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*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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Rubi [A]  time = 0.0311402, antiderivative size = 77, normalized size of antiderivative = 1., 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 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 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 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*}

Mathematica [A]  time = 0.0765913, size = 77, normalized size = 1. \[ \frac{2 d \sqrt{a+b x}}{c \sqrt{c+d x} (a d-b c)}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{3/2}} \]

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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Maple [B]  time = 0.024, size = 243, normalized size = 3.2 \begin{align*}{\frac{1}{ \left ( ad-bc \right ) c} \left ( -\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ) xa{d}^{2}+\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ) xbcd-\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ) acd+\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ) b{c}^{2}+2\,d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac} \right ) \sqrt{bx+a}{\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{dx+c}}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x + a}{\left (d x + c\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}

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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Fricas [B]  time = 3.07168, size = 760, normalized size = 9.87 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \sqrt{a + b x} \left (c + d x\right )^{\frac{3}{2}}}\, dx \end{align*}

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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Giac [B]  time = 3.08445, size = 189, normalized size = 2.45 \begin{align*} -\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 |}} \end{align*}

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))