\(\int \frac {x^2 (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx\) [324]

Optimal result

Integrand size = 22, antiderivative size = 253 \[ \int \frac {x^2 (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {(b c-a d) \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^4 d^2}+\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{96 b^3 d^2}-\frac {(3 b c+13 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 b^2 d^2}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b^2 d}+\frac {(b c-a d)^2 \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{9/2} d^{5/2}} \] Output:

1/64*(-a*d+b*c)*(35*a^2*d^2+10*a*b*c*d+3*b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1 
/2)/b^4/d^2+1/96*(35*a^2*d^2+10*a*b*c*d+3*b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^( 
3/2)/b^3/d^2-1/24*(13*a*d+3*b*c)*(b*x+a)^(1/2)*(d*x+c)^(5/2)/b^2/d^2+1/4*( 
b*x+a)^(3/2)*(d*x+c)^(5/2)/b^2/d+1/64*(-a*d+b*c)^2*(35*a^2*d^2+10*a*b*c*d+ 
3*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(9/2)/d^ 
(5/2)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.76 \[ \int \frac {x^2 (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-105 a^3 d^3+5 a^2 b d^2 (29 c+14 d x)-a b^2 d \left (15 c^2+92 c d x+56 d^2 x^2\right )+b^3 \left (-9 c^3+6 c^2 d x+72 c d^2 x^2+48 d^3 x^3\right )\right )}{192 b^4 d^2}+\frac {(b c-a d)^2 \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{9/2} d^{5/2}} \] Input:

Integrate[(x^2*(c + d*x)^(3/2))/Sqrt[a + b*x],x]
 

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(-105*a^3*d^3 + 5*a^2*b*d^2*(29*c + 14*d*x) - 
 a*b^2*d*(15*c^2 + 92*c*d*x + 56*d^2*x^2) + b^3*(-9*c^3 + 6*c^2*d*x + 72*c 
*d^2*x^2 + 48*d^3*x^3)))/(192*b^4*d^2) + ((b*c - a*d)^2*(3*b^2*c^2 + 10*a* 
b*c*d + 35*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x] 
)])/(64*b^(9/2)*d^(5/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.89, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {101, 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.

\(\displaystyle \int \frac {x^2 (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx\)

\(\Big \downarrow \) 101

\(\displaystyle \frac {\int -\frac {(c+d x)^{3/2} (2 a c+(3 b c+7 a d) x)}{2 \sqrt {a+b x}}dx}{4 b d}+\frac {x \sqrt {a+b x} (c+d x)^{5/2}}{4 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{5/2}}{4 b d}-\frac {\int \frac {(c+d x)^{3/2} (2 a c+(3 b c+7 a d) x)}{\sqrt {a+b x}}dx}{8 b d}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{5/2}}{4 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (7 a d+3 b c)}{3 b d}-\frac {\left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}}dx}{6 b d}}{8 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{5/2}}{4 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (7 a d+3 b c)}{3 b d}-\frac {\left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \left (\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}}{8 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{5/2}}{4 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (7 a d+3 b c)}{3 b d}-\frac {\left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}}{8 b d}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{5/2}}{4 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (7 a d+3 b c)}{3 b d}-\frac {\left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}}{8 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{5/2}}{4 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{5/2} (7 a d+3 b c)}{3 b d}-\frac {\left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}}{8 b d}\)

Input:

Int[(x^2*(c + d*x)^(3/2))/Sqrt[a + b*x],x]
 

Output:

(x*Sqrt[a + b*x]*(c + d*x)^(5/2))/(4*b*d) - (((3*b*c + 7*a*d)*Sqrt[a + b*x 
]*(c + d*x)^(5/2))/(3*b*d) - ((3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2)*((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*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 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(573\) vs. \(2(215)=430\).

