\(\int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx\) [572]

Optimal result

Integrand size = 22, antiderivative size = 227 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}-\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+2 \sqrt {b} d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right ) \]

[Out]

2*d^(5/2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)-1/8*(5*a^3*d^3+15*a^2*b*c*d^2-5*a*b^2*c
^2*d+b^3*c^3)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(5/2)/c^(1/2)-1/12*(5*a*d+b*c)*(d*x+c)^(3
/2)*(b*x+a)^(1/2)/a/x^2-1/3*(d*x+c)^(5/2)*(b*x+a)^(1/2)/x^3+1/8*(-5*a*d+b*c)*(a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(
1/2)/a^2/x

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {99, 154, 163, 65, 223, 212, 95, 214} \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{8 a^2 x}-\frac {\left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+2 \sqrt {b} d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{12 a x^2} \]

[In]

Int[(Sqrt[a + b*x]*(c + d*x)^(5/2))/x^4,x]

[Out]

((b*c - 5*a*d)*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*a^2*x) - ((b*c + 5*a*d)*Sqrt[a + b*x]*(c + d*x)^(3/
2))/(12*a*x^2) - (Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*x^3) - ((b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*
d^3)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*a^(5/2)*Sqrt[c]) + 2*Sqrt[b]*d^(5/2)*ArcTanh
[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])]

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 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 {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}+\frac {1}{3} \int \frac {(c+d x)^{3/2} \left (\frac {1}{2} (b c+5 a d)+3 b d x\right )}{x^3 \sqrt {a+b x}} \, dx \\ & = -\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}+\frac {\int \frac {\sqrt {c+d x} \left (-\frac {3}{4} (b c-5 a d) (b c+a d)+6 a b d^2 x\right )}{x^2 \sqrt {a+b x}} \, dx}{6 a} \\ & = \frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}+\frac {\int \frac {\frac {3}{8} \left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right )+6 a^2 b d^3 x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{6 a^2} \\ & = \frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}+\left (b d^3\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 a^2} \\ & = \frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}+\left (2 d^3\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 )+\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 a^2} \\ & = \frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}-\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+\left (2 d^3\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 ) \\ & = \frac {(b c-5 a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 a^2 x}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 a x^2}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 x^3}-\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+2 \sqrt {b} d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-3 b^2 c^2 x^2+2 a b c x (c+7 d x)+a^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )\right )}{24 a^2 x^3}-\frac {\left (b^3 c^3-5 a b^2 c^2 d+15 a^2 b c d^2+5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 a^{5/2} \sqrt {c}}+2 \sqrt {b} d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right ) \]

[In]

Integrate[(Sqrt[a + b*x]*(c + d*x)^(5/2))/x^4,x]

[Out]

-1/24*(Sqrt[a + b*x]*Sqrt[c + d*x]*(-3*b^2*c^2*x^2 + 2*a*b*c*x*(c + 7*d*x) + a^2*(8*c^2 + 26*c*d*x + 33*d^2*x^
2)))/(a^2*x^3) - ((b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt
[a]*Sqrt[c + d*x])])/(8*a^(5/2)*Sqrt[c]) + 2*Sqrt[b]*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c +
d*x])]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(516\) vs. \(2(183)=366\).

Time = 0.53 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.28

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (48 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{3} x^{3} \sqrt {a c}-15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{3} d^{3} x^{3} \sqrt {b d}-45 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} b c \,d^{2} x^{3} \sqrt {b d}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} d \,x^{3} \sqrt {b d}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{3} c^{3} x^{3} \sqrt {b d}-66 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2} x^{2}-28 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d \,x^{2}+6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2} x^{2}-52 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c d x -4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{2} x -16 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c^{2}\right )}{48 a^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x^{3} \sqrt {b d}\, \sqrt {a c}}\)	\(517\)

[In]

