\(\int \frac {(c+d x)^{3/2} (A+B x+C x^2+D x^3)}{(a+b x)^{3/2}} \, dx\) [139]

Optimal result

Integrand size = 34, antiderivative size = 441 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx=-\frac {\left (315 a^3 d^3 D-35 a^2 b d^2 (8 C d+3 c D)+5 a b^2 d \left (16 c C d+48 B d^2-3 c^2 D\right )+b^3 \left (8 c^2 C d-48 B c d^2-192 A d^3-3 c^3 D\right )\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^5 d^2}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) (c+d x)^{3/2}}{b^4 \sqrt {a+b x}}+\frac {\left (\frac {69 a^2 d D}{b}-2 a (20 C d+9 c D)+b \left (8 c C+16 B d-\frac {3 c^2 D}{d}\right )\right ) \sqrt {a+b x} (c+d x)^{3/2}}{32 b^3 d}+\frac {(8 b C d-3 b c D-21 a d D) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^4 d}+\frac {D (a+b x)^{3/2} (c+d x)^{5/2}}{4 b^3 d}-\frac {(b c-a d) \left (315 a^3 d^3 D-35 a^2 b d^2 (8 C d+3 c D)+5 a b^2 d \left (16 c C d+48 B d^2-3 c^2 D\right )+b^3 \left (8 c^2 C d-48 B c d^2-192 A d^3-3 c^3 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{11/2} d^{5/2}} \] Output:

-1/64*(315*a^3*d^3*D-35*a^2*b*d^2*(8*C*d+3*D*c)+5*a*b^2*d*(48*B*d^2+16*C*c 
*d-3*D*c^2)+b^3*(-192*A*d^3-48*B*c*d^2+8*C*c^2*d-3*D*c^3))*(b*x+a)^(1/2)*( 
d*x+c)^(1/2)/b^5/d^2-2*(A*b^3-a*(B*b^2-C*a*b+D*a^2))*(d*x+c)^(3/2)/b^4/(b* 
x+a)^(1/2)+1/32*(69*a^2*d*D/b-2*a*(20*C*d+9*D*c)+b*(8*C*c+16*B*d-3*c^2*D/d 
))*(b*x+a)^(1/2)*(d*x+c)^(3/2)/b^3/d+1/24*(8*C*b*d-21*D*a*d-3*D*b*c)*(b*x+ 
a)^(3/2)*(d*x+c)^(3/2)/b^4/d+1/4*D*(b*x+a)^(3/2)*(d*x+c)^(5/2)/b^3/d-1/64* 
(-a*d+b*c)*(315*a^3*d^3*D-35*a^2*b*d^2*(8*C*d+3*D*c)+5*a*b^2*d*(48*B*d^2+1 
6*C*c*d-3*D*c^2)+b^3*(-192*A*d^3-48*B*c*d^2+8*C*c^2*d-3*D*c^3))*arctanh(d^ 
(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(11/2)/d^(5/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.27 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.56 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx=\frac {2 (c+d x)^{3/2} \left (-(b c-a d)^3 D \operatorname {Hypergeometric2F1}\left (-\frac {9}{2},-\frac {1}{2},\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )+b (b c-a d)^2 (-C d+3 c D) \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},-\frac {1}{2},\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )-b^2 (b c-a d) \left (-2 c C d+B d^2+3 c^2 D\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {1}{2},\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )+b^3 \left (-c^2 C d+B c d^2-A d^3+c^3 D\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )\right )}{b^4 d^3 \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/2}} \] Input:

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

Output:

(2*(c + d*x)^(3/2)*(-((b*c - a*d)^3*D*Hypergeometric2F1[-9/2, -1/2, 1/2, ( 
d*(a + b*x))/(-(b*c) + a*d)]) + b*(b*c - a*d)^2*(-(C*d) + 3*c*D)*Hypergeom 
etric2F1[-7/2, -1/2, 1/2, (d*(a + b*x))/(-(b*c) + a*d)] - b^2*(b*c - a*d)* 
(-2*c*C*d + B*d^2 + 3*c^2*D)*Hypergeometric2F1[-5/2, -1/2, 1/2, (d*(a + b* 
x))/(-(b*c) + a*d)] + b^3*(-(c^2*C*d) + B*c*d^2 - A*d^3 + c^3*D)*Hypergeom 
etric2F1[-3/2, -1/2, 1/2, (d*(a + b*x))/(-(b*c) + a*d)]))/(b^4*d^3*Sqrt[a 
+ b*x]*((b*(c + d*x))/(b*c - a*d))^(3/2))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 374, normalized size of antiderivative = 0.85, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {2124, 27, 1194, 27, 90, 60, 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.

