3.3.63 \(\int (d+e x)^3 \sqrt {b x+c x^2} \, dx\)

Optimal. Leaf size=210 \[ -\frac {b^2 (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}+\frac {(b+2 c x) \sqrt {b x+c x^2} (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right )}{128 c^4}+\frac {e \left (b x+c x^2\right )^{3/2} \left (35 b^2 e^2+42 c e x (2 c d-b e)-150 b c d e+192 c^2 d^2\right )}{240 c^3}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {742, 779, 612, 620, 206} \begin {gather*} \frac {e \left (b x+c x^2\right )^{3/2} \left (35 b^2 e^2+42 c e x (2 c d-b e)-150 b c d e+192 c^2 d^2\right )}{240 c^3}+\frac {(b+2 c x) \sqrt {b x+c x^2} (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right )}{128 c^4}-\frac {b^2 (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*Sqrt[b*x + c*x^2],x]

[Out]

((2*c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e + 7*b^2*e^2)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^4) + (e*(d + e*x)^2
*(b*x + c*x^2)^(3/2))/(5*c) + (e*(192*c^2*d^2 - 150*b*c*d*e + 35*b^2*e^2 + 42*c*e*(2*c*d - b*e)*x)*(b*x + c*x^
2)^(3/2))/(240*c^3) - (b^2*(2*c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e + 7*b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x +
c*x^2]])/(128*c^(9/2))

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 742

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[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, 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 (d+e x)^3 \sqrt {b x+c x^2} \, dx &=\frac {e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\int (d+e x) \left (\frac {1}{2} d (10 c d-3 b e)+\frac {7}{2} e (2 c d-b e) x\right ) \sqrt {b x+c x^2} \, dx}{5 c}\\ &=\frac {e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {e \left (192 c^2 d^2-150 b c d e+35 b^2 e^2+42 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}+\frac {\left ((2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right )\right ) \int \sqrt {b x+c x^2} \, dx}{32 c^3}\\ &=\frac {(2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {e \left (192 c^2 d^2-150 b c d e+35 b^2 e^2+42 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac {\left (b^2 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right )\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^4}\\ &=\frac {(2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {e \left (192 c^2 d^2-150 b c d e+35 b^2 e^2+42 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac {\left (b^2 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{128 c^4}\\ &=\frac {(2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {e \left (192 c^2 d^2-150 b c d e+35 b^2 e^2+42 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac {b^2 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.49, size = 230, normalized size = 1.10 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {15 b^{3/2} \left (7 b^3 e^3-30 b^2 c d e^2+48 b c^2 d^2 e-32 c^3 d^3\right ) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (-105 b^4 e^3+10 b^3 c e^2 (45 d+7 e x)-4 b^2 c^2 e \left (180 d^2+75 d e x+14 e^2 x^2\right )+48 b c^3 \left (10 d^3+10 d^2 e x+5 d e^2 x^2+e^3 x^3\right )+96 c^4 x \left (10 d^3+20 d^2 e x+15 d e^2 x^2+4 e^3 x^3\right )\right )\right )}{1920 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^4*e^3 + 10*b^3*c*e^2*(45*d + 7*e*x) - 4*b^2*c^2*e*(180*d^2 + 75*d*e*x + 14
*e^2*x^2) + 48*b*c^3*(10*d^3 + 10*d^2*e*x + 5*d*e^2*x^2 + e^3*x^3) + 96*c^4*x*(10*d^3 + 20*d^2*e*x + 15*d*e^2*
x^2 + 4*e^3*x^3)) + (15*b^(3/2)*(-32*c^3*d^3 + 48*b*c^2*d^2*e - 30*b^2*c*d*e^2 + 7*b^3*e^3)*ArcSinh[(Sqrt[c]*S
qrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1920*c^(9/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.11, size = 256, normalized size = 1.22 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-105 b^4 e^3+450 b^3 c d e^2+70 b^3 c e^3 x-720 b^2 c^2 d^2 e-300 b^2 c^2 d e^2 x-56 b^2 c^2 e^3 x^2+480 b c^3 d^3+480 b c^3 d^2 e x+240 b c^3 d e^2 x^2+48 b c^3 e^3 x^3+960 c^4 d^3 x+1920 c^4 d^2 e x^2+1440 c^4 d e^2 x^3+384 c^4 e^3 x^4\right )}{1920 c^4}+\frac {\left (-7 b^5 e^3+30 b^4 c d e^2-48 b^3 c^2 d^2 e+32 b^2 c^3 d^3\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{256 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^3*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[b*x + c*x^2]*(480*b*c^3*d^3 - 720*b^2*c^2*d^2*e + 450*b^3*c*d*e^2 - 105*b^4*e^3 + 960*c^4*d^3*x + 480*b*
c^3*d^2*e*x - 300*b^2*c^2*d*e^2*x + 70*b^3*c*e^3*x + 1920*c^4*d^2*e*x^2 + 240*b*c^3*d*e^2*x^2 - 56*b^2*c^2*e^3
*x^2 + 1440*c^4*d*e^2*x^3 + 48*b*c^3*e^3*x^3 + 384*c^4*e^3*x^4))/(1920*c^4) + ((32*b^2*c^3*d^3 - 48*b^3*c^2*d^
2*e + 30*b^4*c*d*e^2 - 7*b^5*e^3)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(256*c^(9/2))

