + 3 \, B b^{4}\right )} c d^{2} - 15 \, {\left (D a^{3} b - C a^{2} b^{2} + B a b^{3}\right )} d^{3} + 3 \, {\left (D b^{4} c d^{2} - D a b^{3} d^{3}\right )} x^{2} - {\left (4 \, D b^{4} c^{2} d + {\left (D a b^{3} - 5 \, C b^{4}\right )} c d^{2} - 5 \, {\left (D a^{2} b^{2} - C a b^{3}\right )} d^{3}\right )} x\right )} \sqrt {d x + c}\right )}}{15 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )}}\right ] \] Input:

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

Output:

[-1/15*(15*(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*sqrt(b^2*c - a*b*d)*d^3*log 
((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) - 
2*(8*D*b^4*c^3 + 2*(D*a*b^3 - 5*C*b^4)*c^2*d + 5*(D*a^2*b^2 - C*a*b^3 + 3* 
B*b^4)*c*d^2 - 15*(D*a^3*b - C*a^2*b^2 + B*a*b^3)*d^3 + 3*(D*b^4*c*d^2 - D 
*a*b^3*d^3)*x^2 - (4*D*b^4*c^2*d + (D*a*b^3 - 5*C*b^4)*c*d^2 - 5*(D*a^2*b^ 
2 - C*a*b^3)*d^3)*x)*sqrt(d*x + c))/(b^5*c*d^3 - a*b^4*d^4), -2/15*(15*(D* 
a^3 - C*a^2*b + B*a*b^2 - A*b^3)*sqrt(-b^2*c + a*b*d)*d^3*arctan(sqrt(-b^2 
*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (8*D*b^4*c^3 + 2*(D*a*b^3 - 5*C 
*b^4)*c^2*d + 5*(D*a^2*b^2 - C*a*b^3 + 3*B*b^4)*c*d^2 - 15*(D*a^3*b - C*a^ 
2*b^2 + B*a*b^3)*d^3 + 3*(D*b^4*c*d^2 - D*a*b^3*d^3)*x^2 - (4*D*b^4*c^2*d 
+ (D*a*b^3 - 5*C*b^4)*c*d^2 - 5*(D*a^2*b^2 - C*a*b^3)*d^3)*x)*sqrt(d*x + c 
))/(b^5*c*d^3 - a*b^4*d^4)]
 

Sympy [A] (verification not implemented)

Time = 4.29 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.46 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) \sqrt {c+d x}} \, dx=\begin {cases} \frac {2 \left (\frac {D \left (c + d x\right )^{\frac {5}{2}}}{5 b d^{2}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (C b d - D a d - 2 D b c\right )}{3 b^{2} d^{2}} + \frac {\sqrt {c + d x} \left (B b^{2} d^{2} - C a b d^{2} - C b^{2} c d + D a^{2} d^{2} + D a b c d + D b^{2} c^{2}\right )}{b^{3} d^{2}} - \frac {d \left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{4} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\frac {D x^{3}}{3 b} + \frac {x^{2} \left (C b - D a\right )}{2 b^{2}} + \frac {x \left (B b^{2} - C a b + D a^{2}\right )}{b^{3}} - \frac {\left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{3}}}{\sqrt {c}} & \text {otherwise} \end {cases} \] Input:

integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)/(d*x+c)**(1/2),x)
                                                                                    
                                                                                    
 

Output:

Piecewise((2*(D*(c + d*x)**(5/2)/(5*b*d**2) + (c + d*x)**(3/2)*(C*b*d - D* 
a*d - 2*D*b*c)/(3*b**2*d**2) + sqrt(c + d*x)*(B*b**2*d**2 - C*a*b*d**2 - C 
*b**2*c*d + D*a**2*d**2 + D*a*b*c*d + D*b**2*c**2)/(b**3*d**2) - d*(-A*b** 
3 + B*a*b**2 - C*a**2*b + D*a**3)*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/ 
(b**4*sqrt((a*d - b*c)/b)))/d, Ne(d, 0)), ((D*x**3/(3*b) + x**2*(C*b - D*a 
)/(2*b**2) + x*(B*b**2 - C*a*b + D*a**2)/b**3 - (-A*b**3 + B*a*b**2 - C*a* 
*2*b + D*a**3)*Piecewise((x/a, Eq(b, 0)), (log(a + b*x)/b, True))/b**3)/sq 
rt(c), True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.32 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) \sqrt {c+d x}} \, dx=-\frac {2 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{3}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} D b^{4} d^{12} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} D b^{4} c d^{12} + 15 \, \sqrt {d x + c} D b^{4} c^{2} d^{12} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} D a b^{3} d^{13} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} C b^{4} d^{13} + 15 \, \sqrt {d x + c} D a b^{3} c d^{13} - 15 \, \sqrt {d x + c} C b^{4} c d^{13} + 15 \, \sqrt {d x + c} D a^{2} b^{2} d^{14} - 15 \, \sqrt {d x + c} C a b^{3} d^{14} + 15 \, \sqrt {d x + c} B b^{4} d^{14}\right )}}{15 \, b^{5} d^{15}} \] Input:

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

Output:

-2*(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c 
+ a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3) + 2/15*(3*(d*x + c)^(5/2)*D*b^4*d^12 
- 10*(d*x + c)^(3/2)*D*b^4*c*d^12 + 15*sqrt(d*x + c)*D*b^4*c^2*d^12 - 5*(d 
*x + c)^(3/2)*D*a*b^3*d^13 + 5*(d*x + c)^(3/2)*C*b^4*d^13 + 15*sqrt(d*x + 
c)*D*a*b^3*c*d^13 - 15*sqrt(d*x + c)*C*b^4*c*d^13 + 15*sqrt(d*x + c)*D*a^2 
*b^2*d^14 - 15*sqrt(d*x + c)*C*a*b^3*d^14 + 15*sqrt(d*x + c)*B*b^4*d^14)/( 
b^5*d^15)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) \sqrt {c+d x}} \, dx=\frac {-2 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} d^{2}+2 \sqrt {d x +c}\, a^{2} b \,d^{2}-\frac {2 \sqrt {d x +c}\, a \,b^{2} c d}{3}-\frac {2 \sqrt {d x +c}\, a \,b^{2} d^{2} x}{3}+2 \sqrt {d x +c}\, b^{4} d -\frac {4 \sqrt {d x +c}\, b^{3} c^{2}}{15}+\frac {2 \sqrt {d x +c}\, b^{3} c d x}{15}+\frac {2 \sqrt {d x +c}\, b^{3} d^{2} x^{2}}{5}}{b^{4} d^{2}} \] Input:

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

Output:

(2*( - 15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d 
 - b*c)))*a**2*d**2 + 15*sqrt(c + d*x)*a**2*b*d**2 - 5*sqrt(c + d*x)*a*b** 
2*c*d - 5*sqrt(c + d*x)*a*b**2*d**2*x + 15*sqrt(c + d*x)*b**4*d - 2*sqrt(c 
 + d*x)*b**3*c**2 + sqrt(c + d*x)*b**3*c*d*x + 3*sqrt(c + d*x)*b**3*d**2*x 
**2))/(15*b**4*d**2)