\(\int \frac {x (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx\) [310]

Optimal result

Integrand size = 20, antiderivative size = 183 \[ \int \frac {x (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {2 c (a+b x)^{5/2}}{3 d^2 (c+d x)^{3/2}}+\frac {2 (8 b c-3 a d) (a+b x)^{3/2}}{3 d^3 \sqrt {c+d x}}-\frac {5 b (7 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^4}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^3}+\frac {5 \sqrt {b} (7 b c-3 a d) (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{9/2}} \] Output:

2/3*c*(b*x+a)^(5/2)/d^2/(d*x+c)^(3/2)+2/3*(-3*a*d+8*b*c)*(b*x+a)^(3/2)/d^3 
/(d*x+c)^(1/2)-5/4*b*(-3*a*d+7*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^4+1/2*b* 
(b*x+a)^(3/2)*(d*x+c)^(1/2)/d^3+5/4*b^(1/2)*(-3*a*d+7*b*c)*(-a*d+b*c)*arct 
anh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(9/2)
 

Mathematica [C] (verified)

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

Time = 10.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.60 \[ \int \frac {x (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {2 (a+b x)^{7/2} \left (7 c (-b c+a d)+(7 b c-3 a d) (c+d x) \sqrt {\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{2},\frac {9}{2},\frac {d (a+b x)}{-b c+a d}\right )\right )}{21 d (b c-a d)^2 (c+d x)^{3/2}} \] Input:

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

Output:

(2*(a + b*x)^(7/2)*(7*c*(-(b*c) + a*d) + (7*b*c - 3*a*d)*(c + d*x)*Sqrt[(b 
*(c + d*x))/(b*c - a*d)]*Hypergeometric2F1[3/2, 7/2, 9/2, (d*(a + b*x))/(- 
(b*c) + a*d)]))/(21*d*(b*c - a*d)^2*(c + d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {87, 57, 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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, 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 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. \(749\) vs. \(2(145)=290\).

Time = 0.22 (sec) , antiderivative size = 750, normalized size of antiderivative = 4.10

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (45 \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 \,d^{4} x^{2}-150 \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^{2} 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^{3} c^{2} d^{2} x^{2}+12 b^{2} d^{3} x^{3} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+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^{2} b c \,d^{3} x -300 \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^{2} 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^{3} c^{3} d x +54 a b \,d^{3} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}-42 b^{2} c \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+45 \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 \,c^{2} d^{2}-150 \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^{2} 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^{3} c^{4}-48 a^{2} d^{3} x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+316 a b c \,d^{2} x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}-280 b^{2} c^{2} d x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}-32 a^{2} c \,d^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+230 a b \,c^{2} d \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}-210 b^{2} c^{3} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\right )}{24 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \left (x d +c \right )^{\frac {3}{2}} d^{4}}\)	\(750\)

Input:

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

Output:

1/24*(b*x+a)^(1/2)*(45*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*d^4*x^2-150*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^2*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^3*c^2 
*d^2*x^2+12*b^2*d^3*x^3*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+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^2*b*c*d 
^3*x-300*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^2*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^3*c^3*d*x+54*a*b*d^3*x^2*((b*x+a)*(d*x+ 
c))^(1/2)*(d*b)^(1/2)-42*b^2*c*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2) 
+45*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*c^2*d^2-150*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^2*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^3*c^4-48*a^2*d^3*x*((b*x+a)* 
(d*x+c))^(1/2)*(d*b)^(1/2)+316*a*b*c*d^2*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^( 
1/2)-280*b^2*c^2*d*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-32*a^2*c*d^2*((b* 
x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+230*a*b*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(d*b 
)^(1/2)-210*b^2*c^3*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2))/((b*x+a)*(d*x+c)) 
^(1/2)/(d*b)^(1/2)/(d*x+c)^(3/2)/d^4
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (145) = 290\).

Time = 0.35 (sec) , antiderivative size = 619, normalized size of antiderivative = 3.38 \[ \int \frac {x (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\left [\frac {15 \, {\left (7 \, b^{2} c^{4} - 10 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (7 \, b^{2} c^{2} d^{2} - 10 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (7 \, b^{2} c^{3} d - 10 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (6 \, b^{2} d^{3} x^{3} - 105 \, b^{2} c^{3} + 115 \, a b c^{2} d - 16 \, a^{2} c d^{2} - 3 \, {\left (7 \, b^{2} c d^{2} - 9 \, a b d^{3}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d - 79 \, a b c d^{2} + 12 \, a^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}}, -\frac {15 \, {\left (7 \, b^{2} c^{4} - 10 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (7 \, b^{2} c^{2} d^{2} - 10 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (7 \, b^{2} c^{3} d - 10 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) - 2 \, {\left (6 \, b^{2} d^{3} x^{3} - 105 \, b^{2} c^{3} + 115 \, a b c^{2} d - 16 \, a^{2} c d^{2} - 3 \, {\left (7 \, b^{2} c d^{2} - 9 \, a b d^{3}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d - 79 \, a b c d^{2} + 12 \, a^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}}\right ] \] Input:

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

Output:

[1/48*(15*(7*b^2*c^4 - 10*a*b*c^3*d + 3*a^2*c^2*d^2 + (7*b^2*c^2*d^2 - 10* 
a*b*c*d^3 + 3*a^2*d^4)*x^2 + 2*(7*b^2*c^3*d - 10*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^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d 
+ a*b*d^2)*x) + 4*(6*b^2*d^3*x^3 - 105*b^2*c^3 + 115*a*b*c^2*d - 16*a^2*c* 
d^2 - 3*(7*b^2*c*d^2 - 9*a*b*d^3)*x^2 - 2*(70*b^2*c^2*d - 79*a*b*c*d^2 + 1 
2*a^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^6*x^2 + 2*c*d^5*x + c^2*d^4) 
, -1/24*(15*(7*b^2*c^4 - 10*a*b*c^3*d + 3*a^2*c^2*d^2 + (7*b^2*c^2*d^2 - 1 
0*a*b*c*d^3 + 3*a^2*d^4)*x^2 + 2*(7*b^2*c^3*d - 10*a*b*c^2*d^2 + 3*a^2*c*d 
^3)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x 
+ c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) - 2*(6*b^2*d^3*x^ 
3 - 105*b^2*c^3 + 115*a*b*c^2*d - 16*a^2*c*d^2 - 3*(7*b^2*c*d^2 - 9*a*b*d^ 
3)*x^2 - 2*(70*b^2*c^2*d - 79*a*b*c*d^2 + 12*a^2*d^3)*x)*sqrt(b*x + a)*sqr 
t(d*x + c))/(d^6*x^2 + 2*c*d^5*x + c^2*d^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x*(b*x+a)^(5/2)/(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. 404 vs. \(2 (145) = 290\).

Time = 0.21 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.21 \[ \int \frac {x (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx=\frac {{\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{5} c d^{6} {\left | b \right |} - a b^{4} d^{7} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{4} c d^{7} - a b^{3} d^{8}} - \frac {7 \, b^{6} c^{2} d^{5} {\left | b \right |} - 10 \, a b^{5} c d^{6} {\left | b \right |} + 3 \, a^{2} b^{4} d^{7} {\left | b \right |}}{b^{4} c d^{7} - a b^{3} d^{8}}\right )} - \frac {20 \, {\left (7 \, b^{7} c^{3} d^{4} {\left | b \right |} - 17 \, a b^{6} c^{2} d^{5} {\left | b \right |} + 13 \, a^{2} b^{5} c d^{6} {\left | b \right |} - 3 \, a^{3} b^{4} d^{7} {\left | b \right |}\right )}}{b^{4} c d^{7} - a b^{3} d^{8}}\right )} {\left (b x + a\right )} - \frac {15 \, {\left (7 \, b^{8} c^{4} d^{3} {\left | b \right |} - 24 \, a b^{7} c^{3} d^{4} {\left | b \right |} + 30 \, a^{2} b^{6} c^{2} d^{5} {\left | b \right |} - 16 \, a^{3} b^{5} c d^{6} {\left | b \right |} + 3 \, a^{4} b^{4} d^{7} {\left | b \right |}\right )}}{b^{4} c d^{7} - a b^{3} d^{8}}\right )} \sqrt {b x + a}}{12 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {5 \, {\left (7 \, b^{2} c^{2} {\left | b \right |} - 10 \, a b c d {\left | b \right |} + 3 \, a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} d^{4}} \] Input:

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

Output:

1/12*((3*(b*x + a)*(2*(b^5*c*d^6*abs(b) - a*b^4*d^7*abs(b))*(b*x + a)/(b^4 
*c*d^7 - a*b^3*d^8) - (7*b^6*c^2*d^5*abs(b) - 10*a*b^5*c*d^6*abs(b) + 3*a^ 
2*b^4*d^7*abs(b))/(b^4*c*d^7 - a*b^3*d^8)) - 20*(7*b^7*c^3*d^4*abs(b) - 17 
*a*b^6*c^2*d^5*abs(b) + 13*a^2*b^5*c*d^6*abs(b) - 3*a^3*b^4*d^7*abs(b))/(b 
^4*c*d^7 - a*b^3*d^8))*(b*x + a) - 15*(7*b^8*c^4*d^3*abs(b) - 24*a*b^7*c^3 
*d^4*abs(b) + 30*a^2*b^6*c^2*d^5*abs(b) - 16*a^3*b^5*c*d^6*abs(b) + 3*a^4* 
b^4*d^7*abs(b))/(b^4*c*d^7 - a*b^3*d^8))*sqrt(b*x + a)/(b^2*c + (b*x + a)* 
b*d - a*b*d)^(3/2) - 5/4*(7*b^2*c^2*abs(b) - 10*a*b*c*d*abs(b) + 3*a^2*d^2 
*abs(b))*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)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 771, normalized size of antiderivative = 4.21 \[ \int \frac {x (a+b x)^{5/2}}{(c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 128*sqrt(c + d*x)*sqrt(a + b*x)*a**2*c*d**3 - 192*sqrt(c + d*x)*sqrt(a 
 + b*x)*a**2*d**4*x + 920*sqrt(c + d*x)*sqrt(a + b*x)*a*b*c**2*d**2 + 1264 
*sqrt(c + d*x)*sqrt(a + b*x)*a*b*c*d**3*x + 216*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 + 360*sqr 
t(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 + 720*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 + 360*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 - 1200*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sq 
rt(c + d*x))/sqrt(a*d - b*c))*a*b*c**3*d - 2400*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 
 - 1200*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*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)*sqrt(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) + sqr 
t(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**2*c**2*d**2*x**2 + 95*sqrt(d)*sqrt 
(b)*a**2*c**2*d**2 + 190*sqrt(d)*sqrt(b)*a**2*c*d**3*x + 95*sqrt(d)*sqr...