\(\int \frac {\sqrt {c+d x} (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{5/2}} \, dx\) [231]

Optimal result

Integrand size = 41, antiderivative size = 262 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) \sqrt {c+d x}}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {5 c^2 C d-B c d^2+5 A d^3-c^3 D}{12 c^2 d^4 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}-\frac {\left (3 c^2 C d+B c d^2-5 A d^3+9 c^3 D\right ) \sqrt {c+d x}}{8 c^3 d^4 \sqrt {c^2-d^2 x^2}}+\frac {\left (3 c^2 C d+B c d^2-5 A d^3-7 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{8 \sqrt {2} c^{7/2} d^4} \] Output:

1/3*(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)^(3/2) 
-1/12*(5*A*d^3-B*c*d^2+5*C*c^2*d-D*c^3)/c^2/d^4/(d*x+c)^(1/2)/(-d^2*x^2+c^ 
2)^(1/2)-1/8*(-5*A*d^3+B*c*d^2+3*C*c^2*d+9*D*c^3)*(d*x+c)^(1/2)/c^3/d^4/(- 
d^2*x^2+c^2)^(1/2)+1/16*(-5*A*d^3+B*c*d^2+3*C*c^2*d-7*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^(7/2)/d^4
 

Mathematica [A] (verified)

Time = 3.91 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (17 c^5 D+15 A d^5 x^2-c d^4 x (10 A+3 B x)+c^4 d (11 C+2 D x)+c^2 d^3 (-13 A+x (2 B-9 C x))-c^3 d^2 (7 B+x (10 C+27 D x))\right )}{(c-d x)^2 (c+d x)^{3/2}}-3 \sqrt {2} \left (-3 c^2 C d-B c d^2+5 A d^3+7 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{48 c^{7/2} d^4} \] Input:

Integrate[(Sqrt[c + d*x]*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2), 
x]
 

Output:

((-2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(17*c^5*D + 15*A*d^5*x^2 - c*d^4*x*(10*A 
+ 3*B*x) + c^4*d*(11*C + 2*D*x) + c^2*d^3*(-13*A + x*(2*B - 9*C*x)) - c^3* 
d^2*(7*B + x*(10*C + 27*D*x))))/((c - d*x)^2*(c + d*x)^(3/2)) - 3*Sqrt[2]* 
(-3*c^2*C*d - B*c*d^2 + 5*A*d^3 + 7*c^3*D)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c 
 + d*x])/Sqrt[c^2 - d^2*x^2]])/(48*c^(7/2)*d^4)
 

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 284, 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, 671, 467, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {\sqrt {c+d x} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {2 \int \frac {d \left (-5 D c^3+C d c^2+B d^2 c+6 d (C d-c D) x c-5 A d^3\right )}{2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}dx}{d^4}+\frac {12 c D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}}{6 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {c+d x} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\int \frac {-5 D c^3+C d c^2+B d^2 c+6 d (C d-c D) x c-5 A d^3}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}dx}{d^3}+\frac {12 c D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}}{6 c}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {\sqrt {c+d x} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\frac {3 \left (-5 A d^3+B c d^2-7 c^3 D+3 c^2 C d\right ) \int \frac {\sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{4 c}+\frac {5 A d^3-B c d^2+c^3 (-D)+5 c^2 C d}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}}{d^3}+\frac {12 c D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}}{6 c}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {\sqrt {c+d x} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\frac {3 \left (-5 A d^3+B c d^2-7 c^3 D+3 c^2 C d\right ) \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{2 c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}+\frac {5 A d^3-B c d^2+c^3 (-D)+5 c^2 C d}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}}{d^3}+\frac {12 c D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}}{6 c}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {\sqrt {c+d x} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\frac {3 \left (-5 A d^3+B c d^2-7 c^3 D+3 c^2 C d\right ) \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}+\frac {5 A d^3-B c d^2+c^3 (-D)+5 c^2 C d}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}}{d^3}+\frac {12 c D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}}{6 c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {c+d x} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {\frac {3 \left (\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} c^{3/2} d}\right ) \left (-5 A d^3+B c d^2-7 c^3 D+3 c^2 C d\right )}{4 c}+\frac {5 A d^3-B c d^2+c^3 (-D)+5 c^2 C d}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}}{d^3}+\frac {12 c D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}}{6 c}\)

