\(\int \frac {\sqrt {c+d x} (A+B x+C x^2+D x^3)}{(a+b x)^2} \, dx\) [60]

Optimal result

Integrand size = 32, antiderivative size = 228 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^2} \, dx=\frac {2 \left (b^2 B-2 a b C+3 a^2 D\right ) \sqrt {c+d x}}{b^4}-\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt {c+d x}}{b^4 (a+b x)}+\frac {2 (b C d-b c D-2 a d D) (c+d x)^{3/2}}{3 b^3 d^2}+\frac {2 D (c+d x)^{5/2}}{5 b^2 d^2}-\frac {\left (b^3 (2 B c+A d)-a b^2 (4 c C+3 B d)-7 a^3 d D+a^2 b (5 C d+6 c D)\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{9/2} \sqrt {b c-a d}} \] Output:

2*(B*b^2-2*C*a*b+3*D*a^2)*(d*x+c)^(1/2)/b^4-(A*b^3-a*(B*b^2-C*a*b+D*a^2))* 
(d*x+c)^(1/2)/b^4/(b*x+a)+2/3*(C*b*d-2*D*a*d-D*b*c)*(d*x+c)^(3/2)/b^3/d^2+ 
2/5*D*(d*x+c)^(5/2)/b^2/d^2-(b^3*(A*d+2*B*c)-a*b^2*(3*B*d+4*C*c)-7*a^3*d*D 
+a^2*b*(5*C*d+6*D*c))*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(9 
/2)/(-a*d+b*c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^2} \, dx=\frac {\sqrt {c+d x} \left (105 a^3 d^2 D-5 a^2 b d (15 C d+4 c D-14 d D x)-a b^2 \left (4 c^2 D-2 c d (5 C-9 D x)+d^2 \left (-45 B+50 C x+14 D x^2\right )\right )+b^3 \left (-15 A d^2+2 x \left (-2 c^2 D+c d (5 C+D x)+d^2 \left (15 B+5 C x+3 D x^2\right )\right )\right )\right )}{15 b^4 d^2 (a+b x)}+\frac {\left (b^3 (2 B c+A d)-a b^2 (4 c C+3 B d)-7 a^3 d D+a^2 b (5 C d+6 c D)\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{9/2} \sqrt {-b c+a d}} \] Input:

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

Output:

(Sqrt[c + d*x]*(105*a^3*d^2*D - 5*a^2*b*d*(15*C*d + 4*c*D - 14*d*D*x) - a* 
b^2*(4*c^2*D - 2*c*d*(5*C - 9*D*x) + d^2*(-45*B + 50*C*x + 14*D*x^2)) + b^ 
3*(-15*A*d^2 + 2*x*(-2*c^2*D + c*d*(5*C + D*x) + d^2*(15*B + 5*C*x + 3*D*x 
^2)))))/(15*b^4*d^2*(a + b*x)) + ((b^3*(2*B*c + A*d) - a*b^2*(4*c*C + 3*B* 
d) - 7*a^3*d*D + a^2*b*(5*C*d + 6*c*D))*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqr 
t[-(b*c) + a*d]])/(b^(9/2)*Sqrt[-(b*c) + a*d])
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2124, 27, 1192, 1584, 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.

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

\(\Big \downarrow \) 2124

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1192

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

\(\Big \downarrow \) 1584

\(\displaystyle \frac {\int \left (\frac {\left (-7 d D a^3+b (5 C d+6 c D) a^2-b^2 (4 c C+3 B d) a+b^3 (2 B c+A d)\right ) d^2}{b^4}+\frac {2 (b c-a d) D (c+d x)^2}{b^2}+\frac {2 (b c-a d) (b C d-2 a D d-b c D) (c+d x)}{b^3}+\frac {-A c d^3 b^4-2 B c^2 d^2 b^4+a A d^4 b^3+5 a B c d^3 b^3+4 a c^2 C d^2 b^3-3 a^2 B d^4 b^2-9 a^2 c C d^3 b^2-6 a^2 c^2 d^2 D b^2+5 a^3 C d^4 b+13 a^3 c d^3 D b-7 a^4 d^4 D}{b^4 (b c-a d-b (c+d x))}\right )d\sqrt {c+d x}}{d^2 (b c-a d)}-\frac {(c+d x)^{3/2} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{(a+b x) (b c-a d)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {d^2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (-7 a^3 d D+a^2 b (6 c D+5 C d)-a b^2 (3 B d+4 c C)+b^3 (A d+2 B c)\right )}{b^{9/2}}+\frac {d^2 \sqrt {c+d x} \left (-7 a^3 d D+a^2 b (6 c D+5 C d)-a b^2 (3 B d+4 c C)+b^3 (A d+2 B c)\right )}{b^4}+\frac {2 (c+d x)^{3/2} (b c-a d) (-2 a d D-b c D+b C d)}{3 b^3}+\frac {2 D (c+d x)^{5/2} (b c-a d)}{5 b^2}}{d^2 (b c-a d)}-\frac {(c+d x)^{3/2} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{(a+b x) (b c-a d)}\)

