\(\int \frac {\sqrt {a+b x} (A+B x+C x^2+D x^3)}{(c+d x)^{5/2}} \, dx\) [105]

Optimal result

Integrand size = 34, antiderivative size = 264 \[ \int \frac {\sqrt {a+b x} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) (a+b x)^{3/2}}{3 d^3 (b c-a d) (c+d x)^{3/2}}+\frac {2 \left (2 c C d-B d^2-3 c^2 D\right ) \sqrt {a+b x}}{d^4 \sqrt {c+d x}}+\frac {(4 b C d-11 b c D-a d D) \sqrt {a+b x} \sqrt {c+d x}}{4 b d^4}+\frac {D (a+b x)^{3/2} \sqrt {c+d x}}{2 b d^3}-\frac {\left (a^2 d^2 D-2 a b d (2 C d-5 c D)+b^2 \left (20 c C d-8 B d^2-35 c^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{3/2} d^{9/2}} \] Output:

2/3*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(b*x+a)^(3/2)/d^3/(-a*d+b*c)/(d*x+c)^(3/ 
2)+2*(-B*d^2+2*C*c*d-3*D*c^2)*(b*x+a)^(1/2)/d^4/(d*x+c)^(1/2)+1/4*(4*C*b*d 
-D*a*d-11*D*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b/d^4+1/2*D*(b*x+a)^(3/2)*(d* 
x+c)^(1/2)/b/d^3-1/4*(a^2*d^2*D-2*a*b*d*(2*C*d-5*D*c)+b^2*(-8*B*d^2+20*C*c 
*d-35*D*c^2))*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(3/2) 
/d^(9/2)
 

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {a+b x} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {b} \sqrt {d} \sqrt {a+b x} \left (3 a^2 d^2 D (c+d x)^2+b^2 \left (105 c^4 D-8 A d^4 x-20 c^3 d (3 C-7 D x)+2 c d^3 x (16 B-3 x (2 C+D x))+c^2 d^2 (24 B+x (-80 C+21 D x))\right )-2 a b d \left (50 c^3 D+c^2 (-26 C d+69 d D x)+4 c d^2 (2 B+3 x (-3 C+D x))+d^3 \left (4 A-3 x \left (-4 B+2 C x+D x^2\right )\right )\right )\right )-3 (b c-a d) \left (-a^2 d^2 D+2 a b d (2 C d-5 c D)+b^2 \left (-20 c C d+8 B d^2+35 c^2 D\right )\right ) (c+d x)^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{12 b^{3/2} d^{9/2} (-b c+a d) (c+d x)^{3/2}} \] Input:

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

Output:

(Sqrt[b]*Sqrt[d]*Sqrt[a + b*x]*(3*a^2*d^2*D*(c + d*x)^2 + b^2*(105*c^4*D - 
 8*A*d^4*x - 20*c^3*d*(3*C - 7*D*x) + 2*c*d^3*x*(16*B - 3*x*(2*C + D*x)) + 
 c^2*d^2*(24*B + x*(-80*C + 21*D*x))) - 2*a*b*d*(50*c^3*D + c^2*(-26*C*d + 
 69*d*D*x) + 4*c*d^2*(2*B + 3*x*(-3*C + D*x)) + d^3*(4*A - 3*x*(-4*B + 2*C 
*x + D*x^2)))) - 3*(b*c - a*d)*(-(a^2*d^2*D) + 2*a*b*d*(2*C*d - 5*c*D) + b 
^2*(-20*c*C*d + 8*B*d^2 + 35*c^2*D))*(c + d*x)^(3/2)*ArcTanh[(Sqrt[d]*Sqrt 
[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(12*b^(3/2)*d^(9/2)*(-(b*c) + a*d)*(c 
 + d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2124, 27, 1193, 27, 90, 60, 66, 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.

