3.3.10 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^{9/2}} \, dx\)

Optimal. Leaf size=137 \[ \frac {3 c \sqrt {b x+c x^2} (A c+4 b B)}{4 b \sqrt {x}}-\frac {3 c (A c+4 b B) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 \sqrt {b}}-\frac {\left (b x+c x^2\right )^{3/2} (A c+4 b B)}{4 b x^{5/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{2 b x^{9/2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 137, 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, 664, 660, 207} \begin {gather*} -\frac {\left (b x+c x^2\right )^{3/2} (A c+4 b B)}{4 b x^{5/2}}+\frac {3 c \sqrt {b x+c x^2} (A c+4 b B)}{4 b \sqrt {x}}-\frac {3 c (A c+4 b B) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 \sqrt {b}}-\frac {A \left (b x+c x^2\right )^{5/2}}{2 b x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(9/2),x]

[Out]

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

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

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Mathematica [C]  time = 0.03, size = 59, normalized size = 0.43 \begin {gather*} \frac {(x (b+c x))^{5/2} \left (c x^2 (A c+4 b B) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {c x}{b}+1\right )-5 A b^2\right )}{10 b^3 x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(9/2),x]

[Out]

((x*(b + c*x))^(5/2)*(-5*A*b^2 + c*(4*b*B + A*c)*x^2*Hypergeometric2F1[2, 5/2, 7/2, 1 + (c*x)/b]))/(10*b^3*x^(
9/2))

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IntegrateAlgebraic [A]  time = 0.80, size = 90, normalized size = 0.66 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-2 A b-5 A c x-4 b B x+8 B c x^2\right )}{4 x^{5/2}}-\frac {3 \left (A c^2+4 b B c\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{4 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(9/2),x]

[Out]

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

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fricas [A]  time = 0.42, size = 206, normalized size = 1.50 \begin {gather*} \left [\frac {3 \, {\left (4 \, B b c + 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 (8 \, B b c x^{2} - 2 \, A b^{2} - {\left (4 \, B b^{2} + 5 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, b x^{3}}, \frac {3 \, {\left (4 \, B b c + A c^{2}\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (8 \, B b c x^{2} - 2 \, A b^{2} - {\left (4 \, B b^{2} + 5 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, b x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(9/2),x, algorithm="fricas")

[Out]

[1/8*(3*(4*B*b*c + 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*(8*B
*b*c*x^2 - 2*A*b^2 - (4*B*b^2 + 5*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b*x^3), 1/4*(3*(4*B*b*c + A*c^2)*sqrt(
-b)*x^3*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (8*B*b*c*x^2 - 2*A*b^2 - (4*B*b^2 + 5*A*b*c)*x)*sqrt(c*x^
2 + b*x)*sqrt(x))/(b*x^3)]

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giac [A]  time = 0.42, size = 119, normalized size = 0.87 \begin {gather*} \frac {8 \, \sqrt {c x + b} B c^{2} + \frac {3 \, {\left (4 \, B b c^{2} + A c^{3}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c^{2} - 4 \, \sqrt {c x + b} B b^{2} c^{2} + 5 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{3} - 3 \, \sqrt {c x + b} A b c^{3}}{c^{2} x^{2}}}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(9/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 126, normalized size = 0.92 \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 )+12 B b c \,x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-8 \sqrt {c x +b}\, B \sqrt {b}\, c \,x^{2}+5 \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}\, \sqrt {b}\, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/x^(9/2),x)

[Out]

-1/4*((c*x+b)*x)^(1/2)*(3*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*c^2+12*B*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b*c
-8*(c*x+b)^(1/2)*B*b^(1/2)*c*x^2+5*(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)/b^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} {\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)^(3/2)/x^(9/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)*(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 {{\left (c\,x^2+b\,x\right )}^{3/2}\,\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)^(3/2)*(A + B*x))/x^(9/2),x)

[Out]

int(((b*x + c*x^2)^(3/2)*(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 {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \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)**(3/2)/x**(9/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**(9/2), x)

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