3.238 \(\int \frac{A+B x}{\sqrt{x} (b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=97 \[ \frac{\sqrt{x} (2 b B-3 A c)}{b^2 \sqrt{b x+c x^2}}-\frac{(2 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{5/2}}-\frac{A}{b \sqrt{x} \sqrt{b x+c x^2}} \]

[Out]

-(A/(b*Sqrt[x]*Sqrt[b*x + c*x^2])) + ((2*b*B - 3*A*c)*Sqrt[x])/(b^2*Sqrt[b*x + c*x^2]) - ((2*b*B - 3*A*c)*ArcT
anh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(5/2)

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Rubi [A]  time = 0.0752901, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {792, 666, 660, 207} \[ \frac{\sqrt{x} (2 b B-3 A c)}{b^2 \sqrt{b x+c x^2}}-\frac{(2 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{5/2}}-\frac{A}{b \sqrt{x} \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(b*Sqrt[x]*Sqrt[b*x + c*x^2])) + ((2*b*B - 3*A*c)*Sqrt[x])/(b^2*Sqrt[b*x + c*x^2]) - ((2*b*B - 3*A*c)*ArcT
anh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(5/2)

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 666

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((2*c*d - b*e)*(d +
e*x)^m*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*c*d - b*e)*(m + 2*p + 2))/((p + 1)*
(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 207

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

Rubi steps

\begin{align*} \int \frac{A+B x}{\sqrt{x} \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{A}{b \sqrt{x} \sqrt{b x+c x^2}}+\frac{\left (\frac{1}{2} (b B-2 A c)+\frac{1}{2} (b B-A c)\right ) \int \frac{\sqrt{x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{b}\\ &=-\frac{A}{b \sqrt{x} \sqrt{b x+c x^2}}+\frac{(2 b B-3 A c) \sqrt{x}}{b^2 \sqrt{b x+c x^2}}+\frac{(2 b B-3 A c) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{2 b^2}\\ &=-\frac{A}{b \sqrt{x} \sqrt{b x+c x^2}}+\frac{(2 b B-3 A c) \sqrt{x}}{b^2 \sqrt{b x+c x^2}}+\frac{(2 b B-3 A c) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{b^2}\\ &=-\frac{A}{b \sqrt{x} \sqrt{b x+c x^2}}+\frac{(2 b B-3 A c) \sqrt{x}}{b^2 \sqrt{b x+c x^2}}-\frac{(2 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.023136, size = 52, normalized size = 0.54 \[ \frac{x (2 b B-3 A c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c x}{b}+1\right )-A b}{b^2 \sqrt{x} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b) + (2*b*B - 3*A*c)*x*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (c*x)/b])/(b^2*Sqrt[x]*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.021, size = 94, normalized size = 1. \begin{align*}{\frac{1}{cx+b}\sqrt{x \left ( cx+b \right ) } \left ( 3\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}xc-2\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}xb-3\,A\sqrt{b}xc+2\,B{b}^{3/2}x-A{b}^{{\frac{3}{2}}} \right ){x}^{-{\frac{3}{2}}}{b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(c*x^2+b*x)^(3/2)/x^(1/2),x)

[Out]

(x*(c*x+b))^(1/2)*(3*A*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x*c-2*B*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*
x+b)^(1/2)*x*b-3*A*b^(1/2)*x*c+2*B*b^(3/2)*x-A*b^(3/2))/x^(3/2)/(c*x+b)/b^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*sqrt(x)), x)

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Fricas [A]  time = 1.97698, size = 552, normalized size = 5.69 \begin{align*} \left [-\frac{{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{3} +{\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )} \sqrt{b} \log \left (-\frac{c x^{2} + 2 \, b x + 2 \, \sqrt{c x^{2} + b x} \sqrt{b} \sqrt{x}}{x^{2}}\right ) + 2 \,{\left (A b^{2} -{\left (2 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{2 \,{\left (b^{3} c x^{3} + b^{4} x^{2}\right )}}, \frac{{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{3} +{\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) -{\left (A b^{2} -{\left (2 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{b^{3} c x^{3} + b^{4} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(((2*B*b*c - 3*A*c^2)*x^3 + (2*B*b^2 - 3*A*b*c)*x^2)*sqrt(b)*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*s
qrt(b)*sqrt(x))/x^2) + 2*(A*b^2 - (2*B*b^2 - 3*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*c*x^3 + b^4*x^2), (((
2*B*b*c - 3*A*c^2)*x^3 + (2*B*b^2 - 3*A*b*c)*x^2)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - (A*b^2
- (2*B*b^2 - 3*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*c*x^3 + b^4*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x**2+b*x)**(3/2)/x**(1/2),x)

[Out]

Integral((A + B*x)/(sqrt(x)*(x*(b + c*x))**(3/2)), x)

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Giac [A]  time = 1.21979, size = 117, normalized size = 1.21 \begin{align*} \frac{{\left (2 \, B b - 3 \, A c\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{2}} + \frac{2 \,{\left (c x + b\right )} B b - 2 \, B b^{2} - 3 \,{\left (c x + b\right )} A c + 2 \, A b c}{{\left ({\left (c x + b\right )}^{\frac{3}{2}} - \sqrt{c x + b} b\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*b - 3*A*c)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (2*(c*x + b)*B*b - 2*B*b^2 - 3*(c*x + b)*A*c +
2*A*b*c)/(((c*x + b)^(3/2) - sqrt(c*x + b)*b)*b^2)