\(\int \frac {1}{x^3 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx\) [783]

Optimal result

Integrand size = 22, antiderivative size = 268 \[ \int \frac {1}{x^3 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {b \left (15 b^2 c^2-2 a b c d-5 a^2 d^2\right )}{4 a^3 c^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {5 (b c+a d)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}+\frac {d (b c+a d) \left (15 b^2 c^2-22 a b c d+15 a^2 d^2\right ) \sqrt {a+b x}}{4 a^3 c^3 (b c-a d)^2 \sqrt {c+d x}}-\frac {3 \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2} c^{7/2}} \]

[Out]

-3/4*(5*a^2*d^2+6*a*b*c*d+5*b^2*c^2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(7/2)/c^(7/2)+1/4*
b*(-5*a^2*d^2-2*a*b*c*d+15*b^2*c^2)/a^3/c^2/(-a*d+b*c)/(b*x+a)^(1/2)/(d*x+c)^(1/2)-1/2/a/c/x^2/(b*x+a)^(1/2)/(
d*x+c)^(1/2)+5/4*(a*d+b*c)/a^2/c^2/x/(b*x+a)^(1/2)/(d*x+c)^(1/2)+1/4*d*(a*d+b*c)*(15*a^2*d^2-22*a*b*c*d+15*b^2
*c^2)*(b*x+a)^(1/2)/a^3/c^3/(-a*d+b*c)^2/(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 268, 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} \[ \int \frac {1}{x^3 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {5 (a d+b c)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}-\frac {3 \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2} c^{7/2}}+\frac {b \left (-5 a^2 d^2-2 a b c d+15 b^2 c^2\right )}{4 a^3 c^2 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}+\frac {d \sqrt {a+b x} \left (15 a^2 d^2-22 a b c d+15 b^2 c^2\right ) (a d+b c)}{4 a^3 c^3 \sqrt {c+d x} (b c-a d)^2}-\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}} \]

[In]

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

[Out]

(b*(15*b^2*c^2 - 2*a*b*c*d - 5*a^2*d^2))/(4*a^3*c^2*(b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x]) - 1/(2*a*c*x^2*Sq
rt[a + b*x]*Sqrt[c + d*x]) + (5*(b*c + a*d))/(4*a^2*c^2*x*Sqrt[a + b*x]*Sqrt[c + d*x]) + (d*(b*c + a*d)*(15*b^
2*c^2 - 22*a*b*c*d + 15*a^2*d^2)*Sqrt[a + b*x])/(4*a^3*c^3*(b*c - a*d)^2*Sqrt[c + d*x]) - (3*(5*b^2*c^2 + 6*a*
b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(7/2)*c^(7/2))

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 {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}}-\frac {\int \frac {\frac {5}{2} (b c+a d)+3 b d x}{x^2 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx}{2 a c} \\ & = -\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {5 (b c+a d)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}+\frac {\int \frac {\frac {3}{4} \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right )+5 b d (b c+a d) x}{x (a+b x)^{3/2} (c+d x)^{3/2}} \, dx}{2 a^2 c^2} \\ & = \frac {b \left (15 b^2 c^2-2 a b c d-5 a^2 d^2\right )}{4 a^3 c^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {5 (b c+a d)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}+\frac {\int \frac {\frac {3}{8} (b c-a d) \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right )+\frac {1}{4} b d \left (15 b^2 c^2-2 a b c d-5 a^2 d^2\right ) x}{x \sqrt {a+b x} (c+d x)^{3/2}} \, dx}{a^3 c^2 (b c-a d)} \\ & = \frac {b \left (15 b^2 c^2-2 a b c d-5 a^2 d^2\right )}{4 a^3 c^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {5 (b c+a d)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}+\frac {d (b c+a d) \left (15 b^2 c^2-22 a b c d+15 a^2 d^2\right ) \sqrt {a+b x}}{4 a^3 c^3 (b c-a d)^2 \sqrt {c+d x}}-\frac {2 \int -\frac {3 (b c-a d)^2 \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right )}{16 x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{a^3 c^3 (b c-a d)^2} \\ & = \frac {b \left (15 b^2 c^2-2 a b c d-5 a^2 d^2\right )}{4 a^3 c^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {5 (b c+a d)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}+\frac {d (b c+a d) \left (15 b^2 c^2-22 a b c d+15 a^2 d^2\right ) \sqrt {a+b x}}{4 a^3 c^3 (b c-a d)^2 \sqrt {c+d x}}+\frac {\left (3 \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 a^3 c^3} \\ & = \frac {b \left (15 b^2 c^2-2 a b c d-5 a^2 d^2\right )}{4 a^3 c^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {5 (b c+a d)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}+\frac {d (b c+a d) \left (15 b^2 c^2-22 a b c d+15 a^2 d^2\right ) \sqrt {a+b x}}{4 a^3 c^3 (b c-a d)^2 \sqrt {c+d x}}+\frac {\left (3 \left (5 b^2 c^2+6 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 )}{4 a^3 c^3} \\ & = \frac {b \left (15 b^2 c^2-2 a b c d-5 a^2 d^2\right )}{4 a^3 c^2 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {5 (b c+a d)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}+\frac {d (b c+a d) \left (15 b^2 c^2-22 a b c d+15 a^2 d^2\right ) \sqrt {a+b x}}{4 a^3 c^3 (b c-a d)^2 \sqrt {c+d x}}-\frac {3 \left (5 b^2 c^2+6 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 a^{7/2} c^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {15 b^4 c^3 x^2 (c+d x)+a b^3 c^2 x \left (5 c^2-2 c d x-7 d^2 x^2\right )+a^4 d^2 \left (-2 c^2+5 c d x+15 d^2 x^2\right )-a^2 b^2 c \left (2 c^3+5 c^2 d x+10 c d^2 x^2+7 d^3 x^3\right )+a^3 b d \left (4 c^3-5 c^2 d x-2 c d^2 x^2+15 d^3 x^3\right )}{4 a^3 c^3 (b c-a d)^2 x^2 \sqrt {a+b x} \sqrt {c+d x}}-\frac {3 \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{4 a^{7/2} c^{7/2}} \]

