\(\int \frac {\sqrt {c+d x} (A+B x+C x^2+D x^3)}{\sqrt {c^2-d^2 x^2}} \, dx\) [215]

Optimal result

Integrand size = 41, antiderivative size = 184 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \left (49 c^2 C d+35 B c d^2+105 A d^3+27 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{105 d^4 \sqrt {c+d x}}+\frac {2 \left (14 c C d-35 B d^2-33 c^2 D\right ) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{105 d^4}-\frac {2 (7 C d-9 c D) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{35 d^4}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4} \] Output:

-2/105*(105*A*d^3+35*B*c*d^2+49*C*c^2*d+27*D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4 
/(d*x+c)^(1/2)+2/105*(-35*B*d^2+14*C*c*d-33*D*c^2)*(d*x+c)^(1/2)*(-d^2*x^2 
+c^2)^(1/2)/d^4-2/35*(7*C*d-9*D*c)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)/d^4- 
2/7*D*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.53 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \sqrt {c^2-d^2 x^2} \left (48 c^3 D+8 c^2 d (7 C+3 D x)+2 c d^2 (35 B+x (14 C+9 D x))+d^3 (105 A+x (35 B+3 x (7 C+5 D x)))\right )}{105 d^4 \sqrt {c+d x}} \] Input:

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

Output:

(-2*Sqrt[c^2 - d^2*x^2]*(48*c^3*D + 8*c^2*d*(7*C + 3*D*x) + 2*c*d^2*(35*B 
+ x*(14*C + 9*D*x)) + d^3*(105*A + x*(35*B + 3*x*(7*C + 5*D*x)))))/(105*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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int -\frac {d^6 \sqrt {c+d x} \left (-2 D c^3+21 C d c^2+35 A d^3-d \left (-33 D c^2+14 C d c-35 B d^2\right ) x\right )}{2 \sqrt {c^2-d^2 x^2}}dx}{5 d^4}-\frac {2}{5} d \sqrt {c^2-d^2 x^2} (c+d x)^{3/2} (7 C d-9 c D)}{7 d^5}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} d^2 \int \frac {\sqrt {c+d x} \left (-2 D c^3+21 C d c^2+35 A d^3-d \left (-33 D c^2+14 C d c-35 B d^2\right ) x\right )}{\sqrt {c^2-d^2 x^2}}dx-\frac {2}{5} d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (7 C d-9 c D)}{7 d^5}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {\frac {1}{5} d^2 \left (\frac {1}{3} \left (105 A d^3+35 B c d^2+27 c^3 D+49 c^2 C d\right ) \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \left (-35 B d^2-33 c^2 D+14 c C d\right )}{3 d}\right )-\frac {2}{5} d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (7 C d-9 c D)}{7 d^5}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {\frac {1}{5} d^2 \left (\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2} \left (-35 B d^2-33 c^2 D+14 c C d\right )}{3 d}-\frac {2 \sqrt {c^2-d^2 x^2} \left (105 A d^3+35 B c d^2+27 c^3 D+49 c^2 C d\right )}{3 d \sqrt {c+d x}}\right )-\frac {2}{5} d (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (7 C d-9 c D)}{7 d^5}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}\)

Input:

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

Output:

(-2*D*(c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2])/(7*d^4) + ((-2*d*(7*C*d - 9*c*D 
)*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2])/5 + (d^2*((-2*(49*c^2*C*d + 35*B*c* 
d^2 + 105*A*d^3 + 27*c^3*D)*Sqrt[c^2 - d^2*x^2])/(3*d*Sqrt[c + d*x]) + (2* 
(14*c*C*d - 35*B*d^2 - 33*c^2*D)*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2])/(3*d)) 
)/5)/(7*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 = 105, normalized size of antiderivative = 0.57

