\(\int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx\) [257]

Optimal result

Integrand size = 22, antiderivative size = 234 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (b c-a d) (c+d x)^{3/2}}+\frac {4 c (4 b c-3 a d) (a+b x)^{3/2}}{3 d^3 (b c-a d) \sqrt {c+d x}}-\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4 (b c-a d)}+\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d^3}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b} d^{9/2}} \] Output:

2/3*c^2*(b*x+a)^(5/2)/d^2/(-a*d+b*c)/(d*x+c)^(3/2)+4/3*c*(-3*a*d+4*b*c)*(b 
*x+a)^(3/2)/d^3/(-a*d+b*c)/(d*x+c)^(1/2)-1/4*(3*a^2*d^2-30*a*b*c*d+35*b^2* 
c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^4/(-a*d+b*c)+1/2*(b*x+a)^(3/2)*(d*x+c)^ 
(1/2)/d^3+1/4*(3*a^2*d^2-30*a*b*c*d+35*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1 
/2)/b^(1/2)/(d*x+c)^(1/2))/b^(1/2)/d^(9/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.64 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (a d \left (55 c^2+78 c d x+15 d^2 x^2\right )-b \left (105 c^3+140 c^2 d x+21 c d^2 x^2-6 d^3 x^3\right )\right )}{12 d^4 (c+d x)^{3/2}}+\frac {\left (35 b^2 c^2-30 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 \sqrt {b} d^{9/2}} \] Input:

Integrate[(x^2*(a + b*x)^(3/2))/(c + d*x)^(5/2),x]
 

Output:

(Sqrt[a + b*x]*(a*d*(55*c^2 + 78*c*d*x + 15*d^2*x^2) - b*(105*c^3 + 140*c^ 
2*d*x + 21*c*d^2*x^2 - 6*d^3*x^3)))/(12*d^4*(c + d*x)^(3/2)) + ((35*b^2*c^ 
2 - 30*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[ 
a + b*x])])/(4*Sqrt[b]*d^(9/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {100, 27, 87, 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.

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{b c-a d}}{3 d^2 (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 c^2 (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2} (b c-a d)}-\frac {\frac {4 c (a+b x)^{5/2} (4 b c-3 a d)}{\sqrt {c+d x} (b c-a d)}-\frac {\left (3 a^2 d^2-30 a b c d+35 b^2 c^2\right ) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{b c-a d}}{3 d^2 (b c-a d)}\)

Input:

Int[(x^2*(a + b*x)^(3/2))/(c + d*x)^(5/2),x]
 

Output:

(2*c^2*(a + b*x)^(5/2))/(3*d^2*(b*c - a*d)*(c + d*x)^(3/2)) - ((4*c*(4*b*c 
 - 3*a*d)*(a + b*x)^(5/2))/((b*c - a*d)*Sqrt[c + d*x]) - ((35*b^2*c^2 - 30 
*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)))/(b*c - a*d 
))/(3*d^2*(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 87

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

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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. \(675\) vs. \(2(196)=392\).

Time = 0.23 (sec) , antiderivative size = 676, normalized size of antiderivative = 2.89

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (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^{2} d^{4} x^{2}-90 \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 c \,d^{3} x^{2}+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^{2} c^{2} d^{2} x^{2}+12 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, b \,d^{3} x^{3}+18 \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} c \,d^{3} x -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 \,c^{2} d^{2} x +210 \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^{2} c^{3} d x +30 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, a \,d^{3} x^{2}-42 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, b c \,d^{2} x^{2}+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^{2} c^{2} d^{2}-90 \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 \,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^{2} c^{4}+156 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, a c \,d^{2} x -280 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, b \,c^{2} d x +110 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, a \,c^{2} d -210 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, b \,c^{3}\right )}{24 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, d^{4} \left (x d +c \right )^{\frac {3}{2}}}\)	\(676\)