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 497, normalized size = 2.37 \begin {gather*} \left [-\frac {15 \, {\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 30 \, b^{4} c d e^{2} - 7 \, b^{5} e^{3}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (384 \, c^{5} e^{3} x^{4} + 480 \, b c^{4} d^{3} - 720 \, b^{2} c^{3} d^{2} e + 450 \, b^{3} c^{2} d e^{2} - 105 \, b^{4} c e^{3} + 48 \, {\left (30 \, c^{5} d e^{2} + b c^{4} e^{3}\right )} x^{3} + 8 \, {\left (240 \, c^{5} d^{2} e + 30 \, b c^{4} d e^{2} - 7 \, b^{2} c^{3} e^{3}\right )} x^{2} + 10 \, {\left (96 \, c^{5} d^{3} + 48 \, b c^{4} d^{2} e - 30 \, b^{2} c^{3} d e^{2} + 7 \, b^{3} c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{5}}, \frac {15 \, {\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 30 \, b^{4} c d e^{2} - 7 \, b^{5} e^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (384 \, c^{5} e^{3} x^{4} + 480 \, b c^{4} d^{3} - 720 \, b^{2} c^{3} d^{2} e + 450 \, b^{3} c^{2} d e^{2} - 105 \, b^{4} c e^{3} + 48 \, {\left (30 \, c^{5} d e^{2} + b c^{4} e^{3}\right )} x^{3} + 8 \, {\left (240 \, c^{5} d^{2} e + 30 \, b c^{4} d e^{2} - 7 \, b^{2} c^{3} e^{3}\right )} x^{2} + 10 \, {\left (96 \, c^{5} d^{3} + 48 \, b c^{4} d^{2} e - 30 \, b^{2} c^{3} d e^{2} + 7 \, b^{3} c^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3840*(15*(32*b^2*c^3*d^3 - 48*b^3*c^2*d^2*e + 30*b^4*c*d*e^2 - 7*b^5*e^3)*sqrt(c)*log(2*c*x + b + 2*sqrt(c
*x^2 + b*x)*sqrt(c)) - 2*(384*c^5*e^3*x^4 + 480*b*c^4*d^3 - 720*b^2*c^3*d^2*e + 450*b^3*c^2*d*e^2 - 105*b^4*c*
e^3 + 48*(30*c^5*d*e^2 + b*c^4*e^3)*x^3 + 8*(240*c^5*d^2*e + 30*b*c^4*d*e^2 - 7*b^2*c^3*e^3)*x^2 + 10*(96*c^5*
d^3 + 48*b*c^4*d^2*e - 30*b^2*c^3*d*e^2 + 7*b^3*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/c^5, 1/1920*(15*(32*b^2*c^3*d^3
- 48*b^3*c^2*d^2*e + 30*b^4*c*d*e^2 - 7*b^5*e^3)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (384*c^5
*e^3*x^4 + 480*b*c^4*d^3 - 720*b^2*c^3*d^2*e + 450*b^3*c^2*d*e^2 - 105*b^4*c*e^3 + 48*(30*c^5*d*e^2 + b*c^4*e^
3)*x^3 + 8*(240*c^5*d^2*e + 30*b*c^4*d*e^2 - 7*b^2*c^3*e^3)*x^2 + 10*(96*c^5*d^3 + 48*b*c^4*d^2*e - 30*b^2*c^3
*d*e^2 + 7*b^3*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/c^5]