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

\(\Big \downarrow \) 2124

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1193

\(\displaystyle \frac {2 (a+b x)^{3/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 \int \frac {(b c-a d) \sqrt {a+b x} \left (a d (C d-2 c D)-b \left (-8 D c^2+5 C d c-2 B d^2\right )-d (b c-a d) D x\right )}{2 d^3 \sqrt {c+d x}}dx}{b c-a d}+\frac {2 (a+b x)^{3/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{d^3 \sqrt {c+d x}}}{b c-a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 (a+b x)^{3/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\int \frac {\sqrt {a+b x} \left (a d (C d-2 c D)-b \left (-8 D c^2+5 C d c-2 B d^2\right )-d (b c-a d) D x\right )}{\sqrt {c+d x}}dx}{d^3}+\frac {2 (a+b x)^{3/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{d^3 \sqrt {c+d x}}}{b c-a d}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {2 (a+b x)^{3/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {-\frac {\left (a^2 d^2 D-2 a b d (2 C d-5 c D)+b^2 \left (-8 B d^2-35 c^2 D+20 c C d\right )\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 b}-\frac {D (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{2 b}}{d^3}+\frac {2 (a+b x)^{3/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{d^3 \sqrt {c+d x}}}{b c-a d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (a+b x)^{3/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {-\frac {\left (a^2 d^2 D-2 a b d (2 C d-5 c D)+b^2 \left (-8 B d^2-35 c^2 D+20 c C d\right )\right ) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 b}-\frac {D (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{2 b}}{d^3}+\frac {2 (a+b x)^{3/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{d^3 \sqrt {c+d x}}}{b c-a d}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {2 (a+b x)^{3/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {-\frac {\left (a^2 d^2 D-2 a b d (2 C d-5 c D)+b^2 \left (-8 B d^2-35 c^2 D+20 c C d\right )\right ) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 b}-\frac {D (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{2 b}}{d^3}+\frac {2 (a+b x)^{3/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{d^3 \sqrt {c+d x}}}{b c-a d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (a+b x)^{3/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {-\frac {\left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right ) \left (a^2 d^2 D-2 a b d (2 C d-5 c D)+b^2 \left (-8 B d^2-35 c^2 D+20 c C d\right )\right )}{4 b}-\frac {D (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{2 b}}{d^3}+\frac {2 (a+b x)^{3/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{d^3 \sqrt {c+d x}}}{b c-a d}\)

Input:

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

Output:

(2*(A + (c*(c*C*d - B*d^2 - c^2*D))/d^3)*(a + b*x)^(3/2))/(3*(b*c - a*d)*( 
c + d*x)^(3/2)) - ((2*(2*c*C*d - B*d^2 - 3*c^2*D)*(a + b*x)^(3/2))/(d^3*Sq 
rt[c + d*x]) + (-1/2*((b*c - a*d)*D*(a + b*x)^(3/2)*Sqrt[c + d*x])/b - ((a 
^2*d^2*D - 2*a*b*d*(2*C*d - 5*c*D) + b^2*(20*c*C*d - 8*B*d^2 - 35*c^2*D))* 
((Sqrt[a + b*x]*Sqrt[c + d*x])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + 
b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(3/2))))/(4*b))/d^3)/(b*c - a*d 
)
 

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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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 1193

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x 
 + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + 
e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) 
), x] + Simp[1/((m + 1)*(e*f - d*g))   Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex 
pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a 
, b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] &&  !IntegerQ[n] 
&&  !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
 

rule 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2144\) vs. \(2(228)=456\).

