3.3.2 \(\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx\)

Optimal. Leaf size=142 \[ \frac {c^2 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{5/2}}-\frac {c \sqrt {b x+c x^2} (2 b B-A c)}{8 b^2 x^{3/2}}-\frac {\sqrt {b x+c x^2} (2 b B-A c)}{4 b x^{5/2}}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {792, 662, 672, 660, 207} \begin {gather*} \frac {c^2 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{5/2}}-\frac {c \sqrt {b x+c x^2} (2 b B-A c)}{8 b^2 x^{3/2}}-\frac {\sqrt {b x+c x^2} (2 b B-A c)}{4 b x^{5/2}}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^(9/2),x]

[Out]

-((2*b*B - A*c)*Sqrt[b*x + c*x^2])/(4*b*x^(5/2)) - (c*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(8*b^2*x^(3/2)) - (A*(b
*x + c*x^2)^(3/2))/(3*b*x^(9/2)) + (c^2*(2*b*B - A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*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 662

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

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) \sqrt {b x+c x^2}}{x^{9/2}} \, dx &=-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}+\frac {\left (-\frac {9}{2} (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right ) \int \frac {\sqrt {b x+c x^2}}{x^{7/2}} \, dx}{3 b}\\ &=-\frac {(2 b B-A c) \sqrt {b x+c x^2}}{4 b x^{5/2}}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}+\frac {(c (2 b B-A c)) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{8 b}\\ &=-\frac {(2 b B-A c) \sqrt {b x+c x^2}}{4 b x^{5/2}}-\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{8 b^2 x^{3/2}}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}-\frac {\left (c^2 (2 b B-A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{16 b^2}\\ &=-\frac {(2 b B-A c) \sqrt {b x+c x^2}}{4 b x^{5/2}}-\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{8 b^2 x^{3/2}}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}-\frac {\left (c^2 (2 b B-A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{8 b^2}\\ &=-\frac {(2 b B-A c) \sqrt {b x+c x^2}}{4 b x^{5/2}}-\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{8 b^2 x^{3/2}}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}+\frac {c^2 (2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.43 \begin {gather*} -\frac {(x (b+c x))^{3/2} \left (A b^3+c^2 x^3 (2 b B-A c) \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {c x}{b}+1\right )\right )}{3 b^4 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^(9/2),x]

[Out]

-1/3*((x*(b + c*x))^(3/2)*(A*b^3 + c^2*(2*b*B - A*c)*x^3*Hypergeometric2F1[3/2, 3, 5/2, 1 + (c*x)/b]))/(b^4*x^
(9/2))

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IntegrateAlgebraic [A]  time = 0.33, size = 111, normalized size = 0.78 \begin {gather*} \frac {\left (2 b B c^2-A c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{8 b^{5/2}}+\frac {\sqrt {b x+c x^2} \left (-8 A b^2-2 A b c x+3 A c^2 x^2-12 b^2 B x-6 b B c x^2\right )}{24 b^2 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*Sqrt[b*x + c*x^2])/x^(9/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(-8*A*b^2 - 12*b^2*B*x - 2*A*b*c*x - 6*b*B*c*x^2 + 3*A*c^2*x^2))/(24*b^2*x^(7/2)) + ((2*b*B
*c^2 - A*c^3)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(8*b^(5/2))

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fricas [A]  time = 0.43, size = 238, normalized size = 1.68 \begin {gather*} \left [-\frac {3 \, {\left (2 \, B b c^{2} - A c^{3}\right )} \sqrt {b} x^{4} \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 (8 \, A b^{3} + 3 \, {\left (2 \, B b^{2} c - A b c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{3} + A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, b^{3} x^{4}}, -\frac {3 \, {\left (2 \, B b c^{2} - A c^{3}\right )} \sqrt {-b} x^{4} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (8 \, A b^{3} + 3 \, {\left (2 \, B b^{2} c - A b c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{3} + A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, b^{3} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(9/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(2*B*b*c^2 - A*c^3)*sqrt(b)*x^4*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*
(8*A*b^3 + 3*(2*B*b^2*c - A*b*c^2)*x^2 + 2*(6*B*b^3 + A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^4), -1/24*
(3*(2*B*b*c^2 - A*c^3)*sqrt(-b)*x^4*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (8*A*b^3 + 3*(2*B*b^2*c - A*b
*c^2)*x^2 + 2*(6*B*b^3 + A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^4)]

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giac [A]  time = 0.27, size = 128, normalized size = 0.90 \begin {gather*} -\frac {\frac {3 \, {\left (2 \, B b c^{3} - A c^{4}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {6 \, {\left (c x + b\right )}^{\frac {5}{2}} B b c^{3} - 6 \, \sqrt {c x + b} B b^{3} c^{3} - 3 \, {\left (c x + b\right )}^{\frac {5}{2}} A c^{4} + 8 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c^{4} + 3 \, \sqrt {c x + b} A b^{2} c^{4}}{b^{2} c^{3} x^{3}}}{24 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(9/2),x, algorithm="giac")

[Out]

-1/24*(3*(2*B*b*c^3 - A*c^4)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (6*(c*x + b)^(5/2)*B*b*c^3 - 6*sq
rt(c*x + b)*B*b^3*c^3 - 3*(c*x + b)^(5/2)*A*c^4 + 8*(c*x + b)^(3/2)*A*b*c^4 + 3*sqrt(c*x + b)*A*b^2*c^4)/(b^2*
c^3*x^3))/c

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maple [A]  time = 0.09, size = 147, normalized size = 1.04 \begin {gather*} -\frac {\sqrt {\left (c x +b \right ) x}\, \left (3 A \,c^{3} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-6 B b \,c^{2} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-3 \sqrt {c x +b}\, A \sqrt {b}\, c^{2} x^{2}+6 \sqrt {c x +b}\, B \,b^{\frac {3}{2}} c \,x^{2}+2 \sqrt {c x +b}\, A \,b^{\frac {3}{2}} c x +12 \sqrt {c x +b}\, B \,b^{\frac {5}{2}} x +8 \sqrt {c x +b}\, A \,b^{\frac {5}{2}}\right )}{24 \sqrt {c x +b}\, b^{\frac {5}{2}} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^(9/2),x)

[Out]

-1/24*((c*x+b)*x)^(1/2)/b^(5/2)*(3*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*c^3-6*B*arctanh((c*x+b)^(1/2)/b^(1/2))
*x^3*b*c^2-3*A*x^2*c^2*b^(1/2)*(c*x+b)^(1/2)+6*B*x^2*b^(3/2)*c*(c*x+b)^(1/2)+2*A*x*b^(3/2)*c*(c*x+b)^(1/2)+12*
B*x*b^(5/2)*(c*x+b)^(1/2)+8*A*b^(5/2)*(c*x+b)^(1/2))/x^(7/2)/(c*x+b)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c x^{2} + b x} {\left (B x + A\right )}}{x^{\frac {9}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(9/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x)*(B*x + A)/x^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{x^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^(9/2),x)

[Out]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^(9/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{\frac {9}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**(9/2),x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**(9/2), x)

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