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

Optimal. Leaf size=271 \[ -\frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2} (2 c d-b e) \left (3 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{1024 c^5}+\frac{e \left (b x+c x^2\right )^{5/2} \left (21 b^2 e^2+30 c e x (2 c d-b e)-98 b c d e+128 c^2 d^2\right )}{280 c^3}+\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (2 c d-b e) \left (3 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{128 c^4}+\frac{3 b^4 (2 c d-b e) \left (3 b^2 e^2-8 b c d e+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{1024 c^{11/2}}+\frac{e \left (b x+c x^2\right )^{5/2} (d+e x)^2}{7 c} \]

[Out]

(-3*b^2*(2*c*d - b*e)*(8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(1024*c^5) + ((2*c*d
- b*e)*(8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2)*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(128*c^4) + (e*(d + e*x)^2*(b*x +
c*x^2)^(5/2))/(7*c) + (e*(128*c^2*d^2 - 98*b*c*d*e + 21*b^2*e^2 + 30*c*e*(2*c*d - b*e)*x)*(b*x + c*x^2)^(5/2))
/(280*c^3) + (3*b^4*(2*c*d - b*e)*(8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/
(1024*c^(11/2))

________________________________________________________________________________________

Rubi [A]  time = 0.349665, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{3 b^2 (b+2 c x) \sqrt{b x+c x^2} (2 c d-b e) \left (3 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{1024 c^5}+\frac{e \left (b x+c x^2\right )^{5/2} \left (21 b^2 e^2+30 c e x (2 c d-b e)-98 b c d e+128 c^2 d^2\right )}{280 c^3}+\frac{(b+2 c x) \left (b x+c x^2\right )^{3/2} (2 c d-b e) \left (3 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{128 c^4}+\frac{3 b^4 (2 c d-b e) \left (3 b^2 e^2-8 b c d e+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{1024 c^{11/2}}+\frac{e \left (b x+c x^2\right )^{5/2} (d+e x)^2}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(b*x + c*x^2)^(3/2),x]

[Out]

(-3*b^2*(2*c*d - b*e)*(8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(1024*c^5) + ((2*c*d
- b*e)*(8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2)*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(128*c^4) + (e*(d + e*x)^2*(b*x +
c*x^2)^(5/2))/(7*c) + (e*(128*c^2*d^2 - 98*b*c*d*e + 21*b^2*e^2 + 30*c*e*(2*c*d - b*e)*x)*(b*x + c*x^2)^(5/2))
/(280*c^3) + (3*b^4*(2*c*d - b*e)*(8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/
(1024*c^(11/2))

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]

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 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])

