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

Optimal. Leaf size=81 \[ \frac {2 A \sqrt {b x+c x^2}}{\sqrt {x}}-2 A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 81, 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 = {794, 664, 660, 207} \[ \frac {2 A \sqrt {b x+c x^2}}{\sqrt {x}}-2 A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {2 B \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 664

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

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 80, normalized size = 0.99 \[ \frac {2 \sqrt {x} \sqrt {b+c x} \left (\sqrt {b+c x} (3 A c+b B+B c x)-3 A \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{3 c \sqrt {x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[b + c*x]*(Sqrt[b + c*x]*(b*B + 3*A*c + B*c*x) - 3*A*Sqrt[b]*c*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))
/(3*c*Sqrt[x*(b + c*x)])

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fricas [A]  time = 0.71, size = 148, normalized size = 1.83 \[ \left [\frac {3 \, A \sqrt {b} c x \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (B c x + B b + 3 \, A c\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, c x}, \frac {2 \, {\left (3 \, A \sqrt {-b} c x \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (B c x + B b + 3 \, A c\right )} \sqrt {c x^{2} + b x} \sqrt {x}\right )}}{3 \, c x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 103, normalized size = 1.27 \[ \frac {2 \, A b \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {2 \, {\left (3 \, A b c \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + B \sqrt {-b} b^{\frac {3}{2}} + 3 \, A \sqrt {-b} \sqrt {b} c\right )}}{3 \, \sqrt {-b} c} + \frac {2 \, {\left ({\left (c x + b\right )}^{\frac {3}{2}} B c^{2} + 3 \, \sqrt {c x + b} A c^{3}\right )}}{3 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 79, normalized size = 0.98 \[ -\frac {2 \sqrt {\left (c x +b \right ) x}\, \left (3 A \sqrt {b}\, c \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-\sqrt {c x +b}\, B c x -3 \sqrt {c x +b}\, A c -\sqrt {c x +b}\, B b \right )}{3 \sqrt {c x +b}\, c \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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