3.1.67 \(\int x^3 (A+B x) \sqrt {b x+c x^2} \, dx\)

Optimal. Leaf size=200 \[ -\frac {7 b^5 (3 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{11/2}}+\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2} (3 b B-4 A c)}{512 c^5}-\frac {7 b^2 \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{192 c^4}+\frac {7 b x \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{160 c^3}-\frac {x^2 \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{20 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c} \]

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Rubi [A]  time = 0.19, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {794, 670, 640, 612, 620, 206} \begin {gather*} \frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2} (3 b B-4 A c)}{512 c^5}-\frac {7 b^2 \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{192 c^4}-\frac {7 b^5 (3 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{11/2}}+\frac {7 b x \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{160 c^3}-\frac {x^2 \left (b x+c x^2\right )^{3/2} (3 b B-4 A c)}{20 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*b^3*(3*b*B - 4*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^5) - (7*b^2*(3*b*B - 4*A*c)*(b*x + c*x^2)^(3/2))/
(192*c^4) + (7*b*(3*b*B - 4*A*c)*x*(b*x + c*x^2)^(3/2))/(160*c^3) - ((3*b*B - 4*A*c)*x^2*(b*x + c*x^2)^(3/2))/
(20*c^2) + (B*x^3*(b*x + c*x^2)^(3/2))/(6*c) - (7*b^5*(3*b*B - 4*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/
(512*c^(11/2))

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 640

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

Rule 670

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

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

\begin {align*} \int x^3 (A+B x) \sqrt {b x+c x^2} \, dx &=\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}+\frac {\left (3 (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right ) \int x^3 \sqrt {b x+c x^2} \, dx}{6 c}\\ &=-\frac {(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}+\frac {(7 b (3 b B-4 A c)) \int x^2 \sqrt {b x+c x^2} \, dx}{40 c^2}\\ &=\frac {7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac {(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (7 b^2 (3 b B-4 A c)\right ) \int x \sqrt {b x+c x^2} \, dx}{64 c^3}\\ &=-\frac {7 b^2 (3 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{192 c^4}+\frac {7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac {(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}+\frac {\left (7 b^3 (3 b B-4 A c)\right ) \int \sqrt {b x+c x^2} \, dx}{128 c^4}\\ &=\frac {7 b^3 (3 b B-4 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^5}-\frac {7 b^2 (3 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{192 c^4}+\frac {7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac {(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (7 b^5 (3 b B-4 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{1024 c^5}\\ &=\frac {7 b^3 (3 b B-4 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^5}-\frac {7 b^2 (3 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{192 c^4}+\frac {7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac {(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {\left (7 b^5 (3 b B-4 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{512 c^5}\\ &=\frac {7 b^3 (3 b B-4 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^5}-\frac {7 b^2 (3 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{192 c^4}+\frac {7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac {(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {7 b^5 (3 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 166, normalized size = 0.83 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-210 b^4 c (2 A+B x)+56 b^3 c^2 x (5 A+3 B x)-16 b^2 c^3 x^2 (14 A+9 B x)+64 b c^4 x^3 (3 A+2 B x)+256 c^5 x^4 (6 A+5 B x)+315 b^5 B\right )-\frac {105 b^{9/2} (3 b B-4 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{7680 c^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(315*b^5*B - 210*b^4*c*(2*A + B*x) + 64*b*c^4*x^3*(3*A + 2*B*x) + 56*b^3*c^2*x*(5*
A + 3*B*x) + 256*c^5*x^4*(6*A + 5*B*x) - 16*b^2*c^3*x^2*(14*A + 9*B*x)) - (105*b^(9/2)*(3*b*B - 4*A*c)*ArcSinh
[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(7680*c^(11/2))