int((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2*(48*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*b*d^3*x^3*(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^3*d^3*x^
3*(b*d)^(1/2)-45*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*b*c*d^2*x^3*(b*d)^(1/2)+1
5*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^2*c^2*d*x^3*(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^3*c^3*x^3*(b*d)^(1/2)-66*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*
(d*x+c))^(1/2)*a^2*d^2*x^2-28*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d*x^2+6*(b*d)^(1/2)*(a*c)^
(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^2*x^2-52*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*c*d*x-4*(b*d)
^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c^2*x-16*(b*d)^(1/2)*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*c^
2)/((b*x+a)*(d*x+c))^(1/2)/x^3/(b*d)^(1/2)/(a*c)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 1.40 (sec) , antiderivative size = 1245, normalized size of antiderivative = 5.48 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(48*sqrt(b*d)*a^3*c*d^2*x^3*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) + 3*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2
+ 5*a^3*d^3)*sqrt(a*c)*x^3*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*s
qrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*
c^2*d - 33*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c*x^3), -1/96*(9
6*sqrt(-b*d)*a^3*c*d^2*x^3*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)) - 3*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(a*c)*x^3
*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt
(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*c^2*d - 33*a^3*c*d^2)*x^2 +
2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c*x^3), 1/48*(24*sqrt(b*d)*a^3*c*d^2*x^3*lo
g(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) + 3*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(-a*c)*x^3*arctan(
1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d
)*x)) - 2*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*c^2*d - 33*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqr
t(b*x + a)*sqrt(d*x + c))/(a^3*c*x^3), -1/48*(48*sqrt(-b*d)*a^3*c*d^2*x^3*arctan(1/2*(2*b*d*x + b*c + a*d)*sqr
t(-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)) - 3*(b^3*c^3 - 5*a*b^2*c^
2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(-a*c)*x^3*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*s
qrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) + 2*(8*a^3*c^3 - (3*a*b^2*c^3 - 14*a^2*b*c^2*d -
33*a^3*c*d^2)*x^2 + 2*(a^2*b*c^3 + 13*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c*x^3)]

Sympy [F]

\[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\int \frac {\sqrt {a + b x} \left (c + d x\right )^{\frac {5}{2}}}{x^{4}}\, dx \]

[In]

integrate((d*x+c)**(5/2)*(b*x+a)**(1/2)/x**4,x)

[Out]

Integral(sqrt(a + b*x)*(c + d*x)**(5/2)/x**4, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,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. 2265 vs. \(2 (183) = 366\).

Time = 3.49 (sec) , antiderivative size = 2265, normalized size of antiderivative = 9.98 \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/24*(24*sqrt(b*d)*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) + 3*(sqr
t(b*d)*b^4*c^3*abs(b) - 5*sqrt(b*d)*a*b^3*c^2*d*abs(b) + 15*sqrt(b*d)*a^2*b^2*c*d^2*abs(b) + 5*sqrt(b*d)*a^3*b
*d^3*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)*a^2*b) - 2*(3*sqrt(b*d)*b^14*c^8*abs(b) - 32*sqrt(b*d)*a*b^13*c^7*d*abs(b)
+ 96*sqrt(b*d)*a^2*b^12*c^6*d^2*abs(b) - 72*sqrt(b*d)*a^3*b^11*c^5*d^3*abs(b) - 170*sqrt(b*d)*a^4*b^10*c^4*d^4
*abs(b) + 432*sqrt(b*d)*a^5*b^9*c^3*d^5*abs(b) - 408*sqrt(b*d)*a^6*b^8*c^2*d^6*abs(b) + 184*sqrt(b*d)*a^7*b^7*
c*d^7*abs(b) - 33*sqrt(b*d)*a^8*b^6*d^8*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^12*c^7*abs(b) + 111*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^11*c^6*d*abs(b) - 111*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^10*c^5*d^2*abs(b) - 201*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^
9*c^4*d^3*abs(b) + 219*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^8*c^3
*d^4*abs(b) + 309*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^5*b^7*c^2*d^5*
abs(b) - 477*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^6*b^6*c*d^6*abs(b)
+ 165*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^7*b^5*d^7*abs(b) + 30*sqrt
(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^10*c^6*abs(b) - 156*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^9*c^5*d*abs(b) - 198*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^8*c^4*d^2*abs(b) + 312*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^7*c^3*d^3*abs(b) + 114*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b^6*c^2*d^4*abs(b) + 228*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^5*b^5*c*d^5*abs(b) - 330*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^4*a^6*b^4*d^6*abs(b) - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +
a)*b*d - a*b*d))^6*b^8*c^5*abs(b) + 122*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*
d))^6*a*b^7*c^4*d*abs(b) + 396*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2
*b^6*c^3*d^2*abs(b) + 252*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b^5*
c^2*d^3*abs(b) + 338*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^4*b^4*c*d^4
*abs(b) + 330*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^5*b^3*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))^8*b^6*c^4*abs(b) - 60*sqrt(b*d)*
(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a*b^5*c^3*d*abs(b) - 282*sqrt(b*d)*(sqrt(b*d
)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^4*c^2*d^2*abs(b) - 372*sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^3*b^3*c*d^3*abs(b) - 165*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^4*b^2*d^4*abs(b) - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(
b^2*c + (b*x + a)*b*d - a*b*d))^10*b^4*c^3*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))^10*a*b^3*c^2*d*abs(b) + 99*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^10*a^2*b^2*c*d^2*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
))^10*a^3*b*d^3*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 + (sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^3*a^2))/b

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^4} \, dx=\int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2}}{x^4} \,d x \]

[In]

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^4,x)

[Out]

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^4, x)