Input:

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

Output:

-(((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*(c + d*x)^(3/2))/((b*c - a*d)*(a 
+ b*x))) + ((d^2*(b^3*(2*B*c + A*d) - a*b^2*(4*c*C + 3*B*d) - 7*a^3*d*D + 
a^2*b*(5*C*d + 6*c*D))*Sqrt[c + d*x])/b^4 + (2*(b*c - a*d)*(b*C*d - b*c*D 
- 2*a*d*D)*(c + d*x)^(3/2))/(3*b^3) + (2*(b*c - a*d)*D*(c + d*x)^(5/2))/(5 
*b^2) - (d^2*Sqrt[b*c - a*d]*(b^3*(2*B*c + A*d) - a*b^2*(4*c*C + 3*B*d) - 
7*a^3*d*D + a^2*b*(5*C*d + 6*c*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b* 
c - a*d]])/b^(9/2))/(d^2*(b*c - a*d))
 

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 1192

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1)   Subst[Int[x^( 
2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + 
 c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && 
IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
 

rule 1584

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ 
b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 2009

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

rule 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(\frac {-\sqrt {\left (a d -b c \right ) b}\, \left (\left (\left (-\frac {2}{5} D x^{3}-\frac {2}{3} C \,x^{2}+A -2 B x \right ) b^{3}-3 \left (-\frac {14}{45} D x^{2}-\frac {10}{9} C x +B \right ) a \,b^{2}+5 a^{2} \left (-\frac {14 D x}{15}+C \right ) b -7 a^{3} D\right ) d^{2}-\frac {2 \left (\left (\frac {D x}{5}+C \right ) b -2 D a \right ) \left (b x +a \right ) c b d}{3}+\frac {4 D b^{2} c^{2} \left (b x +a \right )}{15}\right ) \sqrt {x d +c}+\left (\left (b^{3} A -3 a \,b^{2} B +5 a^{2} b C -7 a^{3} D\right ) d +2 b c \left (B \,b^{2}-2 C a b +3 D a^{2}\right )\right ) d^{2} \left (b x +a \right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, d^{2} b^{4} \left (b x +a \right )}\)	\(231\)
derivativedivides	\(\frac {\frac {2 \left (\frac {D \left (x d +c \right )^{\frac {5}{2}} b^{2}}{5}+\frac {C \,b^{2} d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {2 D a b d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {D b^{2} c \left (x d +c \right )^{\frac {3}{2}}}{3}+B \,d^{2} b^{2} \sqrt {x d +c}-2 C a \,d^{2} b \sqrt {x d +c}+3 D a^{2} d^{2} \sqrt {x d +c}\right )}{b^{4}}+\frac {2 d^{2} \left (\frac {\left (-\frac {1}{2} b^{3} d A +\frac {1}{2} B a \,b^{2} d -\frac {1}{2} C \,a^{2} b d +\frac {1}{2} a^{3} d D\right ) \sqrt {x d +c}}{\left (x d +c \right ) b +a d -b c}+\frac {\left (b^{3} d A -3 B a \,b^{2} d +2 B \,b^{3} c +5 C \,a^{2} b d -4 a \,b^{2} c C -7 a^{3} d D+6 a^{2} b c D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{4}}}{d^{2}}\)	\(265\)
default	\(\frac {\frac {2 \left (\frac {D \left (x d +c \right )^{\frac {5}{2}} b^{2}}{5}+\frac {C \,b^{2} d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {2 D a b d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {D b^{2} c \left (x d +c \right )^{\frac {3}{2}}}{3}+B \,d^{2} b^{2} \sqrt {x d +c}-2 C a \,d^{2} b \sqrt {x d +c}+3 D a^{2} d^{2} \sqrt {x d +c}\right )}{b^{4}}+\frac {2 d^{2} \left (\frac {\left (-\frac {1}{2} b^{3} d A +\frac {1}{2} B a \,b^{2} d -\frac {1}{2} C \,a^{2} b d +\frac {1}{2} a^{3} d D\right ) \sqrt {x d +c}}{\left (x d +c \right ) b +a d -b c}+\frac {\left (b^{3} d A -3 B a \,b^{2} d +2 B \,b^{3} c +5 C \,a^{2} b d -4 a \,b^{2} c C -7 a^{3} d D+6 a^{2} b c D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{4}}}{d^{2}}\)	\(265\)