Time = 0.23 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.27

method	result	size
default	\(\frac {\sqrt {x d +c}\, \sqrt {b x +a}\, \left (96 b^{3} d^{3} x^{3} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}-112 a \,b^{2} d^{3} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+144 b^{3} c \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+105 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{4} d^{4}-180 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{3} b c \,d^{3}+54 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b^{2} c^{2} d^{2}+12 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{3} c^{3} d +9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{4} c^{4}+140 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, a^{2} b \,d^{3} x -184 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, a \,b^{2} c \,d^{2} x +12 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, b^{3} c^{2} d x -210 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, a^{3} d^{3}+290 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, a^{2} b c \,d^{2}-30 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, a \,b^{2} c^{2} d -18 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, b^{3} c^{3}\right )}{384 d^{2} b^{4} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}}\)	\(574\)

Input:

int(x^2*(d*x+c)^(3/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/384*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(96*b^3*d^3*x^3*((b*x+a)*(d*x+c))^(1/2)* 
(d*b)^(1/2)-112*a*b^2*d^3*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+144*b^3* 
c*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+105*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*d^4-180*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*c* 
d^3+54*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^2*c^2*d^2+12*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^3*c^3*d+9*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^4*c^4+140*((b*x+a)*(d*x+c) 
)^(1/2)*(d*b)^(1/2)*a^2*b*d^3*x-184*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a* 
b^2*c*d^2*x+12*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*b^3*c^2*d*x-210*((b*x+a 
)*(d*x+c))^(1/2)*(d*b)^(1/2)*a^3*d^3+290*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/ 
2)*a^2*b*c*d^2-30*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a*b^2*c^2*d-18*((b*x 
+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*b^3*c^3)/d^2/b^4/((b*x+a)*(d*x+c))^(1/2)/(d 
*b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 546, normalized size of antiderivative = 2.16 \[ \int \frac {x^2 (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\left [\frac {3 \, {\left (3 \, b^{4} c^{4} + 4 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 60 \, a^{3} b c d^{3} + 35 \, a^{4} d^{4}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (48 \, b^{4} d^{4} x^{3} - 9 \, b^{4} c^{3} d - 15 \, a b^{3} c^{2} d^{2} + 145 \, a^{2} b^{2} c d^{3} - 105 \, a^{3} b d^{4} + 8 \, {\left (9 \, b^{4} c d^{3} - 7 \, a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (3 \, b^{4} c^{2} d^{2} - 46 \, a b^{3} c d^{3} + 35 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, b^{5} d^{3}}, -\frac {3 \, {\left (3 \, b^{4} c^{4} + 4 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 60 \, a^{3} b c d^{3} + 35 \, a^{4} d^{4}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (48 \, b^{4} d^{4} x^{3} - 9 \, b^{4} c^{3} d - 15 \, a b^{3} c^{2} d^{2} + 145 \, a^{2} b^{2} c d^{3} - 105 \, a^{3} b d^{4} + 8 \, {\left (9 \, b^{4} c d^{3} - 7 \, a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (3 \, b^{4} c^{2} d^{2} - 46 \, a b^{3} c d^{3} + 35 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, b^{5} d^{3}}\right ] \] Input:

integrate(x^2*(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/768*(3*(3*b^4*c^4 + 4*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 - 60*a^3*b*c*d^3 
 + 35*a^4*d^4)*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*(48*b^4*d^4*x^3 - 9*b^4*c^3*d - 15*a*b^3*c^2*d^2 + 1 
45*a^2*b^2*c*d^3 - 105*a^3*b*d^4 + 8*(9*b^4*c*d^3 - 7*a*b^3*d^4)*x^2 + 2*( 
3*b^4*c^2*d^2 - 46*a*b^3*c*d^3 + 35*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x 
 + c))/(b^5*d^3), -1/384*(3*(3*b^4*c^4 + 4*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^ 
2 - 60*a^3*b*c*d^3 + 35*a^4*d^4)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a* 
d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c* 
d + a*b*d^2)*x)) - 2*(48*b^4*d^4*x^3 - 9*b^4*c^3*d - 15*a*b^3*c^2*d^2 + 14 
5*a^2*b^2*c*d^3 - 105*a^3*b*d^4 + 8*(9*b^4*c*d^3 - 7*a*b^3*d^4)*x^2 + 2*(3 
*b^4*c^2*d^2 - 46*a*b^3*c*d^3 + 35*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x 
+ c))/(b^5*d^3)]
 

Sympy [F]

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

integrate(x**2*(d*x+c)**(3/2)/(b*x+a)**(1/2),x)
 

Output:

Integral(x**2*(c + d*x)**(3/2)/sqrt(a + b*x), x)
 

Maxima [F(-2)]

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

integrate(x^2*(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="maxima")
 

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. 482 vs. \(2 (215) = 430\).

