3.1.69 \(\int x (A+B x) \sqrt {b x+c x^2} \, dx\) [69]

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

[Out]

-1/24*(-6*B*c*x-8*A*c+5*B*b)*(c*x^2+b*x)^(3/2)/c^2-1/64*b^3*(-8*A*c+5*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2)
)/c^(7/2)+1/64*b*(-8*A*c+5*B*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^3

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Rubi [A]
time = 0.03, 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 = {793, 626, 634, 212} \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 212

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

Rule 626

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 634

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 793

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 ) \text {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*}

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Mathematica [A]
time = 0.21, size = 129, normalized size = 1.14 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (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)\right )+\frac {3 b^3 (5 b B-8 A c) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\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^3*(5*b*B - 8*A*c)*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]])/(Sqrt[x]*Sqrt[b + c*x])))/(192*c^(7/2))

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Maple [A]
time = 0.58, size = 186, normalized size = 1.65

method	result	size
risch	\(-\frac {\left (-48 B \,c^{3} x^{3}-64 A \,c^{3} x^{2}-8 B b \,c^{2} x^{2}-16 A b \,c^{2} x +10 B \,b^{2} c x +24 A \,b^{2} c -15 B \,b^{3}\right ) x \left (c x +b \right )}{192 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) A}{16 c^{\frac {5}{2}}}-\frac {5 b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) B}{128 c^{\frac {7}{2}}}\)	\(146\)
default	\(B \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )+A \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (97) = 194\).
time = 0.30, 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

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Fricas [A]
time = 2.83, 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]

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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)

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Giac [A]
time = 0.86, 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)

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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))

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