Rubi steps

\begin{align*} \int (d+e x)^3 \left (b x+c x^2\right )^{3/2} \, dx &=\frac{e (d+e x)^2 \left (b x+c x^2\right )^{5/2}}{7 c}+\frac{\int (d+e x) \left (\frac{1}{2} d (14 c d-5 b e)+\frac{9}{2} e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2} \, dx}{7 c}\\ &=\frac{e (d+e x)^2 \left (b x+c x^2\right )^{5/2}}{7 c}+\frac{e \left (128 c^2 d^2-98 b c d e+21 b^2 e^2+30 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{5/2}}{280 c^3}+\frac{\left ((2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right )\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{16 c^3}\\ &=\frac{(2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right ) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{128 c^4}+\frac{e (d+e x)^2 \left (b x+c x^2\right )^{5/2}}{7 c}+\frac{e \left (128 c^2 d^2-98 b c d e+21 b^2 e^2+30 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{5/2}}{280 c^3}-\frac{\left (3 b^2 (2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right )\right ) \int \sqrt{b x+c x^2} \, dx}{256 c^4}\\ &=-\frac{3 b^2 (2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right ) (b+2 c x) \sqrt{b x+c x^2}}{1024 c^5}+\frac{(2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right ) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{128 c^4}+\frac{e (d+e x)^2 \left (b x+c x^2\right )^{5/2}}{7 c}+\frac{e \left (128 c^2 d^2-98 b c d e+21 b^2 e^2+30 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{5/2}}{280 c^3}+\frac{\left (3 b^4 (2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right )\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{2048 c^5}\\ &=-\frac{3 b^2 (2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right ) (b+2 c x) \sqrt{b x+c x^2}}{1024 c^5}+\frac{(2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right ) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{128 c^4}+\frac{e (d+e x)^2 \left (b x+c x^2\right )^{5/2}}{7 c}+\frac{e \left (128 c^2 d^2-98 b c d e+21 b^2 e^2+30 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{5/2}}{280 c^3}+\frac{\left (3 b^4 (2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 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 )}{1024 c^5}\\ &=-\frac{3 b^2 (2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right ) (b+2 c x) \sqrt{b x+c x^2}}{1024 c^5}+\frac{(2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right ) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{128 c^4}+\frac{e (d+e x)^2 \left (b x+c x^2\right )^{5/2}}{7 c}+\frac{e \left (128 c^2 d^2-98 b c d e+21 b^2 e^2+30 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{5/2}}{280 c^3}+\frac{3 b^4 (2 c d-b e) \left (8 c^2 d^2-8 b c d e+3 b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{1024 c^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.701945, size = 312, normalized size = 1.15 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (28 b^4 c^2 e \left (90 d^2+35 d e x+6 e^2 x^2\right )-16 b^3 c^3 \left (105 d^2 e x+105 d^3+49 d e^2 x^2+9 e^3 x^3\right )+32 b^2 c^4 x \left (42 d^2 e x+35 d^3+21 d e^2 x^2+4 e^3 x^3\right )-210 b^5 c e^2 (7 d+e x)+315 b^6 e^3+128 b c^5 x^2 \left (231 d^2 e x+105 d^3+182 d e^2 x^2+50 e^3 x^3\right )+256 c^6 x^3 \left (84 d^2 e x+35 d^3+70 d e^2 x^2+20 e^3 x^3\right )\right )-\frac{105 b^{7/2} \left (-14 b^2 c d e^2+3 b^3 e^3+24 b c^2 d^2 e-16 c^3 d^3\right ) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{35840 c^{11/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(315*b^6*e^3 - 210*b^5*c*e^2*(7*d + e*x) + 28*b^4*c^2*e*(90*d^2 + 35*d*e*x + 6*e^2
*x^2) + 32*b^2*c^4*x*(35*d^3 + 42*d^2*e*x + 21*d*e^2*x^2 + 4*e^3*x^3) - 16*b^3*c^3*(105*d^3 + 105*d^2*e*x + 49
*d*e^2*x^2 + 9*e^3*x^3) + 256*c^6*x^3*(35*d^3 + 84*d^2*e*x + 70*d*e^2*x^2 + 20*e^3*x^3) + 128*b*c^5*x^2*(105*d
^3 + 231*d^2*e*x + 182*d*e^2*x^2 + 50*e^3*x^3)) - (105*b^(7/2)*(-16*c^3*d^3 + 24*b*c^2*d^2*e - 14*b^2*c*d*e^2
+ 3*b^3*e^3)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(35840*c^(11/2))

