\(\int \frac {(c^2-d^2 x^2)^{3/2} (A+B x+C x^2+D x^3)}{(c+d x)^{5/2}} \, dx\) [207]

Optimal result

Integrand size = 41, antiderivative size = 315 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {4 c \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4 \sqrt {c+d x}}+\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4 (c+d x)^{3/2}}+\frac {2 \left (90 c C d-63 B d^2-113 c^2 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{315 d^4 (c+d x)^{5/2}}-\frac {2 (9 C d-19 c D) \left (c^2-d^2 x^2\right )^{5/2}}{63 d^4 (c+d x)^{3/2}}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}-\frac {4 \sqrt {2} c^{3/2} \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{d^4} \] Output:

4*c*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)^(1/2)+2 
/3*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^(3/2)+2/ 
315*(-63*B*d^2+90*C*c*d-113*D*c^2)*(-d^2*x^2+c^2)^(5/2)/d^4/(d*x+c)^(5/2)- 
2/63*(9*C*d-19*D*c)*(-d^2*x^2+c^2)^(5/2)/d^4/(d*x+c)^(3/2)-2/9*D*(-d^2*x^2 
+c^2)^(5/2)/d^4/(d*x+c)^(1/2)-4*2^(1/2)*c^(3/2)*(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))/d^4
 

Mathematica [A] (verified)

Time = 2.99 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.65 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \left (\frac {\sqrt {c^2-d^2 x^2} \left (-788 c^4 D+4 c^3 d (195 C+59 D x)-6 c^2 d^2 (133 B+x (40 C+23 D x))-d^4 x (105 A+x (63 B+5 x (9 C+7 D x)))+c d^3 (735 A+x (231 B+5 x (27 C+19 D x)))\right )}{\sqrt {c+d x}}+630 \sqrt {2} c^{3/2} \left (-c^2 C d+B c d^2-A d^3+c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )\right )}{315 d^4} \] Input:

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

Output:

(2*((Sqrt[c^2 - d^2*x^2]*(-788*c^4*D + 4*c^3*d*(195*C + 59*D*x) - 6*c^2*d^ 
2*(133*B + x*(40*C + 23*D*x)) - d^4*x*(105*A + x*(63*B + 5*x*(9*C + 7*D*x) 
)) + c*d^3*(735*A + x*(231*B + 5*x*(27*C + 19*D*x)))))/Sqrt[c + d*x] + 630 
*Sqrt[2]*c^(3/2)*(-(c^2*C*d) + B*c*d^2 - A*d^3 + c^3*D)*ArcTanh[(Sqrt[2]*S 
qrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]]))/(315*d^4)
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.90, 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, 672, 466, 466, 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.

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

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {2 \int -\frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((9 C d-19 c D) x^2 d^4+\left (9 B d^2-11 c^2 D\right ) x d^3+\left (9 A d^3-c^3 D\right ) d^2\right )}{2 (c+d x)^{5/2}}dx}{9 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((9 C d-19 c D) x^2 d^4+\left (9 B d^2-11 c^2 D\right ) x d^3+\left (9 A d^3-c^3 D\right ) d^2\right )}{(c+d x)^{5/2}}dx}{9 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int \frac {d^6 \left (-50 D c^3+27 C d c^2-63 A d^3+d \left (-113 D c^2+90 C d c-63 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{2 (c+d x)^{5/2}}dx}{7 d^4}-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 C d-19 c D)}{7 (c+d x)^{3/2}}}{9 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {1}{7} d^2 \int \frac {\left (-50 D c^3+27 C d c^2-63 A d^3+d \left (-113 D c^2+90 C d c-63 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}}dx-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 C d-19 c D)}{7 (c+d x)^{3/2}}}{9 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {-\frac {1}{7} d^2 \left (-63 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-63 B d^2-113 c^2 D+90 c C d\right )}{5 d (c+d x)^{5/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 C d-19 c D)}{7 (c+d x)^{3/2}}}{9 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-\frac {1}{7} d^2 \left (-63 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \left (2 c \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}dx+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{3/2}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-63 B d^2-113 c^2 D+90 c C d\right )}{5 d (c+d x)^{5/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 C d-19 c D)}{7 (c+d x)^{3/2}}}{9 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-\frac {1}{7} d^2 \left (-63 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \left (2 c \left (2 c \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{3/2}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-63 B d^2-113 c^2 D+90 c C d\right )}{5 d (c+d x)^{5/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 C d-19 c D)}{7 (c+d x)^{3/2}}}{9 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {-\frac {1}{7} d^2 \left (-63 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \left (2 c \left (4 c d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{3/2}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-63 B d^2-113 c^2 D+90 c C d\right )}{5 d (c+d x)^{5/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 C d-19 c D)}{7 (c+d x)^{3/2}}}{9 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {1}{7} d^2 \left (-63 \left (2 c \left (\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}-\frac {2 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{d}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^{3/2}}\right ) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-63 B d^2-113 c^2 D+90 c C d\right )}{5 d (c+d x)^{5/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 C d-19 c D)}{7 (c+d x)^{3/2}}}{9 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{9 d^4 \sqrt {c+d x}}\)

