\(\int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx\) [755]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 277 \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}-\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}} \]

[Out]

-1/4*(35*a^2*d^2+10*a*b*c*d+3*b^2*c^2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(5/2)/c^(9/2)+1/
12*d*(-35*a^2*d^2+18*a*b*c*d+9*b^2*c^2)*(b*x+a)^(1/2)/a^2/c^3/(-a*d+b*c)/(d*x+c)^(3/2)-1/2*(b*x+a)^(1/2)/a/c/x
^2/(d*x+c)^(3/2)+1/4*(7*a*d+3*b*c)*(b*x+a)^(1/2)/a^2/c^2/x/(d*x+c)^(3/2)+1/12*d*(105*a^3*d^3-145*a^2*b*c*d^2+1
5*a*b^2*c^2*d+9*b^3*c^3)*(b*x+a)^(1/2)/a^2/c^4/(-a*d+b*c)^2/(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {105, 156, 157, 12, 95, 214} \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\frac {d \sqrt {a+b x} \left (-35 a^2 d^2+18 a b c d+9 b^2 c^2\right )}{12 a^2 c^3 (c+d x)^{3/2} (b c-a d)}+\frac {\sqrt {a+b x} (7 a d+3 b c)}{4 a^2 c^2 x (c+d x)^{3/2}}-\frac {\left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}}+\frac {d \sqrt {a+b x} \left (105 a^3 d^3-145 a^2 b c d^2+15 a b^2 c^2 d+9 b^3 c^3\right )}{12 a^2 c^4 \sqrt {c+d x} (b c-a d)^2}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}} \]

[In]

Int[1/(x^3*Sqrt[a + b*x]*(c + d*x)^(5/2)),x]

[Out]

(d*(9*b^2*c^2 + 18*a*b*c*d - 35*a^2*d^2)*Sqrt[a + b*x])/(12*a^2*c^3*(b*c - a*d)*(c + d*x)^(3/2)) - Sqrt[a + b*
x]/(2*a*c*x^2*(c + d*x)^(3/2)) + ((3*b*c + 7*a*d)*Sqrt[a + b*x])/(4*a^2*c^2*x*(c + d*x)^(3/2)) + (d*(9*b^3*c^3
 + 15*a*b^2*c^2*d - 145*a^2*b*c*d^2 + 105*a^3*d^3)*Sqrt[a + b*x])/(12*a^2*c^4*(b*c - a*d)^2*Sqrt[c + d*x]) - (
(3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(5/2)*c^(
9/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
 c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 156

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 + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

Rule 157

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 + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}-\frac {\int \frac {\frac {1}{2} (3 b c+7 a d)+3 b d x}{x^2 \sqrt {a+b x} (c+d x)^{5/2}} \, dx}{2 a c} \\ & = -\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {\int \frac {\frac {1}{4} \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right )+b d (3 b c+7 a d) x}{x \sqrt {a+b x} (c+d x)^{5/2}} \, dx}{2 a^2 c^2} \\ & = \frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}-\frac {\int \frac {-\frac {3}{8} (b c-a d) \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right )-\frac {1}{4} b d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) x}{x \sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 a^2 c^3 (b c-a d)} \\ & = \frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}+\frac {2 \int \frac {3 (b c-a d)^2 \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right )}{16 x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{3 a^2 c^4 (b c-a d)^2} \\ & = \frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}+\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 a^2 c^4} \\ & = \frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}+\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{4 a^2 c^4} \\ & = \frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}-\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (9 b^3 c^3 x (c+d x)^2-3 a b^2 c^2 (2 c-5 d x) (c+d x)^2+a^2 b c d \left (12 c^3-33 c^2 d x-198 c d^2 x^2-145 d^3 x^3\right )+a^3 d^2 \left (-6 c^3+21 c^2 d x+140 c d^2 x^2+105 d^3 x^3\right )\right )}{12 a^2 c^4 (b c-a d)^2 x^2 (c+d x)^{3/2}}-\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}} \]

[In]

Integrate[1/(x^3*Sqrt[a + b*x]*(c + d*x)^(5/2)),x]

[Out]

(Sqrt[a + b*x]*(9*b^3*c^3*x*(c + d*x)^2 - 3*a*b^2*c^2*(2*c - 5*d*x)*(c + d*x)^2 + a^2*b*c*d*(12*c^3 - 33*c^2*d
*x - 198*c*d^2*x^2 - 145*d^3*x^3) + a^3*d^2*(-6*c^3 + 21*c^2*d*x + 140*c*d^2*x^2 + 105*d^3*x^3)))/(12*a^2*c^4*
(b*c - a*d)^2*x^2*(c + d*x)^(3/2)) - ((3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(S
qrt[a]*Sqrt[c + d*x])])/(4*a^(5/2)*c^(9/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1287\) vs. \(2(239)=478\).