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IntegrateAlgebraic [A]  time = 0.60, size = 177, normalized size = 0.88 \begin {gather*} \frac {7 \left (3 b^6 B-4 A b^5 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{1024 c^{11/2}}+\frac {\sqrt {b x+c x^2} \left (-420 A b^4 c+280 A b^3 c^2 x-224 A b^2 c^3 x^2+192 A b c^4 x^3+1536 A c^5 x^4+315 b^5 B-210 b^4 B c x+168 b^3 B c^2 x^2-144 b^2 B c^3 x^3+128 b B c^4 x^4+1280 B c^5 x^5\right )}{7680 c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(315*b^5*B - 420*A*b^4*c - 210*b^4*B*c*x + 280*A*b^3*c^2*x + 168*b^3*B*c^2*x^2 - 224*A*b^2*
c^3*x^2 - 144*b^2*B*c^3*x^3 + 192*A*b*c^4*x^3 + 128*b*B*c^4*x^4 + 1536*A*c^5*x^4 + 1280*B*c^5*x^5))/(7680*c^5)
+ (7*(3*b^6*B - 4*A*b^5*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(1024*c^(11/2))

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fricas [A]  time = 0.43, size = 349, normalized size = 1.74 \begin {gather*} \left [-\frac {105 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{5} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} + 128 \, {\left (B b c^{5} + 12 \, A c^{6}\right )} x^{4} - 48 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{3} + 56 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{2} - 70 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{15360 \, c^{6}}, \frac {105 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (1280 \, B c^{6} x^{5} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} + 128 \, {\left (B b c^{5} + 12 \, A c^{6}\right )} x^{4} - 48 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{3} + 56 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{2} - 70 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{7680 \, c^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[-1/15360*(105*(3*B*b^6 - 4*A*b^5*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(1280*B*c^6*x^5
+ 315*B*b^5*c - 420*A*b^4*c^2 + 128*(B*b*c^5 + 12*A*c^6)*x^4 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x^3 + 56*(3*B*b^3*
c^3 - 4*A*b^2*c^4)*x^2 - 70*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x))/c^6, 1/7680*(105*(3*B*b^6 - 4*A*
b^5*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (1280*B*c^6*x^5 + 315*B*b^5*c - 420*A*b^4*c^2 + 128
*(B*b*c^5 + 12*A*c^6)*x^4 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x^3 + 56*(3*B*b^3*c^3 - 4*A*b^2*c^4)*x^2 - 70*(3*B*b^
4*c^2 - 4*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x))/c^6]

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giac [A]  time = 0.22, size = 188, normalized size = 0.94 \begin {gather*} \frac {1}{7680} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B x + \frac {B b c^{4} + 12 \, A c^{5}}{c^{5}}\right )} x - \frac {3 \, {\left (3 \, B b^{2} c^{3} - 4 \, A b c^{4}\right )}}{c^{5}}\right )} x + \frac {7 \, {\left (3 \, B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )}}{c^{5}}\right )} x - \frac {35 \, {\left (3 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )}}{c^{5}}\right )} x + \frac {105 \, {\left (3 \, B b^{5} - 4 \, A b^{4} c\right )}}{c^{5}}\right )} + \frac {7 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{1024 \, c^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/7680*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*B*x + (B*b*c^4 + 12*A*c^5)/c^5)*x - 3*(3*B*b^2*c^3 - 4*A*b*c^4)/c^5)*
x + 7*(3*B*b^3*c^2 - 4*A*b^2*c^3)/c^5)*x - 35*(3*B*b^4*c - 4*A*b^3*c^2)/c^5)*x + 105*(3*B*b^5 - 4*A*b^4*c)/c^5
) + 7/1024*(3*B*b^6 - 4*A*b^5*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(11/2)

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maple [A]  time = 0.05, size = 291, normalized size = 1.46 \begin {gather*} \frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,x^{3}}{6 c}+\frac {7 A \,b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {9}{2}}}-\frac {21 B \,b^{6} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{1024 c^{\frac {11}{2}}}-\frac {7 \sqrt {c \,x^{2}+b x}\, A \,b^{3} x}{64 c^{3}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,x^{2}}{5 c}+\frac {21 \sqrt {c \,x^{2}+b x}\, B \,b^{4} x}{256 c^{4}}-\frac {3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B b \,x^{2}}{20 c^{2}}-\frac {7 \sqrt {c \,x^{2}+b x}\, A \,b^{4}}{128 c^{4}}-\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A b x}{40 c^{2}}+\frac {21 \sqrt {c \,x^{2}+b x}\, B \,b^{5}}{512 c^{5}}+\frac {21 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{2} x}{160 c^{3}}+\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,b^{2}}{48 c^{3}}-\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{3}}{64 c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)*(c*x^2+b*x)^(1/2),x)

