\(\int \frac {(c+d x)^{5/2} (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{5/2}} \, dx\) [229]

Optimal result

Integrand size = 41, antiderivative size = 216 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)^{5/2}}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\left (11 c^2 C d+5 B c d^2-A d^3+17 c^3 D\right ) (c+d x)^{3/2}}{6 c^2 d^4 \sqrt {c^2-d^2 x^2}}-\frac {\left (23 c^2 C d+5 B c d^2-A d^3+45 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{6 c^2 d^4 \sqrt {c+d x}}-\frac {2 D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4} \] Output:

1/3*(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^(5/2)/c/d^4/(-d^2*x^2+c^2)^(3/2) 
-1/6*(-A*d^3+5*B*c*d^2+11*C*c^2*d+17*D*c^3)*(d*x+c)^(3/2)/c^2/d^4/(-d^2*x^ 
2+c^2)^(1/2)-1/6*(-A*d^3+5*B*c*d^2+23*C*c^2*d+45*D*c^3)*(-d^2*x^2+c^2)^(1/ 
2)/c^2/d^4/(d*x+c)^(1/2)-2/3*D*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 2.66 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.47 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {c+d x} \left (16 c^3 D+8 c^2 d (C-3 D x)+2 c d^2 (B+3 x (-2 C+D x))+d^3 \left (-A+x \left (-3 B+3 C x+D x^2\right )\right )\right )}{3 d^4 (-c+d x) \sqrt {c^2-d^2 x^2}} \] Input:

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

Output:

(2*Sqrt[c + d*x]*(16*c^3*D + 8*c^2*d*(C - 3*D*x) + 2*c*d^2*(B + 3*x*(-2*C 
+ D*x)) + d^3*(-A + x*(-3*B + 3*C*x + D*x^2))))/(3*d^4*(-c + d*x)*Sqrt[c^2 
 - d^2*x^2])
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {2166, 27, 2166, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2166

\(\displaystyle \frac {(c+d x)^{5/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{3/2} \left (-A d^3+5 B c d^2+17 c^3 D+11 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\int \frac {\sqrt {c+d x} \left (41 D c^3+23 C d c^2+12 d D x c^2+5 B d^2 c-A d^3\right )}{2 d^3 \sqrt {c^2-d^2 x^2}}dx}{c}}{6 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(c+d x)^{5/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{3/2} \left (-A d^3+5 B c d^2+17 c^3 D+11 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\int \frac {\sqrt {c+d x} \left (41 D c^3+23 C d c^2+12 d D x c^2+5 B d^2 c-A d^3\right )}{\sqrt {c^2-d^2 x^2}}dx}{2 c d^3}}{6 c}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {(c+d x)^{5/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{3/2} \left (-A d^3+5 B c d^2+17 c^3 D+11 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\left (-A d^3+5 B c d^2+45 c^3 D+23 c^2 C d\right ) \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx-\frac {8 c^2 D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{d}}{2 c d^3}}{6 c}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {(c+d x)^{5/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{3/2} \left (-A d^3+5 B c d^2+17 c^3 D+11 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {-\frac {2 \sqrt {c^2-d^2 x^2} \left (-A d^3+5 B c d^2+45 c^3 D+23 c^2 C d\right )}{d \sqrt {c+d x}}-\frac {8 c^2 D \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{d}}{2 c d^3}}{6 c}\)

Input:

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

Output:

((c^2*C*d + B*c*d^2 + A*d^3 + c^3*D)*(c + d*x)^(5/2))/(3*c*d^4*(c^2 - d^2* 
x^2)^(3/2)) - (((11*c^2*C*d + 5*B*c*d^2 - A*d^3 + 17*c^3*D)*(c + d*x)^(3/2 
))/(c*d^4*Sqrt[c^2 - d^2*x^2]) - ((-2*(23*c^2*C*d + 5*B*c*d^2 - A*d^3 + 45 
*c^3*D)*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[c + d*x]) - (8*c^2*D*Sqrt[c + d*x]*Sq 
rt[c^2 - d^2*x^2])/d)/(2*c*d^3))/(6*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]
 

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.51

method	result	size
gosper	\(\frac {2 \left (-d x +c \right ) \left (-D x^{3} d^{3}-3 C \,d^{3} x^{2}-6 D c \,d^{2} x^{2}+3 B \,d^{3} x +12 C c \,d^{2} x +24 D c^{2} d x +A \,d^{3}-2 B c \,d^{2}-8 C \,c^{2} d -16 D c^{3}\right ) \left (d x +c \right )^{\frac {5}{2}}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(110\)
orering	\(\frac {2 \left (-d x +c \right ) \left (-D x^{3} d^{3}-3 C \,d^{3} x^{2}-6 D c \,d^{2} x^{2}+3 B \,d^{3} x +12 C c \,d^{2} x +24 D c^{2} d x +A \,d^{3}-2 B c \,d^{2}-8 C \,c^{2} d -16 D c^{3}\right ) \left (d x +c \right )^{\frac {5}{2}}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(110\)
default	\(\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-D x^{3} d^{3}-3 C \,d^{3} x^{2}-6 D c \,d^{2} x^{2}+3 B \,d^{3} x +12 C c \,d^{2} x +24 D c^{2} d x +A \,d^{3}-2 B c \,d^{2}-8 C \,c^{2} d -16 D c^{3}\right )}{3 \sqrt {d x +c}\, \left (-d x +c \right )^{2} d^{4}}\)	\(112\)

