3.16.59 \(\int (b+2 c x) (d+e x) (a+b x+c x^2)^{3/2} \, dx\) [1559]

Optimal. Leaf size=160 \[ -\frac {\left (b^2-4 a c\right )^2 e (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {\left (b^2-4 a c\right ) e (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}+\frac {\left (b^2-4 a c\right )^3 e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{512 c^{7/2}} \]

[Out]

1/96*(-4*a*c+b^2)*e*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c^2+1/30*(10*c*e*x-b*e+12*c*d)*(c*x^2+b*x+a)^(5/2)/c+1/512*(
-4*a*c+b^2)^3*e*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)-1/256*(-4*a*c+b^2)^2*e*(2*c*x+b)*(c
*x^2+b*x+a)^(1/2)/c^3

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Rubi [A]
time = 0.08, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {793, 626, 635, 212} \begin {gather*} \frac {e \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{512 c^{7/2}}-\frac {e \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {e \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {\left (a+b x+c x^2\right )^{5/2} (-b e+12 c d+10 c e x)}{30 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/256*((b^2 - 4*a*c)^2*e*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/c^3 + ((b^2 - 4*a*c)*e*(b + 2*c*x)*(a + b*x + c*x
^2)^(3/2))/(96*c^2) + ((12*c*d - b*e + 10*c*e*x)*(a + b*x + c*x^2)^(5/2))/(30*c) + ((b^2 - 4*a*c)^3*e*ArcTanh[
(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(512*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 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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 (b+2 c x) (d+e x) \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}+\frac {\left (\left (b^2-4 a c\right ) e\right ) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{12 c}\\ &=\frac {\left (b^2-4 a c\right ) e (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}-\frac {\left (\left (b^2-4 a c\right )^2 e\right ) \int \sqrt {a+b x+c x^2} \, dx}{64 c^2}\\ &=-\frac {\left (b^2-4 a c\right )^2 e (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {\left (b^2-4 a c\right ) e (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}+\frac {\left (\left (b^2-4 a c\right )^3 e\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{512 c^3}\\ &=-\frac {\left (b^2-4 a c\right )^2 e (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {\left (b^2-4 a c\right ) e (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}+\frac {\left (\left (b^2-4 a c\right )^3 e\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{256 c^3}\\ &=-\frac {\left (b^2-4 a c\right )^2 e (b+2 c x) \sqrt {a+b x+c x^2}}{256 c^3}+\frac {\left (b^2-4 a c\right ) e (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac {(12 c d-b e+10 c e x) \left (a+b x+c x^2\right )^{5/2}}{30 c}+\frac {\left (b^2-4 a c\right )^3 e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{512 c^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.88, size = 221, normalized size = 1.38 \begin {gather*} \frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (-15 b^5 e+10 b^4 c e x-8 b^3 c^2 e x^2+256 c^5 x^4 (6 d+5 e x)+128 b c^4 x^3 (24 d+19 e x)+48 b^2 c^3 x^2 (32 d+23 e x)+48 a^2 c^2 (32 c d-11 b e+10 c e x)+32 a c \left (5 b^3 e-3 b^2 c e x+3 b c^2 x (32 d+19 e x)+2 c^3 x^2 (48 d+35 e x)\right )\right )-15 \left (b^2-4 a c\right )^3 e \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{7680 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-15*b^5*e + 10*b^4*c*e*x - 8*b^3*c^2*e*x^2 + 256*c^5*x^4*(6*d + 5*e*x) + 128
*b*c^4*x^3*(24*d + 19*e*x) + 48*b^2*c^3*x^2*(32*d + 23*e*x) + 48*a^2*c^2*(32*c*d - 11*b*e + 10*c*e*x) + 32*a*c
*(5*b^3*e - 3*b^2*c*e*x + 3*b*c^2*x*(32*d + 19*e*x) + 2*c^3*x^2*(48*d + 35*e*x))) - 15*(b^2 - 4*a*c)^3*e*Log[b
+ 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(7680*c^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(508\) vs. \(2(138)=276\).
time = 0.90, size = 509, normalized size = 3.18

method	result	size
risch	\(-\frac {\left (-1280 c^{5} e \,x^{5}-2432 b \,c^{4} e \,x^{4}-1536 c^{5} d \,x^{4}-2240 a \,c^{4} e \,x^{3}-1104 b^{2} c^{3} e \,x^{3}-3072 b \,c^{4} d \,x^{3}-1824 a b \,c^{3} e \,x^{2}-3072 a \,c^{4} d \,x^{2}+8 b^{3} c^{2} e \,x^{2}-1536 b^{2} c^{3} d \,x^{2}-480 a^{2} c^{3} e x +96 a \,b^{2} c^{2} e x -3072 a b \,c^{3} d x -10 b^{4} c e x +528 a^{2} b \,c^{2} e -1536 a^{2} c^{3} d -160 a \,b^{3} c e +15 b^{5} e \right ) \sqrt {c \,x^{2}+b x +a}}{3840 c^{3}}-\frac {e \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a^{3}}{8 \sqrt {c}}+\frac {3 e \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a^{2} b^{2}}{32 c^{\frac {3}{2}}}-\frac {3 e \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a \,b^{4}}{128 c^{\frac {5}{2}}}+\frac {e \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{6}}{512 c^{\frac {7}{2}}}\)	\(336\)
default	\(2 c e \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{6 c}-\frac {7 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{12 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{6 c}\right )+\left (b e +2 c d \right ) \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )+b d \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )\)	\(509\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*e*(1/6*x*(c*x^2+b*x+a)^(5/2)/c-7/12*b/c*(1/5*(c*x^2+b*x+a)^(5/2)/c-1/2*b/c*(1/8*(2*c*x+b)*(c*x^2+b*x+a)^(3
/2)/c+3/16*(4*a*c-b^2)/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(
c*x^2+b*x+a)^(1/2)))))-1/6*a/c*(1/8*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c+3/16*(4*a*c-b^2)/c*(1/4*(2*c*x+b)*(c*x^2+b
*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))))+(b*e+2*c*d)*(1/5*(c*x^2+b
*x+a)^(5/2)/c-1/2*b/c*(1/8*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c+3/16*(4*a*c-b^2)/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/
2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))))+b*d*(1/8*(2*c*x+b)*(c*x^2+b*x+a)^(
3/2)/c+3/16*(4*a*c-b^2)/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+
(c*x^2+b*x+a)^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

