3.3.20 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{9/2}} \, dx\)

Optimal. Leaf size=160 \[ -b^{3/2} (5 A c+2 b B) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {b \sqrt {b x+c x^2} (5 A c+2 b B)}{\sqrt {x}}+\frac {\left (b x+c x^2\right )^{5/2} (5 A c+2 b B)}{5 b x^{5/2}}+\frac {\left (b x+c x^2\right )^{3/2} (5 A c+2 b B)}{3 x^{3/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {792, 664, 660, 207} \begin {gather*} -b^{3/2} (5 A c+2 b B) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {\left (b x+c x^2\right )^{5/2} (5 A c+2 b B)}{5 b x^{5/2}}+\frac {\left (b x+c x^2\right )^{3/2} (5 A c+2 b B)}{3 x^{3/2}}+\frac {b \sqrt {b x+c x^2} (5 A c+2 b B)}{\sqrt {x}}-\frac {A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(9/2),x]

[Out]

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

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Mathematica [A]  time = 0.08, size = 118, normalized size = 0.74 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {b+c x} \left (A \left (-15 b^2+70 b c x+10 c^2 x^2\right )+2 B x \left (23 b^2+11 b c x+3 c^2 x^2\right )\right )-15 b^{3/2} x (5 A c+2 b B) \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{15 x^{3/2} \sqrt {b+c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(9/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[b + c*x]*(2*B*x*(23*b^2 + 11*b*c*x + 3*c^2*x^2) + A*(-15*b^2 + 70*b*c*x + 10*c^2*x^2)
) - 15*b^(3/2)*(2*b*B + 5*A*c)*x*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(15*x^(3/2)*Sqrt[b + c*x])

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IntegrateAlgebraic [A]  time = 0.87, size = 113, normalized size = 0.71 \begin {gather*} \left (-5 A b^{3/2} c-2 b^{5/2} B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )+\frac {\sqrt {b x+c x^2} \left (-15 A b^2+70 A b c x+10 A c^2 x^2+46 b^2 B x+22 b B c x^2+6 B c^2 x^3\right )}{15 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(9/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(-15*A*b^2 + 46*b^2*B*x + 70*A*b*c*x + 22*b*B*c*x^2 + 10*A*c^2*x^2 + 6*B*c^2*x^3))/(15*x^(3
/2)) + (-2*b^(5/2)*B - 5*A*b^(3/2)*c)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]]

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fricas [A]  time = 0.43, size = 238, normalized size = 1.49 \begin {gather*} \left [\frac {15 \, {\left (2 \, B b^{2} + 5 \, A b c\right )} \sqrt {b} x^{2} \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 (6 \, B c^{2} x^{3} - 15 \, A b^{2} + 2 \, {\left (11 \, B b c + 5 \, A c^{2}\right )} x^{2} + 2 \, {\left (23 \, B b^{2} + 35 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{30 \, x^{2}}, \frac {15 \, {\left (2 \, B b^{2} + 5 \, A b c\right )} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (6 \, B c^{2} x^{3} - 15 \, A b^{2} + 2 \, {\left (11 \, B b c + 5 \, A c^{2}\right )} x^{2} + 2 \, {\left (23 \, B b^{2} + 35 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{15 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(9/2),x, algorithm="fricas")

[Out]

[1/30*(15*(2*B*b^2 + 5*A*b*c)*sqrt(b)*x^2*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*
(6*B*c^2*x^3 - 15*A*b^2 + 2*(11*B*b*c + 5*A*c^2)*x^2 + 2*(23*B*b^2 + 35*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/x
^2, 1/15*(15*(2*B*b^2 + 5*A*b*c)*sqrt(-b)*x^2*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (6*B*c^2*x^3 - 15*A
*b^2 + 2*(11*B*b*c + 5*A*c^2)*x^2 + 2*(23*B*b^2 + 35*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/x^2]

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giac [A]  time = 0.35, size = 125, normalized size = 0.78 \begin {gather*} \frac {6 \, {\left (c x + b\right )}^{\frac {5}{2}} B c + 10 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c + 30 \, \sqrt {c x + b} B b^{2} c + 10 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{2} + 60 \, \sqrt {c x + b} A b c^{2} - \frac {15 \, \sqrt {c x + b} A b^{2} c}{x} + \frac {15 \, {\left (2 \, B b^{3} c + 5 \, A b^{2} c^{2}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}}}{15 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(9/2),x, algorithm="giac")

[Out]

1/15*(6*(c*x + b)^(5/2)*B*c + 10*(c*x + b)^(3/2)*B*b*c + 30*sqrt(c*x + b)*B*b^2*c + 10*(c*x + b)^(3/2)*A*c^2 +
60*sqrt(c*x + b)*A*b*c^2 - 15*sqrt(c*x + b)*A*b^2*c/x + 15*(2*B*b^3*c + 5*A*b^2*c^2)*arctan(sqrt(c*x + b)/sqr
t(-b))/sqrt(-b))/c

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maple [A]  time = 0.07, size = 162, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {\left (c x +b \right ) x}\, \left (-6 \sqrt {c x +b}\, B \sqrt {b}\, c^{2} x^{3}+75 A \,b^{2} c x \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )+30 B \,b^{3} x \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-10 \sqrt {c x +b}\, A \sqrt {b}\, c^{2} x^{2}-22 \sqrt {c x +b}\, B \,b^{\frac {3}{2}} c \,x^{2}-70 \sqrt {c x +b}\, A \,b^{\frac {3}{2}} c x -46 \sqrt {c x +b}\, B \,b^{\frac {5}{2}} x +15 \sqrt {c x +b}\, A \,b^{\frac {5}{2}}\right )}{15 \sqrt {c x +b}\, \sqrt {b}\, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(9/2),x)

[Out]

-1/15*((c*x+b)*x)^(1/2)*(-6*B*x^3*c^2*(c*x+b)^(1/2)*b^(1/2)-10*(c*x+b)^(1/2)*A*b^(1/2)*c^2*x^2-22*(c*x+b)^(1/2
)*B*b^(3/2)*c*x^2+75*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x*b^2*c-70*(c*x+b)^(1/2)*A*b^(3/2)*c*x+30*B*arctanh((c*x
+b)^(1/2)/b^(1/2))*x*b^3-46*(c*x+b)^(1/2)*B*b^(5/2)*x+15*(c*x+b)^(1/2)*A*b^(5/2))/x^(3/2)/(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*} \frac {2 \, {\left (5 \, {\left (2 \, B b c + A c^{2}\right )} x^{2} + {\left (3 \, B c^{2} x^{2} + B b c x - 2 \, B b^{2}\right )} x + 5 \, {\left (2 \, B b^{2} + A b c\right )} x\right )} \sqrt {c x + b}}{15 \, x} + \int \frac {{\left (A b^{2} + {\left (B b^{2} + 2 \, A b c\right )} x\right )} \sqrt {c x + b}}{x^{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)^(5/2)/x^(9/2),x, algorithm="maxima")

[Out]

2/15*(5*(2*B*b*c + A*c^2)*x^2 + (3*B*c^2*x^2 + B*b*c*x - 2*B*b^2)*x + 5*(2*B*b^2 + A*b*c)*x)*sqrt(c*x + b)/x +
integrate((A*b^2 + (B*b^2 + 2*A*b*c)*x)*sqrt(c*x + b)/x^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 )}^{5/2}\,\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)^(5/2)*(A + B*x))/x^(9/2),x)

[Out]

int(((b*x + c*x^2)^(5/2)*(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 {5}{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)**(5/2)/x**(9/2),x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**(9/2), x)

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