Time = 0.18 (sec) , antiderivative size = 482, normalized size of antiderivative = 1.91 \[ \int \frac {x^2 (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )}}{b^{3}} + \frac {b^{12} c d^{5} - 25 \, a b^{11} d^{6}}{b^{14} d^{6}}\right )} - \frac {5 \, b^{13} c^{2} d^{4} + 14 \, a b^{12} c d^{5} - 163 \, a^{2} b^{11} d^{6}}{b^{14} d^{6}}\right )} + \frac {3 \, {\left (5 \, b^{14} c^{3} d^{3} + 9 \, a b^{13} c^{2} d^{4} + 15 \, a^{2} b^{12} c d^{5} - 93 \, a^{3} b^{11} d^{6}\right )}}{b^{14} d^{6}}\right )} \sqrt {b x + a} + \frac {3 \, {\left (5 \, b^{4} c^{4} + 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 20 \, a^{3} b c d^{3} - 35 \, a^{4} d^{4}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{2} d^{3}}\right )} d {\left | b \right |}}{b^{2}} + \frac {8 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (4 \, b x + 4 \, a + \frac {b c d^{3} - 13 \, a d^{4}}{d^{4}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (b^{2} c^{2} d^{2} + 2 \, a b c d^{3} - 11 \, a^{2} d^{4}\right )}}{d^{4}}\right )} \sqrt {b x + a} - \frac {3 \, {\left (b^{4} c^{3} + a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - 5 \, a^{3} b d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2}}\right )} c {\left | b \right |}}{b^{4}}}{192 \, b} \] Input:

integrate(x^2*(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")
 

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)/b^3 + (b^12*c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^13*c^2*d^4 
+ 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c^3*d^3 + 9* 
a*b^13*c^2*d^4 + 15*a^2*b^12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt(b*x 
 + a) + 3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 
- 35*a^4*d^4)*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^3))*d*abs(b)/b^2 + 8*(sqrt(b^2*c + (b*x + a) 
*b*d - a*b*d)*(2*(4*b*x + 4*a + (b*c*d^3 - 13*a*d^4)/d^4)*(b*x + a) - 3*(b 
^2*c^2*d^2 + 2*a*b*c*d^3 - 11*a^2*d^4)/d^4)*sqrt(b*x + a) - 3*(b^4*c^3 + a 
*b^3*c^2*d + 3*a^2*b^2*c*d^2 - 5*a^3*b*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + 
a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d^2))*c*abs(b)/b^4)/ 
b
 

Mupad [F(-1)]

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

int((x^2*(c + d*x)^(3/2))/(a + b*x)^(1/2),x)
 

Output:

int((x^2*(c + d*x)^(3/2))/(a + b*x)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.86 \[ \int \frac {x^2 (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {-105 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{3} b \,d^{4}+145 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b^{2} c \,d^{3}+70 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b^{2} d^{4} x -15 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{3} c^{2} d^{2}-92 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{3} c \,d^{3} x -56 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{3} d^{4} x^{2}-9 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} c^{3} d +6 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} c^{2} d^{2} x +72 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} c \,d^{3} x^{2}+48 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} d^{4} x^{3}+105 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{4} d^{4}-180 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} b c \,d^{3}+54 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b^{2} c^{2} d^{2}+12 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{3} c^{3} d +9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{4} c^{4}}{192 b^{5} d^{3}} \] Input:

int(x^2*(d*x+c)^(3/2)/(b*x+a)^(1/2),x)
 

Output:

( - 105*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*d**4 + 145*sqrt(c + d*x)*sqrt(a 
 + b*x)*a**2*b**2*c*d**3 + 70*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*d**4*x 
 - 15*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*c**2*d**2 - 92*sqrt(c + d*x)*sqrt 
(a + b*x)*a*b**3*c*d**3*x - 56*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*d**4*x** 
2 - 9*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c**3*d + 6*sqrt(c + d*x)*sqrt(a + b 
*x)*b**4*c**2*d**2*x + 72*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c*d**3*x**2 + 4 
8*sqrt(c + d*x)*sqrt(a + b*x)*b**4*d**4*x**3 + 105*sqrt(d)*sqrt(b)*log((sq 
rt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**4*d**4 - 
180*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sq 
rt(a*d - b*c))*a**3*b*c*d**3 + 54*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b* 
x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b**2*c**2*d**2 + 12*sqrt 
(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - 
 b*c))*a*b**3*c**3*d + 9*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt 
(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**4*c**4)/(192*b**5*d**3)