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

Optimal. Leaf size=77 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}}-\frac {2 d \sqrt {a+b x}}{c \sqrt {c+d x} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {96, 93, 208} \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}}-\frac {2 d \sqrt {a+b x}}{c \sqrt {c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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

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Mathematica [A]  time = 0.08, size = 77, normalized size = 1.00 \[ \frac {2 d \sqrt {a+b x}}{c \sqrt {c+d x} (a d-b c)}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.89, size = 346, normalized size = 4.49 \[ \left [-\frac {4 \, \sqrt {b x + a} \sqrt {d x + c} a c d - {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \sqrt {a c} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right )}{2 \, {\left (a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x\right )}}, -\frac {2 \, \sqrt {b x + a} \sqrt {d x + c} a c d - {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \sqrt {-a c} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right )}{a b c^{4} - a^{2} c^{3} d + {\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.10, size = 140, normalized size = 1.82 \[ -\frac {2 \, \sqrt {b x + a} b^{2} d}{{\left (b c^{2} {\left | b \right |} - a c d {\left | b \right |}\right )} \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {2 \, \sqrt {b d} b \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} c {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(b*x + a)*b^2*d/((b*c^2*abs(b) - a*c*d*abs(b))*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)) - 2*sqrt(b*d)*b*arc
tan(-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)*c*abs(b))

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maple [B]  time = 0.03, size = 243, normalized size = 3.16 \[ \frac {\left (-a \,d^{2} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+b c d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-a c d \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+b \,c^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, d \right ) \sqrt {b x +a}}{\left (a d -b c \right ) \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {d x +c}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x\,\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sqrt {a + b x} \left (c + d x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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