\(\int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} (c^2-d^2 x^2)^{3/2}} \, dx\) [225]

Optimal result

Integrand size = 41, antiderivative size = 264 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {c^2 C d-B c d^2+A d^3-c^3 D}{4 c d^4 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}+\frac {11 c^2 C d-3 B c d^2-5 A d^3-19 c^3 D}{16 c^2 d^4 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}-\frac {\left (c^2 C d-9 B c d^2-15 A d^3-25 c^3 D\right ) \sqrt {c+d x}}{32 c^3 d^4 \sqrt {c^2-d^2 x^2}}+\frac {\left (c^2 C d-9 B c d^2-15 A d^3+39 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{32 \sqrt {2} c^{7/2} d^4} \] Output:

-1/4*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/c/d^4/(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(1/2 
)+1/16*(-5*A*d^3-3*B*c*d^2+11*C*c^2*d-19*D*c^3)/c^2/d^4/(d*x+c)^(1/2)/(-d^ 
2*x^2+c^2)^(1/2)-1/32*(-15*A*d^3-9*B*c*d^2+C*c^2*d-25*D*c^3)*(d*x+c)^(1/2) 
/c^3/d^4/(-d^2*x^2+c^2)^(1/2)+1/64*(-15*A*d^3-9*B*c*d^2+C*c^2*d+39*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 = 3.58 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (5 c^5 D-15 A d^5 x^2-c d^4 x (20 A+9 B x)-c^4 d (13 C+12 D x)+c^2 d^3 (3 A+x (-12 B+C x))-c^3 d^2 (11 B+5 x (4 C+5 D x))\right )}{(c-d x) (c+d x)^{5/2}}+\sqrt {2} \left (c^2 C d-9 B c d^2-15 A d^3+39 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{64 c^{7/2} d^4} \] Input:

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

Output:

((-2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(5*c^5*D - 15*A*d^5*x^2 - c*d^4*x*(20*A + 
 9*B*x) - c^4*d*(13*C + 12*D*x) + c^2*d^3*(3*A + x*(-12*B + C*x)) - c^3*d^ 
2*(11*B + 5*x*(4*C + 5*D*x))))/((c - d*x)*(c + d*x)^(5/2)) + Sqrt[2]*(c^2* 
C*d - 9*B*c*d^2 - 15*A*d^3 + 39*c^3*D)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d 
*x])/Sqrt[c^2 - d^2*x^2]])/(64*c^(7/2)*d^4)
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {2170, 27, 2170, 27, 671, 470, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {\frac {1}{3} d^2 \left (-\frac {\left (-15 A d^3-9 B c d^2+39 c^3 D+c^2 C d\right ) \int \frac {1}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}dx}{8 c}-\frac {3 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )+\frac {2 d (C d-3 c D)}{3 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}}{d^5}+\frac {2 D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {\frac {1}{3} d^2 \left (-\frac {\left (-15 A d^3-9 B c d^2+39 c^3 D+c^2 C d\right ) \left (\frac {3 \int \frac {\sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c}-\frac {3 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )+\frac {2 d (C d-3 c D)}{3 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}}{d^5}+\frac {2 D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {\frac {1}{3} d^2 \left (-\frac {\left (-15 A d^3-9 B c d^2+39 c^3 D+c^2 C d\right ) \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{2 c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c}-\frac {3 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )+\frac {2 d (C d-3 c D)}{3 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}}{d^5}+\frac {2 D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {\frac {1}{3} d^2 \left (-\frac {\left (-15 A d^3-9 B c d^2+39 c^3 D+c^2 C d\right ) \left (\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c}-\frac {3 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )+\frac {2 d (C d-3 c D)}{3 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}}{d^5}+\frac {2 D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{3} d^2 \left (-\frac {\left (\frac {3 \left (\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} c^{3/2} d}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right ) \left (-15 A d^3-9 B c d^2+39 c^3 D+c^2 C d\right )}{8 c}-\frac {3 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\right )+\frac {2 d (C d-3 c D)}{3 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}}{d^5}+\frac {2 D \sqrt {c+d x}}{d^4 \sqrt {c^2-d^2 x^2}}\)

Input:

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

Output:

(2*D*Sqrt[c + d*x])/(d^4*Sqrt[c^2 - d^2*x^2]) + ((2*d*(C*d - 3*c*D))/(3*Sq 
rt[c + d*x]*Sqrt[c^2 - d^2*x^2]) + (d^2*((-3*(c^2*C*d - B*c*d^2 + A*d^3 - 
c^3*D))/(4*c*d*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2]) - ((c^2*C*d - 9*B*c*d^ 
2 - 15*A*d^3 + 39*c^3*D)*(-1/2*1/(c*d*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2]) + 
 (3*(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)))/(8*c)))/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 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 470

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 
2*p + 2)/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; 
 FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + 
 p + 1, 0] && 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 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. \(601\) vs. \(2(232)=464\).

