\(\int \frac {\sqrt {c+d x} (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{3/2}} \, dx\) [223]

Optimal result

Integrand size = 41, antiderivative size = 206 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) \sqrt {c+d x}}{c d^4 \sqrt {c^2-d^2 x^2}}+\frac {2 (3 C d+c D) \sqrt {c^2-d^2 x^2}}{3 d^4 \sqrt {c+d x}}+\frac {2 D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {2} c^{3/2} d^4} \] Output:

(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^(1/2)/c/d^4/(-d^2*x^2+c^2)^(1/2)+2/3 
*(3*C*d+D*c)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)^(1/2)+2/3*D*(d*x+c)^(1/2)*(- 
d^2*x^2+c^2)^(1/2)/d^4-1/2*(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))*2^(1/2)/c^(3/2)/d^4
 

Mathematica [A] (verified)

Time = 2.95 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (3 A d^3+7 c^3 D+c^2 d (9 C-2 D x)+c d^2 (3 B-2 x (3 C+D x))\right )}{(c-d x) \sqrt {c+d x}}+3 \sqrt {2} \left (-c^2 C d+B c d^2-A d^3+c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{6 c^{3/2} d^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2166, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 600

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

\(\Big \downarrow \) 458

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

\(\Big \downarrow \) 471

\(\displaystyle \frac {\sqrt {c+d x} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {-6 d \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}-\frac {4 c \sqrt {c^2-d^2 x^2} (c D+3 C d)}{d \sqrt {c+d x}}}{3 d^3}-\frac {4 c D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4}}{2 c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {c+d x} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right ) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{\sqrt {c} d}-\frac {4 c \sqrt {c^2-d^2 x^2} (c D+3 C d)}{d \sqrt {c+d x}}}{3 d^3}-\frac {4 c D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4}}{2 c}\)

Input:

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

Output:

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

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 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[m, 0]
 

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 = 0.37 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.16

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (3 A \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, d^{3} \sqrt {-d x +c}-3 B \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c \,d^{2} \sqrt {-d x +c}+3 C \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2} d \sqrt {-d x +c}-3 D \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3} \sqrt {-d x +c}+4 D c^{\frac {3}{2}} d^{2} x^{2}+12 C \,c^{\frac {3}{2}} d^{2} x +4 D c^{\frac {5}{2}} d x -6 A \,d^{3} \sqrt {c}-6 B \,c^{\frac {3}{2}} d^{2}-18 C \,c^{\frac {5}{2}} d -14 D c^{\frac {7}{2}}\right )}{6 c^{\frac {3}{2}} \sqrt {d x +c}\, \left (-d x +c \right ) d^{4}}\)	\(238\)

Input:

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

Output:

-1/6*(-d^2*x^2+c^2)^(1/2)/c^(3/2)*(3*A*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/ 
c^(1/2))*2^(1/2)*d^3*(-d*x+c)^(1/2)-3*B*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2) 
/c^(1/2))*2^(1/2)*c*d^2*(-d*x+c)^(1/2)+3*C*arctanh(1/2*(-d*x+c)^(1/2)*2^(1 
/2)/c^(1/2))*2^(1/2)*c^2*d*(-d*x+c)^(1/2)-3*D*arctanh(1/2*(-d*x+c)^(1/2)*2 
^(1/2)/c^(1/2))*2^(1/2)*c^3*(-d*x+c)^(1/2)+4*D*c^(3/2)*d^2*x^2+12*C*c^(3/2 
)*d^2*x+4*D*c^(5/2)*d*x-6*A*d^3*c^(1/2)-6*B*c^(3/2)*d^2-18*C*c^(5/2)*d-14* 
D*c^(7/2))/(d*x+c)^(1/2)/(-d*x+c)/d^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 475, normalized size of antiderivative = 2.31 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (D c^{5} - C c^{4} d + B c^{3} d^{2} - A c^{2} d^{3} - {\left (D c^{3} d^{2} - C c^{2} d^{3} + B c d^{4} - A d^{5}\right )} x^{2}\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 4 \, {\left (2 \, D c^{2} d^{2} x^{2} - 7 \, D c^{4} - 9 \, C c^{3} d - 3 \, B c^{2} d^{2} - 3 \, A c d^{3} + 2 \, {\left (D c^{3} d + 3 \, C c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{12 \, {\left (c^{2} d^{6} x^{2} - c^{4} d^{4}\right )}}, \frac {3 \, \sqrt {2} {\left (D c^{5} - C c^{4} d + B c^{3} d^{2} - A c^{2} d^{3} - {\left (D c^{3} d^{2} - C c^{2} d^{3} + B c d^{4} - A d^{5}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (2 \, D c^{2} d^{2} x^{2} - 7 \, D c^{4} - 9 \, C c^{3} d - 3 \, B c^{2} d^{2} - 3 \, A c d^{3} + 2 \, {\left (D c^{3} d + 3 \, C c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{6 \, {\left (c^{2} d^{6} x^{2} - c^{4} d^{4}\right )}}\right ] \] Input:

