Integrand size = 32, antiderivative size = 174 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{d^3 (b c-a d) \sqrt {c+d x}}+\frac {2 (b C d-2 b c D-a d D) \sqrt {c+d x}}{b^2 d^3}+\frac {2 D (c+d x)^{3/2}}{3 b d^3}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2} (b c-a d)^{3/2}} \] Output:
2*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^3/(-a*d+b*c)/(d*x+c)^(1/2)+2*(C*b*d-D*a* d-2*D*b*c)*(d*x+c)^(1/2)/b^2/d^3+2/3*D*(d*x+c)^(3/2)/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^(5/2)/( -a*d+b*c)^(3/2)
Time = 0.38 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\frac {-6 a^2 d^2 D (c+d x)+2 a b d (c+d x) (3 C d-2 c D+d D x)+2 b^2 \left (-3 A d^3+8 c^3 D+c^2 (-6 C d+4 d D x)+c d^2 (3 B-x (3 C+D x))\right )}{3 b^2 d^3 (-b c+a d) \sqrt {c+d x}}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{5/2} (-b c+a d)^{3/2}} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/((a + b*x)*(c + d*x)^(3/2)),x]
Output:
(-6*a^2*d^2*D*(c + d*x) + 2*a*b*d*(c + d*x)*(3*C*d - 2*c*D + d*D*x) + 2*b^ 2*(-3*A*d^3 + 8*c^3*D + c^2*(-6*C*d + 4*d*D*x) + c*d^2*(3*B - x*(3*C + D*x ))))/(3*b^2*d^3*(-(b*c) + a*d)*Sqrt[c + d*x]) - (2*(A*b^3 - a*(b^2*B - a*b *C + a^2*D))*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(b^(5/2)* (-(b*c) + a*d)^(3/2))
Time = 0.49 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2122, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 2122 |
\(\displaystyle \int \left (\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{b^2 (a+b x) \sqrt {c+d x} (b c-a d)}+\frac {A d^3-B c d^2+c^3 (-D)+c^2 C d}{d^2 (c+d x)^{3/2} (a d-b c)}+\frac {-a d D-b c D+b C d}{b^2 d^2 \sqrt {c+d x}}+\frac {D x}{b d \sqrt {c+d x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2} (b c-a d)^{3/2}}+\frac {2 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d^3 \sqrt {c+d x} (b c-a d)}+\frac {2 \sqrt {c+d x} (-a d D-b c D+b C d)}{b^2 d^3}+\frac {2 D (c+d x)^{3/2}}{3 b d^3}-\frac {2 c D \sqrt {c+d x}}{b d^3}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/((a + b*x)*(c + d*x)^(3/2)),x]
Output:
(2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D))/(d^3*(b*c - a*d)*Sqrt[c + d*x]) - (2*c*D*Sqrt[c + d*x])/(b*d^3) + (2*(b*C*d - b*c*D - a*d*D)*Sqrt[c + d*x])/ (b^2*d^3) + (2*D*(c + d*x)^(3/2))/(3*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^(5/2)*(b*c - a*d)^(3/2))
Int[((Px_)*((c_.) + (d_.)*(x_))^(n_))/((a_.) + (b_.)*(x_)), x_Symbol] :> In t[ExpandIntegrand[1/Sqrt[c + d*x], Px*((c + d*x)^(n + 1/2)/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[n + 1/2, 0]
Time = 0.76 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.93
method | result | size |
pseudoelliptic | \(\frac {\frac {2 \sqrt {x d +c}\, \left (D b d x +3 C b d -3 D a d -5 D b c \right )}{3 b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right ) \sqrt {x d +c}}-\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) b^{2} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\) | \(161\) |
derivativedivides | \(\frac {\frac {2 \left (\frac {D \left (x d +c \right )^{\frac {3}{2}} b}{3}+C d b \sqrt {x d +c}-D a d \sqrt {x d +c}-2 D b c \sqrt {x d +c}\right )}{b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right ) \sqrt {x d +c}}-\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) b^{2} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\) | \(179\) |
default | \(\frac {\frac {2 \left (\frac {D \left (x d +c \right )^{\frac {3}{2}} b}{3}+C d b \sqrt {x d +c}-D a d \sqrt {x d +c}-2 D b c \sqrt {x d +c}\right )}{b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right ) \sqrt {x d +c}}-\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) b^{2} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\) | \(179\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2*(1/3*(d*x+c)^(1/2)*(D*b*d*x+3*C*b*d-3*D*a*d-5*D*b*c)/b^2-(A*d^3-B*c*d^2+ C*c^2*d-D*c^3)/(a*d-b*c)/(d*x+c)^(1/2)-d^3*(A*b^3-B*a*b^2+C*a^2*b-D*a^3)/( a*d-b*c)/b^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2 )))/d^3
Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (156) = 312\).