Time = 0.36 (sec) , antiderivative size = 602, normalized size of antiderivative = 2.28

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-40 A \,c^{\frac {3}{2}} d^{4} x +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}+2 C \,c^{\frac {5}{2}} d^{3} x^{2}-50 D c^{\frac {7}{2}} 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}+9 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}-C \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{4} d -40 C \,c^{\frac {7}{2}} d^{2} x -24 D c^{\frac {9}{2}} d x -39 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}+30 A \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c \,d^{4} x +18 B \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 C \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 -78 D \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{4} d x +9 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}-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}-30 A \sqrt {c}\, d^{5} x^{2}+6 A \,c^{\frac {5}{2}} d^{3}-22 B \,c^{\frac {7}{2}} d^{2}-26 C \,c^{\frac {9}{2}} d -18 B \,c^{\frac {3}{2}} d^{4} x^{2}-39 D \sqrt {-d x +c}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{5}-24 B \,c^{\frac {5}{2}} d^{3} x +10 D c^{\frac {11}{2}}\right )}{64 \left (d x +c \right )^{\frac {5}{2}} \left (-d x +c \right ) d^{4} c^{\frac {7}{2}}}\)	\(602\)

Input:

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

Output:

-1/64/(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2)*(-40*A*c^(3/2)*d^4*x+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+2*C*c^ 
(5/2)*d^3*x^2-50*D*c^(7/2)*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+9*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-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-40*C*c^(7/2)*d^2*x-24*D*c^( 
9/2)*d*x-39*D*(-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*x^2+30*A*(-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+18*B*(-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*C*(-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*x-78*D*(-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*x+9*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-C*(-d*x+c)^(1/2)*arct 
anh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^2*d^3*x^2-30*A*c^(1/2)*d 
^5*x^2+6*A*c^(5/2)*d^3-22*B*c^(7/2)*d^2-26*C*c^(9/2)*d-18*B*c^(3/2)*d^4*x^ 
2-39*D*(-d*x+c)^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)* 
c^5-24*B*c^(5/2)*d^3*x+10*D*c^(11/2))/(-d*x+c)/d^4/c^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 767, normalized size of antiderivative = 2.91 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\left [\frac {\sqrt {2} {\left (39 \, D c^{7} + C c^{6} d - 9 \, B c^{5} d^{2} - 15 \, A c^{4} d^{3} - {\left (39 \, D c^{3} d^{4} + C c^{2} d^{5} - 9 \, B c d^{6} - 15 \, A d^{7}\right )} x^{4} - 2 \, {\left (39 \, D c^{4} d^{3} + C c^{3} d^{4} - 9 \, B c^{2} d^{5} - 15 \, A c d^{6}\right )} x^{3} + 2 \, {\left (39 \, D c^{6} d + C c^{5} d^{2} - 9 \, B c^{4} d^{3} - 15 \, A c^{3} 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 (5 \, D c^{6} - 13 \, C c^{5} d - 11 \, B c^{4} d^{2} + 3 \, A c^{3} d^{3} - {\left (25 \, D c^{4} d^{2} - C c^{3} d^{3} + 9 \, B c^{2} d^{4} + 15 \, A c d^{5}\right )} x^{2} - 4 \, {\left (3 \, D c^{5} d + 5 \, C c^{4} d^{2} + 3 \, 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}}{128 \, {\left (c^{4} d^{8} x^{4} + 2 \, c^{5} d^{7} x^{3} - 2 \, c^{7} d^{5} x - c^{8} d^{4}\right )}}, \frac {\sqrt {2} {\left (39 \, D c^{7} + C c^{6} d - 9 \, B c^{5} d^{2} - 15 \, A c^{4} d^{3} - {\left (39 \, D c^{3} d^{4} + C c^{2} d^{5} - 9 \, B c d^{6} - 15 \, A d^{7}\right )} x^{4} - 2 \, {\left (39 \, D c^{4} d^{3} + C c^{3} d^{4} - 9 \, B c^{2} d^{5} - 15 \, A c d^{6}\right )} x^{3} + 2 \, {\left (39 \, D c^{6} d + C c^{5} d^{2} - 9 \, B c^{4} d^{3} - 15 \, A c^{3} 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 (5 \, D c^{6} - 13 \, C c^{5} d - 11 \, B c^{4} d^{2} + 3 \, A c^{3} d^{3} - {\left (25 \, D c^{4} d^{2} - C c^{3} d^{3} + 9 \, B c^{2} d^{4} + 15 \, A c d^{5}\right )} x^{2} - 4 \, {\left (3 \, D c^{5} d + 5 \, C c^{4} d^{2} + 3 \, 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}}{64 \, {\left (c^{4} d^{8} x^{4} + 2 \, c^{5} d^{7} x^{3} - 2 \, c^{7} d^{5} x - c^{8} 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)^(3/2),x, algori 
thm="fricas")
 

Output:

[1/128*(sqrt(2)*(39*D*c^7 + C*c^6*d - 9*B*c^5*d^2 - 15*A*c^4*d^3 - (39*D*c 
^3*d^4 + C*c^2*d^5 - 9*B*c*d^6 - 15*A*d^7)*x^4 - 2*(39*D*c^4*d^3 + C*c^3*d 
^4 - 9*B*c^2*d^5 - 15*A*c*d^6)*x^3 + 2*(39*D*c^6*d + C*c^5*d^2 - 9*B*c^4*d 
^3 - 15*A*c^3*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 
*(5*D*c^6 - 13*C*c^5*d - 11*B*c^4*d^2 + 3*A*c^3*d^3 - (25*D*c^4*d^2 - C*c^ 
3*d^3 + 9*B*c^2*d^4 + 15*A*c*d^5)*x^2 - 4*(3*D*c^5*d + 5*C*c^4*d^2 + 3*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^5*d^7*x^3 - 2*c^7*d^5*x - c^8*d^4), 1/64*(sqrt(2)*(39*D*c^7 + C*c^6* 
d - 9*B*c^5*d^2 - 15*A*c^4*d^3 - (39*D*c^3*d^4 + C*c^2*d^5 - 9*B*c*d^6 - 1 
5*A*d^7)*x^4 - 2*(39*D*c^4*d^3 + C*c^3*d^4 - 9*B*c^2*d^5 - 15*A*c*d^6)*x^3 
 + 2*(39*D*c^6*d + C*c^5*d^2 - 9*B*c^4*d^3 - 15*A*c^3*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*(5*D*c^6 - 13*C*c^5*d - 11*B*c^4*d^2 + 3*A*c^3*d^3 - (25*D*c^4*d^2 - 
C*c^3*d^3 + 9*B*c^2*d^4 + 15*A*c*d^5)*x^2 - 4*(3*D*c^5*d + 5*C*c^4*d^2 + 3 
*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^5*d^7*x^3 - 2*c^7*d^5*x - c^8*d^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {\frac {\sqrt {2} {\left (39 \, D c^{3} + C c^{2} d - 9 \, B c d^{2} - 15 \, A d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{3} d^{3}} - \frac {16 \, {\left (D c^{3} + C c^{2} d + B c d^{2} + A d^{3}\right )}}{\sqrt {-d x + c} c^{3} d^{3}} - \frac {2 \, {\left (17 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{3} - 30 \, \sqrt {-d x + c} D c^{4} - 9 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c^{2} d + 14 \, \sqrt {-d x + c} C c^{3} d + {\left (-d x + c\right )}^{\frac {3}{2}} B c d^{2} + 2 \, \sqrt {-d x + c} B c^{2} d^{2} + 7 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d^{3} - 18 \, \sqrt {-d x + c} A c d^{3}\right )}}{{\left (d x + c\right )}^{2} c^{3} d^{3}}}{64 \, d} \] Input:

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

Output:

-1/64*(sqrt(2)*(39*D*c^3 + C*c^2*d - 9*B*c*d^2 - 15*A*d^3)*arctan(1/2*sqrt 
(2)*sqrt(-d*x + c)/sqrt(-c))/(sqrt(-c)*c^3*d^3) - 16*(D*c^3 + C*c^2*d + B* 
c*d^2 + A*d^3)/(sqrt(-d*x + c)*c^3*d^3) - 2*(17*(-d*x + c)^(3/2)*D*c^3 - 3 
0*sqrt(-d*x + c)*D*c^4 - 9*(-d*x + c)^(3/2)*C*c^2*d + 14*sqrt(-d*x + c)*C* 
c^3*d + (-d*x + c)^(3/2)*B*c*d^2 + 2*sqrt(-d*x + c)*B*c^2*d^2 + 7*(-d*x + 
c)^(3/2)*A*d^3 - 18*sqrt(-d*x + c)*A*c*d^3)/((d*x + c)^2*c^3*d^3))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 728, normalized size of antiderivative = 2.76 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int((D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(3/2),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 + 30*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sq 
rt(2))*a*c*d**3*x + 15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - s 
qrt(c)*sqrt(2))*a*d**4*x**2 + 9*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - 
 d*x) - sqrt(c)*sqrt(2))*b*c**3*d + 18*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(s 
qrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**2*d**2*x + 9*sqrt(c)*sqrt(c - d*x)*sq 
rt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c*d**3*x**2 - 40*sqrt(c)*sqrt 
(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c**5 - 80*sqrt(c)*s 
qrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c**4*d*x - 40*sq 
rt(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 - 30*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqr 
t(c)*sqrt(2))*a*c*d**3*x - 15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d 
*x) + sqrt(c)*sqrt(2))*a*d**4*x**2 - 9*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(s 
qrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**3*d - 18*sqrt(c)*sqrt(c - d*x)*sqrt(2 
)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**2*d**2*x - 9*sqrt(c)*sqrt(c - 
d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c*d**3*x**2 + 40*sqrt( 
c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*c**5 + 80*sq 
rt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*c**4*d*x 
+ 40*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))...