________________________________________________________________________________________

Maple [B]  time = 0.056, size = 629, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/64*d^3*b^3/c^2*(c*x^2+b*x)^(1/2)+3/128*d^3*b^4/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/7*e^3*x^
2*(c*x^2+b*x)^(5/2)/c+3/40*e^3*b^2/c^3*(c*x^2+b*x)^(5/2)-3/128*e^3*b^4/c^4*(c*x^2+b*x)^(3/2)+9/1024*e^3*b^6/c^
5*(c*x^2+b*x)^(1/2)-9/2048*e^3*b^7/c^(11/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))-21/512*d*e^2*b^5/c^4*(c*
x^2+b*x)^(1/2)+21/1024*d*e^2*b^6/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))-3/28*e^3*b/c^2*x*(c*x^2+b*x
)^(5/2)-3/64*e^3*b^3/c^3*x*(c*x^2+b*x)^(3/2)+9/512*e^3*b^5/c^4*(c*x^2+b*x)^(1/2)*x+7/64*d*e^2*b^3/c^3*(c*x^2+b
*x)^(3/2)-9/256*d^2*e*b^5/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/2*d*e^2*x*(c*x^2+b*x)^(5/2)/c-7/
20*d*e^2*b/c^2*(c*x^2+b*x)^(5/2)+9/128*d^2*e*b^4/c^3*(c*x^2+b*x)^(1/2)+1/4*d^3*x*(c*x^2+b*x)^(3/2)-3/32*d^3*b^
2/c*(c*x^2+b*x)^(1/2)*x-3/16*d^2*e*b^2/c^2*(c*x^2+b*x)^(3/2)+3/5*d^2*e*(c*x^2+b*x)^(5/2)/c+1/8*d^3/c*(c*x^2+b*
x)^(3/2)*b+7/32*d*e^2*b^2/c^2*x*(c*x^2+b*x)^(3/2)-21/256*d*e^2*b^4/c^3*(c*x^2+b*x)^(1/2)*x-3/8*d^2*e*b/c*x*(c*
x^2+b*x)^(3/2)+9/64*d^2*e*b^3/c^2*(c*x^2+b*x)^(1/2)*x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.40864, size = 1586, normalized size = 5.85 \begin{align*} \left [-\frac{105 \,{\left (16 \, b^{4} c^{3} d^{3} - 24 \, b^{5} c^{2} d^{2} e + 14 \, b^{6} c d e^{2} - 3 \, b^{7} e^{3}\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (5120 \, c^{7} e^{3} x^{6} - 1680 \, b^{3} c^{4} d^{3} + 2520 \, b^{4} c^{3} d^{2} e - 1470 \, b^{5} c^{2} d e^{2} + 315 \, b^{6} c e^{3} + 1280 \,{\left (14 \, c^{7} d e^{2} + 5 \, b c^{6} e^{3}\right )} x^{5} + 128 \,{\left (168 \, c^{7} d^{2} e + 182 \, b c^{6} d e^{2} + b^{2} c^{5} e^{3}\right )} x^{4} + 16 \,{\left (560 \, c^{7} d^{3} + 1848 \, b c^{6} d^{2} e + 42 \, b^{2} c^{5} d e^{2} - 9 \, b^{3} c^{4} e^{3}\right )} x^{3} + 56 \,{\left (240 \, b c^{6} d^{3} + 24 \, b^{2} c^{5} d^{2} e - 14 \, b^{3} c^{4} d e^{2} + 3 \, b^{4} c^{3} e^{3}\right )} x^{2} + 70 \,{\left (16 \, b^{2} c^{5} d^{3} - 24 \, b^{3} c^{4} d^{2} e + 14 \, b^{4} c^{3} d e^{2} - 3 \, b^{5} c^{2} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{71680 \, c^{6}}, -\frac{105 \,{\left (16 \, b^{4} c^{3} d^{3} - 24 \, b^{5} c^{2} d^{2} e + 14 \, b^{6} c d e^{2} - 3 \, b^{7} e^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (5120 \, c^{7} e^{3} x^{6} - 1680 \, b^{3} c^{4} d^{3} + 2520 \, b^{4} c^{3} d^{2} e - 1470 \, b^{5} c^{2} d e^{2} + 315 \, b^{6} c e^{3} + 1280 \,{\left (14 \, c^{7} d e^{2} + 5 \, b c^{6} e^{3}\right )} x^{5} + 128 \,{\left (168 \, c^{7} d^{2} e + 182 \, b c^{6} d e^{2} + b^{2} c^{5} e^{3}\right )} x^{4} + 16 \,{\left (560 \, c^{7} d^{3} + 1848 \, b c^{6} d^{2} e + 42 \, b^{2} c^{5} d e^{2} - 9 \, b^{3} c^{4} e^{3}\right )} x^{3} + 56 \,{\left (240 \, b c^{6} d^{3} + 24 \, b^{2} c^{5} d^{2} e - 14 \, b^{3} c^{4} d e^{2} + 3 \, b^{4} c^{3} e^{3}\right )} x^{2} + 70 \,{\left (16 \, b^{2} c^{5} d^{3} - 24 \, b^{3} c^{4} d^{2} e + 14 \, b^{4} c^{3} d e^{2} - 3 \, b^{5} c^{2} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{35840 \, c^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/71680*(105*(16*b^4*c^3*d^3 - 24*b^5*c^2*d^2*e + 14*b^6*c*d*e^2 - 3*b^7*e^3)*sqrt(c)*log(2*c*x + b - 2*sqrt
