\(\int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx\) [237]

Optimal result

Integrand size = 22, antiderivative size = 258 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx=3 d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}+\frac {d (7 b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 c}-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}-\frac {3 \sqrt {c} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a}}+\frac {3 \sqrt {d} \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b}} \] Output:

3*d*(a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)+1/4*d*(5*a*d+7*b*c)*(b*x+a)^(1/2 
)*(d*x+c)^(3/2)/c-1/4*(5*a*d+3*b*c)*(b*x+a)^(1/2)*(d*x+c)^(5/2)/c/x-1/2*(b 
*x+a)^(3/2)*(d*x+c)^(5/2)/x^2-3/4*c^(1/2)*(5*a^2*d^2+10*a*b*c*d+b^2*c^2)*a 
rctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(1/2)+3/4*d^(1/2)*(a 
^2*d^2+10*a*b*c*d+5*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c) 
^(1/2))/b^(1/2)
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx=\frac {1}{4} \left (\frac {\sqrt {a+b x} \sqrt {c+d x} \left (b x \left (-5 c^2+9 c d x+2 d^2 x^2\right )+a \left (-2 c^2-9 c d x+5 d^2 x^2\right )\right )}{x^2}-\frac {3 \sqrt {c} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{\sqrt {a}}+\frac {3 \sqrt {d} \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {b}}\right ) \] Input:

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

Output:

((Sqrt[a + b*x]*Sqrt[c + d*x]*(b*x*(-5*c^2 + 9*c*d*x + 2*d^2*x^2) + a*(-2* 
c^2 - 9*c*d*x + 5*d^2*x^2)))/x^2 - (3*Sqrt[c]*(b^2*c^2 + 10*a*b*c*d + 5*a^ 
2*d^2)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/Sqrt[a] + 
 (3*Sqrt[d]*(5*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d 
*x])/(Sqrt[d]*Sqrt[a + b*x])])/Sqrt[b])/4
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {108, 27, 166, 27, 171, 27, 171, 27, 175, 66, 104, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b x} (c+d x)^{3/2} (3 b c+5 a d+8 b d x)}{2 x^2}dx-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/2*((a + b*x)^(3/2)*(c + d*x)^(5/2))/x^2 + (-(((3*b*c + 5*a*d)*Sqrt[a + 
b*x]*(c + d*x)^(5/2))/(c*x)) + (2*d*(7*b*c + 5*a*d)*Sqrt[a + b*x]*(c + d*x 
)^(3/2) + 3*c*(8*d*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x] - (2*Sqrt[c]*(b 
^2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]* 
Sqrt[c + d*x])])/Sqrt[a] + (2*Sqrt[d]*(5*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*A 
rcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b]))/(2*c))/ 
4
 

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 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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 108

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

rule 166

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - 
a*f)*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* 
c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h 
)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
 

rule 171

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegersQ[2*m, 2*n, 2*p]
 

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

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. \(558\) vs. \(2(208)=416\).

Time = 0.23 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.17

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {x d +c}\, \left (3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} d^{3} x^{2} \sqrt {a c}+30 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a b c \,d^{2} x^{2} \sqrt {a c}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{2} c^{2} d \,x^{2} \sqrt {a c}-15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} c \,d^{2} x^{2} \sqrt {d b}-30 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a b \,c^{2} d \,x^{2} \sqrt {d b}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) b^{2} c^{3} x^{2} \sqrt {d b}+4 b \,d^{2} x^{3} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}+10 a \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}+18 b c d \,x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}-18 a c d x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}-10 b \,c^{2} x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}-4 a \,c^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\right )}{8 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x^{2} \sqrt {d b}\, \sqrt {a c}}\)	\(559\)

Input:

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

Output:

