\(\int \frac {x^2 (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx\) [242]

Optimal result

Integrand size = 22, antiderivative size = 253 \[ \int \frac {x^2 (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=-\frac {(b c-a d) \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d^4}+\frac {\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) (a+b x)^{3/2} \sqrt {c+d x}}{96 b^2 d^3}-\frac {(7 b c+9 a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b^2 d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b^2 d}+\frac {(b c-a d)^2 \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{9/2}} \] Output:

-1/64*(-a*d+b*c)*(3*a^2*d^2+10*a*b*c*d+35*b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^( 
1/2)/b^2/d^4+1/96*(3*a^2*d^2+10*a*b*c*d+35*b^2*c^2)*(b*x+a)^(3/2)*(d*x+c)^ 
(1/2)/b^2/d^3-1/24*(9*a*d+7*b*c)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b^2/d^2+1/4*( 
b*x+a)^(7/2)*(d*x+c)^(1/2)/b^2/d+1/64*(-a*d+b*c)^2*(3*a^2*d^2+10*a*b*c*d+3 
5*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2)/d^ 
(9/2)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.75 \[ \int \frac {x^2 (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-9 a^3 d^3+3 a^2 b d^2 (-5 c+2 d x)+a b^2 d \left (145 c^2-92 c d x+72 d^2 x^2\right )+b^3 \left (-105 c^3+70 c^2 d x-56 c d^2 x^2+48 d^3 x^3\right )\right )}{192 b^2 d^4}+\frac {(b c-a d)^2 \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{64 b^{5/2} d^{9/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(-9*a^3*d^3 + 3*a^2*b*d^2*(-5*c + 2*d*x) + a* 
b^2*d*(145*c^2 - 92*c*d*x + 72*d^2*x^2) + b^3*(-105*c^3 + 70*c^2*d*x - 56* 
c*d^2*x^2 + 48*d^3*x^3)))/(192*b^2*d^4) + ((b*c - a*d)^2*(35*b^2*c^2 + 10* 
a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x 
])])/(64*b^(5/2)*d^(9/2))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 225, 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.

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(x*(a + b*x)^(5/2)*Sqrt[c + d*x])/(4*b*d) - (((7*b*c + 3*a*d)*(a + b*x)^(5 
/2)*Sqrt[c + d*x])/(3*b*d) - ((35*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*(((a + 
 b*x)^(3/2)*Sqrt[c + d*x])/(2*d) - (3*(b*c - a*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*d)))/(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 = 0.24 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.27

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 546, normalized size of antiderivative = 2.16 \[ \int \frac {x^2 (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (35 \, b^{4} c^{4} - 60 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a^{3} b c d^{3} + 3 \, a^{4} d^{4}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (48 \, b^{4} d^{4} x^{3} - 105 \, b^{4} c^{3} d + 145 \, a b^{3} c^{2} d^{2} - 15 \, a^{2} b^{2} c d^{3} - 9 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 9 \, a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 46 \, a b^{3} c d^{3} + 3 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, b^{3} d^{5}}, -\frac {3 \, {\left (35 \, b^{4} c^{4} - 60 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a^{3} b c d^{3} + 3 \, a^{4} d^{4}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (48 \, b^{4} d^{4} x^{3} - 105 \, b^{4} c^{3} d + 145 \, a b^{3} c^{2} d^{2} - 15 \, a^{2} b^{2} c d^{3} - 9 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 9 \, a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 46 \, a b^{3} c d^{3} + 3 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, b^{3} d^{5}}\right ] \] Input:

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

Output:

[1/768*(3*(35*b^4*c^4 - 60*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^ 
3 + 3*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 - 105*b^4*c^3*d + 145*a*b^3*c^2*d^2 
- 15*a^2*b^2*c*d^3 - 9*a^3*b*d^4 - 8*(7*b^4*c*d^3 - 9*a*b^3*d^4)*x^2 + 2*( 
35*b^4*c^2*d^2 - 46*a*b^3*c*d^3 + 3*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x 
 + c))/(b^3*d^5), -1/384*(3*(35*b^4*c^4 - 60*a*b^3*c^3*d + 18*a^2*b^2*c^2* 
d^2 + 4*a^3*b*c*d^3 + 3*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 - 105*b^4*c^3*d + 145*a*b^3*c^2*d^2 - 
 15*a^2*b^2*c*d^3 - 9*a^3*b*d^4 - 8*(7*b^4*c*d^3 - 9*a*b^3*d^4)*x^2 + 2*(3 
5*b^4*c^2*d^2 - 46*a*b^3*c*d^3 + 3*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x 
+ c))/(b^3*d^5)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Time = 0.15 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.15 \[ \int \frac {x^2 (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )}}{b^{3} d} - \frac {7 \, b^{7} c d^{5} + 9 \, a b^{6} d^{6}}{b^{9} d^{7}}\right )} + \frac {35 \, b^{8} c^{2} d^{4} + 10 \, a b^{7} c d^{5} + 3 \, a^{2} b^{6} d^{6}}{b^{9} d^{7}}\right )} - \frac {3 \, {\left (35 \, b^{9} c^{3} d^{3} - 25 \, a b^{8} c^{2} d^{4} - 7 \, a^{2} b^{7} c d^{5} - 3 \, a^{3} b^{6} d^{6}\right )}}{b^{9} d^{7}}\right )} \sqrt {b x + a} - \frac {3 \, {\left (35 \, b^{4} c^{4} - 60 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a^{3} b c d^{3} + 3 \, a^{4} d^{4}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{2} d^{4}}\right )} b}{192 \, {\left | b \right |}} \] Input:

integrate(x^2*(b*x+a)^(3/2)/(d*x+c)^(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*d) - (7*b^7*c*d^5 + 9*a*b^6*d^6)/(b^9*d^7)) + (35*b^8*c^2*d^4 
 + 10*a*b^7*c*d^5 + 3*a^2*b^6*d^6)/(b^9*d^7)) - 3*(35*b^9*c^3*d^3 - 25*a*b 
^8*c^2*d^4 - 7*a^2*b^7*c*d^5 - 3*a^3*b^6*d^6)/(b^9*d^7))*sqrt(b*x + a) - 3 
*(35*b^4*c^4 - 60*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 + 3*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^4))*b/abs(b)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.86 \[ \int \frac {x^2 (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx=\frac {-9 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{3} b \,d^{4}-15 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b^{2} c \,d^{3}+6 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b^{2} d^{4} x +145 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{3} c^{2} d^{2}-92 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{3} c \,d^{3} x +72 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{3} d^{4} x^{2}-105 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} c^{3} d +70 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} c^{2} d^{2} x -56 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} c \,d^{3} x^{2}+48 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{4} d^{4} x^{3}+9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{4} d^{4}+12 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} b c \,d^{3}+54 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b^{2} c^{2} d^{2}-180 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{3} c^{3} d +105 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{4} c^{4}}{192 b^{3} d^{5}} \] Input:

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

Output:

( - 9*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b*d**4 - 15*sqrt(c + d*x)*sqrt(a + 
b*x)*a**2*b**2*c*d**3 + 6*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**2*d**4*x + 1 
45*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 + 72*sqrt(c + d*x)*sqrt(a + b*x)*a*b**3*d**4*x**2 - 
 105*sqrt(c + d*x)*sqrt(a + b*x)*b**4*c**3*d + 70*sqrt(c + d*x)*sqrt(a + b 
*x)*b**4*c**2*d**2*x - 56*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 + 9*sqrt(d)*sqrt(b)*log((sqrt 
(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**4*d**4 + 12 
*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt( 
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 - 180*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 + 105*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**3*d**5)