Time = 0.10 (sec) , antiderivative size = 834, normalized size of antiderivative = 4.79 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx =\text {Too large to display} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(3/2),x, algorithm="fricas")
Output:
[1/3*(3*((D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*d^4*x + (D*a^3 - C*a^2*b + B* a*b^2 - A*b^3)*c*d^3)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d + 2*sqr t(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) - 2*(8*D*b^4*c^4 + 3*A*a*b^3*d^ 4 - 2*(5*D*a*b^3 + 3*C*b^4)*c^3*d - (D*a^2*b^2 - 9*C*a*b^3 - 3*B*b^4)*c^2* d^2 + 3*(D*a^3*b - C*a^2*b^2 - B*a*b^3 - A*b^4)*c*d^3 - (D*b^4*c^2*d^2 - 2 *D*a*b^3*c*d^3 + D*a^2*b^2*d^4)*x^2 + (4*D*b^4*c^3*d - (5*D*a*b^3 + 3*C*b^ 4)*c^2*d^2 - 2*(D*a^2*b^2 - 3*C*a*b^3)*c*d^3 + 3*(D*a^3*b - C*a^2*b^2)*d^4 )*x)*sqrt(d*x + c))/(b^5*c^3*d^3 - 2*a*b^4*c^2*d^4 + a^2*b^3*c*d^5 + (b^5* c^2*d^4 - 2*a*b^4*c*d^5 + a^2*b^3*d^6)*x), -2/3*(3*((D*a^3 - C*a^2*b + B*a *b^2 - A*b^3)*d^4*x + (D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*c*d^3)*sqrt(-b^2 *c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (8* D*b^4*c^4 + 3*A*a*b^3*d^4 - 2*(5*D*a*b^3 + 3*C*b^4)*c^3*d - (D*a^2*b^2 - 9 *C*a*b^3 - 3*B*b^4)*c^2*d^2 + 3*(D*a^3*b - C*a^2*b^2 - B*a*b^3 - A*b^4)*c* d^3 - (D*b^4*c^2*d^2 - 2*D*a*b^3*c*d^3 + D*a^2*b^2*d^4)*x^2 + (4*D*b^4*c^3 *d - (5*D*a*b^3 + 3*C*b^4)*c^2*d^2 - 2*(D*a^2*b^2 - 3*C*a*b^3)*c*d^3 + 3*( D*a^3*b - C*a^2*b^2)*d^4)*x)*sqrt(d*x + c))/(b^5*c^3*d^3 - 2*a*b^4*c^2*d^4 + a^2*b^3*c*d^5 + (b^5*c^2*d^4 - 2*a*b^4*c*d^5 + a^2*b^3*d^6)*x)]
Time = 9.98 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.51 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {D \left (c + d x\right )^{\frac {3}{2}}}{3 b d^{2}} + \frac {- A d^{3} + B c d^{2} - C c^{2} d + D c^{3}}{d^{2} \sqrt {c + d x} \left (a d - b c\right )} + \frac {\sqrt {c + d x} \left (C b d - D a d - 2 D b c\right )}{b^{2} d^{2}} + \frac {d \left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{3} \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\frac {D x^{3}}{3 b} + \frac {x^{2} \left (C b - D a\right )}{2 b^{2}} + \frac {x \left (B b^{2} - C a b + D a^{2}\right )}{b^{3}} - \frac {\left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{3}}}{c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)/(d*x+c)**(3/2),x)
Output:
Piecewise((2*(D*(c + d*x)**(3/2)/(3*b*d**2) + (-A*d**3 + B*c*d**2 - C*c**2 *d + D*c**3)/(d**2*sqrt(c + d*x)*(a*d - b*c)) + sqrt(c + d*x)*(C*b*d - D*a *d - 2*D*b*c)/(b**2*d**2) + d*(-A*b**3 + B*a*b**2 - C*a**2*b + D*a**3)*ata n(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(b**3*sqrt((a*d - b*c)/b)*(a*d - b*c) ))/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)/c**(3/2), True))
Exception generated. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(3/2),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
Time = 0.13 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{3} c - a b^{2} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{{\left (b c d^{3} - a d^{4}\right )} \sqrt {d x + c}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} D b^{2} d^{6} - 6 \, \sqrt {d x + c} D b^{2} c d^{6} - 3 \, \sqrt {d x + c} D a b d^{7} + 3 \, \sqrt {d x + c} C b^{2} d^{7}\right )}}{3 \, b^{3} d^{9}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(3/2),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))/((b^3*c - a*b^2*d)*sqrt(-b^2*c + a*b*d)) - 2*(D*c^3 - C*c^2*d + B*c*d^2 - A*d^3)/((b*c*d^3 - a*d^4)*sqrt(d*x + c)) + 2/3*((d*x + c)^(3/2)* D*b^2*d^6 - 6*sqrt(d*x + c)*D*b^2*c*d^6 - 3*sqrt(d*x + c)*D*a*b*d^7 + 3*sq rt(d*x + c)*C*b^2*d^7)/(b^3*d^9)
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{\left (a+b\,x\right )\,{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:
int((A + B*x + C*x^2 + x^3*D)/((a + b*x)*(c + d*x)^(3/2)),x)
Output:
int((A + B*x + C*x^2 + x^3*D)/((a + b*x)*(c + d*x)^(3/2)), x)
Time = 0.14 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\frac {2 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} d^{2}-2 a^{2} b c \,d^{2}-2 a^{2} b \,d^{3} x -2 a \,b^{3} d^{2}+\frac {2 a \,b^{2} c^{2} d}{3}+\frac {4 a \,b^{2} c \,d^{2} x}{3}+\frac {2 a \,b^{2} d^{3} x^{2}}{3}+2 b^{4} c d +\frac {4 b^{3} c^{3}}{3}+\frac {2 b^{3} c^{2} d x}{3}-\frac {2 b^{3} c \,d^{2} x^{2}}{3}}{\sqrt {d x +c}\, b^{3} d^{2} \left (a d -b c \right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(3/2),x)
Output:
(2*(3*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b )*sqrt(a*d - b*c)))*a**2*d**2 - 3*a**2*b*c*d**2 - 3*a**2*b*d**3*x - 3*a*b* *3*d**2 + a*b**2*c**2*d + 2*a*b**2*c*d**2*x + a*b**2*d**3*x**2 + 3*b**4*c* d + 2*b**3*c**3 + b**3*c**2*d*x - b**3*c*d**2*x**2))/(3*sqrt(c + d*x)*b**3 *d**2*(a*d - b*c))