\(\int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx\) [621]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {99, 154, 159, 163, 65, 223, 212, 95, 214} \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx=-\frac {3 \sqrt {c} \left (5 a^2 d^2+10 a b c d+b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a}}+\frac {3 \sqrt {d} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b}}-\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2} (5 a d+3 b c)}{4 c x}+\frac {d \sqrt {a+b x} (c+d x)^{3/2} (5 a d+7 b c)}{4 c}+3 d \sqrt {a+b x} \sqrt {c+d x} (a d+b c) \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[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] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))), x] - Dist[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 154

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

Rule 159

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[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] /; FreeQ[{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 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

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]

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{2 x^2}+\frac {1}{2} \int \frac {\sqrt {a+b x} (c+d x)^{3/2} \left (\frac {1}{2} (3 b c+5 a d)+4 b d x\right )}{x^2} \, dx \\ & = -\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 {\int \frac {(c+d x)^{3/2} \left (\frac {3}{4} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )+b d (7 b c+5 a d) x\right )}{x \sqrt {a+b x}} \, dx}{2 c} \\ & = \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 {\int \frac {\sqrt {c+d x} \left (\frac {3}{2} b c \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )+12 b^2 c d (b c+a d) x\right )}{x \sqrt {a+b x}} \, dx}{4 b c} \\ & = 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 {\int \frac {\frac {3}{2} b^2 c^2 \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )+\frac {3}{2} b^2 c d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{4 b^2 c} \\ & = 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 {1}{8} \left (3 d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {1}{8} \left (3 c \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, 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 {\left (3 d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{4 b}+\frac {1}{4} \left (3 c \left (b^2 c^2+10 a b c d+5 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right ) \\ & = 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 ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a}}+\frac {\left (3 d \left (5 b^2 c^2+10 a b c d+a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{4 b} \\ & = 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 ) \tanh ^{-1}\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 ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx=\frac {1}{4} \left (\frac {\sqrt {a+b x} \sqrt {c+d x} \left (b x \left (-5 c^2+9 c d x+2 d^2 x^2\right )+a \left (-2 c^2-9 c d x+5 d^2 x^2\right )\right )}{x^2}-\frac {3 \sqrt {c} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{\sqrt {a}}+\frac {3 \sqrt {d} \left (5 b^2 c^2+10 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {b}}\right ) \]

[In]

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

[Out]

((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])])/Sqr
t[b])/4

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(558\) vs. \(2(208)=416\).

Time = 0.55 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.17

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 1.69 (sec) , antiderivative size = 1185, normalized size of antiderivative = 4.59 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[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 + 1
0*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)*sqr
t(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)*sqrt(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(b*x + a)*sqrt(d*x + c))/x^2, 1/8*(3*(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)*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)) + 2*(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]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more 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 = 1.22 (sec) , antiderivative size = 1247, normalized size of antiderivative = 4.83 \[ \int \frac {(a+b x)^{3/2} (c+d x)^{5/2}}{x^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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))*
log((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*sq
rt(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) - sqr
t(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)*sqrt(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 + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^3*c^2*d^2*ab
s(b) + 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^2*c*d^3*abs(b) - 5
*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^3*c^3*abs(b) - 18*sqrt(b*d)*(sq
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^2*c^2*d*abs(b) - 9*sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b*c*d^2*abs(b))/(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 -
2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt
(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^
2)/b

Mupad [F(-1)]

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

[In]

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

[Out]

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