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

Optimal. Leaf size=121 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2} c^{3/2}}+\frac {2 b}{a \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}+\frac {2 d \sqrt {a+b x} (a d+b c)}{a c \sqrt {c+d x} (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {104, 152, 12, 93, 208} \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2} c^{3/2}}+\frac {2 b}{a \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}+\frac {2 d \sqrt {a+b x} (a d+b c)}{a c \sqrt {c+d x} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 104

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] &&
IntegersQ[2*m, 2*n, 2*p]

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

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Mathematica [A]  time = 0.14, size = 104, normalized size = 0.86 \[ \frac {2 \left (a^2 d^2+a b d^2 x+b^2 c (c+d x)\right )}{a c \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.24, size = 719, normalized size = 5.94 \[ \left [\frac {{\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x\right )} \sqrt {a c} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (a b^{2} c^{3} + a^{3} c d^{2} + {\left (a b^{2} c^{2} d + a^{2} b c d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a^{3} b^{2} c^{5} - 2 \, a^{4} b c^{4} d + a^{5} c^{3} d^{2} + {\left (a^{2} b^{3} c^{4} d - 2 \, a^{3} b^{2} c^{3} d^{2} + a^{4} b c^{2} d^{3}\right )} x^{2} + {\left (a^{2} b^{3} c^{5} - a^{3} b^{2} c^{4} d - a^{4} b c^{3} d^{2} + a^{5} c^{2} d^{3}\right )} x\right )}}, \frac {{\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x\right )} \sqrt {-a c} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (a b^{2} c^{3} + a^{3} c d^{2} + {\left (a b^{2} c^{2} d + a^{2} b c d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{a^{3} b^{2} c^{5} - 2 \, a^{4} b c^{4} d + a^{5} c^{3} d^{2} + {\left (a^{2} b^{3} c^{4} d - 2 \, a^{3} b^{2} c^{3} d^{2} + a^{4} b c^{2} d^{3}\right )} x^{2} + {\left (a^{2} b^{3} c^{5} - a^{3} b^{2} c^{4} d - a^{4} b c^{3} d^{2} + a^{5} c^{2} d^{3}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^2 + (b^3*c^3 - a*b^2*
c^2*d - a^2*b*c*d^2 + a^3*d^3)*x)*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*(a*b^2*c^3 + a^3*c*d^
2 + (a*b^2*c^2*d + a^2*b*c*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b^2*c^5 - 2*a^4*b*c^4*d + a^5*c^3*d^2 + (
a^2*b^3*c^4*d - 2*a^3*b^2*c^3*d^2 + a^4*b*c^2*d^3)*x^2 + (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), ((a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^2 + (b^3*c^3 -
a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x)*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*(a*b^2*c^3 + a^3*c*d^2 + (a*b^2*c^2*d + a^
2*b*c*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b^2*c^5 - 2*a^4*b*c^4*d + a^5*c^3*d^2 + (a^2*b^3*c^4*d - 2*a^3
*b^2*c^3*d^2 + a^4*b*c^2*d^3)*x^2 + (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)]

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giac [B]  time = 3.98, size = 241, normalized size = 1.99 \[ \frac {2 \, \sqrt {b x + a} b^{2} d^{2}}{{\left (b^{2} c^{3} {\left | b \right |} - 2 \, a b c^{2} d {\left | b \right |} + a^{2} c d^{2} {\left | b \right |}\right )} \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {4 \, \sqrt {b d} b^{3}}{{\left (a b c {\left | b \right |} - a^{2} d {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} - \frac {2 \, \sqrt {b d} b \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} a c {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*x + a)*b^2*d^2/((b^2*c^3*abs(b) - 2*a*b*c^2*d*abs(b) + a^2*c*d^2*abs(b))*sqrt(b^2*c + (b*x + a)*b*d -
a*b*d)) + 4*sqrt(b*d)*b^3/((a*b*c*abs(b) - a^2*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)) - 2*sqrt(b*d)*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*c*abs(b))

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maple [B]  time = 0.03, size = 638, normalized size = 5.27 \[ \frac {-a^{2} b \,d^{3} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+2 a \,b^{2} c \,d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-b^{3} c^{2} d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-a^{3} d^{3} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+a^{2} b c \,d^{2} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+a \,b^{2} c^{2} d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-b^{3} c^{3} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-a^{3} c \,d^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+2 a^{2} b \,c^{2} d \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-a \,b^{2} c^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b \,d^{2} x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b^{2} c d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} d^{2}+2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b^{2} c^{2}}{\sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (a d -b c \right )^{2} \sqrt {a c}\, \sqrt {b x +a}\, \sqrt {d x +c}\, a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-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*d^3+2*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^2*c*d^2-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*b^3*c^2*d-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*a^3*d^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*a^2*b*c*d^2+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*a*b^2*c^2*d-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*b^3*c^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)*a^3*c*d^2+2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*
x+a)*(d*x+c))^(1/2))/x)*a^2*b*c^2*d-ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*a*b^2*c^3+
2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*a*b*d^2+2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*b^2*c*d+2*((b*x+a)*(d*
x+c))^(1/2)*(a*c)^(1/2)*a^2*d^2+2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*b^2*c^2)/c/a/((b*x+a)*(d*x+c))^(1/2)/(a*
d-b*c)^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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(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 = -0.01 \[ \int \frac {1}{x\,{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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