Input:

int(x^2*(b*x+a)^(3/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/24*(b*x+a)^(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^2*d^4*x^2-90*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*c*d^3*x^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))*b^2*c^2*d^2*x 
^2+12*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*b*d^3*x^3+18*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*c*d^3*x-180*l 
n(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*c^2*d^2*x+210*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^2*c^3*d*x+30*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a* 
d^3*x^2-42*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*b*c*d^2*x^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^2*c^2*d^2 
-90*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*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^2*c^4+156*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a*c* 
d^2*x-280*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*b*c^2*d*x+110*((b*x+a)*(d*x+ 
c))^(1/2)*(d*b)^(1/2)*a*c^2*d-210*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*b*c^ 
3)/(d*b)^(1/2)/((b*x+a)*(d*x+c))^(1/2)/d^4/(d*x+c)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 594, normalized size of antiderivative = 2.54 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\left [\frac {3 \, {\left (35 \, b^{2} c^{4} - 30 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (6 \, b^{2} d^{4} x^{3} - 105 \, b^{2} c^{3} d + 55 \, a b c^{2} d^{2} - 3 \, {\left (7 \, b^{2} c d^{3} - 5 \, a b d^{4}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d^{2} - 39 \, a b c d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b d^{7} x^{2} + 2 \, b c d^{6} x + b c^{2} d^{5}\right )}}, -\frac {3 \, {\left (35 \, b^{2} c^{4} - 30 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (6 \, b^{2} d^{4} x^{3} - 105 \, b^{2} c^{3} d + 55 \, a b c^{2} d^{2} - 3 \, {\left (7 \, b^{2} c d^{3} - 5 \, a b d^{4}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d^{2} - 39 \, a b c d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (b d^{7} x^{2} + 2 \, b c d^{6} x + b c^{2} d^{5}\right )}}\right ] \] Input:

integrate(x^2*(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/48*(3*(35*b^2*c^4 - 30*a*b*c^3*d + 3*a^2*c^2*d^2 + (35*b^2*c^2*d^2 - 30 
*a*b*c*d^3 + 3*a^2*d^4)*x^2 + 2*(35*b^2*c^3*d - 30*a*b*c^2*d^2 + 3*a^2*c*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*(6*b^2*d^4*x^3 - 105*b^2*c^3*d + 55*a*b*c^2*d^2 - 3*(7*b^2*c 
*d^3 - 5*a*b*d^4)*x^2 - 2*(70*b^2*c^2*d^2 - 39*a*b*c*d^3)*x)*sqrt(b*x + a) 
*sqrt(d*x + c))/(b*d^7*x^2 + 2*b*c*d^6*x + b*c^2*d^5), -1/24*(3*(35*b^2*c^ 
4 - 30*a*b*c^3*d + 3*a^2*c^2*d^2 + (35*b^2*c^2*d^2 - 30*a*b*c*d^3 + 3*a^2* 
d^4)*x^2 + 2*(35*b^2*c^3*d - 30*a*b*c^2*d^2 + 3*a^2*c*d^3)*x)*sqrt(-b*d)*a 
rctan(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*(6*b^2*d^4*x^3 - 105*b^2 
*c^3*d + 55*a*b*c^2*d^2 - 3*(7*b^2*c*d^3 - 5*a*b*d^4)*x^2 - 2*(70*b^2*c^2* 
d^2 - 39*a*b*c*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^7*x^2 + 2*b*c*d^6 
*x + b*c^2*d^5)]
 

Sympy [F]

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

integrate(x**2*(b*x+a)**(3/2)/(d*x+c)**(5/2),x)
 

Output:

Integral(x**2*(a + b*x)**(3/2)/(c + d*x)**(5/2), x)
 

Maxima [F(-2)]

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

integrate(x^2*(b*x+a)^(3/2)/(d*x+c)^(5/2),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. 393 vs. \(2 (196) = 392\).

