3.1.69 \(\int x (A+B x) \sqrt {b x+c x^2} \, dx\)

Optimal. Leaf size=113 \[ -\frac {b^3 (5 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}+\frac {b (b+2 c x) \sqrt {b x+c x^2} (5 b B-8 A c)}{64 c^3}-\frac {\left (b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {779, 612, 620, 206} \begin {gather*} -\frac {b^3 (5 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}+\frac {b (b+2 c x) \sqrt {b x+c x^2} (5 b B-8 A c)}{64 c^3}-\frac {\left (b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*Sqrt[b*x + c*x^2],x]

[Out]

(b*(5*b*B - 8*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) - ((5*b*B - 8*A*c - 6*B*c*x)*(b*x + c*x^2)^(3/2))/(
24*c^2) - (b^3*(5*b*B - 8*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(7/2))

Rule 206

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

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int x (A+B x) \sqrt {b x+c x^2} \, dx &=-\frac {(5 b B-8 A c-6 B c x) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {(b (5 b B-8 A c)) \int \sqrt {b x+c x^2} \, dx}{16 c^2}\\ &=\frac {b (5 b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {(5 b B-8 A c-6 B c x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac {\left (b^3 (5 b B-8 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^3}\\ &=\frac {b (5 b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {(5 b B-8 A c-6 B c x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac {\left (b^3 (5 b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c^3}\\ &=\frac {b (5 b B-8 A c) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {(5 b B-8 A c-6 B c x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac {b^3 (5 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 128, normalized size = 1.13 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-2 b^2 c (12 A+5 B x)+8 b c^2 x (2 A+B x)+16 c^3 x^2 (4 A+3 B x)+15 b^3 B\right )-\frac {3 b^{5/2} (5 b B-8 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{192 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(A + B*x)*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3*B + 8*b*c^2*x*(2*A + B*x) + 16*c^3*x^2*(4*A + 3*B*x) - 2*b^2*c*(12*A + 5*B
*x)) - (3*b^(5/2)*(5*b*B - 8*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(7/
2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.42, size = 129, normalized size = 1.14 \begin {gather*} \frac {\left (5 b^4 B-8 A b^3 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{128 c^{7/2}}+\frac {\sqrt {b x+c x^2} \left (-24 A b^2 c+16 A b c^2 x+64 A c^3 x^2+15 b^3 B-10 b^2 B c x+8 b B c^2 x^2+48 B c^3 x^3\right )}{192 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*(A + B*x)*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[b*x + c*x^2]*(15*b^3*B - 24*A*b^2*c - 10*b^2*B*c*x + 16*A*b*c^2*x + 8*b*B*c^2*x^2 + 64*A*c^3*x^2 + 48*B*
c^3*x^3))/(192*c^3) + ((5*b^4*B - 8*A*b^3*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(128*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 253, normalized size = 2.24 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(48*B*c^4*x^3 + 15*B
*b^3*c - 24*A*b^2*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^2 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/1
92*(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (48*B*c^4*x^3 + 15*B*b^3*c - 2
4*A*b^2*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^2 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/c^4]

________________________________________________________________________________________

giac [A]  time = 0.43, size = 132, normalized size = 1.17 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, B x + \frac {B b c^{2} + 8 \, A c^{3}}{c^{3}}\right )} x - \frac {5 \, B b^{2} c - 8 \, A b c^{2}}{c^{3}}\right )} x + \frac {3 \, {\left (5 \, B b^{3} - 8 \, A b^{2} c\right )}}{c^{3}}\right )} + \frac {{\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*x + (B*b*c^2 + 8*A*c^3)/c^3)*x - (5*B*b^2*c - 8*A*b*c^2)/c^3)*x + 3*(5*B*b^
3 - 8*A*b^2*c)/c^3) + 1/128*(5*B*b^4 - 8*A*b^3*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(
7/2)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 201, normalized size = 1.78 \begin {gather*} \frac {A \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {5}{2}}}-\frac {5 B \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {7}{2}}}-\frac {\sqrt {c \,x^{2}+b x}\, A b x}{4 c}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{2} x}{32 c^{2}}-\frac {\sqrt {c \,x^{2}+b x}\, A \,b^{2}}{8 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{3}}{64 c^{3}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B x}{4 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A}{3 c}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B b}{24 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x+A)*(c*x^2+b*x)^(1/2),x)

[Out]

1/4*B*x*(c*x^2+b*x)^(3/2)/c-5/24*B*b/c^2*(c*x^2+b*x)^(3/2)+5/32*B*b^2/c^2*x*(c*x^2+b*x)^(1/2)+5/64*B*b^3/c^3*(
c*x^2+b*x)^(1/2)-5/128*B*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/3*A*(c*x^2+b*x)^(3/2)/c-1/4*A
*b/c*x*(c*x^2+b*x)^(1/2)-1/8*A*b^2/c^2*(c*x^2+b*x)^(1/2)+1/16*A*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)
^(1/2))

________________________________________________________________________________________

maxima [B]  time = 0.90, size = 198, normalized size = 1.75 \begin {gather*} \frac {5 \, \sqrt {c x^{2} + b x} B b^{2} x}{32 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B x}{4 \, c} - \frac {\sqrt {c x^{2} + b x} A b x}{4 \, c} - \frac {5 \, B b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {A b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} B b^{3}}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{24 \, c^{2}} - \frac {\sqrt {c x^{2} + b x} A b^{2}}{8 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{3 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

5/32*sqrt(c*x^2 + b*x)*B*b^2*x/c^2 + 1/4*(c*x^2 + b*x)^(3/2)*B*x/c - 1/4*sqrt(c*x^2 + b*x)*A*b*x/c - 5/128*B*b
^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 1/16*A*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(
c))/c^(5/2) + 5/64*sqrt(c*x^2 + b*x)*B*b^3/c^3 - 5/24*(c*x^2 + b*x)^(3/2)*B*b/c^2 - 1/8*sqrt(c*x^2 + b*x)*A*b^
2/c^2 + 1/3*(c*x^2 + b*x)^(3/2)*A/c

________________________________________________________________________________________

mupad [B]  time = 1.52, size = 165, normalized size = 1.46 \begin {gather*} \frac {A\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}+\frac {B\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,B\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}+\frac {A\,b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x + c*x^2)^(1/2)*(A + B*x),x)

[Out]

(A*(b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2) + (B*x*(b*x + c*x^2)^(3/2))/(4*c) - (5*B*b*((b^
3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2)) + ((b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2
*b*c*x))/(24*c^2)))/(8*c) + (A*b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {x \left (b + c x\right )} \left (A + B x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x**2+b*x)**(1/2),x)

[Out]

Integral(x*sqrt(x*(b + c*x))*(A + B*x), x)

________________________________________________________________________________________