\(\int \frac {\sqrt {c+d x} (A+B x+C x^2+D x^3)}{(a+b x)^{5/2}} \, dx\) [147]

Optimal result

Integrand size = 34, antiderivative size = 263 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{5/2}} \, dx=-\frac {2 \left (b^2 B-2 a b C+3 a^2 D\right ) \sqrt {c+d x}}{b^4 \sqrt {a+b x}}+\frac {(4 b C d-b c D-11 a d D) \sqrt {a+b x} \sqrt {c+d x}}{4 b^4 d}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) (c+d x)^{3/2}}{3 b^3 (b c-a d) (a+b x)^{3/2}}+\frac {D \sqrt {a+b x} (c+d x)^{3/2}}{2 b^3 d}+\frac {\left (35 a^2 d^2 D-10 a b d (2 C d+c D)+b^2 \left (4 c C d+8 B d^2-c^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{9/2} d^{3/2}} \] Output:

-2*(B*b^2-2*C*a*b+3*D*a^2)*(d*x+c)^(1/2)/b^4/(b*x+a)^(1/2)+1/4*(4*C*b*d-11 
*D*a*d-D*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^4/d-2/3*(A*b^3-a*(B*b^2-C*a*b+ 
D*a^2))*(d*x+c)^(3/2)/b^3/(-a*d+b*c)/(b*x+a)^(3/2)+1/2*D*(b*x+a)^(1/2)*(d* 
x+c)^(3/2)/b^3/d+1/4*(35*a^2*d^2*D-10*a*b*d*(2*C*d+D*c)+b^2*(8*B*d^2+4*C*c 
*d-D*c^2))*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(9/2)/d^ 
(3/2)
 

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{5/2}} \, dx=\frac {\sqrt {c+d x} \left (105 a^4 d^2 D-8 A b^4 d (c+d x)-20 a^3 b d (3 C d+5 c D-7 d D x)+3 b^4 c x (-8 B d+4 C d x+D x (c+2 d x))-2 a b^3 \left (8 B d (c-2 d x)+3 x \left (-c^2 D+4 c d (-3 C+D x)+d^2 x (2 C+D x)\right )\right )+a^2 b^2 \left (3 c^2 D+2 c d (26 C-69 D x)+d^2 (24 B+x (-80 C+21 D x))\right )\right )}{12 b^4 d (b c-a d) (a+b x)^{3/2}}+\frac {\left (35 a^2 d^2 D-10 a b d (2 C d+c D)+b^2 \left (4 c C d+8 B d^2-c^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 b^{9/2} d^{3/2}} \] Input:

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

Output:

(Sqrt[c + d*x]*(105*a^4*d^2*D - 8*A*b^4*d*(c + d*x) - 20*a^3*b*d*(3*C*d + 
5*c*D - 7*d*D*x) + 3*b^4*c*x*(-8*B*d + 4*C*d*x + D*x*(c + 2*d*x)) - 2*a*b^ 
3*(8*B*d*(c - 2*d*x) + 3*x*(-(c^2*D) + 4*c*d*(-3*C + D*x) + d^2*x*(2*C + D 
*x))) + a^2*b^2*(3*c^2*D + 2*c*d*(26*C - 69*D*x) + d^2*(24*B + x*(-80*C + 
21*D*x)))))/(12*b^4*d*(b*c - a*d)*(a + b*x)^(3/2)) + ((35*a^2*d^2*D - 10*a 
*b*d*(2*C*d + c*D) + b^2*(4*c*C*d + 8*B*d^2 - c^2*D))*ArcTanh[(Sqrt[b]*Sqr 
t[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(4*b^(9/2)*d^(3/2))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.05, 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.

