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

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

[Out]

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

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Rubi [A]  time = 0.0432569, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {78, 63, 217, 206} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{b} d^{3/2}}-\frac{2 c \sqrt{a+b x}}{d \sqrt{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.167007, size = 110, normalized size = 1.43 \[ \frac{2 \left (b c \sqrt{d} \sqrt{a+b x}-(b c-a d)^{3/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )\right )}{b d^{3/2} \sqrt{c+d x} (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.019, size = 251, normalized size = 3.3 \begin{align*}{\frac{1}{d \left ( ad-bc \right ) }\sqrt{bx+a} \left ( \ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) xa{d}^{2}-\ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) xbcd+\ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) acd-\ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) b{c}^{2}+2\,c\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.56558, size = 744, normalized size = 9.66 \begin{align*} \left [-\frac{4 \, \sqrt{b x + a} \sqrt{d x + c} b c d -{\left (b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x\right )} \sqrt{b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \,{\left (2 \, b d x + b c + a d\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right )}{2 \,{\left (b^{2} c^{2} d^{2} - a b c d^{3} +{\left (b^{2} c d^{3} - a b d^{4}\right )} x\right )}}, -\frac{2 \, \sqrt{b x + a} \sqrt{d x + c} b c d +{\left (b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x\right )} \sqrt{-b d} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (b^{2} d^{2} x^{2} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right )}{b^{2} c^{2} d^{2} - a b c d^{3} +{\left (b^{2} c d^{3} - a b d^{4}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)*b*c*d - (b*c^2 - a*c*d + (b*c*d - a*d^2)*x)*sqrt(b*d)*log(8*b^2*d^2*x^2 +
b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d +
a*b*d^2)*x))/(b^2*c^2*d^2 - a*b*c*d^3 + (b^2*c*d^3 - a*b*d^4)*x), -(2*sqrt(b*x + a)*sqrt(d*x + c)*b*c*d + (b*c
^2 - a*c*d + (b*c*d - a*d^2)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x
+ c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)))/(b^2*c^2*d^2 - a*b*c*d^3 + (b^2*c*d^3 - a*b*d^4)*x)]

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

[Out]

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

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Giac [A]  time = 1.27627, size = 146, normalized size = 1.9 \begin{align*} -\frac{2 \,{\left (\frac{\sqrt{b x + a} b^{3} c{\left | b \right |}}{{\left (b^{3} c d - a b^{2} d^{2}\right )} \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}} + \frac{{\left | b \right |} \log \left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} d}\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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