Time = 0.53 (sec) , antiderivative size = 2145, normalized size of antiderivative = 8.12

method	result	size
default	\(\text {Expression too large to display}\)	\(2145\)

Input:

int((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x,method=_RETURNVERBOS 
E)
 

Output:

-1/24*(6*D*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/ 
(d*b)^(1/2))*a^3*c*d^4*x-12*C*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d 
*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b*c^2*d^3+72*C*ln(1/2*(2*b*d*x+2*((b*x 
+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^2*c^3*d^2+27*D*ln 
(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))* 
a^2*b*c^3*d^2-12*D*a*b*d^4*x^3*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+12*D*b^ 
2*c*d^3*x^3*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+200*D*a*b*c^3*d*(d*b)^(1/2 
)*((b*x+a)*(d*x+c))^(1/2)+72*C*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*( 
d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^2*c*d^4*x^2+27*D*ln(1/2*(2*b*d*x+2*(( 
b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b*c*d^4*x^2-13 
5*D*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^( 
1/2))*a*b^2*c^2*d^3*x^2-48*B*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d* 
b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^2*c*d^4*x-48*B*b^2*c^2*d^2*(d*b)^(1/2)* 
((b*x+a)*(d*x+c))^(1/2)+120*C*b^2*c^3*d*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2 
)-6*D*a^2*c^2*d^2*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-24*B*ln(1/2*(2*b*d*x 
+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^2*d^5*x^2 
+24*B*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b) 
^(1/2))*b^3*c*d^4*x^2-12*C*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b) 
^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b*d^5*x^2-120*C*ln(1/2*(2*b*d*x+2*((b*x+a 
)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^3*c^3*d^2*x-24*C*a...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 673 vs. \(2 (227) = 454\).

Time = 12.26 (sec) , antiderivative size = 1360, normalized size of antiderivative = 5.15 \[ \int \frac {\sqrt {a+b x} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x, algorithm="fr 
icas")
 

Output:

[1/48*(3*(35*D*b^3*c^5 - 5*(9*D*a*b^2 + 4*C*b^3)*c^4*d + (9*D*a^2*b + 24*C 
*a*b^2 + 8*B*b^3)*c^3*d^2 + (D*a^3 - 4*C*a^2*b - 8*B*a*b^2)*c^2*d^3 + (35* 
D*b^3*c^3*d^2 - 5*(9*D*a*b^2 + 4*C*b^3)*c^2*d^3 + (9*D*a^2*b + 24*C*a*b^2 
+ 8*B*b^3)*c*d^4 + (D*a^3 - 4*C*a^2*b - 8*B*a*b^2)*d^5)*x^2 + 2*(35*D*b^3* 
c^4*d - 5*(9*D*a*b^2 + 4*C*b^3)*c^3*d^2 + (9*D*a^2*b + 24*C*a*b^2 + 8*B*b^ 
3)*c^2*d^3 + (D*a^3 - 4*C*a^2*b - 8*B*a*b^2)*c*d^4)*x)*sqrt(b*d)*log(8*b^2 
*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b* 
d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(105*D*b^3*c 
^4*d - 16*B*a*b^2*c*d^4 - 8*A*a*b^2*d^5 - 20*(5*D*a*b^2 + 3*C*b^3)*c^3*d^2 
 + (3*D*a^2*b + 52*C*a*b^2 + 24*B*b^3)*c^2*d^3 - 6*(D*b^3*c*d^4 - D*a*b^2* 
d^5)*x^3 + 3*(7*D*b^3*c^2*d^3 - 4*(2*D*a*b^2 + C*b^3)*c*d^4 + (D*a^2*b + 4 
*C*a*b^2)*d^5)*x^2 + 2*(70*D*b^3*c^3*d^2 - (69*D*a*b^2 + 40*C*b^3)*c^2*d^3 
 + (3*D*a^2*b + 36*C*a*b^2 + 16*B*b^3)*c*d^4 - 4*(3*B*a*b^2 + A*b^3)*d^5)* 
x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*c^3*d^5 - a*b^2*c^2*d^6 + (b^3*c*d^7 
- a*b^2*d^8)*x^2 + 2*(b^3*c^2*d^6 - a*b^2*c*d^7)*x), -1/24*(3*(35*D*b^3*c^ 
5 - 5*(9*D*a*b^2 + 4*C*b^3)*c^4*d + (9*D*a^2*b + 24*C*a*b^2 + 8*B*b^3)*c^3 
*d^2 + (D*a^3 - 4*C*a^2*b - 8*B*a*b^2)*c^2*d^3 + (35*D*b^3*c^3*d^2 - 5*(9* 
D*a*b^2 + 4*C*b^3)*c^2*d^3 + (9*D*a^2*b + 24*C*a*b^2 + 8*B*b^3)*c*d^4 + (D 
*a^3 - 4*C*a^2*b - 8*B*a*b^2)*d^5)*x^2 + 2*(35*D*b^3*c^4*d - 5*(9*D*a*b^2 
+ 4*C*b^3)*c^3*d^2 + (9*D*a^2*b + 24*C*a*b^2 + 8*B*b^3)*c^2*d^3 + (D*a^...
 

