3.3.11 \(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^{11/2}} \, dx\)

Optimal. Leaf size=142 \[ -\frac {c^2 (6 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2} (6 b B-A c)}{12 b x^{7/2}}-\frac {c \sqrt {b x+c x^2} (6 b B-A c)}{8 b x^{3/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{3 b x^{11/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {792, 662, 660, 207} \begin {gather*} -\frac {c^2 (6 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2} (6 b B-A c)}{12 b x^{7/2}}-\frac {c \sqrt {b x+c x^2} (6 b B-A c)}{8 b x^{3/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{3 b x^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(11/2),x]

[Out]

-(c*(6*b*B - A*c)*Sqrt[b*x + c*x^2])/(8*b*x^(3/2)) - ((6*b*B - A*c)*(b*x + c*x^2)^(3/2))/(12*b*x^(7/2)) - (A*(
b*x + c*x^2)^(5/2))/(3*b*x^(11/2)) - (c^2*(6*b*B - A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*b^(3/
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 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^{11/2}} \, dx &=-\frac {A \left (b x+c x^2\right )^{5/2}}{3 b x^{11/2}}+\frac {\left (-\frac {11}{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^{9/2}} \, dx}{3 b}\\ &=-\frac {(6 b B-A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{3 b x^{11/2}}+\frac {(c (6 b B-A c)) \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx}{8 b}\\ &=-\frac {c (6 b B-A c) \sqrt {b x+c x^2}}{8 b x^{3/2}}-\frac {(6 b B-A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{3 b x^{11/2}}+\frac {\left (c^2 (6 b B-A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{16 b}\\ &=-\frac {c (6 b B-A c) \sqrt {b x+c x^2}}{8 b x^{3/2}}-\frac {(6 b B-A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{3 b x^{11/2}}+\frac {\left (c^2 (6 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}\\ &=-\frac {c (6 b B-A c) \sqrt {b x+c x^2}}{8 b x^{3/2}}-\frac {(6 b B-A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{7/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{3 b x^{11/2}}-\frac {c^2 (6 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 107, normalized size = 0.75 \begin {gather*} -\frac {(b+c x) \left (A \left (8 b^2+14 b c x+3 c^2 x^2\right )+6 b B x (2 b+5 c x)\right )+3 c^2 x^3 \sqrt {\frac {c x}{b}+1} (6 b B-A c) \tanh ^{-1}\left (\sqrt {\frac {c x}{b}+1}\right )}{24 b x^{5/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(11/2),x]

[Out]

-1/24*((b + c*x)*(6*b*B*x*(2*b + 5*c*x) + A*(8*b^2 + 14*b*c*x + 3*c^2*x^2)) + 3*c^2*(6*b*B - A*c)*x^3*Sqrt[1 +
(c*x)/b]*ArcTanh[Sqrt[1 + (c*x)/b]])/(b*x^(5/2)*Sqrt[x*(b + c*x)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.95, size = 110, normalized size = 0.77 \begin {gather*} \frac {\left (A c^3-6 b B c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{8 b^{3/2}}+\frac {\sqrt {b x+c x^2} \left (-8 A b^2-14 A b c x-3 A c^2 x^2-12 b^2 B x-30 b B c x^2\right )}{24 b x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(11/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(-8*A*b^2 - 12*b^2*B*x - 14*A*b*c*x - 30*b*B*c*x^2 - 3*A*c^2*x^2))/(24*b*x^(7/2)) + ((-6*b*
B*c^2 + A*c^3)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(8*b^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 239, normalized size = 1.68 \begin {gather*} \left [-\frac {3 \, {\left (6 \, 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 (10 \, B b^{2} c + A b c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{3} + 7 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, b^{2} x^{4}}, \frac {3 \, {\left (6 \, 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 (10 \, B b^{2} c + A b c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{3} + 7 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, b^{2} 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)^(3/2)/x^(11/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(6*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*(10*B*b^2*c + A*b*c^2)*x^2 + 2*(6*B*b^3 + 7*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^4), 1/2
4*(3*(6*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*(10*B*b^2*c +
A*b*c^2)*x^2 + 2*(6*B*b^3 + 7*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^4)]

________________________________________________________________________________________

giac [A]  time = 0.32, size = 145, normalized size = 1.02 \begin {gather*} \frac {\frac {3 \, {\left (6 \, B b c^{3} - A c^{4}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {30 \, {\left (c x + b\right )}^{\frac {5}{2}} B b c^{3} - 48 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} c^{3} + 18 \, \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 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)^(3/2)/x^(11/2),x, algorithm="giac")

[Out]

1/24*(3*(6*B*b*c^3 - A*c^4)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b) - (30*(c*x + b)^(5/2)*B*b*c^3 - 48*(c*
x + b)^(3/2)*B*b^2*c^3 + 18*sqrt(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*c^3*x^3))/c

________________________________________________________________________________________

maple [A]  time = 0.07, 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 )-18 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}-30 \sqrt {c x +b}\, B \,b^{\frac {3}{2}} c \,x^{2}-14 \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 {3}{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)^(3/2)/x^(11/2),x)

[Out]

1/24*((c*x+b)*x)^(1/2)/b^(3/2)*(3*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*c^3-18*B*arctanh((c*x+b)^(1/2)/b^(1/2))
*x^3*b*c^2-3*(c*x+b)^(1/2)*A*b^(1/2)*c^2*x^2-30*(c*x+b)^(1/2)*B*b^(3/2)*c*x^2-14*(c*x+b)^(1/2)*A*b^(3/2)*c*x-1
2*(c*x+b)^(1/2)*B*b^(5/2)*x-8*(c*x+b)^(1/2)*A*b^(5/2))/x^(7/2)/(c*x+b)^(1/2)

________________________________________________________________________________________

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 {11}{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^(11/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)*(B*x + A)/x^(11/2), x)

________________________________________________________________________________________

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^{11/2}} \,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^(11/2),x)

[Out]

int(((b*x + c*x^2)^(3/2)*(A + B*x))/x^(11/2), x)

________________________________________________________________________________________

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 {11}{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**(11/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**(11/2), x)

________________________________________________________________________________________