\(\int \frac {A+B x+C x^2+D x^3}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx\) [216]

Optimal result

Integrand size = 41, antiderivative size = 202 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \left (10 c C d-15 B d^2-17 c^2 D\right ) \sqrt {c^2-d^2 x^2}}{15 d^4 \sqrt {c+d x}}-\frac {2 (5 C d-7 c D) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{15 d^4}-\frac {2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}-\frac {\sqrt {2} \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c} d^4} \] Output:

2/15*(-15*B*d^2+10*C*c*d-17*D*c^2)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)^(1/2)- 
2/15*(5*C*d-7*D*c)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/d^4-2/5*D*(d*x+c)^(3 
/2)*(-d^2*x^2+c^2)^(1/2)/d^4-2^(1/2)*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*arctanh 
(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2))/c^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\frac {-\frac {2 \sqrt {c^2-d^2 x^2} \left (13 c^2 D-c d (5 C+D x)+d^2 (15 B+x (5 C+3 D x))\right )}{\sqrt {c+d x}}+\frac {15 \sqrt {2} \left (-c^2 C d+B c d^2-A d^3+c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c}}}{15 d^4} \] Input:

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

Output:

((-2*Sqrt[c^2 - d^2*x^2]*(13*c^2*D - c*d*(5*C + D*x) + d^2*(15*B + x*(5*C 
+ 3*D*x))))/Sqrt[c + d*x] + (15*Sqrt[2]*(-(c^2*C*d) + B*c*d^2 - A*d^3 + c^ 
3*D)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/Sqrt[c] 
)/(15*d^4)
 

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2170, 27, 2170, 27, 600, 458, 471, 221}

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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int -\frac {d^6 \left (2 D c^3+5 C d c^2+15 A d^3-d \left (-17 D c^2+10 C d c-15 B d^2\right ) x\right )}{2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{3 d^4}-\frac {2}{3} d \sqrt {c+d x} \sqrt {c^2-d^2 x^2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} d^2 \int \frac {2 D c^3+5 C d c^2+15 A d^3-d \left (-17 D c^2+10 C d c-15 B d^2\right ) x}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx-\frac {2}{3} d \sqrt {c+d x} \sqrt {c^2-d^2 x^2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}\)

\(\Big \downarrow \) 600

\(\displaystyle \frac {\frac {1}{3} d^2 \left (15 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx-\left (-15 B d^2-17 c^2 D+10 c C d\right ) \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx\right )-\frac {2}{3} d \sqrt {c+d x} \sqrt {c^2-d^2 x^2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {\frac {1}{3} d^2 \left (15 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c^2-d^2 x^2} \left (-15 B d^2-17 c^2 D+10 c C d\right )}{d \sqrt {c+d x}}\right )-\frac {2}{3} d \sqrt {c+d x} \sqrt {c^2-d^2 x^2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {\frac {1}{3} d^2 \left (30 d \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}+\frac {2 \sqrt {c^2-d^2 x^2} \left (-15 B d^2-17 c^2 D+10 c C d\right )}{d \sqrt {c+d x}}\right )-\frac {2}{3} d \sqrt {c+d x} \sqrt {c^2-d^2 x^2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {2 \sqrt {c^2-d^2 x^2} \left (-15 B d^2-17 c^2 D+10 c C d\right )}{d \sqrt {c+d x}}-\frac {15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right ) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{\sqrt {c} d}\right )-\frac {2}{3} d \sqrt {c+d x} \sqrt {c^2-d^2 x^2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}\)

Input:

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

Output:

(-2*D*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2])/(5*d^4) + ((-2*d*(5*C*d - 7*c*D 
)*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2])/3 + (d^2*((2*(10*c*C*d - 15*B*d^2 - 1 
7*c^2*D)*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[c + d*x]) - (15*Sqrt[2]*(c^2*C*d - B 
*c*d^2 + A*d^3 - c^3*D)*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[ 
c + d*x])])/(Sqrt[c]*d)))/3)/(5*d^5)
 

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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 458

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

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

rule 600

Int[((A_.) + (B_.)*(x_))/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2] 
), x_Symbol] :> Simp[B/d   Int[Sqrt[c + d*x]/Sqrt[a + b*x^2], x], x] - Simp 
[(B*c - A*d)/d   Int[1/(Sqrt[c + d*x]*Sqrt[a + b*x^2]), x], x] /; FreeQ[{a, 
 b, c, d, A, B}, x] && NegQ[b/a]
 

rule 2170

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + 
 p + q)*(d + e*x)^(q - 2)*(a*e - b*d*x), x], x], x] /; NeQ[m + q + 2*p + 1, 
 0]] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 
0] &&  !IGtQ[m, 0]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.21

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (6 D d^{2} x^{2} \sqrt {-d x +c}\, \sqrt {c}+15 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{3}-15 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{2}+15 C \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d +10 C \,d^{2} x \sqrt {-d x +c}\, \sqrt {c}-15 D \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}-2 D c^{\frac {3}{2}} d x \sqrt {-d x +c}+30 d^{2} B \sqrt {-d x +c}\, \sqrt {c}-10 C \,c^{\frac {3}{2}} d \sqrt {-d x +c}+26 D c^{\frac {5}{2}} \sqrt {-d x +c}\right )}{15 \sqrt {d x +c}\, \sqrt {-d x +c}\, d^{4} \sqrt {c}}\)	\(245\)