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

\(\Big \downarrow \) 2124

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1193

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {\frac {\left (35 a^2 d^2 D-10 a b d (c D+2 C d)+b^2 \left (8 B d^2+c^2 (-D)+4 c C d\right )\right ) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 d}+\frac {D \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}{2 d}}{b^3}-\frac {2 (c+d x)^{3/2} \left (3 a^2 D-2 a b C+b^2 B\right )}{b^3 \sqrt {a+b x}}}{b c-a d}-\frac {2 (c+d x)^{3/2} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\frac {\left (35 a^2 d^2 D-10 a b d (c D+2 C d)+b^2 \left (8 B d^2+c^2 (-D)+4 c C d\right )\right ) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 d}+\frac {D \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}{2 d}}{b^3}-\frac {2 (c+d x)^{3/2} \left (3 a^2 D-2 a b C+b^2 B\right )}{b^3 \sqrt {a+b x}}}{b c-a d}-\frac {2 (c+d x)^{3/2} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {\frac {\left (35 a^2 d^2 D-10 a b d (c D+2 C d)+b^2 \left (8 B d^2+c^2 (-D)+4 c C d\right )\right ) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 d}+\frac {D \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}{2 d}}{b^3}-\frac {2 (c+d x)^{3/2} \left (3 a^2 D-2 a b C+b^2 B\right )}{b^3 \sqrt {a+b x}}}{b c-a d}-\frac {2 (c+d x)^{3/2} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {\left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right ) \left (35 a^2 d^2 D-10 a b d (c D+2 C d)+b^2 \left (8 B d^2+c^2 (-D)+4 c C d\right )\right )}{4 d}+\frac {D \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}{2 d}}{b^3}-\frac {2 (c+d x)^{3/2} \left (3 a^2 D-2 a b C+b^2 B\right )}{b^3 \sqrt {a+b x}}}{b c-a d}-\frac {2 (c+d x)^{3/2} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^{3/2} (b c-a d)}\)

Input:

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

Output:

(-2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*(c + d*x)^(3/2))/(3*b^3*(b*c - a*d 
)*(a + b*x)^(3/2)) + ((-2*(b^2*B - 2*a*b*C + 3*a^2*D)*(c + d*x)^(3/2))/(b^ 
3*Sqrt[a + b*x]) + (((b*c - a*d)*D*Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*d) + 
((35*a^2*d^2*D - 10*a*b*d*(2*C*d + c*D) + b^2*(4*c*C*d + 8*B*d^2 - c^2*D)) 
*((Sqrt[a + b*x]*Sqrt[c + d*x])/b + ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + 
 b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(3/2)*Sqrt[d])))/(4*d))/b^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. \(2(227)=454\).

Time = 0.54 (sec) , antiderivative size = 2145, normalized size of antiderivative = 8.16

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

Input:

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

Output:

1/24*(-104*C*a^2*b^2*c*d*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+200*D*a^3*b*c 
*d*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(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^4*c*d^2*x^2-135*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^3*c*d^2*x^2-12*D*a*b^3*c^2*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+32*B*a* 
b^3*c*d*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+105*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^5*d^3-42*D*a^2*b^2*d 
^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-64*B*a*b^3*d^2*x*((b*x+a)*(d*x+ 
c))^(1/2)*(d*b)^(1/2)+48*B*b^4*c*d*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+1 
60*C*a^2*b^2*d^2*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-280*D*a^3*b*d^2*x*( 
(b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+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*b^4*c^2*d*x^2+144*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^3* 
c*d^2*x-24*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^4*c^2*d*x-270*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*b^2*c*d^2*x+54*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^3*c^2*d* 
x-210*D*a^4*d^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+3*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))*b^5*c^3*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)^(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 694 vs. \(2 (226) = 452\).

