3.221 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{11/2}} \, dx\)

Optimal. Leaf size=172 \[ \frac {5 c \sqrt {b x+c x^2} (3 A c+4 b B)}{4 \sqrt {x}}-\frac {5}{4} \sqrt {b} c (3 A c+4 b B) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )-\frac {\left (b x+c x^2\right )^{5/2} (3 A c+4 b B)}{4 b x^{7/2}}+\frac {5 c \left (b x+c x^2\right )^{3/2} (3 A c+4 b B)}{12 b x^{3/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}} \]

[Out]

5/12*c*(3*A*c+4*B*b)*(c*x^2+b*x)^(3/2)/b/x^(3/2)-1/4*(3*A*c+4*B*b)*(c*x^2+b*x)^(5/2)/b/x^(7/2)-1/2*A*(c*x^2+b*
x)^(7/2)/b/x^(11/2)-5/4*c*(3*A*c+4*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))*b^(1/2)+5/4*c*(3*A*c+4*B*b)
*(c*x^2+b*x)^(1/2)/x^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {\left (b x+c x^2\right )^{5/2} (3 A c+4 b B)}{4 b x^{7/2}}+\frac {5 c \left (b x+c x^2\right )^{3/2} (3 A c+4 b B)}{12 b x^{3/2}}+\frac {5 c \sqrt {b x+c x^2} (3 A c+4 b B)}{4 \sqrt {x}}-\frac {5}{4} \sqrt {b} c (3 A c+4 b B) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(11/2),x]

[Out]

(5*c*(4*b*B + 3*A*c)*Sqrt[b*x + c*x^2])/(4*Sqrt[x]) + (5*c*(4*b*B + 3*A*c)*(b*x + c*x^2)^(3/2))/(12*b*x^(3/2))
- ((4*b*B + 3*A*c)*(b*x + c*x^2)^(5/2))/(4*b*x^(7/2)) - (A*(b*x + c*x^2)^(7/2))/(2*b*x^(11/2)) - (5*Sqrt[b]*c
*(4*b*B + 3*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/4

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 )^{5/2}}{x^{11/2}} \, dx &=-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac {\left (-\frac {11}{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^{9/2}} \, dx}{2 b}\\ &=-\frac {(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac {(5 c (4 b B+3 A c)) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{5/2}} \, dx}{8 b}\\ &=\frac {5 c (4 b B+3 A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{3/2}}-\frac {(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac {1}{8} (5 c (4 b B+3 A c)) \int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx\\ &=\frac {5 c (4 b B+3 A c) \sqrt {b x+c x^2}}{4 \sqrt {x}}+\frac {5 c (4 b B+3 A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{3/2}}-\frac {(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac {1}{8} (5 b c (4 b B+3 A c)) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx\\ &=\frac {5 c (4 b B+3 A c) \sqrt {b x+c x^2}}{4 \sqrt {x}}+\frac {5 c (4 b B+3 A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{3/2}}-\frac {(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac {1}{4} (5 b 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 )\\ &=\frac {5 c (4 b B+3 A c) \sqrt {b x+c x^2}}{4 \sqrt {x}}+\frac {5 c (4 b B+3 A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{3/2}}-\frac {(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}-\frac {5}{4} \sqrt {b} c (4 b B+3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 67, normalized size = 0.39 \[ \frac {(b+c x)^3 \sqrt {x (b+c x)} \left (c x^2 (3 A c+4 b B) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {c x}{b}+1\right )-7 A b^2\right )}{14 b^3 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(11/2),x]

[Out]

((b + c*x)^3*Sqrt[x*(b + c*x)]*(-7*A*b^2 + c*(4*b*B + 3*A*c)*x^2*Hypergeometric2F1[2, 7/2, 9/2, 1 + (c*x)/b]))
/(14*b^3*x^(5/2))