Input:

Int[(Sqrt[c + d*x]*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2),x]
 

Output:

((c^2*C*d + B*c*d^2 + A*d^3 + c^3*D)*Sqrt[c + d*x])/(3*c*d^4*(c^2 - d^2*x^ 
2)^(3/2)) - ((12*c*D*Sqrt[c + d*x])/(d^4*Sqrt[c^2 - d^2*x^2]) + ((5*c^2*C* 
d - B*c*d^2 + 5*A*d^3 - c^3*D)/(2*c*d*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2]) + 
 (3*(3*c^2*C*d + B*c*d^2 - 5*A*d^3 - 7*c^3*D)*(Sqrt[c + d*x]/(c*d*Sqrt[c^2 
 - d^2*x^2]) - ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x]) 
]/(Sqrt[2]*c^(3/2)*d)))/(4*c))/d^3)/(6*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 467

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

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 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, 
e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 0]
 

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 = 454, normalized size of antiderivative = 1.73

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-15 A \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, d^{5} x^{2}+3 B \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c \,d^{4} x^{2}+9 C \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2} d^{3} x^{2}-21 D \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3} d^{2} x^{2}+15 A \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2} d^{3}+30 A \sqrt {c}\, d^{5} x^{2}-3 B \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3} d^{2}-6 B \,c^{\frac {3}{2}} d^{4} x^{2}-9 C \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{4} d -18 C \,c^{\frac {5}{2}} d^{3} x^{2}+21 D \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{5}-54 D c^{\frac {7}{2}} d^{2} x^{2}-20 A \,c^{\frac {3}{2}} d^{4} x +4 B \,c^{\frac {5}{2}} d^{3} x -20 C \,c^{\frac {7}{2}} d^{2} x +4 D c^{\frac {9}{2}} d x -26 A \,c^{\frac {5}{2}} d^{3}-14 B \,c^{\frac {7}{2}} d^{2}+22 C \,c^{\frac {9}{2}} d +34 D c^{\frac {11}{2}}\right )}{48 \left (d x +c \right )^{\frac {3}{2}} \left (-d x +c \right )^{2} d^{4} c^{\frac {7}{2}}}\)	\(454\)

Input:

int((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-1/48*(-d^2*x^2+c^2)^(1/2)*(-15*A*(-d*x+c)^(1/2)*arctanh(1/2*(-d*x+c)^(1/2 
)*2^(1/2)/c^(1/2))*2^(1/2)*d^5*x^2+3*B*(-d*x+c)^(1/2)*arctanh(1/2*(-d*x+c) 
^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c*d^4*x^2+9*C*(-d*x+c)^(1/2)*arctanh(1/2*( 
-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^2*d^3*x^2-21*D*(-d*x+c)^(1/2)*arc 
tanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^3*d^2*x^2+15*A*(-d*x+c) 
^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^2*d^3+30*A*c^ 
(1/2)*d^5*x^2-3*B*(-d*x+c)^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2 
))*2^(1/2)*c^3*d^2-6*B*c^(3/2)*d^4*x^2-9*C*(-d*x+c)^(1/2)*arctanh(1/2*(-d* 
x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^4*d-18*C*c^(5/2)*d^3*x^2+21*D*(-d*x+ 
c)^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^5-54*D*c^(7 
/2)*d^2*x^2-20*A*c^(3/2)*d^4*x+4*B*c^(5/2)*d^3*x-20*C*c^(7/2)*d^2*x+4*D*c^ 
(9/2)*d*x-26*A*c^(5/2)*d^3-14*B*c^(7/2)*d^2+22*C*c^(9/2)*d+34*D*c^(11/2))/ 
(d*x+c)^(3/2)/(-d*x+c)^2/d^4/c^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 677, normalized size of antiderivative = 2.58 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (7 \, D c^{7} - 3 \, C c^{6} d - B c^{5} d^{2} + 5 \, A c^{4} d^{3} + {\left (7 \, D c^{3} d^{4} - 3 \, C c^{2} d^{5} - B c d^{6} + 5 \, A d^{7}\right )} x^{4} - 2 \, {\left (7 \, D c^{5} d^{2} - 3 \, C c^{4} d^{3} - B c^{3} d^{4} + 5 \, A c^{2} d^{5}\right )} x^{2}\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, {\left (17 \, D c^{6} + 11 \, C c^{5} d - 7 \, B c^{4} d^{2} - 13 \, A c^{3} d^{3} - 3 \, {\left (9 \, D c^{4} d^{2} + 3 \, C c^{3} d^{3} + B c^{2} d^{4} - 5 \, A c d^{5}\right )} x^{2} + 2 \, {\left (D c^{5} d - 5 \, C c^{4} d^{2} + B c^{3} d^{3} - 5 \, A c^{2} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{96 \, {\left (c^{4} d^{8} x^{4} - 2 \, c^{6} d^{6} x^{2} + c^{8} d^{4}\right )}}, \frac {3 \, \sqrt {2} {\left (7 \, D c^{7} - 3 \, C c^{6} d - B c^{5} d^{2} + 5 \, A c^{4} d^{3} + {\left (7 \, D c^{3} d^{4} - 3 \, C c^{2} d^{5} - B c d^{6} + 5 \, A d^{7}\right )} x^{4} - 2 \, {\left (7 \, D c^{5} d^{2} - 3 \, C c^{4} d^{3} - B c^{3} d^{4} + 5 \, A c^{2} d^{5}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) - 2 \, {\left (17 \, D c^{6} + 11 \, C c^{5} d - 7 \, B c^{4} d^{2} - 13 \, A c^{3} d^{3} - 3 \, {\left (9 \, D c^{4} d^{2} + 3 \, C c^{3} d^{3} + B c^{2} d^{4} - 5 \, A c d^{5}\right )} x^{2} + 2 \, {\left (D c^{5} d - 5 \, C c^{4} d^{2} + B c^{3} d^{3} - 5 \, A c^{2} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{48 \, {\left (c^{4} d^{8} x^{4} - 2 \, c^{6} d^{6} x^{2} + c^{8} d^{4}\right )}}\right ] \] Input:

integrate((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori 
thm="fricas")
 

Output:

[1/96*(3*sqrt(2)*(7*D*c^7 - 3*C*c^6*d - B*c^5*d^2 + 5*A*c^4*d^3 + (7*D*c^3 
*d^4 - 3*C*c^2*d^5 - B*c*d^6 + 5*A*d^7)*x^4 - 2*(7*D*c^5*d^2 - 3*C*c^4*d^3 
 - B*c^3*d^4 + 5*A*c^2*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*(17*D*c^6 + 11*C*c^5*d - 7*B*c^4*d^2 - 13*A*c^3*d^3 - 3*(9*D*c 
^4*d^2 + 3*C*c^3*d^3 + B*c^2*d^4 - 5*A*c*d^5)*x^2 + 2*(D*c^5*d - 5*C*c^4*d 
^2 + B*c^3*d^3 - 5*A*c^2*d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^4* 
d^8*x^4 - 2*c^6*d^6*x^2 + c^8*d^4), 1/48*(3*sqrt(2)*(7*D*c^7 - 3*C*c^6*d - 
 B*c^5*d^2 + 5*A*c^4*d^3 + (7*D*c^3*d^4 - 3*C*c^2*d^5 - B*c*d^6 + 5*A*d^7) 
*x^4 - 2*(7*D*c^5*d^2 - 3*C*c^4*d^3 - B*c^3*d^4 + 5*A*c^2*d^5)*x^2)*sqrt(- 
c)*arctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + 
 c^2)) - 2*(17*D*c^6 + 11*C*c^5*d - 7*B*c^4*d^2 - 13*A*c^3*d^3 - 3*(9*D*c^ 
4*d^2 + 3*C*c^3*d^3 + B*c^2*d^4 - 5*A*c*d^5)*x^2 + 2*(D*c^5*d - 5*C*c^4*d^ 
2 + B*c^3*d^3 - 5*A*c^2*d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^4*d 
^8*x^4 - 2*c^6*d^6*x^2 + c^8*d^4)]
 

Sympy [F]

\[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\sqrt {c + d x} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((d*x+c)**(1/2)*(D*x**3+C*x**2+B*x+A)/(-d**2*x**2+c**2)**(5/2),x)
 

Output:

Integral(sqrt(c + d*x)*(A + B*x + C*x**2 + D*x**3)/(-(-c + d*x)*(c + d*x)) 
**(5/2), x)
 