________________________________________________________________________________________

giac [A]  time = 0.21, size = 250, normalized size = 1.19 \begin {gather*} \frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x e^{3} + \frac {30 \, c^{4} d e^{2} + b c^{3} e^{3}}{c^{4}}\right )} x + \frac {240 \, c^{4} d^{2} e + 30 \, b c^{3} d e^{2} - 7 \, b^{2} c^{2} e^{3}}{c^{4}}\right )} x + \frac {5 \, {\left (96 \, c^{4} d^{3} + 48 \, b c^{3} d^{2} e - 30 \, b^{2} c^{2} d e^{2} + 7 \, b^{3} c e^{3}\right )}}{c^{4}}\right )} x + \frac {15 \, {\left (32 \, b c^{3} d^{3} - 48 \, b^{2} c^{2} d^{2} e + 30 \, b^{3} c d e^{2} - 7 \, b^{4} e^{3}\right )}}{c^{4}}\right )} + \frac {{\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 30 \, b^{4} c d e^{2} - 7 \, b^{5} e^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*x*e^3 + (30*c^4*d*e^2 + b*c^3*e^3)/c^4)*x + (240*c^4*d^2*e + 30*b*c^3*d*e
^2 - 7*b^2*c^2*e^3)/c^4)*x + 5*(96*c^4*d^3 + 48*b*c^3*d^2*e - 30*b^2*c^2*d*e^2 + 7*b^3*c*e^3)/c^4)*x + 15*(32*
b*c^3*d^3 - 48*b^2*c^2*d^2*e + 30*b^3*c*d*e^2 - 7*b^4*e^3)/c^4) + 1/256*(32*b^2*c^3*d^3 - 48*b^3*c^2*d^2*e + 3
0*b^4*c*d*e^2 - 7*b^5*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 444, normalized size = 2.11 \begin {gather*} \frac {7 b^{5} e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {9}{2}}}-\frac {15 b^{4} d \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {7}{2}}}+\frac {3 b^{3} d^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {5}{2}}}-\frac {b^{2} d^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}-\frac {7 \sqrt {c \,x^{2}+b x}\, b^{3} e^{3} x}{64 c^{3}}+\frac {15 \sqrt {c \,x^{2}+b x}\, b^{2} d \,e^{2} x}{32 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x}\, b \,d^{2} e x}{4 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} e^{3} x^{2}}{5 c}+\frac {\sqrt {c \,x^{2}+b x}\, d^{3} x}{2}-\frac {7 \sqrt {c \,x^{2}+b x}\, b^{4} e^{3}}{128 c^{4}}+\frac {15 \sqrt {c \,x^{2}+b x}\, b^{3} d \,e^{2}}{64 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x}\, b^{2} d^{2} e}{8 c^{2}}+\frac {\sqrt {c \,x^{2}+b x}\, b \,d^{3}}{4 c}-\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b \,e^{3} x}{40 c^{2}}+\frac {3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} d \,e^{2} x}{4 c}+\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{2} e^{3}}{48 c^{3}}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b d \,e^{2}}{8 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} d^{2} e}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(c*x^2+b*x)^(1/2),x)

[Out]

1/5*e^3*x^2*(c*x^2+b*x)^(3/2)/c-7/40*e^3*b/c^2*x*(c*x^2+b*x)^(3/2)+7/48*e^3*b^2/c^3*(c*x^2+b*x)^(3/2)-7/64*e^3
*b^3/c^3*x*(c*x^2+b*x)^(1/2)-7/128*e^3*b^4/c^4*(c*x^2+b*x)^(1/2)+7/256*e^3*b^5/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+
(c*x^2+b*x)^(1/2))+3/4*d*e^2*x*(c*x^2+b*x)^(3/2)/c-5/8*d*e^2*b/c^2*(c*x^2+b*x)^(3/2)+15/32*d*e^2*b^2/c^2*x*(c*
x^2+b*x)^(1/2)+15/64*d*e^2*b^3/c^3*(c*x^2+b*x)^(1/2)-15/128*d*e^2*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*
x)^(1/2))+d^2*e*(c*x^2+b*x)^(3/2)/c-3/4*d^2*e*b/c*x*(c*x^2+b*x)^(1/2)-3/8*d^2*e*b^2/c^2*(c*x^2+b*x)^(1/2)+3/16
*d^2*e*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/2*d^3*x*(c*x^2+b*x)^(1/2)+1/4*d^3/c*(c*x^2+b*x)
^(1/2)*b-1/8*d^3*b^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