method	result	size
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (15 D x^{3} d^{3}+21 C \,d^{3} x^{2}+18 D c \,d^{2} x^{2}+35 B \,d^{3} x +28 C c \,d^{2} x +24 D c^{2} d x +105 A \,d^{3}+70 B c \,d^{2}+56 C \,c^{2} d +48 D c^{3}\right )}{105 \sqrt {d x +c}\, d^{4}}\)	\(105\)
gosper	\(-\frac {2 \left (-d x +c \right ) \left (15 D x^{3} d^{3}+21 C \,d^{3} x^{2}+18 D c \,d^{2} x^{2}+35 B \,d^{3} x +28 C c \,d^{2} x +24 D c^{2} d x +105 A \,d^{3}+70 B c \,d^{2}+56 C \,c^{2} d +48 D c^{3}\right ) \sqrt {d x +c}}{105 d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(111\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (15 D x^{3} d^{3}+21 C \,d^{3} x^{2}+18 D c \,d^{2} x^{2}+35 B \,d^{3} x +28 C c \,d^{2} x +24 D c^{2} d x +105 A \,d^{3}+70 B c \,d^{2}+56 C \,c^{2} d +48 D c^{3}\right ) \sqrt {d x +c}}{105 d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(111\)

Input:

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

Output:

-2/105/(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)*(15*D*d^3*x^3+21*C*d^3*x^2+18*D* 
c*d^2*x^2+35*B*d^3*x+28*C*c*d^2*x+24*D*c^2*d*x+105*A*d^3+70*B*c*d^2+56*C*c 
^2*d+48*D*c^3)/d^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \, {\left (15 \, D d^{3} x^{3} + 48 \, D c^{3} + 56 \, C c^{2} d + 70 \, B c d^{2} + 105 \, A d^{3} + 3 \, {\left (6 \, D c d^{2} + 7 \, C d^{3}\right )} x^{2} + {\left (24 \, D c^{2} d + 28 \, C c d^{2} + 35 \, B d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{105 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

-2/105*(15*D*d^3*x^3 + 48*D*c^3 + 56*C*c^2*d + 70*B*c*d^2 + 105*A*d^3 + 3* 
(6*D*c*d^2 + 7*C*d^3)*x^2 + (24*D*c^2*d + 28*C*c*d^2 + 35*B*d^3)*x)*sqrt(- 
d^2*x^2 + c^2)*sqrt(d*x + c)/(d^5*x + c*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \, {\left (d x - c\right )} A}{\sqrt {-d x + c} d} + \frac {2 \, {\left (d^{2} x^{2} + c d x - 2 \, c^{2}\right )} B}{3 \, \sqrt {-d x + c} d^{2}} + \frac {2 \, {\left (3 \, d^{3} x^{3} + c d^{2} x^{2} + 4 \, c^{2} d x - 8 \, c^{3}\right )} C}{15 \, \sqrt {-d x + c} d^{3}} + \frac {2 \, {\left (5 \, d^{4} x^{4} + c d^{3} x^{3} + 2 \, c^{2} d^{2} x^{2} + 8 \, c^{3} d x - 16 \, c^{4}\right )} D}{35 \, \sqrt {-d x + c} d^{4}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} D + 63 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D c - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{2} + 105 \, \sqrt {-d x + c} D c^{3} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} C d - 70 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c d + 105 \, \sqrt {-d x + c} C c^{2} d - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} B d^{2} + 105 \, \sqrt {-d x + c} B c d^{2} + 105 \, \sqrt {-d x + c} A d^{3}\right )}}{105 \, d^{4}} \] Input:

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

Output:

-2/105*(15*(d*x - c)^3*sqrt(-d*x + c)*D + 63*(d*x - c)^2*sqrt(-d*x + c)*D* 
c - 105*(-d*x + c)^(3/2)*D*c^2 + 105*sqrt(-d*x + c)*D*c^3 + 21*(d*x - c)^2 
*sqrt(-d*x + c)*C*d - 70*(-d*x + c)^(3/2)*C*c*d + 105*sqrt(-d*x + c)*C*c^2 
*d - 35*(-d*x + c)^(3/2)*B*d^2 + 105*sqrt(-d*x + c)*B*c*d^2 + 105*sqrt(-d* 
x + c)*A*d^3)/d^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \sqrt {-d x +c}\, \left (-15 d^{3} x^{3}-39 c \,d^{2} x^{2}-35 b \,d^{2} x -52 c^{2} d x -105 a \,d^{2}-70 b c d -104 c^{3}\right )}{105 d^{3}} \] Input:

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

Output:

(2*sqrt(c - d*x)*( - 105*a*d**2 - 70*b*c*d - 35*b*d**2*x - 104*c**3 - 52*c 
**2*d*x - 39*c*d**2*x**2 - 15*d**3*x**3))/(105*d**3)