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

\(\Big \downarrow \) 2124

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1194

\(\displaystyle \frac {\frac {\int \frac {(c+d x)^{3/2} \left (-\frac {35 d^2 D a^3}{b}+2 d (20 C d+3 c D) a^2-b \left (3 D c^2+8 C d c+40 B d^2\right ) a+8 b^2 d (B c+4 A d)+(b c-a d) (8 b C d-21 a D d-3 b c D) x\right )}{2 \sqrt {a+b x}}dx}{4 b^2 d}+\frac {D (a+b x)^{3/2} (c+d x)^{5/2} (b c-a d)}{4 b^3 d}}{b c-a d}-\frac {2 (c+d x)^{5/2} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {(c+d x)^{3/2} \left (-\frac {35 d^2 D a^3}{b}+2 d (20 C d+3 c D) a^2-b \left (3 D c^2+8 C d c+40 B d^2\right ) a+8 b^2 d (B c+4 A d)+(b c-a d) (8 b C d-21 a D d-3 b c D) x\right )}{\sqrt {a+b x}}dx}{8 b^2 d}+\frac {D (a+b x)^{3/2} (c+d x)^{5/2} (b c-a d)}{4 b^3 d}}{b c-a d}-\frac {2 (c+d x)^{5/2} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d) (-21 a d D-3 b c D+8 b C d)}{3 b d}-\frac {\left (315 a^3 d^3 D-35 a^2 b d^2 (3 c D+8 C d)+5 a b^2 d \left (48 B d^2-3 c^2 D+16 c C d\right )+b^3 \left (-192 A d^3-48 B c d^2-3 c^3 D+8 c^2 C d\right )\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}}dx}{6 b d}}{8 b^2 d}+\frac {D (a+b x)^{3/2} (c+d x)^{5/2} (b c-a d)}{4 b^3 d}}{b c-a d}-\frac {2 (c+d x)^{5/2} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d) (-21 a d D-3 b c D+8 b C d)}{3 b d}-\frac {\left (315 a^3 d^3 D-35 a^2 b d^2 (3 c D+8 C d)+5 a b^2 d \left (48 B d^2-3 c^2 D+16 c C d\right )+b^3 \left (-192 A d^3-48 B c d^2-3 c^3 D+8 c^2 C d\right )\right ) \left (\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}}{8 b^2 d}+\frac {D (a+b x)^{3/2} (c+d x)^{5/2} (b c-a d)}{4 b^3 d}}{b c-a d}-\frac {2 (c+d x)^{5/2} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d) (-21 a d D-3 b c D+8 b C d)}{3 b d}-\frac {\left (315 a^3 d^3 D-35 a^2 b d^2 (3 c D+8 C d)+5 a b^2 d \left (48 B d^2-3 c^2 D+16 c C d\right )+b^3 \left (-192 A d^3-48 B c d^2-3 c^3 D+8 c^2 C d\right )\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}}{8 b^2 d}+\frac {D (a+b x)^{3/2} (c+d x)^{5/2} (b c-a d)}{4 b^3 d}}{b c-a d}-\frac {2 (c+d x)^{5/2} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d) (-21 a d D-3 b c D+8 b C d)}{3 b d}-\frac {\left (315 a^3 d^3 D-35 a^2 b d^2 (3 c D+8 C d)+5 a b^2 d \left (48 B d^2-3 c^2 D+16 c C d\right )+b^3 \left (-192 A d^3-48 B c d^2-3 c^3 D+8 c^2 C d\right )\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}}{8 b^2 d}+\frac {D (a+b x)^{3/2} (c+d x)^{5/2} (b c-a d)}{4 b^3 d}}{b c-a d}-\frac {2 (c+d x)^{5/2} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-a d) (-21 a d D-3 b c D+8 b C d)}{3 b d}-\frac {\left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right ) \left (315 a^3 d^3 D-35 a^2 b d^2 (3 c D+8 C d)+5 a b^2 d \left (48 B d^2-3 c^2 D+16 c C d\right )+b^3 \left (-192 A d^3-48 B c d^2-3 c^3 D+8 c^2 C d\right )\right )}{6 b d}}{8 b^2 d}+\frac {D (a+b x)^{3/2} (c+d x)^{5/2} (b c-a d)}{4 b^3 d}}{b c-a d}-\frac {2 (c+d x)^{5/2} \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{\sqrt {a+b x} (b c-a d)}\)