________________________________________________________________________________________

maxima [B]  time = 1.46, size = 439, normalized size = 2.09 \begin {gather*} \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} e^{3} x^{2}}{5 \, c} + \frac {1}{2} \, \sqrt {c x^{2} + b x} d^{3} x - \frac {3 \, \sqrt {c x^{2} + b x} b d^{2} e x}{4 \, c} + \frac {15 \, \sqrt {c x^{2} + b x} b^{2} d e^{2} x}{32 \, c^{2}} + \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} d e^{2} x}{4 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} b^{3} e^{3} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b e^{3} x}{40 \, c^{2}} - \frac {b^{2} d^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} + \frac {3 \, b^{3} d^{2} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} - \frac {15 \, b^{4} d e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {7 \, b^{5} e^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} + \frac {\sqrt {c x^{2} + b x} b d^{3}}{4 \, c} - \frac {3 \, \sqrt {c x^{2} + b x} b^{2} d^{2} e}{8 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} d^{2} e}{c} + \frac {15 \, \sqrt {c x^{2} + b x} b^{3} d e^{2}}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b d e^{2}}{8 \, c^{2}} - \frac {7 \, \sqrt {c x^{2} + b x} b^{4} e^{3}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} e^{3}}{48 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(c*x^2 + b*x)^(3/2)*e^3*x^2/c + 1/2*sqrt(c*x^2 + b*x)*d^3*x - 3/4*sqrt(c*x^2 + b*x)*b*d^2*e*x/c + 15/32*sq
rt(c*x^2 + b*x)*b^2*d*e^2*x/c^2 + 3/4*(c*x^2 + b*x)^(3/2)*d*e^2*x/c - 7/64*sqrt(c*x^2 + b*x)*b^3*e^3*x/c^3 - 7
/40*(c*x^2 + b*x)^(3/2)*b*e^3*x/c^2 - 1/8*b^2*d^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + 3/16*
b^3*d^2*e*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 15/128*b^4*d*e^2*log(2*c*x + b + 2*sqrt(c*x^2
+ b*x)*sqrt(c))/c^(7/2) + 7/256*b^5*e^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) + 1/4*sqrt(c*x^2
+ b*x)*b*d^3/c - 3/8*sqrt(c*x^2 + b*x)*b^2*d^2*e/c^2 + (c*x^2 + b*x)^(3/2)*d^2*e/c + 15/64*sqrt(c*x^2 + b*x)*
b^3*d*e^2/c^3 - 5/8*(c*x^2 + b*x)^(3/2)*b*d*e^2/c^2 - 7/128*sqrt(c*x^2 + b*x)*b^4*e^3/c^4 + 7/48*(c*x^2 + b*x)
^(3/2)*b^2*e^3/c^3

________________________________________________________________________________________

mupad [B]  time = 0.92, size = 361, normalized size = 1.72 \begin {gather*} d^3\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {7\,b\,e^3\,\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 {e^3\,x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}-\frac {b^2\,d^3\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}+\frac {3\,d\,e^2\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {15\,b\,d\,e^2\,\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 {3\,b^3\,d^2\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {d^2\,e\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{8\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)*(d + e*x)^3,x)

[Out]

d^3*(b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (7*b*e^3*((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) + (e^3*x^2*(b*x + c*x^2)^(3/2))/(5*c) - (b^2*d^3*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/
2)))/(8*c^(3/2)) + (3*d*e^2*x*(b*x + c*x^2)^(3/2))/(4*c) - (15*b*d*e^2*((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) + (3*b^3*
d^2*e*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2)) + (d^2*e*(b*x + c*x^2)^(1/2)*(8*c^2*x^2 -
3*b^2 + 2*b*c*x))/(8*c^2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {x \left (b + c x\right )} \left (d + e x\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(c*x**2+b*x)**(1/2),x)

[Out]

Integral(sqrt(x*(b + c*x))*(d + e*x)**3, x)

________________________________________________________________________________________