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(3*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^3*x^2*(a*c)^(1/2)+30*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^ 
2*x^2*(a*c)^(1/2)+15*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*x^2*(a*c)^(1/2)-15*ln((a*d*x+b*c*x+2*(a*c 
)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a^2*c*d^2*x^2*(d*b)^(1/2)-30*ln( 
(a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a*b*c^2*d*x^2 
*(d*b)^(1/2)-3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c 
)/x)*b^2*c^3*x^2*(d*b)^(1/2)+4*b*d^2*x^3*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/ 
2)*(a*c)^(1/2)+10*a*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2 
)+18*b*c*d*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)-18*a*c*d*x* 
((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)-10*b*c^2*x*((b*x+a)*(d*x+c 
))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)-4*a*c^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/ 
2)*(a*c)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/x^2/(d*b)^(1/2)/(a*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 1.54 (sec) , antiderivative size = 1185, normalized size of antiderivative = 4.59 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/16*(3*(5*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*x^2*sqrt(d/b)*log(8*b^2*d^2*x^ 
2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x 
 + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 3*(b^2*c^2 + 10 
*a*b*c*d + 5*a^2*d^2)*x^2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d 
+ a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + 
c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(2*b*d^2*x^3 - 2*a*c^2 + 
(9*b*c*d + 5*a*d^2)*x^2 - (5*b*c^2 + 9*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + 
c))/x^2, -1/16*(6*(5*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*x^2*sqrt(-d/b)*arctan 
(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x 
^2 + a*c*d + (b*c*d + a*d^2)*x)) - 3*(b^2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*x^ 
2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^ 
2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^ 
2 + a^2*c*d)*x)/x^2) - 4*(2*b*d^2*x^3 - 2*a*c^2 + (9*b*c*d + 5*a*d^2)*x^2 
- (5*b*c^2 + 9*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/x^2, 1/16*(6*(b^2*c^ 
2 + 10*a*b*c*d + 5*a^2*d^2)*x^2*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d) 
*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a 
*c*d)*x)) + 3*(5*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*x^2*sqrt(d/b)*log(8*b^2*d 
^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqr 
t(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*b*d^2 
*x^3 - 2*a*c^2 + (9*b*c*d + 5*a*d^2)*x^2 - (5*b*c^2 + 9*a*c*d)*x)*sqrt(...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(3/2)*(d*x+c)^(5/2)/x^3,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. 1247 vs. \(2 (208) = 416\).

Time = 0.93 (sec) , antiderivative size = 1247, normalized size of antiderivative = 4.83 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*d^2*abs(b)/b + 3*( 
3*b*c*d^3*abs(b) + a*d^4*abs(b))/(b*d^2))*sqrt(b*x + a) - 3*(5*sqrt(b*d)*b 
^2*c^2*abs(b) + 10*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b*d)*a^2*d^2*abs(b))*lo 
g((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b - 6 
*(sqrt(b*d)*b^3*c^3*abs(b) + 10*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 5*sqrt(b*d) 
*a^2*b*c*d^2*abs(b))*arctan(-1/2*(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(-a*b*c*d)*b))/(sqrt(-a*b* 
c*d)*b) - 4*(5*sqrt(b*d)*b^9*c^6*abs(b) - 11*sqrt(b*d)*a*b^8*c^5*d*abs(b) 
- 6*sqrt(b*d)*a^2*b^7*c^4*d^2*abs(b) + 34*sqrt(b*d)*a^3*b^6*c^3*d^3*abs(b) 
 - 31*sqrt(b*d)*a^4*b^5*c^2*d^4*abs(b) + 9*sqrt(b*d)*a^5*b^4*c*d^5*abs(b) 
- 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b 
*d))^2*b^7*c^5*abs(b) - 8*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c 
+ (b*x + a)*b*d - a*b*d))^2*a*b^6*c^4*d*abs(b) + 34*sqrt(b*d)*(sqrt(b*d)*s 
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^5*c^3*d^2*abs( 
b) + 16*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - 
a*b*d))^2*a^3*b^4*c^2*d^3*abs(b) - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - 
 sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^3*c*d^4*abs(b) + 15*sqrt(b*d 
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^5*c^ 
4*abs(b) + 37*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)* 
b*d - a*b*d))^4*a*b^4*c^3*d*abs(b) + 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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