Time = 0.20 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.68 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {{\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{6} c d^{6} - a b^{5} d^{7}\right )} {\left (b x + a\right )}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}} - \frac {7 \, b^{7} c^{2} d^{5} - 6 \, a b^{6} c d^{6} - a^{2} b^{5} d^{7}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}}\right )} - \frac {4 \, {\left (35 \, b^{8} c^{3} d^{4} - 65 \, a b^{7} c^{2} d^{5} + 33 \, a^{2} b^{6} c d^{6} - 3 \, a^{3} b^{5} d^{7}\right )}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (35 \, b^{9} c^{4} d^{3} - 100 \, a b^{8} c^{3} d^{4} + 98 \, a^{2} b^{7} c^{2} d^{5} - 36 \, a^{3} b^{6} c d^{6} + 3 \, a^{4} b^{5} d^{7}\right )}}{b^{4} c d^{7} {\left | b \right |} - a b^{3} d^{8} {\left | b \right |}}\right )} \sqrt {b x + a}}{12 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {{\left (35 \, b^{3} c^{2} - 30 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} d^{4} {\left | b \right |}} \] Input:

integrate(x^2*(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

1/12*((3*(b*x + a)*(2*(b^6*c*d^6 - a*b^5*d^7)*(b*x + a)/(b^4*c*d^7*abs(b) 
- a*b^3*d^8*abs(b)) - (7*b^7*c^2*d^5 - 6*a*b^6*c*d^6 - a^2*b^5*d^7)/(b^4*c 
*d^7*abs(b) - a*b^3*d^8*abs(b))) - 4*(35*b^8*c^3*d^4 - 65*a*b^7*c^2*d^5 + 
33*a^2*b^6*c*d^6 - 3*a^3*b^5*d^7)/(b^4*c*d^7*abs(b) - a*b^3*d^8*abs(b)))*( 
b*x + a) - 3*(35*b^9*c^4*d^3 - 100*a*b^8*c^3*d^4 + 98*a^2*b^7*c^2*d^5 - 36 
*a^3*b^6*c*d^6 + 3*a^4*b^5*d^7)/(b^4*c*d^7*abs(b) - a*b^3*d^8*abs(b)))*sqr 
t(b*x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 1/4*(35*b^3*c^2 - 30*a* 
b^2*c*d + 3*a^2*b*d^2)*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^4*abs(b))
 

Mupad [F(-1)]

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

int((x^2*(a + b*x)^(3/2))/(c + d*x)^(5/2),x)
 

Output:

int((x^2*(a + b*x)^(3/2))/(c + d*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.99 (sec) , antiderivative size = 732, normalized size of antiderivative = 3.13 \[ \int \frac {x^2 (a+b x)^{3/2}}{(c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

int(x^2*(b*x+a)^(3/2)/(d*x+c)^(5/2),x)
 

Output:

(440*sqrt(c + d*x)*sqrt(a + b*x)*a*b*c**2*d**2 + 624*sqrt(c + d*x)*sqrt(a 
+ b*x)*a*b*c*d**3*x + 120*sqrt(c + d*x)*sqrt(a + b*x)*a*b*d**4*x**2 - 840* 
sqrt(c + d*x)*sqrt(a + b*x)*b**2*c**3*d - 1120*sqrt(c + d*x)*sqrt(a + b*x) 
*b**2*c**2*d**2*x - 168*sqrt(c + d*x)*sqrt(a + b*x)*b**2*c*d**3*x**2 + 48* 
sqrt(c + d*x)*sqrt(a + b*x)*b**2*d**4*x**3 + 72*sqrt(d)*sqrt(b)*log((sqrt( 
d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*c**2*d**2 
+ 144*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/ 
sqrt(a*d - b*c))*a**2*c*d**3*x + 72*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + 
b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*d**4*x**2 - 720*sqrt(d 
)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b 
*c))*a*b*c**3*d - 1440*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b 
)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b*c**2*d**2*x - 720*sqrt(d)*sqrt(b)*lo 
g((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b*c*d 
**3*x**2 + 840*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c 
 + d*x))/sqrt(a*d - b*c))*b**2*c**4 + 1680*sqrt(d)*sqrt(b)*log((sqrt(d)*sq 
rt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**2*c**3*d*x + 840* 
sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a 
*d - b*c))*b**2*c**2*d**2*x**2 + 15*sqrt(d)*sqrt(b)*a**2*c**2*d**2 + 30*sq 
rt(d)*sqrt(b)*a**2*c*d**3*x + 15*sqrt(d)*sqrt(b)*a**2*d**4*x**2 - 190*sqrt 
(d)*sqrt(b)*a*b*c**3*d - 380*sqrt(d)*sqrt(b)*a*b*c**2*d**2*x - 190*sqrt...