3.3.85 \(\int (d+e x)^3 \sqrt {b x+c x^2} \, dx\) [285]

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

[Out]

1/5*e*(e*x+d)^2*(c*x^2+b*x)^(3/2)/c+1/240*e*(192*c^2*d^2-150*b*c*d*e+35*b^2*e^2+42*c*e*(-b*e+2*c*d)*x)*(c*x^2+
b*x)^(3/2)/c^3-1/128*b^2*(-b*e+2*c*d)*(7*b^2*e^2-16*b*c*d*e+16*c^2*d^2)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c
^(9/2)+1/128*(-b*e+2*c*d)*(7*b^2*e^2-16*b*c*d*e+16*c^2*d^2)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^4

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Rubi [A]
time = 0.17, 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 = {756, 793, 626, 634, 212} \begin {gather*} -\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} \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 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 756

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 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 (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 ) \text {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*}

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Mathematica [A]
time = 0.36, size = 231, normalized size = 1.10 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\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 )-\frac {15 b^2 \left (-32 c^3 d^3+48 b c^2 d^2 e-30 b^2 c d e^2+7 b^3 e^3\right ) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\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^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)*Log[-(Sqrt[c]*Sqrt[x])
+ Sqrt[b + c*x]])/(Sqrt[x]*Sqrt[b + c*x])))/(1920*c^(9/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(384\) vs. \(2(190)=380\).
time = 0.46, size = 385, normalized size = 1.83

method	result	size
risch	\(-\frac {\left (-384 c^{4} e^{3} x^{4}-48 b \,c^{3} e^{3} x^{3}-1440 c^{4} d \,e^{2} x^{3}+56 b^{2} c^{2} e^{3} x^{2}-240 b \,c^{3} d \,e^{2} x^{2}-1920 c^{4} d^{2} e \,x^{2}-70 b^{3} c \,e^{3} x +300 b^{2} c^{2} d \,e^{2} x -480 b \,c^{3} d^{2} e x -960 c^{4} d^{3} x +105 b^{4} e^{3}-450 b^{3} c d \,e^{2}+720 b^{2} c^{2} d^{2} e -480 c^{3} b \,d^{3}\right ) x \left (c x +b \right )}{1920 c^{4} \sqrt {x \left (c x +b \right )}}+\frac {7 b^{5} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) e^{3}}{256 c^{\frac {9}{2}}}-\frac {15 b^{4} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) d \,e^{2}}{128 c^{\frac {7}{2}}}+\frac {3 b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) d^{2} e}{16 c^{\frac {5}{2}}}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) d^{3}}{8 c^{\frac {3}{2}}}\)	\(321\)
default	\(e^{3} \left (\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )+3 d \,e^{2} \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )+3 d^{2} e \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )+d^{3} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )\)	\(385\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^3*(1/5*x^2*(c*x^2+b*x)^(3/2)/c-7/10*b/c*(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))))))+3*d*e^2*(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)))))+3*d^2*e*(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))))+d^3*(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. 432 vs. \(2 (197) = 394\).
time = 0.28, size = 432, normalized size = 2.06 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + b x} d^{3} x - \frac {3 \, \sqrt {c x^{2} + b x} b d^{2} x e}{4 \, c} - \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 {\sqrt {c x^{2} + b x} b d^{3}}{4 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{2} e^{3}}{5 \, c} + \frac {15 \, \sqrt {c x^{2} + b x} b^{2} d x e^{2}}{32 \, c^{2}} + \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} d x e^{2}}{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 \, 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 \, \sqrt {c x^{2} + b x} b^{3} x e^{3}}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x e^{3}}{40 \, c^{2}} + \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 \, 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 {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]

1/2*sqrt(c*x^2 + b*x)*d^3*x - 3/4*sqrt(c*x^2 + b*x)*b*d^2*x*e/c - 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) + 1/4*sqrt(c*x^2 +
b*x)*b*d^3/c + 1/5*(c*x^2 + b*x)^(3/2)*x^2*e^3/c + 15/32*sqrt(c*x^2 + b*x)*b^2*d*x*e^2/c^2 + 3/4*(c*x^2 + b*x)
^(3/2)*d*x*e^2/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 - 15/128*b^4*d*e^2*log(2*
c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 7/64*sqrt(c*x^2 + b*x)*b^3*x*e^3/c^3 - 7/40*(c*x^2 + b*x)^(3/
2)*b*x*e^3/c^2 + 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 + 7/256*b^5*e^3*l
og(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) - 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

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Fricas [A]
time = 2.17, size = 471, normalized size = 2.24 \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 (960 \, c^{5} d^{3} x + 480 \, b c^{4} d^{3} + {\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 56 \, b^{2} c^{3} x^{2} + 70 \, b^{3} c^{2} x - 105 \, b^{4} c\right )} e^{3} + 30 \, {\left (48 \, c^{5} d x^{3} + 8 \, b c^{4} d x^{2} - 10 \, b^{2} c^{3} d x + 15 \, b^{3} c^{2} d\right )} e^{2} + 240 \, {\left (8 \, c^{5} d^{2} x^{2} + 2 \, b c^{4} d^{2} x - 3 \, b^{2} c^{3} d^{2}\right )} e\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 (960 \, c^{5} d^{3} x + 480 \, b c^{4} d^{3} + {\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 56 \, b^{2} c^{3} x^{2} + 70 \, b^{3} c^{2} x - 105 \, b^{4} c\right )} e^{3} + 30 \, {\left (48 \, c^{5} d x^{3} + 8 \, b c^{4} d x^{2} - 10 \, b^{2} c^{3} d x + 15 \, b^{3} c^{2} d\right )} e^{2} + 240 \, {\left (8 \, c^{5} d^{2} x^{2} + 2 \, b c^{4} d^{2} x - 3 \, b^{2} c^{3} d^{2}\right )} e\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*(960*c^5*d^3*x + 480*b*c^4*d^3 + (384*c^5*x^4 + 48*b*c^4*x^3 - 56*b^2*c^3*x^2 + 70*b^
3*c^2*x - 105*b^4*c)*e^3 + 30*(48*c^5*d*x^3 + 8*b*c^4*d*x^2 - 10*b^2*c^3*d*x + 15*b^3*c^2*d)*e^2 + 240*(8*c^5*
d^2*x^2 + 2*b*c^4*d^2*x - 3*b^2*c^3*d^2)*e)*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)) + (960*c^5*d^3*x + 480*b*c
^4*d^3 + (384*c^5*x^4 + 48*b*c^4*x^3 - 56*b^2*c^3*x^2 + 70*b^3*c^2*x - 105*b^4*c)*e^3 + 30*(48*c^5*d*x^3 + 8*b
*c^4*d*x^2 - 10*b^2*c^3*d*x + 15*b^3*c^2*d)*e^2 + 240*(8*c^5*d^2*x^2 + 2*b*c^4*d^2*x - 3*b^2*c^3*d^2)*e)*sqrt(
c*x^2 + b*x))/c^5]

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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)

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Giac [A]
time = 2.01, 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)

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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)

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