Input:

int((D*x^3+C*x^2+B*x+A)/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-1/15*(-d^2*x^2+c^2)^(1/2)*(6*D*d^2*x^2*(-d*x+c)^(1/2)*c^(1/2)+15*A*2^(1/2 
)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^3-15*B*2^(1/2)*arctanh(1/2 
*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^2+15*C*2^(1/2)*arctanh(1/2*(-d*x+c)^( 
1/2)*2^(1/2)/c^(1/2))*c^2*d+10*C*d^2*x*(-d*x+c)^(1/2)*c^(1/2)-15*D*2^(1/2) 
*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3-2*D*c^(3/2)*d*x*(-d*x+c)^ 
(1/2)+30*d^2*B*(-d*x+c)^(1/2)*c^(1/2)-10*C*c^(3/2)*d*(-d*x+c)^(1/2)+26*D*c 
^(5/2)*(-d*x+c)^(1/2))/(d*x+c)^(1/2)/(-d*x+c)^(1/2)/d^4/c^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.05 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\left [-\frac {4 \, {\left (3 \, D d^{2} x^{2} + 13 \, D c^{2} - 5 \, C c d + 15 \, B d^{2} - {\left (D c d - 5 \, C d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} - \frac {15 \, \sqrt {2} {\left (D c^{4} - C c^{3} d + B c^{2} d^{2} - A c d^{3} + {\left (D c^{3} d - C c^{2} d^{2} + B c d^{3} - A d^{4}\right )} x\right )} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{\sqrt {c}}}{30 \, {\left (d^{5} x + c d^{4}\right )}}, \frac {15 \, \sqrt {2} {\left (D c^{4} - C c^{3} d + B c^{2} d^{2} - A c d^{3} + {\left (D c^{3} d - C c^{2} d^{2} + B c d^{3} - A d^{4}\right )} x\right )} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} c \sqrt {-\frac {1}{c}}}{d^{2} x^{2} - c^{2}}\right ) - 2 \, {\left (3 \, D d^{2} x^{2} + 13 \, D c^{2} - 5 \, C c d + 15 \, B d^{2} - {\left (D c d - 5 \, C d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{15 \, {\left (d^{5} x + c d^{4}\right )}}\right ] \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2),x, algori 
thm="fricas")
                                                                                    
                                                                                    
 

Output:

[-1/30*(4*(3*D*d^2*x^2 + 13*D*c^2 - 5*C*c*d + 15*B*d^2 - (D*c*d - 5*C*d^2) 
*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c) - 15*sqrt(2)*(D*c^4 - C*c^3*d + B*c 
^2*d^2 - A*c*d^3 + (D*c^3*d - C*c^2*d^2 + B*c*d^3 - A*d^4)*x)*log(-(d^2*x^ 
2 - 2*c*d*x - 2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(c) - 3*c^2 
)/(d^2*x^2 + 2*c*d*x + c^2))/sqrt(c))/(d^5*x + c*d^4), 1/15*(15*sqrt(2)*(D 
*c^4 - C*c^3*d + B*c^2*d^2 - A*c*d^3 + (D*c^3*d - C*c^2*d^2 + B*c*d^3 - A* 
d^4)*x)*sqrt(-1/c)*arctan(sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*c*sqr 
t(-1/c)/(d^2*x^2 - c^2)) - 2*(3*D*d^2*x^2 + 13*D*c^2 - 5*C*c*d + 15*B*d^2 
- (D*c*d - 5*C*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^5*x + c*d^4) 
]
 

Sympy [F]

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

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

Output:

Integral((A + B*x + C*x**2 + D*x**3)/(sqrt(-(-c + d*x)*(c + d*x))*sqrt(c + 
 d*x)), x)
 

Maxima [F]

\[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\int { \frac {D x^{3} + C x^{2} + B x + A}{\sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}} \,d x } \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2),x, algori 
thm="maxima")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)/(sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)), 
x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=-\frac {6 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D - 10 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c + 30 \, \sqrt {-d x + c} D c^{2} - 10 \, {\left (-d x + c\right )}^{\frac {3}{2}} C d + 30 \, \sqrt {-d x + c} B d^{2} + \frac {15 \, \sqrt {2} {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}}}{15 \, d^{4}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2),x, algori 
thm="giac")
 

Output:

-1/15*(6*(d*x - c)^2*sqrt(-d*x + c)*D - 10*(-d*x + c)^(3/2)*D*c + 30*sqrt( 
-d*x + c)*D*c^2 - 10*(-d*x + c)^(3/2)*C*d + 30*sqrt(-d*x + c)*B*d^2 + 15*s 
qrt(2)*(D*c^3 - C*c^2*d + B*c*d^2 - A*d^3)*arctan(1/2*sqrt(2)*sqrt(-d*x + 
c)/sqrt(-c))/sqrt(-c))/d^4
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{\sqrt {c^2-d^2\,x^2}\,\sqrt {c+d\,x}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x+C x^2+D x^3}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}} \, dx=\frac {-60 \sqrt {-d x +c}\, b c d -32 \sqrt {-d x +c}\, c^{3}-16 \sqrt {-d x +c}\, c^{2} d x -12 \sqrt {-d x +c}\, c \,d^{2} x^{2}+15 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{2}-15 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c d -15 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{2}+15 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c d}{30 c \,d^{3}} \] Input:

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

Output:

( - 60*sqrt(c - d*x)*b*c*d - 32*sqrt(c - d*x)*c**3 - 16*sqrt(c - d*x)*c**2 
*d*x - 12*sqrt(c - d*x)*c*d**2*x**2 + 15*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) 
 - sqrt(c)*sqrt(2))*a*d**2 - 15*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c 
)*sqrt(2))*b*c*d - 15*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2)) 
*a*d**2 + 15*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c*d)/( 
30*c*d**3)