\(\int \frac {\sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3)}{\sqrt {c+d x}} \, dx\) [196]

Optimal result

Integrand size = 41, antiderivative size = 184 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=-\frac {2 \left (33 c^2 C d-21 B c d^2+105 A d^3-13 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{315 d^4 (c+d x)^{3/2}}+\frac {2 \left (18 c C d-21 B d^2-23 c^2 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{105 d^4 \sqrt {c+d x}}-\frac {2 (3 C d-5 c D) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{21 d^4}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^4} \] Output:

-2/315*(105*A*d^3-21*B*c*d^2+33*C*c^2*d-13*D*c^3)*(-d^2*x^2+c^2)^(3/2)/d^4 
/(d*x+c)^(3/2)+2/105*(-21*B*d^2+18*C*c*d-23*D*c^2)*(-d^2*x^2+c^2)^(3/2)/d^ 
4/(d*x+c)^(1/2)-2/21*(3*C*d-5*D*c)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(3/2)/d^4- 
2/9*D*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(3/2)/d^4
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=-\frac {2 (c-d x) \sqrt {c^2-d^2 x^2} \left (16 c^3 D+24 c^2 d (C+D x)+6 c d^2 (7 B+x (6 C+5 D x))+d^3 (105 A+x (63 B+5 x (9 C+7 D x)))\right )}{315 d^4 \sqrt {c+d x}} \] Input:

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

Output:

(-2*(c - d*x)*Sqrt[c^2 - d^2*x^2]*(16*c^3*D + 24*c^2*d*(C + D*x) + 6*c*d^2 
*(7*B + x*(6*C + 5*D*x)) + d^3*(105*A + x*(63*B + 5*x*(9*C + 7*D*x)))))/(3 
15*d^4*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {2170, 27, 2170, 27, 672, 458}

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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int -\frac {d^6 \left (2 D c^3+3 C d c^2+21 A d^3-d \left (-23 D c^2+18 C d c-21 B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}}{2 \sqrt {c+d x}}dx}{7 d^4}-\frac {2}{7} d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} (3 C d-5 c D)}{3 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} d^2 \int \frac {\left (2 D c^3+3 C d c^2+21 A d^3-d \left (-23 D c^2+18 C d c-21 B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}dx-\frac {2}{7} d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} (3 C d-5 c D)}{3 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^4}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {\frac {1}{7} d^2 \left (\frac {1}{5} \left (105 A d^3-21 B c d^2-13 c^3 D+33 c^2 C d\right ) \int \frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}dx+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (-21 B d^2-23 c^2 D+18 c C d\right )}{5 d \sqrt {c+d x}}\right )-\frac {2}{7} d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} (3 C d-5 c D)}{3 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {\frac {1}{7} d^2 \left (\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (-21 B d^2-23 c^2 D+18 c C d\right )}{5 d \sqrt {c+d x}}-\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (105 A d^3-21 B c d^2-13 c^3 D+33 c^2 C d\right )}{15 d (c+d x)^{3/2}}\right )-\frac {2}{7} d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} (3 C d-5 c D)}{3 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d^4}\)

Input:

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

Output:

(-2*D*(c + d*x)^(3/2)*(c^2 - d^2*x^2)^(3/2))/(9*d^4) + ((-2*d*(3*C*d - 5*c 
*D)*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2))/7 + (d^2*((-2*(33*c^2*C*d - 21*B* 
c*d^2 + 105*A*d^3 - 13*c^3*D)*(c^2 - d^2*x^2)^(3/2))/(15*d*(c + d*x)^(3/2) 
) + (2*(18*c*C*d - 21*B*d^2 - 23*c^2*D)*(c^2 - d^2*x^2)^(3/2))/(5*d*Sqrt[c 
 + d*x])))/7)/(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 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 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 = 0.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.60

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (35 D x^{3} d^{3}+45 C \,d^{3} x^{2}+30 D c \,d^{2} x^{2}+63 B \,d^{3} x +36 C c \,d^{2} x +24 D c^{2} d x +105 A \,d^{3}+42 B c \,d^{2}+24 C \,c^{2} d +16 D c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{315 d^{4} \sqrt {d x +c}}\)	\(111\)
default	\(-\frac {2 \left (-d x +c \right ) \left (35 D x^{3} d^{3}+45 C \,d^{3} x^{2}+30 D c \,d^{2} x^{2}+63 B \,d^{3} x +36 C c \,d^{2} x +24 D c^{2} d x +105 A \,d^{3}+42 B c \,d^{2}+24 C \,c^{2} d +16 D c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{315 d^{4} \sqrt {d x +c}}\)	\(111\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (35 D x^{3} d^{3}+45 C \,d^{3} x^{2}+30 D c \,d^{2} x^{2}+63 B \,d^{3} x +36 C c \,d^{2} x +24 D c^{2} d x +105 A \,d^{3}+42 B c \,d^{2}+24 C \,c^{2} d +16 D c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{315 d^{4} \sqrt {d x +c}}\)	\(111\)