Input:

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

Output:

(-2*(A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*(c + d*x)^(5/2))/((b*c - a*d)*Sq 
rt[a + b*x]) + (((b*c - a*d)*D*(a + b*x)^(3/2)*(c + d*x)^(5/2))/(4*b^3*d) 
+ (((b*c - a*d)*(8*b*C*d - 3*b*c*D - 21*a*d*D)*Sqrt[a + b*x]*(c + d*x)^(5/ 
2))/(3*b*d) - ((315*a^3*d^3*D - 35*a^2*b*d^2*(8*C*d + 3*c*D) + 5*a*b^2*d*( 
16*c*C*d + 48*B*d^2 - 3*c^2*D) + b^3*(8*c^2*C*d - 48*B*c*d^2 - 192*A*d^3 - 
 3*c^3*D))*((Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b) + (3*(b*c - a*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*b)))/(6*b*d))/(8*b^2*d))/( 
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 1194

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x 
)^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Simp[1/(g*e^(2*p)*(m + n + 2 
*p + 1))   Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^( 
2*p)*(a + b*x + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p) 
*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ 
[p, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && NeQ[m + n + 2*p + 1, 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. \(2427\) vs. \(2(399)=798\).

Time = 0.51 (sec) , antiderivative size = 2428, normalized size of antiderivative = 5.51

method	result	size
default	\(\text {Expression too large to display}\)	\(2428\)

Input:

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

Output:

-1/384*(d*x+c)^(1/2)*(264*D*a*b^3*c*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^ 
(1/2)+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))*b^5*c^3*d*x-945*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^4*b*d^4*x-576*A*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^3+1890*D*a^ 
4*d^3*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-9*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^4*x+576*A*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* 
d^4-720*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^3*b^2*d^4+840*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^4*b*d^4-9*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^4+864*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^2*b^3* 
c*d^3-144*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^4*c^2*d^2-1080*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^3*b^2*c*d^3+216*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^2*d^2 
+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^3*d+1260*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^4*b*c*d^3-270*D*ln(1/2*(2*b*d*x+2*((b*x+...
 

Fricas [A] (verification not implemented)

