3.8.58 \(\int \frac {1}{x^3 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx\)
Optimal. Leaf size=268 \[ \frac {5 (a d+b c)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}-\frac {3 \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2} c^{7/2}}+\frac {b \left (-5 a^2 d^2-2 a b c d+15 b^2 c^2\right )}{4 a^3 c^2 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}+\frac {d \sqrt {a+b x} \left (15 a^2 d^2-22 a b c d+15 b^2 c^2\right ) (a d+b c)}{4 a^3 c^3 \sqrt {c+d x} (b c-a d)^2}-\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}} \]
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Rubi [A] time = 0.26, antiderivative size = 268, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used =
{103, 151, 152, 12, 93, 208} \begin {gather*} \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 {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 {3 \left (5 a^2 d^2+6 a b c d+5 b^2 c^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}}+\frac {5 (a d+b c)}{4 a^2 c^2 x \sqrt {a+b x} \sqrt {c+d x}}-\frac {1}{2 a c x^2 \sqrt {a+b x} \sqrt {c+d x}} \end {gather*}
Antiderivative was successfully verified.
[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 93
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 103
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] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])
Rule 151
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] && IntegerQ[m]
Rule 152
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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin {align*} \int \frac {1}{x^3 (a+b x)^{3/2} (c+d x)^{3/2}} \, dx &=-\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 ) \operatorname {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*}
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Mathematica [A] time = 0.29, size = 254, normalized size = 0.95 \begin {gather*} -\frac {3 \left (5 a^2 d^2+6 a b c d+5 b^2 c^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}}-\frac {a^4 d^2 \left (2 c^2-5 c d x-15 d^2 x^2\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 )+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 b^3 c^2 x \left (-5 c^2+2 c d x+7 d^2 x^2\right )-15 b^4 c^3 x^2 (c+d x)}{4 a^3 c^3 x^2 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[1/(x^3*(a + b*x)^(3/2)*(c + d*x)^(3/2)),x]
[Out]
-1/4*(-15*b^4*c^3*x^2*(c + d*x) + a^4*d^2*(2*c^2 - 5*c*d*x - 15*d^2*x^2) + a*b^3*c^2*x*(-5*c^2 + 2*c*d*x + 7*d
^2*x^2) + a^3*b*d*(-4*c^3 + 5*c^2*d*x + 2*c*d^2*x^2 - 15*d^3*x^3) + 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*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)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(7/2)*c^(7/2))
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IntegrateAlgebraic [A] time = 0.50, size = 399, normalized size = 1.49 \begin {gather*} \frac {\sqrt {a+b x} \left (\frac {15 a^5 d^4 (c+d x)^2}{(a+b x)^2}-\frac {25 a^4 c d^4 (c+d x)}{a+b x}-\frac {12 a^4 b c d^3 (c+d x)^2}{(a+b x)^2}-\frac {6 a^3 b^2 c^2 d^2 (c+d x)^2}{(a+b x)^2}+\frac {20 a^3 b c^2 d^3 (c+d x)}{a+b x}+8 a^3 c^2 d^4+\frac {8 a^2 b^4 c^3 (c+d x)^3}{(a+b x)^3}+\frac {20 a^2 b^3 c^3 d (c+d x)^2}{(a+b x)^2}-\frac {6 a^2 b^2 c^3 d^2 (c+d x)}{a+b x}+\frac {15 b^4 c^5 (c+d x)}{a+b x}-\frac {25 a b^4 c^4 (c+d x)^2}{(a+b x)^2}-\frac {12 a b^3 c^4 d (c+d x)}{a+b x}\right )}{4 a^3 c^3 \sqrt {c+d x} (a d-b c)^2 \left (\frac {a (c+d x)}{a+b x}-c\right )^2}-\frac {3 \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{4 a^{7/2} c^{7/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[1/(x^3*(a + b*x)^(3/2)*(c + d*x)^(3/2)),x]
[Out]
(Sqrt[a + b*x]*(8*a^3*c^2*d^4 + (15*b^4*c^5*(c + d*x))/(a + b*x) - (12*a*b^3*c^4*d*(c + d*x))/(a + b*x) - (6*a
^2*b^2*c^3*d^2*(c + d*x))/(a + b*x) + (20*a^3*b*c^2*d^3*(c + d*x))/(a + b*x) - (25*a^4*c*d^4*(c + d*x))/(a + b
*x) - (25*a*b^4*c^4*(c + d*x)^2)/(a + b*x)^2 + (20*a^2*b^3*c^3*d*(c + d*x)^2)/(a + b*x)^2 - (6*a^3*b^2*c^2*d^2
*(c + d*x)^2)/(a + b*x)^2 - (12*a^4*b*c*d^3*(c + d*x)^2)/(a + b*x)^2 + (15*a^5*d^4*(c + d*x)^2)/(a + b*x)^2 +
(8*a^2*b^4*c^3*(c + d*x)^3)/(a + b*x)^3))/(4*a^3*c^3*(-(b*c) + a*d)^2*Sqrt[c + d*x]*(-c + (a*(c + d*x))/(a + b
*x))^2) - (3*(5*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(4*
a^(7/2)*c^(7/2))
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fricas [B] time = 10.12, size = 1244, normalized size = 4.64
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[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)]
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giac [B] time = 57.94, size = 1204, normalized size = 4.49
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[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))
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maple [B] time = 0.04, size = 1372, normalized size = 5.12
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x^3/(b*x+a)^(3/2)/(d*x+c)^(3/2),x)
[Out]
-1/8/a^3/c^3*(15*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^5*d^5+15*ln((a*d*x+b*c*
x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*b^5*c^5-10*x*a*b^3*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+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a*b^4*c^5-30*x^2*a^4*d^4*(a*c)^(1/2)
*((b*x+a)*(d*x+c))^(1/2)-30*x^2*b^4*c^4*(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)+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+2*(a*c)^(1/2)*((b*x
+a)*(d*x+c))^(1/2))/x)*x^4*a^4*b*d^5+15*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*b^
5*c^4*d+15*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a^5*c*d^4-8*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+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*a^3*b^2*c*
d^4-6*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*a^2*b^3*c^2*d^3-12*ln((a*d*x+b*c*x+2
*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*a*b^4*c^3*d^2+3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a
)*(d*x+c))^(1/2))/x)*x^3*a^4*b*c*d^4-18*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^
3*b^2*c^2*d^3-18*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^2*b^3*c^3*d^2+3*ln((a*d
*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a*b^4*c^4*d-12*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2
)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a^4*b*c^2*d^3-6*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))
/x)*x^2*a^3*b^2*c^3*d^2-12*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a^2*b^3*c^4*d-3
0*x^3*a^3*b*d^4*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-30*x^3*b^4*c^3*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*x*
a^4*c*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+14*x^3*a^2*b^2*c*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+14*x^3*
a*b^3*c^2*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*x^2*a^3*b*c*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+20*x^2
*a^2*b^2*c^2*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*x^2*a*b^3*c^3*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+10*
x*a^3*b*c^2*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+10*x*a^2*b^2*c^3*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/((
b*x+a)*(d*x+c))^(1/2)/(a*d-b*c)^2/x^2/(a*c)^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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 details)Is a*d-b*c zero or nonzero?
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^3\,{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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