[In]

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

[Out]

(15*b^4*c^3*x^2*(c + d*x) + a*b^3*c^2*x*(5*c^2 - 2*c*d*x - 7*d^2*x^2) + a^4*d^2*(-2*c^2 + 5*c*d*x + 15*d^2*x^2
) - a^2*b^2*c*(2*c^3 + 5*c^2*d*x + 10*c*d^2*x^2 + 7*d^3*x^3) + a^3*b*d*(4*c^3 - 5*c^2*d*x - 2*c*d^2*x^2 + 15*d
^3*x^3))/(4*a^3*c^3*(b*c - a*d)^2*x^2*Sqrt[a + b*x]*Sqrt[c + d*x]) - (3*(5*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*Ar
cTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(4*a^(7/2)*c^(7/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1371\) vs. \(2(230)=460\).

Time = 0.60 (sec) , antiderivative size = 1372, normalized size of antiderivative = 5.12

method	result	size
default	\(\text {Expression too large to display}\)	\(1372\)

[In]

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

[Out]

-1/8/c^3/a^3*(-30*a^4*d^4*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-30*b^4*c^4*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+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^5*d^5*x^3+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)*b^5*c^5*x^3+14*a^2*b^2*c*d^3*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2
)+14*a*b^3*c^2*d^2*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*a^3*b*c*d^3*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/
2)+20*a^2*b^2*c^2*d^2*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*a*b^3*c^3*d*x^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^
(1/2)+10*a^3*b*c^2*d^2*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+10*a^2*b^2*c^3*d*x*(a*c)^(1/2)*((b*x+a)*(d*x+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^4*b*d^5*x^4+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)*b^5*c^4*d*x^4+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^5*c*d^4*x^2+4*a^2*b^2*c^4*(a*c)^(1/2)*((b*x+a)*(d*x+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*b^4*c^5*x^2-10*a^4*c*d^3*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*a*b^3*
c^4*x*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-8*a^3*b*c^3*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-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^3*b^2*c*d^4*x^4-6*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^3*c^2*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^4*c^3*d^2*x^4+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)*a^4*b*c*d^4*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)*a^3*b^2*c^2*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)*a^2*b^3*c^3*d^2*x^3+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)*a*b^4*c^4*d*x^3-12*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^4*b*c^2*d^3*x^2
-6*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^2*c^3*d^2*x^2-12*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^2*b^3*c^4*d*x^2-30*a^3*b*d^4*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(
1/2)-30*b^4*c^3*d*x^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*a^4*c^2*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/(
(b*x+a)*(d*x+c))^(1/2)/x^2/(a*c)^(1/2)/(a*d-b*c)^2/(b*x+a)^(1/2)/(d*x+c)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 612 vs. \(2 (230) = 460\).