Input:

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

Output:

(-2*D*(c^2 - d^2*x^2)^(5/2))/(9*d^4*Sqrt[c + d*x]) + ((-2*d*(9*C*d - 19*c* 
D)*(c^2 - d^2*x^2)^(5/2))/(7*(c + d*x)^(3/2)) - (d^2*((-2*(90*c*C*d - 63*B 
*d^2 - 113*c^2*D)*(c^2 - d^2*x^2)^(5/2))/(5*d*(c + d*x)^(5/2)) - 63*(c^2*C 
*d - B*c*d^2 + A*d^3 - c^3*D)*((2*(c^2 - d^2*x^2)^(3/2))/(3*d*(c + d*x)^(3 
/2)) + 2*c*((2*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[c + d*x]) - (2*Sqrt[2]*Sqrt[c] 
*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])])/d))))/7)/(9 
*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 466

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 0]) && NeQ[n + 2*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 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(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] && NeQ[m + 2*p + 2, 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 = 380, normalized size of antiderivative = 1.21

method	result	size
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (35 D d^{4} x^{4} \sqrt {-d x +c}+45 C \,d^{4} x^{3} \sqrt {-d x +c}-95 D c \,d^{3} x^{3} \sqrt {-d x +c}+630 A \,c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{3}-630 B \,c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{2}+63 B \,d^{4} x^{2} \sqrt {-d x +c}+630 C \,c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d -135 C c \,d^{3} x^{2} \sqrt {-d x +c}-630 D c^{\frac {9}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )+138 D c^{2} d^{2} x^{2} \sqrt {-d x +c}+105 A \,d^{4} x \sqrt {-d x +c}-231 B c \,d^{3} x \sqrt {-d x +c}+240 C \,c^{2} d^{2} x \sqrt {-d x +c}-236 D c^{3} d x \sqrt {-d x +c}-735 A c \,d^{3} \sqrt {-d x +c}+798 B \,c^{2} d^{2} \sqrt {-d x +c}-780 C \,c^{3} d \sqrt {-d x +c}+788 D c^{4} \sqrt {-d x +c}\right )}{315 \sqrt {d x +c}\, \sqrt {-d x +c}\, d^{4}}\)	\(380\)

Input:

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

Output:

-2/315*(-d^2*x^2+c^2)^(1/2)*(35*D*d^4*x^4*(-d*x+c)^(1/2)+45*C*d^4*x^3*(-d* 
x+c)^(1/2)-95*D*c*d^3*x^3*(-d*x+c)^(1/2)+630*A*c^(3/2)*2^(1/2)*arctanh(1/2 
*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^3-630*B*c^(5/2)*2^(1/2)*arctanh(1/2*(-d 
*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^2+63*B*d^4*x^2*(-d*x+c)^(1/2)+630*C*c^(7/2) 
*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d-135*C*c*d^3*x^2*(-d 
*x+c)^(1/2)-630*D*c^(9/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/ 
2))+138*D*c^2*d^2*x^2*(-d*x+c)^(1/2)+105*A*d^4*x*(-d*x+c)^(1/2)-231*B*c*d^ 
3*x*(-d*x+c)^(1/2)+240*C*c^2*d^2*x*(-d*x+c)^(1/2)-236*D*c^3*d*x*(-d*x+c)^( 
1/2)-735*A*c*d^3*(-d*x+c)^(1/2)+798*B*c^2*d^2*(-d*x+c)^(1/2)-780*C*c^3*d*( 
-d*x+c)^(1/2)+788*D*c^4*(-d*x+c)^(1/2))/(d*x+c)^(1/2)/(-d*x+c)^(1/2)/d^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 570, normalized size of antiderivative = 1.81 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\left [\frac {2 \, {\left (315 \, \sqrt {2} {\left (D c^{5} - C c^{4} d + B c^{3} d^{2} - A c^{2} d^{3} + {\left (D c^{4} d - C c^{3} d^{2} + B c^{2} d^{3} - A c d^{4}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - {\left (35 \, D d^{4} x^{4} + 788 \, D c^{4} - 780 \, C c^{3} d + 798 \, B c^{2} d^{2} - 735 \, A c d^{3} - 5 \, {\left (19 \, D c d^{3} - 9 \, C d^{4}\right )} x^{3} + 3 \, {\left (46 \, D c^{2} d^{2} - 45 \, C c d^{3} + 21 \, B d^{4}\right )} x^{2} - {\left (236 \, D c^{3} d - 240 \, C c^{2} d^{2} + 231 \, B c d^{3} - 105 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}\right )}}{315 \, {\left (d^{5} x + c d^{4}\right )}}, -\frac {2 \, {\left (630 \, \sqrt {2} {\left (D c^{5} - C c^{4} d + B c^{3} d^{2} - A c^{2} d^{3} + {\left (D c^{4} d - C c^{3} d^{2} + B c^{2} d^{3} - A c d^{4}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + {\left (35 \, D d^{4} x^{4} + 788 \, D c^{4} - 780 \, C c^{3} d + 798 \, B c^{2} d^{2} - 735 \, A c d^{3} - 5 \, {\left (19 \, D c d^{3} - 9 \, C d^{4}\right )} x^{3} + 3 \, {\left (46 \, D c^{2} d^{2} - 45 \, C c d^{3} + 21 \, B d^{4}\right )} x^{2} - {\left (236 \, D c^{3} d - 240 \, C c^{2} d^{2} + 231 \, B c d^{3} - 105 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}\right )}}{315 \, {\left (d^{5} x + c d^{4}\right )}}\right ] \] Input:

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

Output:

[2/315*(315*sqrt(2)*(D*c^5 - C*c^4*d + B*c^3*d^2 - A*c^2*d^3 + (D*c^4*d - 
C*c^3*d^2 + B*c^2*d^3 - A*c*d^4)*x)*sqrt(c)*log(-(d^2*x^2 - 2*c*d*x - 2*sq 
rt(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)) - (35*D*d^4*x^4 + 788*D*c^4 - 780*C*c^3*d + 798*B*c^2*d^2 - 735 
*A*c*d^3 - 5*(19*D*c*d^3 - 9*C*d^4)*x^3 + 3*(46*D*c^2*d^2 - 45*C*c*d^3 + 2 
1*B*d^4)*x^2 - (236*D*c^3*d - 240*C*c^2*d^2 + 231*B*c*d^3 - 105*A*d^4)*x)* 
sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^5*x + c*d^4), -2/315*(630*sqrt(2)*( 
D*c^5 - C*c^4*d + B*c^3*d^2 - A*c^2*d^3 + (D*c^4*d - C*c^3*d^2 + B*c^2*d^3 
 - A*c*d^4)*x)*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)) + (35*D*d^4*x^4 + 788*D*c^4 - 780*C*c^3*d + 79 
8*B*c^2*d^2 - 735*A*c*d^3 - 5*(19*D*c*d^3 - 9*C*d^4)*x^3 + 3*(46*D*c^2*d^2 
 - 45*C*c*d^3 + 21*B*d^4)*x^2 - (236*D*c^3*d - 240*C*c^2*d^2 + 231*B*c*d^3 
 - 105*A*d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^5*x + c*d^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.93 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} D + 45 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} D c + 63 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D c^{2} + 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{3} + 630 \, \sqrt {-d x + c} D c^{4} + 45 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} C d - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c^{2} d - 630 \, \sqrt {-d x + c} C c^{3} d + 63 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B d^{2} + 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c d^{2} + 630 \, \sqrt {-d x + c} B c^{2} d^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d^{3} - 630 \, \sqrt {-d x + c} A c d^{3} + \frac {630 \, \sqrt {2} {\left (D c^{5} - C c^{4} d + B c^{3} d^{2} - A c^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}}\right )}}{315 \, d^{4}} \] Input:

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

Output:

-2/315*(35*(d*x - c)^4*sqrt(-d*x + c)*D + 45*(d*x - c)^3*sqrt(-d*x + c)*D* 
c + 63*(d*x - c)^2*sqrt(-d*x + c)*D*c^2 + 105*(-d*x + c)^(3/2)*D*c^3 + 630 
*sqrt(-d*x + c)*D*c^4 + 45*(d*x - c)^3*sqrt(-d*x + c)*C*d - 105*(-d*x + c) 
^(3/2)*C*c^2*d - 630*sqrt(-d*x + c)*C*c^3*d + 63*(d*x - c)^2*sqrt(-d*x + c 
)*B*d^2 + 105*(-d*x + c)^(3/2)*B*c*d^2 + 630*sqrt(-d*x + c)*B*c^2*d^2 - 10 
5*(-d*x + c)^(3/2)*A*d^3 - 630*sqrt(-d*x + c)*A*c*d^3 + 630*sqrt(2)*(D*c^5 
 - C*c^4*d + B*c^3*d^2 - A*c^2*d^3)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt 
(-c))/sqrt(-c))/d^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.78 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {\frac {14 \sqrt {-d x +c}\, a c \,d^{2}}{3}-\frac {2 \sqrt {-d x +c}\, a \,d^{3} x}{3}-\frac {76 \sqrt {-d x +c}\, b \,c^{2} d}{15}+\frac {22 \sqrt {-d x +c}\, b c \,d^{2} x}{15}-\frac {2 \sqrt {-d x +c}\, b \,d^{3} x^{2}}{5}-\frac {16 \sqrt {-d x +c}\, c^{4}}{315}-\frac {8 \sqrt {-d x +c}\, c^{3} d x}{315}-\frac {2 \sqrt {-d x +c}\, c^{2} d^{2} x^{2}}{105}+\frac {20 \sqrt {-d x +c}\, c \,d^{3} x^{3}}{63}-\frac {2 \sqrt {-d x +c}\, d^{4} x^{4}}{9}+4 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a c \,d^{2}-4 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{2} d -\frac {16 \sqrt {c}\, \sqrt {2}\, a c \,d^{2}}{3}+\frac {104 \sqrt {c}\, \sqrt {2}\, b \,c^{2} d}{15}+\frac {184 \sqrt {c}\, \sqrt {2}\, c^{4}}{315}}{d^{3}} \] Input:

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

Output:

(2*(735*sqrt(c - d*x)*a*c*d**2 - 105*sqrt(c - d*x)*a*d**3*x - 798*sqrt(c - 
 d*x)*b*c**2*d + 231*sqrt(c - d*x)*b*c*d**2*x - 63*sqrt(c - d*x)*b*d**3*x* 
*2 - 8*sqrt(c - d*x)*c**4 - 4*sqrt(c - d*x)*c**3*d*x - 3*sqrt(c - d*x)*c** 
2*d**2*x**2 + 50*sqrt(c - d*x)*c*d**3*x**3 - 35*sqrt(c - d*x)*d**4*x**4 + 
630*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c* 
d**2 - 630*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2 
))*b*c**2*d - 840*sqrt(c)*sqrt(2)*a*c*d**2 + 1092*sqrt(c)*sqrt(2)*b*c**2*d 
 + 92*sqrt(c)*sqrt(2)*c**4))/(315*d**3)