Input:

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

Output:

-2/315*(-d*x+c)*(35*D*d^3*x^3+45*C*d^3*x^2+30*D*c*d^2*x^2+63*B*d^3*x+36*C* 
c*d^2*x+24*D*c^2*d*x+105*A*d^3+42*B*c*d^2+24*C*c^2*d+16*D*c^3)*(-d^2*x^2+c 
^2)^(1/2)/d^4/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (35 \, D d^{4} x^{4} - 16 \, D c^{4} - 24 \, C c^{3} d - 42 \, B c^{2} d^{2} - 105 \, A c d^{3} - 5 \, {\left (D c d^{3} - 9 \, C d^{4}\right )} x^{3} - 3 \, {\left (2 \, D c^{2} d^{2} + 3 \, C c d^{3} - 21 \, B d^{4}\right )} x^{2} - {\left (8 \, D c^{3} d + 12 \, C c^{2} d^{2} + 21 \, B c d^{3} - 105 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{315 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

2/315*(35*D*d^4*x^4 - 16*D*c^4 - 24*C*c^3*d - 42*B*c^2*d^2 - 105*A*c*d^3 - 
 5*(D*c*d^3 - 9*C*d^4)*x^3 - 3*(2*D*c^2*d^2 + 3*C*c*d^3 - 21*B*d^4)*x^2 - 
(8*D*c^3*d + 12*C*c^2*d^2 + 21*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]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (d x - c\right )} \sqrt {-d x + c} A}{3 \, d} + \frac {2 \, {\left (3 \, d^{2} x^{2} - c d x - 2 \, c^{2}\right )} \sqrt {-d x + c} B}{15 \, d^{2}} + \frac {2 \, {\left (15 \, d^{3} x^{3} - 3 \, c d^{2} x^{2} - 4 \, c^{2} d x - 8 \, c^{3}\right )} \sqrt {-d x + c} C}{105 \, d^{3}} + \frac {2 \, {\left (35 \, d^{4} x^{4} - 5 \, c d^{3} x^{3} - 6 \, c^{2} d^{2} x^{2} - 8 \, c^{3} d x - 16 \, c^{4}\right )} \sqrt {-d x + c} D}{315 \, d^{4}} \] Input:

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

Output:

2/3*(d*x - c)*sqrt(-d*x + c)*A/d + 2/15*(3*d^2*x^2 - c*d*x - 2*c^2)*sqrt(- 
d*x + c)*B/d^2 + 2/105*(15*d^3*x^3 - 3*c*d^2*x^2 - 4*c^2*d*x - 8*c^3)*sqrt 
(-d*x + c)*C/d^3 + 2/315*(35*d^4*x^4 - 5*c*d^3*x^3 - 6*c^2*d^2*x^2 - 8*c^3 
*d*x - 16*c^4)*sqrt(-d*x + c)*D/d^4
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=-\frac {2 \, {\left (105 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d^{3} - 21 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} c\right )} B d^{2} - 3 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 42 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2}\right )} C d - {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 135 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 189 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3}\right )} D\right )}}{315 \, d^{4}} \] Input:

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

Output:

-2/315*(105*(-d*x + c)^(3/2)*A*d^3 - 21*(3*(d*x - c)^2*sqrt(-d*x + c) - 5* 
(-d*x + c)^(3/2)*c)*B*d^2 - 3*(15*(d*x - c)^3*sqrt(-d*x + c) + 42*(d*x - c 
)^2*sqrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2)*C*d - (35*(d*x - c)^4*sqrt 
(-d*x + c) + 135*(d*x - c)^3*sqrt(-d*x + c)*c + 189*(d*x - c)^2*sqrt(-d*x 
+ c)*c^2 - 105*(-d*x + c)^(3/2)*c^3)*D)/d^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {-d x +c}\, \left (35 d^{4} x^{4}+40 c \,d^{3} x^{3}+63 b \,d^{3} x^{2}-15 c^{2} d^{2} x^{2}+105 a \,d^{3} x -21 b c \,d^{2} x -20 c^{3} d x -105 a c \,d^{2}-42 b \,c^{2} d -40 c^{4}\right )}{315 d^{3}} \] Input:

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

Output:

(2*sqrt(c - d*x)*( - 105*a*c*d**2 + 105*a*d**3*x - 42*b*c**2*d - 21*b*c*d* 
*2*x + 63*b*d**3*x**2 - 40*c**4 - 20*c**3*d*x - 15*c**2*d**2*x**2 + 40*c*d 
**3*x**3 + 35*d**4*x**4))/(315*d**3)