Time = 0.59 (sec) , antiderivative size = 1288, normalized size of antiderivative = 4.65

method result size
default \(\text {Expression too large to display}\) \(1288\)

[In]

int(1/x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/24*(b*x+a)^(1/2)/a^2/c^4*(210*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^4*c*d^5*x^3
-360*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*b*c^2*d^4*x^3+108*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^2*c^3*d^3*x^3+18*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^4*c^5*d*x^3+12*a^3*c^3*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+12*a*b^2*c^5*(a*c)^(1/2
)*((b*x+a)*(d*x+c))^(1/2)+9*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^4*c^4*d^2*x^4+10
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^4*d^6*x^4+9*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^4*c^6*x^2-210*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*d^5*x^3-18*(a*c)^(1/
2)*((b*x+a)*(d*x+c))^(1/2)*b^3*c^5*x-18*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^3*c^3*d^2*x^3-280*(a*c)^(1/2)*((
b*x+a)*(d*x+c))^(1/2)*a^3*c*d^4*x^2-36*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^3*c^4*d*x^2-42*(a*c)^(1/2)*((b*x+
a)*(d*x+c))^(1/2)*a^3*c^2*d^3*x+24*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^3*c^4*d
^2*x^3-180*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*b*c^3*d^3*x^2+54*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^2*c^4*d^2*x^2+105*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^4*c^2*d^4*x^2+290*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c*d^4*x^3-30*(a*c)^(
1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c^2*d^3*x^3+396*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c^2*d^3*x^2-48*(a
*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c^3*d^2*x^2+66*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b*c^3*d^2*x-6*(
a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c^4*d*x-24*a^2*b*c^4*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-180*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*b*c*d^5*x^4+54*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^2*c^2*d^4*x^4+12*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^3*c^3*d^3*x^4+12*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^3*c^5*d*x^2)/
(a*d-b*c)^2/(a*c)^(1/2)/x^2/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(3/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 583 vs. \(2 (239) = 478\).

Time = 1.57 (sec) , antiderivative size = 1186, normalized size of antiderivative = 4.28 \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(3*((3*b^4*c^4*d^2 + 4*a*b^3*c^3*d^3 + 18*a^2*b^2*c^2*d^4 - 60*a^3*b*c*d^5 + 35*a^4*d^6)*x^4 + 2*(3*b^4*
c^5*d + 4*a*b^3*c^4*d^2 + 18*a^2*b^2*c^3*d^3 - 60*a^3*b*c^2*d^4 + 35*a^4*c*d^5)*x^3 + (3*b^4*c^6 + 4*a*b^3*c^5
*d + 18*a^2*b^2*c^4*d^2 - 60*a^3*b*c^3*d^3 + 35*a^4*c^2*d^4)*x^2)*sqrt(a*c)*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*(6*a^2*b^2*c^6 - 12*a^3*b*c^5*d + 6*a^4*c^4*d^2 - (9*a*b^3*c^4*d^2 + 15*a^2*b^2*c^3*d^3 - 145*a^3*b
*c^2*d^4 + 105*a^4*c*d^5)*x^3 - 2*(9*a*b^3*c^5*d + 12*a^2*b^2*c^4*d^2 - 99*a^3*b*c^3*d^3 + 70*a^4*c^2*d^4)*x^2
 - 3*(3*a*b^3*c^6 + a^2*b^2*c^5*d - 11*a^3*b*c^4*d^2 + 7*a^4*c^3*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((a^3*b^
2*c^7*d^2 - 2*a^4*b*c^6*d^3 + a^5*c^5*d^4)*x^4 + 2*(a^3*b^2*c^8*d - 2*a^4*b*c^7*d^2 + a^5*c^6*d^3)*x^3 + (a^3*
b^2*c^9 - 2*a^4*b*c^8*d + a^5*c^7*d^2)*x^2), 1/24*(3*((3*b^4*c^4*d^2 + 4*a*b^3*c^3*d^3 + 18*a^2*b^2*c^2*d^4 -
60*a^3*b*c*d^5 + 35*a^4*d^6)*x^4 + 2*(3*b^4*c^5*d + 4*a*b^3*c^4*d^2 + 18*a^2*b^2*c^3*d^3 - 60*a^3*b*c^2*d^4 +
35*a^4*c*d^5)*x^3 + (3*b^4*c^6 + 4*a*b^3*c^5*d + 18*a^2*b^2*c^4*d^2 - 60*a^3*b*c^3*d^3 + 35*a^4*c^2*d^4)*x^2)*
sqrt(-a*c)*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*(6*a^2*b^2*c^6 - 12*a^3*b*c^5*d + 6*a^4*c^4*d^2 - (9*a*b^3*c^4*d^2 + 15*a^2*b^2*c^
3*d^3 - 145*a^3*b*c^2*d^4 + 105*a^4*c*d^5)*x^3 - 2*(9*a*b^3*c^5*d + 12*a^2*b^2*c^4*d^2 - 99*a^3*b*c^3*d^3 + 70
*a^4*c^2*d^4)*x^2 - 3*(3*a*b^3*c^6 + a^2*b^2*c^5*d - 11*a^3*b*c^4*d^2 + 7*a^4*c^3*d^3)*x)*sqrt(b*x + a)*sqrt(d
*x + c))/((a^3*b^2*c^7*d^2 - 2*a^4*b*c^6*d^3 + a^5*c^5*d^4)*x^4 + 2*(a^3*b^2*c^8*d - 2*a^4*b*c^7*d^2 + a^5*c^6
*d^3)*x^3 + (a^3*b^2*c^9 - 2*a^4*b*c^8*d + a^5*c^7*d^2)*x^2)]

