3.3.31 \(\int \frac {A+B x}{x^{5/2} \sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=105 \[ \frac {c (4 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{5/2}}-\frac {\sqrt {b x+c x^2} (4 b B-3 A c)}{4 b^2 x^{3/2}}-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 105, 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, 672, 660, 207} \begin {gather*} -\frac {\sqrt {b x+c x^2} (4 b B-3 A c)}{4 b^2 x^{3/2}}+\frac {c (4 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{5/2}}-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(5/2)*Sqrt[b*x + c*x^2]),x]

[Out]

-(A*Sqrt[b*x + c*x^2])/(2*b*x^(5/2)) - ((4*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(4*b^2*x^(3/2)) + (c*(4*b*B - 3*A*c
)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(4*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 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && 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}{x^{5/2} \sqrt {b x+c x^2}} \, dx &=-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}}+\frac {\left (-\frac {5}{2} (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right ) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{2 b}\\ &=-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}}-\frac {(4 b B-3 A c) \sqrt {b x+c x^2}}{4 b^2 x^{3/2}}-\frac {(c (4 b B-3 A c)) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{8 b^2}\\ &=-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}}-\frac {(4 b B-3 A c) \sqrt {b x+c x^2}}{4 b^2 x^{3/2}}-\frac {(c (4 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 )}{4 b^2}\\ &=-\frac {A \sqrt {b x+c x^2}}{2 b x^{5/2}}-\frac {(4 b B-3 A c) \sqrt {b x+c x^2}}{4 b^2 x^{3/2}}+\frac {c (4 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 92, normalized size = 0.88 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (c x^2 (4 b B-3 A c) \tanh ^{-1}\left (\sqrt {\frac {c x}{b}+1}\right )+b \sqrt {\frac {c x}{b}+1} (-2 A b+3 A c x-4 b B x)\right )}{4 b^3 x^{5/2} \sqrt {\frac {c x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(5/2)*Sqrt[b*x + c*x^2]),x]

[Out]

(Sqrt[x*(b + c*x)]*(b*(-2*A*b - 4*b*B*x + 3*A*c*x)*Sqrt[1 + (c*x)/b] + c*(4*b*B - 3*A*c)*x^2*ArcTanh[Sqrt[1 +
(c*x)/b]]))/(4*b^3*x^(5/2)*Sqrt[1 + (c*x)/b])

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IntegrateAlgebraic [A]  time = 0.23, size = 87, normalized size = 0.83 \begin {gather*} \frac {\left (4 b B c-3 A c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{4 b^{5/2}}+\frac {\sqrt {b x+c x^2} (-2 A b+3 A c x-4 b B x)}{4 b^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(5/2)*Sqrt[b*x + c*x^2]),x]

[Out]

((-2*A*b - 4*b*B*x + 3*A*c*x)*Sqrt[b*x + c*x^2])/(4*b^2*x^(5/2)) + ((4*b*B*c - 3*A*c^2)*ArcTanh[(Sqrt[b]*Sqrt[
x])/Sqrt[b*x + c*x^2]])/(4*b^(5/2))

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fricas [A]  time = 0.42, size = 188, normalized size = 1.79 \begin {gather*} \left [-\frac {{\left (4 \, B b c - 3 \, A c^{2}\right )} \sqrt {b} x^{3} \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 (2 \, A b^{2} + {\left (4 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, b^{3} x^{3}}, -\frac {{\left (4 \, B b c - 3 \, A c^{2}\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (2 \, A b^{2} + {\left (4 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, b^{3} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*((4*B*b*c - 3*A*c^2)*sqrt(b)*x^3*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(2*
A*b^2 + (4*B*b^2 - 3*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^3), -1/4*((4*B*b*c - 3*A*c^2)*sqrt(-b)*x^3*ar
ctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (2*A*b^2 + (4*B*b^2 - 3*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x
^3)]

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giac [A]  time = 0.26, size = 111, normalized size = 1.06 \begin {gather*} -\frac {\frac {{\left (4 \, B b c^{2} - 3 \, A c^{3}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c^{2} - 4 \, \sqrt {c x + b} B b^{2} c^{2} - 3 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{3} + 5 \, \sqrt {c x + b} A b c^{3}}{b^{2} c^{2} x^{2}}}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-1/4*((4*B*b*c^2 - 3*A*c^3)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (4*(c*x + b)^(3/2)*B*b*c^2 - 4*sqr
t(c*x + b)*B*b^2*c^2 - 3*(c*x + b)^(3/2)*A*c^3 + 5*sqrt(c*x + b)*A*b*c^3)/(b^2*c^2*x^2))/c

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maple [A]  time = 0.07, size = 109, normalized size = 1.04 \begin {gather*} -\frac {\sqrt {\left (c x +b \right ) x}\, \left (3 A \,c^{2} x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-4 B b c \,x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-3 \sqrt {c x +b}\, A \sqrt {b}\, c x +4 \sqrt {c x +b}\, B \,b^{\frac {3}{2}} x +2 \sqrt {c x +b}\, A \,b^{\frac {3}{2}}\right )}{4 \sqrt {c x +b}\, b^{\frac {5}{2}} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(5/2)/(c*x^2+b*x)^(1/2),x)

[Out]

-1/4*((c*x+b)*x)^(1/2)/b^(5/2)*(3*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*c^2-4*B*arctanh((c*x+b)^(1/2)/b^(1/2))*
x^2*b*c-3*(c*x+b)^(1/2)*A*b^(1/2)*c*x+4*(c*x+b)^(1/2)*B*b^(3/2)*x+2*(c*x+b)^(1/2)*A*b^(3/2))/x^(5/2)/(c*x+b)^(
1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{\sqrt {c x^{2} + b x} x^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/(sqrt(c*x^2 + b*x)*x^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{x^{5/2}\,\sqrt {c\,x^2+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(5/2)*(b*x + c*x^2)^(1/2)),x)

[Out]

int((A + B*x)/(x^(5/2)*(b*x + c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{\frac {5}{2}} \sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(5/2)/(c*x**2+b*x)**(1/2),x)

[Out]

Integral((A + B*x)/(x**(5/2)*sqrt(x*(b + c*x))), x)

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