Time = 1.58 (sec) , antiderivative size = 1244, normalized size of antiderivative = 4.64 \[ \int \frac {1}{x^3 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/16*(3*((5*b^5*c^4*d - 4*a*b^4*c^3*d^2 - 2*a^2*b^3*c^2*d^3 - 4*a^3*b^2*c*d^4 + 5*a^4*b*d^5)*x^4 + (5*b^5*c^5
+ a*b^4*c^4*d - 6*a^2*b^3*c^3*d^2 - 6*a^3*b^2*c^2*d^3 + a^4*b*c*d^4 + 5*a^5*d^5)*x^3 + (5*a*b^4*c^5 - 4*a^2*b
^3*c^4*d - 2*a^3*b^2*c^3*d^2 - 4*a^4*b*c^2*d^3 + 5*a^5*c*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*(2*a^3*b^2*c^5 - 4*a^4*b*c^4*d + 2*a^5*c^3*d^2 - (15*a*b^4*c^4*d - 7*a^2*b^3*c^3*d^2 - 7*a^3*b^2*c
^2*d^3 + 15*a^4*b*c*d^4)*x^3 - (15*a*b^4*c^5 - 2*a^2*b^3*c^4*d - 10*a^3*b^2*c^3*d^2 - 2*a^4*b*c^2*d^3 + 15*a^5
*c*d^4)*x^2 - 5*(a^2*b^3*c^5 - a^3*b^2*c^4*d - a^4*b*c^3*d^2 + a^5*c^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((
a^4*b^3*c^6*d - 2*a^5*b^2*c^5*d^2 + a^6*b*c^4*d^3)*x^4 + (a^4*b^3*c^7 - a^5*b^2*c^6*d - a^6*b*c^5*d^2 + a^7*c^
4*d^3)*x^3 + (a^5*b^2*c^7 - 2*a^6*b*c^6*d + a^7*c^5*d^2)*x^2), 1/8*(3*((5*b^5*c^4*d - 4*a*b^4*c^3*d^2 - 2*a^2*
b^3*c^2*d^3 - 4*a^3*b^2*c*d^4 + 5*a^4*b*d^5)*x^4 + (5*b^5*c^5 + a*b^4*c^4*d - 6*a^2*b^3*c^3*d^2 - 6*a^3*b^2*c^
2*d^3 + a^4*b*c*d^4 + 5*a^5*d^5)*x^3 + (5*a*b^4*c^5 - 4*a^2*b^3*c^4*d - 2*a^3*b^2*c^3*d^2 - 4*a^4*b*c^2*d^3 +
5*a^5*c*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*(2*a^3*b^2*c^5 - 4*a^4*b*c^4*d + 2*a^5*c^3*d^2 - (15*a*b^4*c^4*d
- 7*a^2*b^3*c^3*d^2 - 7*a^3*b^2*c^2*d^3 + 15*a^4*b*c*d^4)*x^3 - (15*a*b^4*c^5 - 2*a^2*b^3*c^4*d - 10*a^3*b^2*
c^3*d^2 - 2*a^4*b*c^2*d^3 + 15*a^5*c*d^4)*x^2 - 5*(a^2*b^3*c^5 - a^3*b^2*c^4*d - a^4*b*c^3*d^2 + a^5*c^2*d^3)*
x)*sqrt(b*x + a)*sqrt(d*x + c))/((a^4*b^3*c^6*d - 2*a^5*b^2*c^5*d^2 + a^6*b*c^4*d^3)*x^4 + (a^4*b^3*c^7 - a^5*
b^2*c^6*d - a^6*b*c^5*d^2 + a^7*c^4*d^3)*x^3 + (a^5*b^2*c^7 - 2*a^6*b*c^6*d + a^7*c^5*d^2)*x^2)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x^3 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/x^3/(b*x+a)^(3/2)/(d*x+c)^(3/2),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. 1204 vs. \(2 (230) = 460\).

Time = 3.37 (sec) , antiderivative size = 1204, normalized size of antiderivative = 4.49 \[ \int \frac {1}{x^3 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

2*sqrt(b*x + a)*b^2*d^4/((b^2*c^5*abs(b) - 2*a*b*c^4*d*abs(b) + a^2*c^3*d^2*abs(b))*sqrt(b^2*c + (b*x + a)*b*d
- a*b*d)) + 4*sqrt(b*d)*b^5/((a^3*b*c*abs(b) - a^4*d*abs(b))*(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)) - 3/4*(5*sqrt(b*d)*b^4*c^2 + 6*sqrt(b*d)*a*b^3*c*d + 5*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^3*b*c^3*abs(b)) + 1/2*(7*sqrt(b*d)*b^10*c^5 - 21*sqrt(b*d)*a*b^9*c^4*d + 14*sqrt(
b*d)*a^2*b^8*c^3*d^2 + 14*sqrt(b*d)*a^3*b^7*c^2*d^3 - 21*sqrt(b*d)*a^4*b^6*c*d^4 + 7*sqrt(b*d)*a^5*b^5*d^5 - 2
1*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 - 4*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 - 4*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^5*c*d^3 - 21*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 + 21*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
+ 35*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 + 35*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 + 21*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 - 7*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 - 7*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^3*c^3*abs(b))

Mupad [F(-1)]

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

[In]

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

[Out]

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