Time = 5.51 (sec) , antiderivative size = 1454, normalized size of antiderivative = 3.30 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/768*(3*(3*D*a*b^4*c^4 + 4*(3*D*a^2*b^3 - 2*C*a*b^4)*c^3*d + 6*(15*D*a^ 
3*b^2 - 12*C*a^2*b^3 + 8*B*a*b^4)*c^2*d^2 - 12*(35*D*a^4*b - 30*C*a^3*b^2 
+ 24*B*a^2*b^3 - 16*A*a*b^4)*c*d^3 + (315*D*a^5 - 280*C*a^4*b + 240*B*a^3* 
b^2 - 192*A*a^2*b^3)*d^4 + (3*D*b^5*c^4 + 4*(3*D*a*b^4 - 2*C*b^5)*c^3*d + 
6*(15*D*a^2*b^3 - 12*C*a*b^4 + 8*B*b^5)*c^2*d^2 - 12*(35*D*a^3*b^2 - 30*C* 
a^2*b^3 + 24*B*a*b^4 - 16*A*b^5)*c*d^3 + (315*D*a^4*b - 280*C*a^3*b^2 + 24 
0*B*a^2*b^3 - 192*A*a*b^4)*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)*sqr 
t(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(48*D*b^5*d^4*x^4 - 9*D*a*b^4*c^ 
3*d - 3*(13*D*a^2*b^3 - 8*C*a*b^4)*c^2*d^2 + (945*D*a^3*b^2 - 800*C*a^2*b^ 
3 + 624*B*a*b^4 - 384*A*b^5)*c*d^3 - 3*(315*D*a^4*b - 280*C*a^3*b^2 + 240* 
B*a^2*b^3 - 192*A*a*b^4)*d^4 + 8*(9*D*b^5*c*d^3 - (9*D*a*b^4 - 8*C*b^5)*d^ 
4)*x^3 + 2*(3*D*b^5*c^2*d^2 - 2*(33*D*a*b^4 - 28*C*b^5)*c*d^3 + (63*D*a^2* 
b^3 - 56*C*a*b^4 + 48*B*b^5)*d^4)*x^2 - (9*D*b^5*c^3*d + 3*(11*D*a*b^4 - 8 
*C*b^5)*c^2*d^2 - (357*D*a^2*b^3 - 304*C*a*b^4 + 240*B*b^5)*c*d^3 + (315*D 
*a^3*b^2 - 280*C*a^2*b^3 + 240*B*a*b^4 - 192*A*b^5)*d^4)*x)*sqrt(b*x + a)* 
sqrt(d*x + c))/(b^7*d^3*x + a*b^6*d^3), -1/384*(3*(3*D*a*b^4*c^4 + 4*(3*D* 
a^2*b^3 - 2*C*a*b^4)*c^3*d + 6*(15*D*a^3*b^2 - 12*C*a^2*b^3 + 8*B*a*b^4)*c 
^2*d^2 - 12*(35*D*a^4*b - 30*C*a^3*b^2 + 24*B*a^2*b^3 - 16*A*a*b^4)*c*d^3 
+ (315*D*a^5 - 280*C*a^4*b + 240*B*a^3*b^2 - 192*A*a^2*b^3)*d^4 + (3*D*...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((d*x+c)^(3/2)*(D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),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 [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 793, normalized size of antiderivative = 1.80 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{(a+b x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/192*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b* 
x + a)*D*d*abs(b)/b^7 + (9*D*b^28*c*d^6*abs(b) - 33*D*a*b^27*d^7*abs(b) + 
8*C*b^28*d^7*abs(b))/(b^34*d^6)) + (3*D*b^29*c^2*d^5*abs(b) - 174*D*a*b^28 
*c*d^6*abs(b) + 56*C*b^29*c*d^6*abs(b) + 315*D*a^2*b^27*d^7*abs(b) - 152*C 
*a*b^28*d^7*abs(b) + 48*B*b^29*d^7*abs(b))/(b^34*d^6)) - 3*(3*D*b^30*c^3*d 
^4*abs(b) + 15*D*a*b^29*c^2*d^5*abs(b) - 8*C*b^30*c^2*d^5*abs(b) - 279*D*a 
^2*b^28*c*d^6*abs(b) + 176*C*a*b^29*c*d^6*abs(b) - 80*B*b^30*c*d^6*abs(b) 
+ 325*D*a^3*b^27*d^7*abs(b) - 232*C*a^2*b^28*d^7*abs(b) + 144*B*a*b^29*d^7 
*abs(b) - 64*A*b^30*d^7*abs(b))/(b^34*d^6))*sqrt(b*x + a) + 4*(D*a^3*b^2*c 
^2*d*abs(b) - C*a^2*b^3*c^2*d*abs(b) + B*a*b^4*c^2*d*abs(b) - A*b^5*c^2*d* 
abs(b) - 2*D*a^4*b*c*d^2*abs(b) + 2*C*a^3*b^2*c*d^2*abs(b) - 2*B*a^2*b^3*c 
*d^2*abs(b) + 2*A*a*b^4*c*d^2*abs(b) + D*a^5*d^3*abs(b) - C*a^4*b*d^3*abs( 
b) + B*a^3*b^2*d^3*abs(b) - A*a^2*b^3*d^3*abs(b))/((b^2*c - a*b*d - (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 
) - 1/128*(3*D*b^4*c^4*abs(b) + 12*D*a*b^3*c^3*d*abs(b) - 8*C*b^4*c^3*d*ab 
s(b) + 90*D*a^2*b^2*c^2*d^2*abs(b) - 72*C*a*b^3*c^2*d^2*abs(b) + 48*B*b^4* 
c^2*d^2*abs(b) - 420*D*a^3*b*c*d^3*abs(b) + 360*C*a^2*b^2*c*d^3*abs(b) - 2 
88*B*a*b^3*c*d^3*abs(b) + 192*A*b^4*c*d^3*abs(b) + 315*D*a^4*d^4*abs(b) - 
280*C*a^3*b*d^4*abs(b) + 240*B*a^2*b^2*d^4*abs(b) - 192*A*a*b^3*d^4*abs(b) 
)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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