Input:

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

Output:

2/3*(-d*x+c)*(-D*d^3*x^3-3*C*d^3*x^2-6*D*c*d^2*x^2+3*B*d^3*x+12*C*c*d^2*x+ 
24*D*c^2*d*x+A*d^3-2*B*c*d^2-8*C*c^2*d-16*D*c^3)*(d*x+c)^(5/2)/d^4/(-d^2*x 
^2+c^2)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.62 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (D d^{3} x^{3} + 16 \, D c^{3} + 8 \, C c^{2} d + 2 \, B c d^{2} - A d^{3} + 3 \, {\left (2 \, D c d^{2} + C d^{3}\right )} x^{2} - 3 \, {\left (8 \, D c^{2} d + 4 \, C c d^{2} + B d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{3 \, {\left (d^{7} x^{3} - c d^{6} x^{2} - c^{2} d^{5} x + c^{3} d^{4}\right )}} \] Input:

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

Output:

-2/3*(D*d^3*x^3 + 16*D*c^3 + 8*C*c^2*d + 2*B*c*d^2 - A*d^3 + 3*(2*D*c*d^2 
+ C*d^3)*x^2 - 3*(8*D*c^2*d + 4*C*c*d^2 + B*d^3)*x)*sqrt(-d^2*x^2 + c^2)*s 
qrt(d*x + c)/(d^7*x^3 - c*d^6*x^2 - c^2*d^5*x + c^3*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.72 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, d x - 2 \, c\right )} B}{3 \, {\left (d^{3} x - c d^{2}\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (3 \, d^{2} x^{2} - 12 \, c d x + 8 \, c^{2}\right )} C}{3 \, {\left (d^{4} x - c d^{3}\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (d^{3} x^{3} + 6 \, c d^{2} x^{2} - 24 \, c^{2} d x + 16 \, c^{3}\right )} D}{3 \, {\left (d^{5} x - c d^{4}\right )} \sqrt {-d x + c}} - \frac {2 \, A}{3 \, {\left (d^{2} x - c d\right )} \sqrt {-d x + c}} \] Input:

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

Output:

-2/3*(3*d*x - 2*c)*B/((d^3*x - c*d^2)*sqrt(-d*x + c)) + 2/3*(3*d^2*x^2 - 1 
2*c*d*x + 8*c^2)*C/((d^4*x - c*d^3)*sqrt(-d*x + c)) + 2/3*(d^3*x^3 + 6*c*d 
^2*x^2 - 24*c^2*d*x + 16*c^3)*D/((d^5*x - c*d^4)*sqrt(-d*x + c)) - 2/3*A/( 
(d^2*x - c*d)*sqrt(-d*x + c))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.63 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (\frac {9 \, {\left (d x - c\right )} D c^{2} + D c^{3} + 6 \, {\left (d x - c\right )} C c d + C c^{2} d + 3 \, {\left (d x - c\right )} B d^{2} + B c d^{2} + A d^{3}}{{\left (d x - c\right )} \sqrt {-d x + c} d^{3}} - \frac {{\left (-d x + c\right )}^{\frac {3}{2}} D d^{6} - 9 \, \sqrt {-d x + c} D c d^{6} - 3 \, \sqrt {-d x + c} C d^{7}}{d^{9}}\right )}}{3 \, d} \] Input:

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

Output:

-2/3*((9*(d*x - c)*D*c^2 + D*c^3 + 6*(d*x - c)*C*c*d + C*c^2*d + 3*(d*x - 
c)*B*d^2 + B*c*d^2 + A*d^3)/((d*x - c)*sqrt(-d*x + c)*d^3) - ((-d*x + c)^( 
3/2)*D*d^6 - 9*sqrt(-d*x + c)*D*c*d^6 - 3*sqrt(-d*x + c)*C*d^7)/d^9)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.32 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-\frac {2}{3} d^{3} x^{3}-6 c \,d^{2} x^{2}+2 b \,d^{2} x +24 c^{2} d x +\frac {2}{3} a \,d^{2}-\frac {4}{3} b c d -16 c^{3}}{\sqrt {-d x +c}\, d^{3} \left (-d x +c \right )} \] Input:

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

Output:

(2*(a*d**2 - 2*b*c*d + 3*b*d**2*x - 24*c**3 + 36*c**2*d*x - 9*c*d**2*x**2 
- d**3*x**3))/(3*sqrt(c - d*x)*d**3*(c - d*x))