Maxima [F]

\[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} \sqrt {d x + c}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),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)^(5/2), 
x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left (7 \, D c^{3} - 3 \, C c^{2} d - B c d^{2} + 5 \, A d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{3} d^{3}} + \frac {6 \, {\left (\sqrt {-d x + c} D c^{3} - \sqrt {-d x + c} C c^{2} d + \sqrt {-d x + c} B c d^{2} - \sqrt {-d x + c} A d^{3}\right )}}{{\left (d x + c\right )} c^{3} d^{3}} - \frac {8 \, {\left (6 \, {\left (d x - c\right )} D c^{3} + D c^{4} + 3 \, {\left (d x - c\right )} C c^{2} d + C c^{3} d + B c^{2} d^{2} - 3 \, {\left (d x - c\right )} A d^{3} + A c d^{3}\right )}}{{\left (d x - c\right )} \sqrt {-d x + c} c^{3} d^{3}}}{48 \, d} \] Input:

integrate((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori 
thm="giac")
 

Output:

1/48*(3*sqrt(2)*(7*D*c^3 - 3*C*c^2*d - B*c*d^2 + 5*A*d^3)*arctan(1/2*sqrt( 
2)*sqrt(-d*x + c)/sqrt(-c))/(sqrt(-c)*c^3*d^3) + 6*(sqrt(-d*x + c)*D*c^3 - 
 sqrt(-d*x + c)*C*c^2*d + sqrt(-d*x + c)*B*c*d^2 - sqrt(-d*x + c)*A*d^3)/( 
(d*x + c)*c^3*d^3) - 8*(6*(d*x - c)*D*c^3 + D*c^4 + 3*(d*x - c)*C*c^2*d + 
C*c^3*d + B*c^2*d^2 - 3*(d*x - c)*A*d^3 + A*c*d^3)/((d*x - c)*sqrt(-d*x + 
c)*c^3*d^3))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\sqrt {c+d\,x}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c^2-d^2\,x^2\right )}^{5/2}} \,d x \] Input:

int(((c + d*x)^(1/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2),x)
 

Output:

int(((c + d*x)^(1/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.99 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,c^{2} d^{2}-15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{4} x^{2}-3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{3} d +3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c \,d^{3} x^{2}+12 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c^{5}-12 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c^{3} d^{2} x^{2}-15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,c^{2} d^{2}+15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{4} x^{2}+3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{3} d -3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c \,d^{3} x^{2}-12 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c^{5}+12 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c^{3} d^{2} x^{2}+52 a \,c^{3} d^{2}+40 a \,c^{2} d^{3} x -60 a c \,d^{4} x^{2}+28 b \,c^{4} d -8 b \,c^{3} d^{2} x +12 b \,c^{2} d^{3} x^{2}-112 c^{6}+32 c^{5} d x +144 c^{4} d^{2} x^{2}}{96 \sqrt {-d x +c}\, c^{4} d^{3} \left (-d^{2} x^{2}+c^{2}\right )} \] Input:

int((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x)
 

Output:

(15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a*c 
**2*d**2 - 15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sq 
rt(2))*a*d**4*x**2 - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - s 
qrt(c)*sqrt(2))*b*c**3*d + 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d* 
x) - sqrt(c)*sqrt(2))*b*c*d**3*x**2 + 12*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log 
(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c**5 - 12*sqrt(c)*sqrt(c - d*x)*sqrt(2)* 
log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c**3*d**2*x**2 - 15*sqrt(c)*sqrt(c - 
d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*c**2*d**2 + 15*sqrt(c) 
*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**4*x**2 + 
3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c** 
3*d - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2)) 
*b*c*d**3*x**2 - 12*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt 
(c)*sqrt(2))*c**5 + 12*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + s 
qrt(c)*sqrt(2))*c**3*d**2*x**2 + 52*a*c**3*d**2 + 40*a*c**2*d**3*x - 60*a* 
c*d**4*x**2 + 28*b*c**4*d - 8*b*c**3*d**2*x + 12*b*c**2*d**3*x**2 - 112*c* 
*6 + 32*c**5*d*x + 144*c**4*d**2*x**2)/(96*sqrt(c - d*x)*c**4*d**3*(c**2 - 
 d**2*x**2))