(c*x^2 + b*x)*sqrt(c)) - 2*(5120*c^7*e^3*x^6 - 1680*b^3*c^4*d^3 + 2520*b^4*c^3*d^2*e - 1470*b^5*c^2*d*e^2 + 31
5*b^6*c*e^3 + 1280*(14*c^7*d*e^2 + 5*b*c^6*e^3)*x^5 + 128*(168*c^7*d^2*e + 182*b*c^6*d*e^2 + b^2*c^5*e^3)*x^4
+ 16*(560*c^7*d^3 + 1848*b*c^6*d^2*e + 42*b^2*c^5*d*e^2 - 9*b^3*c^4*e^3)*x^3 + 56*(240*b*c^6*d^3 + 24*b^2*c^5*
d^2*e - 14*b^3*c^4*d*e^2 + 3*b^4*c^3*e^3)*x^2 + 70*(16*b^2*c^5*d^3 - 24*b^3*c^4*d^2*e + 14*b^4*c^3*d*e^2 - 3*b
^5*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/c^6, -1/35840*(105*(16*b^4*c^3*d^3 - 24*b^5*c^2*d^2*e + 14*b^6*c*d*e^2 - 3*b
^7*e^3)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (5120*c^7*e^3*x^6 - 1680*b^3*c^4*d^3 + 2520*b^4*c^
3*d^2*e - 1470*b^5*c^2*d*e^2 + 315*b^6*c*e^3 + 1280*(14*c^7*d*e^2 + 5*b*c^6*e^3)*x^5 + 128*(168*c^7*d^2*e + 18
2*b*c^6*d*e^2 + b^2*c^5*e^3)*x^4 + 16*(560*c^7*d^3 + 1848*b*c^6*d^2*e + 42*b^2*c^5*d*e^2 - 9*b^3*c^4*e^3)*x^3
+ 56*(240*b*c^6*d^3 + 24*b^2*c^5*d^2*e - 14*b^3*c^4*d*e^2 + 3*b^4*c^3*e^3)*x^2 + 70*(16*b^2*c^5*d^3 - 24*b^3*c
^4*d^2*e + 14*b^4*c^3*d*e^2 - 3*b^5*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/c^6]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (x \left (b + c x\right )\right )^{\frac{3}{2}} \left (d + e x\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.34195, size = 493, normalized size = 1.82 \begin{align*} \frac{1}{35840} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \,{\left (4 \, c x e^{3} + \frac{14 \, c^{7} d e^{2} + 5 \, b c^{6} e^{3}}{c^{6}}\right )} x + \frac{168 \, c^{7} d^{2} e + 182 \, b c^{6} d e^{2} + b^{2} c^{5} e^{3}}{c^{6}}\right )} x + \frac{560 \, c^{7} d^{3} + 1848 \, b c^{6} d^{2} e + 42 \, b^{2} c^{5} d e^{2} - 9 \, b^{3} c^{4} e^{3}}{c^{6}}\right )} x + \frac{7 \,{\left (240 \, b c^{6} d^{3} + 24 \, b^{2} c^{5} d^{2} e - 14 \, b^{3} c^{4} d e^{2} + 3 \, b^{4} c^{3} e^{3}\right )}}{c^{6}}\right )} x + \frac{35 \,{\left (16 \, b^{2} c^{5} d^{3} - 24 \, b^{3} c^{4} d^{2} e + 14 \, b^{4} c^{3} d e^{2} - 3 \, b^{5} c^{2} e^{3}\right )}}{c^{6}}\right )} x - \frac{105 \,{\left (16 \, b^{3} c^{4} d^{3} - 24 \, b^{4} c^{3} d^{2} e + 14 \, b^{5} c^{2} d e^{2} - 3 \, b^{6} c e^{3}\right )}}{c^{6}}\right )} - \frac{3 \,{\left (16 \, b^{4} c^{3} d^{3} - 24 \, b^{5} c^{2} d^{2} e + 14 \, b^{6} c d e^{2} - 3 \, b^{7} e^{3}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{2048 \, c^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35840*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*(4*c*x*e^3 + (14*c^7*d*e^2 + 5*b*c^6*e^3)/c^6)*x + (168*c^7*d^2*e +
182*b*c^6*d*e^2 + b^2*c^5*e^3)/c^6)*x + (560*c^7*d^3 + 1848*b*c^6*d^2*e + 42*b^2*c^5*d*e^2 - 9*b^3*c^4*e^3)/c^
6)*x + 7*(240*b*c^6*d^3 + 24*b^2*c^5*d^2*e - 14*b^3*c^4*d*e^2 + 3*b^4*c^3*e^3)/c^6)*x + 35*(16*b^2*c^5*d^3 - 2
4*b^3*c^4*d^2*e + 14*b^4*c^3*d*e^2 - 3*b^5*c^2*e^3)/c^6)*x - 105*(16*b^3*c^4*d^3 - 24*b^4*c^3*d^2*e + 14*b^5*c
^2*d*e^2 - 3*b^6*c*e^3)/c^6) - 3/2048*(16*b^4*c^3*d^3 - 24*b^5*c^2*d^2*e + 14*b^6*c*d*e^2 - 3*b^7*e^3)*log(abs
(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(11/2)