\(\int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx\) [217]

Optimal result

Integrand size = 41, antiderivative size = 216 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{2 c d^4 (c+d x)^{3/2}}-\frac {2 (3 C d-5 c D) \sqrt {c^2-d^2 x^2}}{3 d^4 \sqrt {c+d x}}-\frac {2 D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4}+\frac {\left (7 c^2 C d-3 B c d^2-A d^3-11 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{2 \sqrt {2} c^{3/2} d^4} \] Output:

-1/2*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(1/2)/c/d^4/(d*x+c)^(3/2 
)-2/3*(3*C*d-5*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+1/4*(-A*d^3-3*B*c*d^2+7*C*c^2*d-11*D*c^3)*ar 
ctanh(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 = 0.97 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.75 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (-3 A d^3+19 c^3 D-3 c^2 d (5 C-4 D x)+c d^2 (3 B-4 x (3 C+D x))\right )}{(c+d x)^{3/2}}-3 \sqrt {2} \left (-7 c^2 C d+3 B c d^2+A d^3+11 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{12 c^{3/2} d^4} \] Input:

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

Output:

((2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(-3*A*d^3 + 19*c^3*D - 3*c^2*d*(5*C - 4*D* 
x) + c*d^2*(3*B - 4*x*(3*C + D*x))))/(c + d*x)^(3/2) - 3*Sqrt[2]*(-7*c^2*C 
*d + 3*B*c*d^2 + A*d^3 + 11*c^3*D)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x]) 
/Sqrt[c^2 - d^2*x^2]])/(12*c^(3/2)*d^4)
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2170, 27, 2170, 27, 671, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {-3 d^2 \left (\frac {\left (-A d^3-3 B c d^2-11 c^3 D+7 c^2 C d\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{4 c}+\frac {\sqrt {c^2-d^2 x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (c+d x)^{3/2}}\right )-\frac {2 d \sqrt {c^2-d^2 x^2} (3 C d-5 c D)}{\sqrt {c+d x}}}{3 d^5}-\frac {2 D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {-3 d^2 \left (\frac {d \left (-A d^3-3 B c d^2-11 c^3 D+7 c^2 C d\right ) \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{2 c}+\frac {\sqrt {c^2-d^2 x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (c+d x)^{3/2}}\right )-\frac {2 d \sqrt {c^2-d^2 x^2} (3 C d-5 c D)}{\sqrt {c+d x}}}{3 d^5}-\frac {2 D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-3 d^2 \left (\frac {\sqrt {c^2-d^2 x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (c+d x)^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right ) \left (-A d^3-3 B c d^2-11 c^3 D+7 c^2 C d\right )}{2 \sqrt {2} c^{3/2} d}\right )-\frac {2 d \sqrt {c^2-d^2 x^2} (3 C d-5 c D)}{\sqrt {c+d x}}}{3 d^5}-\frac {2 D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4}\)

Input:

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

Output:

(-2*D*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2])/(3*d^4) + ((-2*d*(3*C*d - 5*c*D)* 
Sqrt[c^2 - d^2*x^2])/Sqrt[c + d*x] - 3*d^2*(((c^2*C*d - B*c*d^2 + A*d^3 - 
c^3*D)*Sqrt[c^2 - d^2*x^2])/(2*c*d*(c + d*x)^(3/2)) - ((7*c^2*C*d - 3*B*c* 
d^2 - A*d^3 - 11*c^3*D)*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[ 
c + d*x])])/(2*Sqrt[2]*c^(3/2)*d)))/(3*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 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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(377\) vs. \(2(184)=368\).