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Fricas [A]
time = 3.89, size = 551, normalized size = 3.44 \begin {gather*} \left [-\frac {15 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {c} e \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (1536 \, c^{6} d x^{4} + 3072 \, b c^{5} d x^{3} + 3072 \, a b c^{4} d x + 1536 \, a^{2} c^{4} d + 1536 \, {\left (b^{2} c^{4} + 2 \, a c^{5}\right )} d x^{2} + {\left (1280 \, c^{6} x^{5} + 2432 \, b c^{5} x^{4} - 15 \, b^{5} c + 160 \, a b^{3} c^{2} - 528 \, a^{2} b c^{3} + 16 \, {\left (69 \, b^{2} c^{4} + 140 \, a c^{5}\right )} x^{3} - 8 \, {\left (b^{3} c^{3} - 228 \, a b c^{4}\right )} x^{2} + 2 \, {\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} x\right )} e\right )} \sqrt {c x^{2} + b x + a}}{15360 \, c^{4}}, -\frac {15 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) e - 2 \, {\left (1536 \, c^{6} d x^{4} + 3072 \, b c^{5} d x^{3} + 3072 \, a b c^{4} d x + 1536 \, a^{2} c^{4} d + 1536 \, {\left (b^{2} c^{4} + 2 \, a c^{5}\right )} d x^{2} + {\left (1280 \, c^{6} x^{5} + 2432 \, b c^{5} x^{4} - 15 \, b^{5} c + 160 \, a b^{3} c^{2} - 528 \, a^{2} b c^{3} + 16 \, {\left (69 \, b^{2} c^{4} + 140 \, a c^{5}\right )} x^{3} - 8 \, {\left (b^{3} c^{3} - 228 \, a b c^{4}\right )} x^{2} + 2 \, {\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} x\right )} e\right )} \sqrt {c x^{2} + b x + a}}{7680 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15360*(15*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)*e*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sq
rt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(1536*c^6*d*x^4 + 3072*b*c^5*d*x^3 + 3072*a*b*c^4*d*x + 1
536*a^2*c^4*d + 1536*(b^2*c^4 + 2*a*c^5)*d*x^2 + (1280*c^6*x^5 + 2432*b*c^5*x^4 - 15*b^5*c + 160*a*b^3*c^2 - 5
28*a^2*b*c^3 + 16*(69*b^2*c^4 + 140*a*c^5)*x^3 - 8*(b^3*c^3 - 228*a*b*c^4)*x^2 + 2*(5*b^4*c^2 - 48*a*b^2*c^3 +
240*a^2*c^4)*x)*e)*sqrt(c*x^2 + b*x + a))/c^4, -1/7680*(15*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*s
qrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c))*e - 2*(1536*c^6*d*x^4 +
3072*b*c^5*d*x^3 + 3072*a*b*c^4*d*x + 1536*a^2*c^4*d + 1536*(b^2*c^4 + 2*a*c^5)*d*x^2 + (1280*c^6*x^5 + 2432*
b*c^5*x^4 - 15*b^5*c + 160*a*b^3*c^2 - 528*a^2*b*c^3 + 16*(69*b^2*c^4 + 140*a*c^5)*x^3 - 8*(b^3*c^3 - 228*a*b*
c^4)*x^2 + 2*(5*b^4*c^2 - 48*a*b^2*c^3 + 240*a^2*c^4)*x)*e)*sqrt(c*x^2 + b*x + a))/c^4]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b + 2 c x\right ) \left (d + e x\right ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (143) = 286\).
time = 4.60, size = 294, normalized size = 1.84 \begin {gather*} \frac {1}{3840} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, c^{2} x e + \frac {12 \, c^{7} d + 19 \, b c^{6} e}{c^{5}}\right )} x + \frac {192 \, b c^{6} d + 69 \, b^{2} c^{5} e + 140 \, a c^{6} e}{c^{5}}\right )} x + \frac {192 \, b^{2} c^{5} d + 384 \, a c^{6} d - b^{3} c^{4} e + 228 \, a b c^{5} e}{c^{5}}\right )} x + \frac {1536 \, a b c^{5} d + 5 \, b^{4} c^{3} e - 48 \, a b^{2} c^{4} e + 240 \, a^{2} c^{5} e}{c^{5}}\right )} x + \frac {1536 \, a^{2} c^{5} d - 15 \, b^{5} c^{2} e + 160 \, a b^{3} c^{3} e - 528 \, a^{2} b c^{4} e}{c^{5}}\right )} - \frac {{\left (b^{6} e - 12 \, a b^{4} c e + 48 \, a^{2} b^{2} c^{2} e - 64 \, a^{3} c^{3} e\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{512 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3840*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(10*c^2*x*e + (12*c^7*d + 19*b*c^6*e)/c^5)*x + (192*b*c^6*d + 69*b^2*
c^5*e + 140*a*c^6*e)/c^5)*x + (192*b^2*c^5*d + 384*a*c^6*d - b^3*c^4*e + 228*a*b*c^5*e)/c^5)*x + (1536*a*b*c^5
*d + 5*b^4*c^3*e - 48*a*b^2*c^4*e + 240*a^2*c^5*e)/c^5)*x + (1536*a^2*c^5*d - 15*b^5*c^2*e + 160*a*b^3*c^3*e -
528*a^2*b*c^4*e)/c^5) - 1/512*(b^6*e - 12*a*b^4*c*e + 48*a^2*b^2*c^2*e - 64*a^3*c^3*e)*log(abs(-2*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (b+2\,c\,x\right )\,\left (d+e\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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