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

Output:

[1/12*(3*sqrt(2)*(D*c^5 - C*c^4*d + B*c^3*d^2 - A*c^2*d^3 - (D*c^3*d^2 - C 
*c^2*d^3 + B*c*d^4 - A*d^5)*x^2)*sqrt(c)*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)) + 4*(2*D*c^2*d^2*x^2 - 7*D*c^4 - 9*C*c^3*d - 3*B*c^2*d^2 - 3*A*c*d 
^3 + 2*(D*c^3*d + 3*C*c^2*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^2 
*d^6*x^2 - c^4*d^4), 1/6*(3*sqrt(2)*(D*c^5 - C*c^4*d + B*c^3*d^2 - A*c^2*d 
^3 - (D*c^3*d^2 - C*c^2*d^3 + B*c*d^4 - A*d^5)*x^2)*sqrt(-c)*arctan(1/2*sq 
rt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) + 2*(2*D* 
c^2*d^2*x^2 - 7*D*c^4 - 9*C*c^3*d - 3*B*c^2*d^2 - 3*A*c*d^3 + 2*(D*c^3*d + 
 3*C*c^2*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^2*d^6*x^2 - c^4*d^ 
4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c d^{3}} - \frac {6 \, {\left (D c^{3} + C c^{2} d + B c d^{2} + A d^{3}\right )}}{\sqrt {-d x + c} c d^{3}} + \frac {4 \, {\left ({\left (-d x + c\right )}^{\frac {3}{2}} D d^{6} - 3 \, \sqrt {-d x + c} D c d^{6} - 3 \, \sqrt {-d x + c} C d^{7}\right )}}{d^{9}}}{6 \, d} \] Input:

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

Output:

-1/6*(3*sqrt(2)*(D*c^3 - C*c^2*d + B*c*d^2 - A*d^3)*arctan(1/2*sqrt(2)*sqr 
t(-d*x + c)/sqrt(-c))/(sqrt(-c)*c*d^3) - 6*(D*c^3 + C*c^2*d + B*c*d^2 + A* 
d^3)/(sqrt(-d*x + c)*c*d^3) + 4*((-d*x + c)^(3/2)*D*d^6 - 3*sqrt(-d*x + c) 
*D*c*d^6 - 3*sqrt(-d*x + c)*C*d^7)/d^9)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{2}-3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c d -3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{2}+3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c d +12 a c \,d^{2}+12 b \,c^{2} d +64 c^{4}-32 c^{3} d x -8 c^{2} d^{2} x^{2}}{12 \sqrt {-d x +c}\, c^{2} d^{3}} \] Input:

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

Output:

(3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a*d* 
*2 - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))* 
b*c*d - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2 
))*a*d**2 + 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sq 
rt(2))*b*c*d + 12*a*c*d**2 + 12*b*c**2*d + 64*c**4 - 32*c**3*d*x - 8*c**2* 
d**2*x**2)/(12*sqrt(c - d*x)*c**2*d**3)