\(\int \frac {(c+d x)^{3/2} (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{3/2}} \, dx\) [222]

Optimal result

Integrand size = 41, antiderivative size = 192 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)^{3/2}}{c d^4 \sqrt {c^2-d^2 x^2}}+\frac {\left (55 c^2 C d+45 B c d^2+15 A d^3+69 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{15 c d^4 \sqrt {c+d x}}+\frac {2 (5 C d+3 c D) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{15 d^4}+\frac {2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4} \] Output:

(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^(3/2)/c/d^4/(-d^2*x^2+c^2)^(1/2)+1/1 
5*(15*A*d^3+45*B*c*d^2+55*C*c^2*d+69*D*c^3)*(-d^2*x^2+c^2)^(1/2)/c/d^4/(d* 
x+c)^(1/2)+2/15*(5*C*d+3*D*c)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/d^4+2/5*D 
*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.55 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {c^2-d^2 x^2} \left (-48 c^3 D-8 c^2 d (5 C-3 D x)+2 c d^2 (-15 B+x (10 C+3 D x))+d^3 \left (-15 A+x \left (15 B+5 C x+3 D x^2\right )\right )\right )}{15 d^4 (-c+d x) \sqrt {c+d x}} \] Input:

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

Output:

(2*Sqrt[c^2 - d^2*x^2]*(-48*c^3*D - 8*c^2*d*(5*C - 3*D*x) + 2*c*d^2*(-15*B 
 + x*(10*C + 3*D*x)) + d^3*(-15*A + x*(15*B + 5*C*x + 3*D*x^2))))/(15*d^4* 
(-c + d*x)*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 207, 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 = {2166, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {-\frac {2 \int -\frac {d \sqrt {c+d x} \left (21 D c^3+15 C d c^2+15 B d^2 c+2 d (5 C d+3 c D) x c+5 A d^3\right )}{2 \sqrt {c^2-d^2 x^2}}dx}{5 d^4}-\frac {4 c D \sqrt {c^2-d^2 x^2} (c+d x)^{3/2}}{5 d^4}}{2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\int \frac {\sqrt {c+d x} \left (21 D c^3+15 C d c^2+15 B d^2 c+2 d (5 C d+3 c D) x c+5 A d^3\right )}{\sqrt {c^2-d^2 x^2}}dx}{5 d^3}-\frac {4 c D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}}{2 c}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {1}{3} \left (15 A d^3+45 B c d^2+69 c^3 D+55 c^2 C d\right ) \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx-\frac {4 c \sqrt {c+d x} \sqrt {c^2-d^2 x^2} (3 c D+5 C d)}{3 d}}{5 d^3}-\frac {4 c D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}}{2 c}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {(c+d x)^{3/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {-\frac {2 \sqrt {c^2-d^2 x^2} \left (15 A d^3+45 B c d^2+69 c^3 D+55 c^2 C d\right )}{3 d \sqrt {c+d x}}-\frac {4 c \sqrt {c+d x} \sqrt {c^2-d^2 x^2} (3 c D+5 C d)}{3 d}}{5 d^3}-\frac {4 c D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4}}{2 c}\)

Input:

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

Output:

((c^2*C*d + B*c*d^2 + A*d^3 + c^3*D)*(c + d*x)^(3/2))/(c*d^4*Sqrt[c^2 - d^ 
2*x^2]) - ((-4*c*D*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2])/(5*d^4) + ((-2*(55 
*c^2*C*d + 45*B*c*d^2 + 15*A*d^3 + 69*c^3*D)*Sqrt[c^2 - d^2*x^2])/(3*d*Sqr 
t[c + d*x]) - (4*c*(5*C*d + 3*c*D)*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2])/(3*d 
))/(5*d^3))/(2*c)
 

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 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[m, 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 = 0.58