Time = 11.67 (sec) , antiderivative size = 1402, normalized size of antiderivative = 5.33 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/48*(3*(D*a^2*b^3*c^3 + (9*D*a^3*b^2 - 4*C*a^2*b^3)*c^2*d - (45*D*a^4*b 
 - 24*C*a^3*b^2 + 8*B*a^2*b^3)*c*d^2 + (35*D*a^5 - 20*C*a^4*b + 8*B*a^3*b^ 
2)*d^3 + (D*b^5*c^3 + (9*D*a*b^4 - 4*C*b^5)*c^2*d - (45*D*a^2*b^3 - 24*C*a 
*b^4 + 8*B*b^5)*c*d^2 + (35*D*a^3*b^2 - 20*C*a^2*b^3 + 8*B*a*b^4)*d^3)*x^2 
 + 2*(D*a*b^4*c^3 + (9*D*a^2*b^3 - 4*C*a*b^4)*c^2*d - (45*D*a^3*b^2 - 24*C 
*a^2*b^3 + 8*B*a*b^4)*c*d^2 + (35*D*a^4*b - 20*C*a^3*b^2 + 8*B*a^2*b^3)*d^ 
3)*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*(3*D*a^2*b^3*c^2*d - 4*(25*D*a^3*b^2 - 13*C*a^2*b^3 + 4*B*a*b 
^4 + 2*A*b^5)*c*d^2 + 3*(35*D*a^4*b - 20*C*a^3*b^2 + 8*B*a^2*b^3)*d^3 + 6* 
(D*b^5*c*d^2 - D*a*b^4*d^3)*x^3 + 3*(D*b^5*c^2*d - 4*(2*D*a*b^4 - C*b^5)*c 
*d^2 + (7*D*a^2*b^3 - 4*C*a*b^4)*d^3)*x^2 + 2*(3*D*a*b^4*c^2*d - 3*(23*D*a 
^2*b^3 - 12*C*a*b^4 + 4*B*b^5)*c*d^2 + 2*(35*D*a^3*b^2 - 20*C*a^2*b^3 + 8* 
B*a*b^4 - 2*A*b^5)*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^6*c*d^2 - a 
^3*b^5*d^3 + (b^8*c*d^2 - a*b^7*d^3)*x^2 + 2*(a*b^7*c*d^2 - a^2*b^6*d^3)*x 
), 1/24*(3*(D*a^2*b^3*c^3 + (9*D*a^3*b^2 - 4*C*a^2*b^3)*c^2*d - (45*D*a^4* 
b - 24*C*a^3*b^2 + 8*B*a^2*b^3)*c*d^2 + (35*D*a^5 - 20*C*a^4*b + 8*B*a^3*b 
^2)*d^3 + (D*b^5*c^3 + (9*D*a*b^4 - 4*C*b^5)*c^2*d - (45*D*a^2*b^3 - 24*C* 
a*b^4 + 8*B*b^5)*c*d^2 + (35*D*a^3*b^2 - 20*C*a^2*b^3 + 8*B*a*b^4)*d^3)*x^ 
2 + 2*(D*a*b^4*c^3 + (9*D*a^2*b^3 - 4*C*a*b^4)*c^2*d - (45*D*a^3*b^2 - ...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((d*x+c)^(1/2)*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(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. 1305 vs. \(2 (226) = 452\).

Time = 0.42 (sec) , antiderivative size = 1305, normalized size of antiderivative = 4.96 \[ \int \frac {\sqrt {c+d x} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*D*abs(b 
)/b^6 + (D*b^12*c*d*abs(b) - 13*D*a*b^11*d^2*abs(b) + 4*C*b^12*d^2*abs(b)) 
/(b^17*d^2)) + 1/8*(D*b^2*c^2*abs(b) + 10*D*a*b*c*d*abs(b) - 4*C*b^2*c*d*a 
bs(b) - 35*D*a^2*d^2*abs(b) + 20*C*a*b*d^2*abs(b) - 8*B*b^2*d^2*abs(b))*lo 
g((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt 
(b*d)*b^5*d) - 4/3*(9*D*a^2*b^5*c^3*d*abs(b) - 6*C*a*b^6*c^3*d*abs(b) + 3* 
B*b^7*c^3*d*abs(b) - 28*D*a^3*b^4*c^2*d^2*abs(b) + 19*C*a^2*b^5*c^2*d^2*ab 
s(b) - 10*B*a*b^6*c^2*d^2*abs(b) + A*b^7*c^2*d^2*abs(b) + 29*D*a^4*b^3*c*d 
^3*abs(b) - 20*C*a^3*b^4*c*d^3*abs(b) + 11*B*a^2*b^5*c*d^3*abs(b) - 2*A*a* 
b^6*c*d^3*abs(b) - 10*D*a^5*b^2*d^4*abs(b) + 7*C*a^4*b^3*d^4*abs(b) - 4*B* 
a^3*b^4*d^4*abs(b) + A*a^2*b^5*d^4*abs(b) - 18*(sqrt(b*d)*sqrt(b*x + a) - 
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*D*a^2*b^3*c^2*d*abs(b) + 12*(sqrt(b 
*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*C*a*b^4*c^2*d*a 
bs(b) - 6*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^ 
2*B*b^5*c^2*d*abs(b) + 36*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a 
)*b*d - a*b*d))^2*D*a^3*b^2*c*d^2*abs(b) - 24*(sqrt(b*d)*sqrt(b*x + a) - s 
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*C*a^2*b^3*c*d^2*abs(b) + 12*(sqrt(b* 
d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*B*a*b^4*c*d^2*ab 
s(b) - 18*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^ 
2*D*a^4*b*d^3*abs(b) + 12*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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