Input:

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

Output:

(-((a*d-b*c)*b)^(1/2)*(((-2/5*D*x^3-2/3*C*x^2+A-2*B*x)*b^3-3*(-14/45*D*x^2 
-10/9*C*x+B)*a*b^2+5*a^2*(-14/15*D*x+C)*b-7*a^3*D)*d^2-2/3*((1/5*D*x+C)*b- 
2*D*a)*(b*x+a)*c*b*d+4/15*D*b^2*c^2*(b*x+a))*(d*x+c)^(1/2)+((A*b^3-3*B*a*b 
^2+5*C*a^2*b-7*D*a^3)*d+2*b*c*(B*b^2-2*C*a*b+3*D*a^2))*d^2*(b*x+a)*arctan( 
b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)))/((a*d-b*c)*b)^(1/2)/d^2/b^4/(b*x+a)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (206) = 412\).

Time = 0.11 (sec) , antiderivative size = 1045, normalized size of antiderivative = 4.58 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/30*(15*(2*(3*D*a^3*b - 2*C*a^2*b^2 + B*a*b^3)*c*d^2 - (7*D*a^4 - 5*C*a^ 
3*b + 3*B*a^2*b^2 - A*a*b^3)*d^3 + (2*(3*D*a^2*b^2 - 2*C*a*b^3 + B*b^4)*c* 
d^2 - (7*D*a^3*b - 5*C*a^2*b^2 + 3*B*a*b^3 - A*b^4)*d^3)*x)*sqrt(b^2*c - a 
*b*d)*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*(4*D*a*b^4*c^3 + 2*(8*D*a^2*b^3 - 5*C*a*b^4)*c^2*d - 5*(25*D*a^ 
3*b^2 - 17*C*a^2*b^3 + 9*B*a*b^4 - 3*A*b^5)*c*d^2 + 15*(7*D*a^4*b - 5*C*a^ 
3*b^2 + 3*B*a^2*b^3 - A*a*b^4)*d^3 - 6*(D*b^5*c*d^2 - D*a*b^4*d^3)*x^3 - 2 
*(D*b^5*c^2*d - (8*D*a*b^4 - 5*C*b^5)*c*d^2 + (7*D*a^2*b^3 - 5*C*a*b^4)*d^ 
3)*x^2 + 2*(2*D*b^5*c^3 + (7*D*a*b^4 - 5*C*b^5)*c^2*d - (44*D*a^2*b^3 - 30 
*C*a*b^4 + 15*B*b^5)*c*d^2 + 5*(7*D*a^3*b^2 - 5*C*a^2*b^3 + 3*B*a*b^4)*d^3 
)*x)*sqrt(d*x + c))/(a*b^6*c*d^2 - a^2*b^5*d^3 + (b^7*c*d^2 - a*b^6*d^3)*x 
), 1/15*(15*(2*(3*D*a^3*b - 2*C*a^2*b^2 + B*a*b^3)*c*d^2 - (7*D*a^4 - 5*C* 
a^3*b + 3*B*a^2*b^2 - A*a*b^3)*d^3 + (2*(3*D*a^2*b^2 - 2*C*a*b^3 + B*b^4)* 
c*d^2 - (7*D*a^3*b - 5*C*a^2*b^2 + 3*B*a*b^3 - A*b^4)*d^3)*x)*sqrt(-b^2*c 
+ a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (4*D*a 
*b^4*c^3 + 2*(8*D*a^2*b^3 - 5*C*a*b^4)*c^2*d - 5*(25*D*a^3*b^2 - 17*C*a^2* 
b^3 + 9*B*a*b^4 - 3*A*b^5)*c*d^2 + 15*(7*D*a^4*b - 5*C*a^3*b^2 + 3*B*a^2*b 
^3 - A*a*b^4)*d^3 - 6*(D*b^5*c*d^2 - D*a*b^4*d^3)*x^3 - 2*(D*b^5*c^2*d - ( 
8*D*a*b^4 - 5*C*b^5)*c*d^2 + (7*D*a^2*b^3 - 5*C*a*b^4)*d^3)*x^2 + 2*(2*D*b 
^5*c^3 + (7*D*a*b^4 - 5*C*b^5)*c^2*d - (44*D*a^2*b^3 - 30*C*a*b^4 + 15*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(b*x+a)^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 = 299, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^2} \, dx=\frac {{\left (6 \, D a^{2} b c - 4 \, C a b^{2} c + 2 \, B b^{3} c - 7 \, D a^{3} d + 5 \, C a^{2} b d - 3 \, B a b^{2} d + A b^{3} d\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{4}} + \frac {\sqrt {d x + c} D a^{3} d - \sqrt {d x + c} C a^{2} b d + \sqrt {d x + c} B a b^{2} d - \sqrt {d x + c} A b^{3} d}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{4}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} D b^{8} d^{8} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} D b^{8} c d^{8} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} D a b^{7} d^{9} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} C b^{8} d^{9} + 45 \, \sqrt {d x + c} D a^{2} b^{6} d^{10} - 30 \, \sqrt {d x + c} C a b^{7} d^{10} + 15 \, \sqrt {d x + c} B b^{8} d^{10}\right )}}{15 \, b^{10} d^{10}} \] Input:

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

Output:

(6*D*a^2*b*c - 4*C*a*b^2*c + 2*B*b^3*c - 7*D*a^3*d + 5*C*a^2*b*d - 3*B*a*b 
^2*d + A*b^3*d)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c 
+ a*b*d)*b^4) + (sqrt(d*x + c)*D*a^3*d - sqrt(d*x + c)*C*a^2*b*d + sqrt(d* 
x + c)*B*a*b^2*d - sqrt(d*x + c)*A*b^3*d)/(((d*x + c)*b - b*c + a*d)*b^4) 
+ 2/15*(3*(d*x + c)^(5/2)*D*b^8*d^8 - 5*(d*x + c)^(3/2)*D*b^8*c*d^8 - 10*( 
d*x + c)^(3/2)*D*a*b^7*d^9 + 5*(d*x + c)^(3/2)*C*b^8*d^9 + 45*sqrt(d*x + c 
)*D*a^2*b^6*d^10 - 30*sqrt(d*x + c)*C*a*b^7*d^10 + 15*sqrt(d*x + c)*B*b^8* 
d^10)/(b^10*d^10)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.96 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^2} \, dx=\frac {-105 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{3} d^{2}+60 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} b c d -105 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} b \,d^{2} x -30 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a \,b^{3} d +60 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a \,b^{2} c d x -30 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{4} d x +105 \sqrt {d x +c}\, a^{3} b \,d^{2}-95 \sqrt {d x +c}\, a^{2} b^{2} c d +70 \sqrt {d x +c}\, a^{2} b^{2} d^{2} x +30 \sqrt {d x +c}\, a \,b^{4} d +6 \sqrt {d x +c}\, a \,b^{3} c^{2}-68 \sqrt {d x +c}\, a \,b^{3} c d x -14 \sqrt {d x +c}\, a \,b^{3} d^{2} x^{2}+30 \sqrt {d x +c}\, b^{5} d x +6 \sqrt {d x +c}\, b^{4} c^{2} x +12 \sqrt {d x +c}\, b^{4} c d \,x^{2}+6 \sqrt {d x +c}\, b^{4} d^{2} x^{3}}{15 b^{5} d \left (b x +a \right )} \] Input:

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

Output:

( - 105*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - 
 b*c)))*a**3*d**2 + 60*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqr 
t(b)*sqrt(a*d - b*c)))*a**2*b*c*d - 105*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt 
(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**2*b*d**2*x - 30*sqrt(b)*sqrt(a* 
d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a*b**3*d + 60*s 
qrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a 
*b**2*c*d*x - 30*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*s 
qrt(a*d - b*c)))*b**4*d*x + 105*sqrt(c + d*x)*a**3*b*d**2 - 95*sqrt(c + d* 
x)*a**2*b**2*c*d + 70*sqrt(c + d*x)*a**2*b**2*d**2*x + 30*sqrt(c + d*x)*a* 
b**4*d + 6*sqrt(c + d*x)*a*b**3*c**2 - 68*sqrt(c + d*x)*a*b**3*c*d*x - 14* 
sqrt(c + d*x)*a*b**3*d**2*x**2 + 30*sqrt(c + d*x)*b**5*d*x + 6*sqrt(c + d* 
x)*b**4*c**2*x + 12*sqrt(c + d*x)*b**4*c*d*x**2 + 6*sqrt(c + d*x)*b**4*d** 
2*x**3)/(15*b**5*d*(a + b*x))