Sympy [F]

\[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\int \frac {1}{x^{3} \sqrt {a + b x} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/x**3/(d*x+c)**(5/2)/(b*x+a)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(a + b*x)*(c + d*x)**(5/2)), x)

Maxima [F]

\[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\int { \frac {1}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {5}{2}} x^{3}} \,d x } \]

[In]

integrate(1/x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x + a)*(d*x + c)^(5/2)*x^3), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1234 vs. \(2 (239) = 478\).

Time = 1.08 (sec) , antiderivative size = 1234, normalized size of antiderivative = 4.45 \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-2/3*sqrt(b*x + a)*((11*b^4*c^5*d^5*abs(b) - 9*a*b^3*c^4*d^6*abs(b))*(b*x + a)/(b^4*c^10*d - 2*a*b^3*c^9*d^2 +
 a^2*b^2*c^8*d^3) + 3*(4*b^5*c^6*d^4*abs(b) - 7*a*b^4*c^5*d^5*abs(b) + 3*a^2*b^3*c^4*d^6*abs(b))/(b^4*c^10*d -
 2*a*b^3*c^9*d^2 + a^2*b^2*c^8*d^3))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 1/4*(3*sqrt(b*d)*b^4*c^2 + 10*sqr
t(b*d)*a*b^3*c*d + 35*sqrt(b*d)*a^2*b^2*d^2)*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*c^4*abs(b)) + 1/2*(3*sqrt(b*d)*b^10*c
^5 - sqrt(b*d)*a*b^9*c^4*d - 26*sqrt(b*d)*a^2*b^8*c^3*d^2 + 54*sqrt(b*d)*a^3*b^7*c^2*d^3 - 41*sqrt(b*d)*a^4*b^
6*c*d^4 + 11*sqrt(b*d)*a^5*b^5*d^5 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d
))^2*b^8*c^4 - 28*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^7*c^3*d + 50
*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^6*c^2*d^2 + 20*sqrt(b*d)*(s
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^5*c*d^3 - 33*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^4*d^4 + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2
*c + (b*x + a)*b*d - a*b*d))^4*b^6*c^3 + 39*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^5*c^2*d + 31*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^4
*c*d^2 + 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^3*b^3*d^3 - 3*sqrt(b
*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^4*c^2 - 10*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^3*c*d - 11*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^2*d^2)/((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*a^2*c^4*abs(b))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx=\int \frac {1}{x^3\,\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2}} \,d x \]

[In]

int(1/(x^3*(a + b*x)^(1/2)*(c + d*x)^(5/2)),x)

[Out]

int(1/(x^3*(a + b*x)^(1/2)*(c + d*x)^(5/2)), x)