[Out]

1/6*B*x^3*(c*x^2+b*x)^(3/2)/c-3/20*B*b/c^2*x^2*(c*x^2+b*x)^(3/2)+21/160*B*b^2/c^3*x*(c*x^2+b*x)^(3/2)-7/64*B*b
^3/c^4*(c*x^2+b*x)^(3/2)+21/256*B*b^4/c^4*(c*x^2+b*x)^(1/2)*x+21/512*B*b^5/c^5*(c*x^2+b*x)^(1/2)-21/1024*B*b^6
/c^(11/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/5*A*x^2*(c*x^2+b*x)^(3/2)/c-7/40*A*b/c^2*x*(c*x^2+b*x)^(
3/2)+7/48*A*b^2/c^3*(c*x^2+b*x)^(3/2)-7/64*A*b^3/c^3*(c*x^2+b*x)^(1/2)*x-7/128*A*b^4/c^4*(c*x^2+b*x)^(1/2)+7/2
56*A*b^5/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 0.90, size = 288, normalized size = 1.44 \begin {gather*} \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B x^{3}}{6 \, c} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b x^{2}}{20 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A x^{2}}{5 \, c} + \frac {21 \, \sqrt {c x^{2} + b x} B b^{4} x}{256 \, c^{4}} + \frac {21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2} x}{160 \, c^{3}} - \frac {7 \, \sqrt {c x^{2} + b x} A b^{3} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b x}{40 \, c^{2}} - \frac {21 \, B b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {11}{2}}} + \frac {7 \, A b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} + \frac {21 \, \sqrt {c x^{2} + b x} B b^{5}}{512 \, c^{5}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{3}}{64 \, c^{4}} - \frac {7 \, \sqrt {c x^{2} + b x} A b^{4}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{2}}{48 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

1/6*(c*x^2 + b*x)^(3/2)*B*x^3/c - 3/20*(c*x^2 + b*x)^(3/2)*B*b*x^2/c^2 + 1/5*(c*x^2 + b*x)^(3/2)*A*x^2/c + 21/
256*sqrt(c*x^2 + b*x)*B*b^4*x/c^4 + 21/160*(c*x^2 + b*x)^(3/2)*B*b^2*x/c^3 - 7/64*sqrt(c*x^2 + b*x)*A*b^3*x/c^
3 - 7/40*(c*x^2 + b*x)^(3/2)*A*b*x/c^2 - 21/1024*B*b^6*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(11/2) +
7/256*A*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) + 21/512*sqrt(c*x^2 + b*x)*B*b^5/c^5 - 7/64*
(c*x^2 + b*x)^(3/2)*B*b^3/c^4 - 7/128*sqrt(c*x^2 + b*x)*A*b^4/c^4 + 7/48*(c*x^2 + b*x)^(3/2)*A*b^2/c^3

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mupad [B]  time = 1.82, size = 267, normalized size = 1.34 \begin {gather*} \frac {3\,B\,b\,\left (\frac {7\,b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,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}\right )}{10\,c}-\frac {x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}\right )}{4\,c}-\frac {7\,A\,b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,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}\right )}{10\,c}+\frac {A\,x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}+\frac {B\,x^3\,{\left (c\,x^2+b\,x\right )}^{3/2}}{6\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x + c*x^2)^(1/2)*(A + B*x),x)

[Out]

(3*B*b*((7*b*((x*(b*x + c*x^2)^(3/2))/(4*c) - (5*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)))/(10*c) - (x^2*(b*x + c*x^2)
^(3/2))/(5*c)))/(4*c) - (7*A*b*((x*(b*x + c*x^2)^(3/2))/(4*c) - (5*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)))/(10*c) +
(A*x^2*(b*x + c*x^2)^(3/2))/(5*c) + (B*x^3*(b*x + c*x^2)^(3/2))/(6*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \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**3*(B*x+A)*(c*x**2+b*x)**(1/2),x)

[Out]

Integral(x**3*sqrt(x*(b + c*x))*(A + B*x), x)

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