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

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

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [C]  time = 0.02, 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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fricas [A]  time = 0.98, size = 248, normalized size = 2.56 \[ \left [-\frac {{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{3} + {\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (A b^{2} - {\left (2 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{2 \, {\left (b^{3} c x^{3} + b^{4} x^{2}\right )}}, \frac {{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{3} + {\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (A b^{2} - {\left (2 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{b^{3} c x^{3} + b^{4} x^{2}}\right ] \]

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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giac [A]  time = 0.24, size = 87, normalized size = 0.90 \[ \frac {{\left (2 \, B b - 3 \, A c\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {2 \, {\left (c x + b\right )} B b - 2 \, B b^{2} - 3 \, {\left (c x + b\right )} A c + 2 \, A b c}{{\left ({\left (c x + b\right )}^{\frac {3}{2}} - \sqrt {c x + b} b\right )} b^{2}} \]

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)

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maple [A]  time = 0.09, size = 94, normalized size = 0.97 \[ \frac {\sqrt {\left (c x +b \right ) x}\, \left (3 \sqrt {c x +b}\, A c x \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-2 \sqrt {c x +b}\, B b x \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-3 A \sqrt {b}\, c x +2 B \,b^{\frac {3}{2}} x -A \,b^{\frac {3}{2}}\right )}{\left (c x +b \right ) b^{\frac {5}{2}} x^{\frac {3}{2}}} \]

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]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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