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

Optimal. Leaf size=113 \[ \frac{\sqrt{a+b x} (b c-3 a d)}{a c^2 \sqrt{c+d x}}-\frac{(b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{5/2}}-\frac{(a+b x)^{3/2}}{a c x \sqrt{c+d x}} \]

[Out]

((b*c - 3*a*d)*Sqrt[a + b*x])/(a*c^2*Sqrt[c + d*x]) - (a + b*x)^(3/2)/(a*c*x*Sqrt[c + d*x]) - ((b*c - 3*a*d)*A
rcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(5/2))

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Rubi [A]  time = 0.0500735, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \[ \frac{\sqrt{a+b x} (b c-3 a d)}{a c^2 \sqrt{c+d x}}-\frac{(b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{5/2}}-\frac{(a+b x)^{3/2}}{a c x \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - 3*a*d)*Sqrt[a + b*x])/(a*c^2*Sqrt[c + d*x]) - (a + b*x)^(3/2)/(a*c*x*Sqrt[c + d*x]) - ((b*c - 3*a*d)*A
rcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(5/2))

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 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[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{\sqrt{a+b x}}{x^2 (c+d x)^{3/2}} \, dx &=-\frac{(a+b x)^{3/2}}{a c x \sqrt{c+d x}}-\frac{\left (-\frac{b c}{2}+\frac{3 a d}{2}\right ) \int \frac{\sqrt{a+b x}}{x (c+d x)^{3/2}} \, dx}{a c}\\ &=\frac{(b c-3 a d) \sqrt{a+b x}}{a c^2 \sqrt{c+d x}}-\frac{(a+b x)^{3/2}}{a c x \sqrt{c+d x}}+\frac{(b c-3 a d) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 c^2}\\ &=\frac{(b c-3 a d) \sqrt{a+b x}}{a c^2 \sqrt{c+d x}}-\frac{(a+b x)^{3/2}}{a c x \sqrt{c+d x}}+\frac{(b c-3 a d) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{c^2}\\ &=\frac{(b c-3 a d) \sqrt{a+b x}}{a c^2 \sqrt{c+d x}}-\frac{(a+b x)^{3/2}}{a c x \sqrt{c+d x}}-\frac{(b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0658191, size = 83, normalized size = 0.73 \[ \frac{(3 a d-b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{\sqrt{a} c^{5/2}}-\frac{\sqrt{a+b x} (c+3 d x)}{c^2 x \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a + b*x]*(c + 3*d*x))/(c^2*x*Sqrt[c + d*x])) + ((-(b*c) + 3*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt
[a]*Sqrt[c + d*x])])/(Sqrt[a]*c^(5/2))

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Maple [B]  time = 0.021, size = 267, normalized size = 2.4 \begin{align*}{\frac{1}{2\,{c}^{2}x}\sqrt{bx+a} \left ( 3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}a{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 ){x}^{2}bcd+3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) xacd-\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac \right ) } \right ) xb{c}^{2}-6\,xd\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }-2\,c\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) } \right ){\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((b*x+a)^(1/2)/x^2/(d*x+c)^(3/2),x)

[Out]

1/2*(b*x+a)^(1/2)/c^2*(3*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^2*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^2*b*c*d+3*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*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)*x*b*c^2-6*x*d*(a*c)
^(1/2)*((b*x+a)*(d*x+c))^(1/2)-2*c*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x/(a*c)^(1/2)/((b*x+a)*(d*x+c))^(1/2)/
(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((b*x+a)^(1/2)/x^2/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.40593, size = 753, normalized size = 6.66 \begin{align*} \left [-\frac{{\left ({\left (b c d - 3 \, a d^{2}\right )} x^{2} +{\left (b c^{2} - 3 \, a c d\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 (3 \, a c d x + a c^{2}\right )} \sqrt{b x + a} \sqrt{d x + c}}{4 \,{\left (a c^{3} d x^{2} + a c^{4} x\right )}}, \frac{{\left ({\left (b c d - 3 \, a d^{2}\right )} x^{2} +{\left (b c^{2} - 3 \, a c d\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 (3 \, a c d x + a c^{2}\right )} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (a c^{3} d x^{2} + a c^{4} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError