\(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{11/2}} \, dx\) [221]

Optimal result

Integrand size = 24, antiderivative size = 172 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{11/2}} \, dx=\frac {5 c (4 b B+3 A c) \sqrt {b x+c x^2}}{4 \sqrt {x}}+\frac {5 c (4 b B+3 A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{3/2}}-\frac {(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}-\frac {5}{4} \sqrt {b} c (4 b B+3 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right ) \]

[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)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {806, 676, 678, 674, 213} \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{11/2}} \, dx=-\frac {5}{4} \sqrt {b} c (3 A c+4 b B) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )+\frac {5 c \sqrt {b x+c x^2} (3 A c+4 b B)}{4 \sqrt {x}}-\frac {\left (b x+c x^2\right )^{5/2} (3 A c+4 b B)}{4 b x^{7/2}}+\frac {5 c \left (b x+c x^2\right )^{3/2} (3 A c+4 b B)}{12 b x^{3/2}}-\frac {A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}} \]

[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 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 674

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 676

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 678

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 806

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*} \text {integral}& = -\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)) \text {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*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{11/2}} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {b+c x} \left (-3 A \left (2 b^2+9 b c x-8 c^2 x^2\right )+4 B x \left (-3 b^2+14 b c x+2 c^2 x^2\right )\right )-15 \sqrt {b} c (4 b B+3 A c) x^2 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{12 x^{5/2} \sqrt {b+c x}} \]

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[b + c*x]*(-3*A*(2*b^2 + 9*b*c*x - 8*c^2*x^2) + 4*B*x*(-3*b^2 + 14*b*c*x + 2*c^2*x^2))
- 15*Sqrt[b]*c*(4*b*B + 3*A*c)*x^2*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(12*x^(5/2)*Sqrt[b + c*x])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.69

method	result	size
risch	\(-\frac {b \left (c x +b \right ) \left (9 A c x +4 B b x +2 A b \right )}{4 x^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}+\frac {c \left (\frac {16 B \left (c x +b \right )^{\frac {3}{2}}}{3}+16 A c \sqrt {c x +b}+32 B b \sqrt {c x +b}-10 \sqrt {b}\, \left (3 A c +4 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )\right ) \sqrt {c x +b}\, \sqrt {x}}{8 \sqrt {x \left (c x +b \right )}}\)	\(118\)
default	\(-\frac {\sqrt {x \left (c x +b \right )}\, \left (-8 B \,c^{2} x^{3} \sqrt {b}\, \sqrt {c x +b}+45 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{2} x^{2}-24 A \,c^{2} x^{2} \sqrt {c x +b}\, \sqrt {b}+60 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b^{2} c \,x^{2}-56 B \,b^{\frac {3}{2}} c \,x^{2} \sqrt {c x +b}+27 A \,b^{\frac {3}{2}} c x \sqrt {c x +b}+12 B \,b^{\frac {5}{2}} x \sqrt {c x +b}+6 A \,b^{\frac {5}{2}} \sqrt {c x +b}\right )}{12 x^{\frac {5}{2}} \sqrt {c x +b}\, \sqrt {b}}\)	\(167\)

[In]

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

[Out]

-1/4*b*(c*x+b)*(9*A*c*x+4*B*b*x+2*A*b)/x^(3/2)/(x*(c*x+b))^(1/2)+1/8*c*(16/3*B*(c*x+b)^(3/2)+16*A*c*(c*x+b)^(1
/2)+32*B*b*(c*x+b)^(1/2)-10*b^(1/2)*(3*A*c+4*B*b)*arctanh((c*x+b)^(1/2)/b^(1/2)))*(c*x+b)^(1/2)*x^(1/2)/(x*(c*
x+b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{11/2}} \, dx=\left [\frac {15 \, {\left (4 \, B b c + 3 \, A c^{2}\right )} \sqrt {b} x^{3} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (8 \, B c^{2} x^{3} - 6 \, A b^{2} + 8 \, {\left (7 \, B b c + 3 \, A c^{2}\right )} x^{2} - 3 \, {\left (4 \, B b^{2} + 9 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, x^{3}}, \frac {15 \, {\left (4 \, B b c + 3 \, A c^{2}\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (8 \, B c^{2} x^{3} - 6 \, A b^{2} + 8 \, {\left (7 \, B b c + 3 \, A c^{2}\right )} x^{2} - 3 \, {\left (4 \, B b^{2} + 9 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{12 \, x^{3}}\right ] \]

[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]

Sympy [F]

\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{11/2}} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{x^{\frac {11}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{11/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{x^{\frac {11}{2}}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{11/2}} \, dx=\frac {8 \, {\left (c x + b\right )}^{\frac {3}{2}} B c^{2} + 48 \, \sqrt {c x + b} B b c^{2} + 24 \, \sqrt {c x + b} A c^{3} + \frac {15 \, {\left (4 \, B b^{2} c^{2} + 3 \, A b c^{3}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {3 \, {\left (4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} c^{2} - 4 \, \sqrt {c x + b} B b^{3} c^{2} + 9 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c^{3} - 7 \, \sqrt {c x + b} A b^{2} c^{3}\right )}}{c^{2} x^{2}}}{12 \, c} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{11/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^{11/2}} \,d x \]

[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)