Time = 0.36 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.75

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (3 A \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, d^{4} x +9 B \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c \,d^{3} x -21 C \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2} d^{2} x +33 D \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3} d x +8 D c^{\frac {3}{2}} d^{2} x^{2} \sqrt {-d x +c}+3 A \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c \,d^{3}+9 B \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2} d^{2}-21 C \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3} d +24 C \sqrt {-d x +c}\, c^{\frac {3}{2}} d^{2} x +33 D \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{4}-24 D \sqrt {-d x +c}\, c^{\frac {5}{2}} d x +6 A \sqrt {-d x +c}\, \sqrt {c}\, d^{3}-6 B \sqrt {-d x +c}\, c^{\frac {3}{2}} d^{2}+30 C \sqrt {-d x +c}\, c^{\frac {5}{2}} d -38 D \sqrt {-d x +c}\, c^{\frac {7}{2}}\right )}{12 c^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} \sqrt {-d x +c}\, d^{4}}\)	\(378\)

Input:

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

Output:

-1/12*(-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^4*x+9*B*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^ 
(1/2)*c*d^3*x-21*C*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^2 
*d^2*x+33*D*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^3*d*x+8* 
D*c^(3/2)*d^2*x^2*(-d*x+c)^(1/2)+3*A*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^ 
(1/2))*2^(1/2)*c*d^3+9*B*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/ 
2)*c^2*d^2-21*C*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^3*d+ 
24*C*(-d*x+c)^(1/2)*c^(3/2)*d^2*x+33*D*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/ 
c^(1/2))*2^(1/2)*c^4-24*D*(-d*x+c)^(1/2)*c^(5/2)*d*x+6*A*(-d*x+c)^(1/2)*c^ 
(1/2)*d^3-6*B*(-d*x+c)^(1/2)*c^(3/2)*d^2+30*C*(-d*x+c)^(1/2)*c^(5/2)*d-38* 
D*(-d*x+c)^(1/2)*c^(7/2))/(d*x+c)^(3/2)/(-d*x+c)^(1/2)/d^4
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 563, normalized size of antiderivative = 2.61 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (11 \, D c^{5} - 7 \, C c^{4} d + 3 \, B c^{3} d^{2} + A c^{2} d^{3} + {\left (11 \, D c^{3} d^{2} - 7 \, C c^{2} d^{3} + 3 \, B c d^{4} + A d^{5}\right )} x^{2} + 2 \, {\left (11 \, D c^{4} d - 7 \, C c^{3} d^{2} + 3 \, B c^{2} d^{3} + A c d^{4}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, {\left (4 \, D c^{2} d^{2} x^{2} - 19 \, D c^{4} + 15 \, C c^{3} d - 3 \, B c^{2} d^{2} + 3 \, A c d^{3} - 12 \, {\left (D c^{3} d - C c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{24 \, {\left (c^{2} d^{6} x^{2} + 2 \, c^{3} d^{5} x + c^{4} d^{4}\right )}}, \frac {3 \, \sqrt {2} {\left (11 \, D c^{5} - 7 \, C c^{4} d + 3 \, B c^{3} d^{2} + A c^{2} d^{3} + {\left (11 \, D c^{3} d^{2} - 7 \, C c^{2} d^{3} + 3 \, B c d^{4} + A d^{5}\right )} x^{2} + 2 \, {\left (11 \, D c^{4} d - 7 \, C c^{3} d^{2} + 3 \, B c^{2} d^{3} + A c d^{4}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) - 2 \, {\left (4 \, D c^{2} d^{2} x^{2} - 19 \, D c^{4} + 15 \, C c^{3} d - 3 \, B c^{2} d^{2} + 3 \, A c d^{3} - 12 \, {\left (D c^{3} d - C c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{12 \, {\left (c^{2} d^{6} x^{2} + 2 \, c^{3} d^{5} x + c^{4} d^{4}\right )}}\right ] \] Input:

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

Output:

[1/24*(3*sqrt(2)*(11*D*c^5 - 7*C*c^4*d + 3*B*c^3*d^2 + A*c^2*d^3 + (11*D*c 
^3*d^2 - 7*C*c^2*d^3 + 3*B*c*d^4 + A*d^5)*x^2 + 2*(11*D*c^4*d - 7*C*c^3*d^ 
2 + 3*B*c^2*d^3 + A*c*d^4)*x)*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*(4*D*c^2*d^2*x^2 - 19*D*c^4 + 15*C*c^3*d - 3*B*c^2*d^2 + 3*A*c*d^ 
3 - 12*(D*c^3*d - C*c^2*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^2*d 
^6*x^2 + 2*c^3*d^5*x + c^4*d^4), 1/12*(3*sqrt(2)*(11*D*c^5 - 7*C*c^4*d + 3 
*B*c^3*d^2 + A*c^2*d^3 + (11*D*c^3*d^2 - 7*C*c^2*d^3 + 3*B*c*d^4 + A*d^5)* 
x^2 + 2*(11*D*c^4*d - 7*C*c^3*d^2 + 3*B*c^2*d^3 + A*c*d^4)*x)*sqrt(-c)*arc 
tan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) 
 - 2*(4*D*c^2*d^2*x^2 - 19*D*c^4 + 15*C*c^3*d - 3*B*c^2*d^2 + 3*A*c*d^3 - 
12*(D*c^3*d - C*c^2*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^2*d^6*x 
^2 + 2*c^3*d^5*x + c^4*d^4)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Maxima [F]

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

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {8 \, {\left (-d x + c\right )}^{\frac {3}{2}} D + 24 \, \sqrt {-d x + c} D c - 24 \, \sqrt {-d x + c} C d + \frac {3 \, \sqrt {2} {\left (11 \, D c^{3} - 7 \, C c^{2} d + 3 \, B c d^{2} + A d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {6 \, {\left (\sqrt {-d x + c} D c^{3} - \sqrt {-d x + c} C c^{2} d + \sqrt {-d x + c} B c d^{2} - \sqrt {-d x + c} A d^{3}\right )}}{{\left (d x + c\right )} c}}{12 \, d^{4}} \] Input:

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

Output:

1/12*(8*(-d*x + c)^(3/2)*D + 24*sqrt(-d*x + c)*D*c - 24*sqrt(-d*x + c)*C*d 
 + 3*sqrt(2)*(11*D*c^3 - 7*C*c^2*d + 3*B*c*d^2 + A*d^3)*arctan(1/2*sqrt(2) 
*sqrt(-d*x + c)/sqrt(-c))/(sqrt(-c)*c) + 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))/d^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.75 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \, dx=\frac {-12 \sqrt {-d x +c}\, a c \,d^{2}+12 \sqrt {-d x +c}\, b \,c^{2} d +16 \sqrt {-d x +c}\, c^{4}-16 \sqrt {-d x +c}\, c^{2} d^{2} x^{2}+3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a c \,d^{2}+3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x +9 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} d +9 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x +12 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c^{4}+12 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c^{3} d x -3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a c \,d^{2}-3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x -9 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} d -9 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x -12 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c^{4}-12 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c^{3} d x}{24 c^{2} d^{3} \left (d x +c \right )} \] Input:

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

Output:

( - 12*sqrt(c - d*x)*a*c*d**2 + 12*sqrt(c - d*x)*b*c**2*d + 16*sqrt(c - d* 
x)*c**4 - 16*sqrt(c - d*x)*c**2*d**2*x**2 + 3*sqrt(c)*sqrt(2)*log(sqrt(c - 
 d*x) - sqrt(c)*sqrt(2))*a*c*d**2 + 3*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - 
sqrt(c)*sqrt(2))*a*d**3*x + 9*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)* 
sqrt(2))*b*c**2*d + 9*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2)) 
*b*c*d**2*x + 12*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c**4 
 + 12*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c**3*d*x - 3*sq 
rt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*c*d**2 - 3*sqrt(c)*sq 
rt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**3*x - 9*sqrt(c)*sqrt(2)*lo 
g(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**2*d - 9*sqrt(c)*sqrt(2)*log(sqrt(c 
 - d*x) + sqrt(c)*sqrt(2))*b*c*d**2*x - 12*sqrt(c)*sqrt(2)*log(sqrt(c - d* 
x) + sqrt(c)*sqrt(2))*c**4 - 12*sqrt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c 
)*sqrt(2))*c**3*d*x)/(24*c**2*d**3*(c + d*x))