Sympy [F]

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

integrate((b*x+a)**(1/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**(5/2),x)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x, algorithm="ma 
xima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 654 vs. \(2 (227) = 454\).

Time = 0.24 (sec) , antiderivative size = 654, normalized size of antiderivative = 2.48 \[ \int \frac {\sqrt {a+b x} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x, algorithm="gi 
ac")
 

Output:

1/12*((3*(b*x + a)*(2*(D*b^5*c*d^6*abs(b) - D*a*b^4*d^7*abs(b))*(b*x + a)/ 
(b^6*c*d^7 - a*b^5*d^8) - (7*D*b^6*c^2*d^5*abs(b) - 2*D*a*b^5*c*d^6*abs(b) 
 - 4*C*b^6*c*d^6*abs(b) - 5*D*a^2*b^4*d^7*abs(b) + 4*C*a*b^5*d^7*abs(b))/( 
b^6*c*d^7 - a*b^5*d^8)) - 4*(35*D*b^7*c^3*d^4*abs(b) - 45*D*a*b^6*c^2*d^5* 
abs(b) - 20*C*b^7*c^2*d^5*abs(b) + 9*D*a^2*b^5*c*d^6*abs(b) + 24*C*a*b^6*c 
*d^6*abs(b) + 8*B*b^7*c*d^6*abs(b) + 3*D*a^3*b^4*d^7*abs(b) - 6*C*a^2*b^5* 
d^7*abs(b) - 6*B*a*b^6*d^7*abs(b) - 2*A*b^7*d^7*abs(b))/(b^6*c*d^7 - a*b^5 
*d^8))*(b*x + a) - 3*(35*D*b^8*c^4*d^3*abs(b) - 80*D*a*b^7*c^3*d^4*abs(b) 
- 20*C*b^8*c^3*d^4*abs(b) + 54*D*a^2*b^6*c^2*d^5*abs(b) + 44*C*a*b^7*c^2*d 
^5*abs(b) + 8*B*b^8*c^2*d^5*abs(b) - 8*D*a^3*b^5*c*d^6*abs(b) - 28*C*a^2*b 
^6*c*d^6*abs(b) - 16*B*a*b^7*c*d^6*abs(b) - D*a^4*b^4*d^7*abs(b) + 4*C*a^3 
*b^5*d^7*abs(b) + 8*B*a^2*b^6*d^7*abs(b))/(b^6*c*d^7 - a*b^5*d^8))*sqrt(b* 
x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 1/4*(35*D*b^2*c^2*abs(b) - 
10*D*a*b*c*d*abs(b) - 20*C*b^2*c*d*abs(b) - D*a^2*d^2*abs(b) + 4*C*a*b*d^2 
*abs(b) + 8*B*b^2*d^2*abs(b))*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2* 
c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^4)
 

Mupad [F(-1)]

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

int(((a + b*x)^(1/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(5/2),x)
 

Output:

int(((a + b*x)^(1/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(5/2), x)
 

Reduce [F]

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

int((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x)
 

Output:

int((b*x+a)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x)