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fricas [A]  time = 0.73, size = 238, normalized size = 1.38 \[ \left [\frac {15 \, {\left (4 \, B b c + 3 \, A c^{2}\right )} \sqrt {b} x^{3} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (8 \, B c^{2} x^{3} - 6 \, A b^{2} + 8 \, {\left (7 \, B b c + 3 \, A c^{2}\right )} x^{2} - 3 \, {\left (4 \, B b^{2} + 9 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, x^{3}}, \frac {15 \, {\left (4 \, B b c + 3 \, A c^{2}\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (8 \, B c^{2} x^{3} - 6 \, A b^{2} + 8 \, {\left (7 \, B b c + 3 \, A c^{2}\right )} x^{2} - 3 \, {\left (4 \, B b^{2} + 9 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{12 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(11/2),x, algorithm="fricas")

[Out]

[1/24*(15*(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*
(8*B*c^2*x^3 - 6*A*b^2 + 8*(7*B*b*c + 3*A*c^2)*x^2 - 3*(4*B*b^2 + 9*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/x^3,
1/12*(15*(4*B*b*c + 3*A*c^2)*sqrt(-b)*x^3*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (8*B*c^2*x^3 - 6*A*b^2
+ 8*(7*B*b*c + 3*A*c^2)*x^2 - 3*(4*B*b^2 + 9*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/x^3]

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giac [A]  time = 0.32, size = 155, normalized size = 0.90 \[ \frac {8 \, {\left (c x + b\right )}^{\frac {3}{2}} B c^{2} + 48 \, \sqrt {c x + b} B b c^{2} + 24 \, \sqrt {c x + b} A c^{3} + \frac {15 \, {\left (4 \, B b^{2} c^{2} + 3 \, A b c^{3}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {3 \, {\left (4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} c^{2} - 4 \, \sqrt {c x + b} B b^{3} c^{2} + 9 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c^{3} - 7 \, \sqrt {c x + b} A b^{2} c^{3}\right )}}{c^{2} x^{2}}}{12 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(11/2),x, algorithm="giac")

[Out]

1/12*(8*(c*x + b)^(3/2)*B*c^2 + 48*sqrt(c*x + b)*B*b*c^2 + 24*sqrt(c*x + b)*A*c^3 + 15*(4*B*b^2*c^2 + 3*A*b*c^
3)*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) - 3*(4*(c*x + b)^(3/2)*B*b^2*c^2 - 4*sqrt(c*x + b)*B*b^3*c^2 + 9*(c
*x + b)^(3/2)*A*b*c^3 - 7*sqrt(c*x + b)*A*b^2*c^3)/(c^2*x^2))/c

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maple [A]  time = 0.06, size = 167, normalized size = 0.97 \[ -\frac {\sqrt {\left (c x +b \right ) x}\, \left (45 A b \,c^{2} x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )+60 B \,b^{2} c \,x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-8 \sqrt {c x +b}\, B \sqrt {b}\, c^{2} x^{3}-24 \sqrt {c x +b}\, A \sqrt {b}\, c^{2} x^{2}-56 \sqrt {c x +b}\, B \,b^{\frac {3}{2}} c \,x^{2}+27 \sqrt {c x +b}\, A \,b^{\frac {3}{2}} c x +12 \sqrt {c x +b}\, B \,b^{\frac {5}{2}} x +6 \sqrt {c x +b}\, A \,b^{\frac {5}{2}}\right )}{12 \sqrt {c x +b}\, \sqrt {b}\, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(11/2),x)

[Out]

-1/12*((c*x+b)*x)^(1/2)*(-8*(c*x+b)^(1/2)*B*b^(1/2)*c^2*x^3+45*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b*c^2-24*(
c*x+b)^(1/2)*A*b^(1/2)*c^2*x^2+60*B*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b^2*c-56*(c*x+b)^(1/2)*B*b^(3/2)*c*x^2+
27*(c*x+b)^(1/2)*A*b^(3/2)*c*x+12*(c*x+b)^(1/2)*B*b^(5/2)*x+6*(c*x+b)^(1/2)*A*b^(5/2))/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 \[ \frac {2}{3} \, {\left (B c^{2} x + B b c\right )} \sqrt {c x + b} + \int \frac {{\left (A b^{2} + {\left (2 \, B b c + A c^{2}\right )} x^{2} + {\left (B b^{2} + 2 \, A b c\right )} x\right )} \sqrt {c x + b}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(11/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^{11/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(11/2),x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(11/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{x^{\frac {11}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**(11/2),x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**(11/2), x)

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