method	result	size
gosper	\(\frac {2 \left (-d x +c \right ) \left (-3 D x^{3} d^{3}-5 C \,d^{3} x^{2}-6 D c \,d^{2} x^{2}-15 B \,d^{3} x -20 C c \,d^{2} x -24 D c^{2} d x +15 A \,d^{3}+30 B c \,d^{2}+40 C \,c^{2} d +48 D c^{3}\right ) \left (d x +c \right )^{\frac {3}{2}}}{15 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(111\)
orering	\(\frac {2 \left (-d x +c \right ) \left (-3 D x^{3} d^{3}-5 C \,d^{3} x^{2}-6 D c \,d^{2} x^{2}-15 B \,d^{3} x -20 C c \,d^{2} x -24 D c^{2} d x +15 A \,d^{3}+30 B c \,d^{2}+40 C \,c^{2} d +48 D c^{3}\right ) \left (d x +c \right )^{\frac {3}{2}}}{15 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(111\)
default	\(\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-3 D x^{3} d^{3}-5 C \,d^{3} x^{2}-6 D c \,d^{2} x^{2}-15 B \,d^{3} x -20 C c \,d^{2} x -24 D c^{2} d x +15 A \,d^{3}+30 B c \,d^{2}+40 C \,c^{2} d +48 D c^{3}\right )}{15 \sqrt {d x +c}\, \left (-d x +c \right ) d^{4}}\)	\(113\)

Input:

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

Output:

2/15*(-d*x+c)*(-3*D*d^3*x^3-5*C*d^3*x^2-6*D*c*d^2*x^2-15*B*d^3*x-20*C*c*d^ 
2*x-24*D*c^2*d*x+15*A*d^3+30*B*c*d^2+40*C*c^2*d+48*D*c^3)*(d*x+c)^(3/2)/d^ 
4/(-d^2*x^2+c^2)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.61 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (3 \, D d^{3} x^{3} - 48 \, D c^{3} - 40 \, C c^{2} d - 30 \, B c d^{2} - 15 \, A d^{3} + {\left (6 \, D c d^{2} + 5 \, C d^{3}\right )} x^{2} + {\left (24 \, D c^{2} d + 20 \, C c d^{2} + 15 \, B d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{15 \, {\left (d^{6} x^{2} - c^{2} d^{4}\right )}} \] Input:

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

Output:

2/15*(3*D*d^3*x^3 - 48*D*c^3 - 40*C*c^2*d - 30*B*c*d^2 - 15*A*d^3 + (6*D*c 
*d^2 + 5*C*d^3)*x^2 + (24*D*c^2*d + 20*C*c*d^2 + 15*B*d^3)*x)*sqrt(-d^2*x^ 
2 + c^2)*sqrt(d*x + c)/(d^6*x^2 - c^2*d^4)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.58 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (d x - 2 \, c\right )} B}{\sqrt {-d x + c} d^{2}} + \frac {2 \, A}{\sqrt {-d x + c} d} - \frac {2 \, {\left (d^{2} x^{2} + 4 \, c d x - 8 \, c^{2}\right )} C}{3 \, \sqrt {-d x + c} d^{3}} - \frac {2 \, {\left (d^{3} x^{3} + 2 \, c d^{2} x^{2} + 8 \, c^{2} d x - 16 \, c^{3}\right )} D}{5 \, \sqrt {-d x + c} d^{4}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.76 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {15 \, {\left (D c^{3} + C c^{2} d + B c d^{2} + A d^{3}\right )}}{\sqrt {-d x + c} d^{3}} + \frac {3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D d^{12} - 15 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c d^{12} + 45 \, \sqrt {-d x + c} D c^{2} d^{12} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} C d^{13} + 30 \, \sqrt {-d x + c} C c d^{13} + 15 \, \sqrt {-d x + c} B d^{14}}{d^{15}}\right )}}{15 \, d} \] Input:

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

Output:

2/15*(15*(D*c^3 + C*c^2*d + B*c*d^2 + A*d^3)/(sqrt(-d*x + c)*d^3) + (3*(d* 
x - c)^2*sqrt(-d*x + c)*D*d^12 - 15*(-d*x + c)^(3/2)*D*c*d^12 + 45*sqrt(-d 
*x + c)*D*c^2*d^12 - 5*(-d*x + c)^(3/2)*C*d^13 + 30*sqrt(-d*x + c)*C*c*d^1 
3 + 15*sqrt(-d*x + c)*B*d^14)/d^15)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.32 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-\frac {2}{5} d^{3} x^{3}-\frac {22}{15} c \,d^{2} x^{2}-2 b \,d^{2} x -\frac {88}{15} c^{2} d x +2 a \,d^{2}+4 b c d +\frac {176}{15} c^{3}}{\sqrt {-d x +c}\, d^{3}} \] Input:

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

Output:

(2*(15*a*d**2 + 30*b*c*d - 15*b*d**2*x + 88*c**3 - 44*c**2*d*x - 11*c*d**2 
*x**2 - 3*d**3*x**3))/(15*sqrt(c - d*x)*d**3)