3.1559 \(\int (b+2 c x) (d+e x) (a+b x+c x^2)^{3/2} \, dx\)

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

[Out]

-((b^2 - 4*a*c)^2*e*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(256*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))

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Rubi [A]  time = 0.123079, antiderivative size = 160, normalized size of antiderivative = 1., 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 = {779, 612, 621, 206} \[ -\frac{e \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{256 c^3}+\frac{e \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{96 c^2}+\frac{e \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{512 c^{7/2}}+\frac{\left (a+b x+c x^2\right )^{5/2} (-b e+12 c d+10 c e x)}{30 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)^2*e*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(256*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 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 621

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 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 (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 ) \operatorname{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*}

Mathematica [A]  time = 0.232304, size = 147, normalized size = 0.92 \[ \frac{e \left (b^2-4 a c\right ) \left (2 \sqrt{c} (b+2 c x) \sqrt{a+x (b+c x)} \left (4 c \left (5 a+2 c x^2\right )-3 b^2+8 b c x\right )+3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )}{1536 c^{7/2}}+\frac{(a+x (b+c x))^{5/2} (2 c (6 d+5 e x)-b e)}{30 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + x*(b + c*x))^(5/2)*(-(b*e) + 2*c*(6*d + 5*e*x)))/(30*c) + ((b^2 - 4*a*c)*e*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a
+ x*(b + c*x)]*(-3*b^2 + 8*b*c*x + 4*c*(5*a + 2*c*x^2)) + 3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sq
rt[a + x*(b + c*x)])]))/(1536*c^(7/2))

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Maple [B]  time = 0.007, size = 401, normalized size = 2.5 \begin{align*}{\frac{{b}^{2}exa}{16\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{2}e{a}^{2}}{32}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{3\,e{b}^{4}a}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{2\,d}{5} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}}+{\frac{ex}{3} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}}+{\frac{e{b}^{6}}{512}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}-{\frac{abe}{24\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{e{a}^{2}b}{16\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{e{a}^{3}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{{b}^{2}ex}{48\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{e{b}^{4}x}{128\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{{b}^{3}ea}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{be}{30\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}}+{\frac{{b}^{3}e}{96\,{c}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{e{b}^{5}}{256\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{aex}{12} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{e{a}^{2}x}{8}\sqrt{c{x}^{2}+bx+a}} \end{align*}

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)

[Out]

1/16/c*e*b^2*(c*x^2+b*x+a)^(1/2)*x*a+3/32/c^(3/2)*e*b^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-3/128/
c^(5/2)*e*b^4*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+2/5*(c*x^2+b*x+a)^(5/2)*d+1/3*e*x*(c*x^2+b*x+a)^(5
/2)+1/512/c^(7/2)*e*b^6*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/24/c*e*a*(c*x^2+b*x+a)^(3/2)*b-1/16/c*e*
a^2*(c*x^2+b*x+a)^(1/2)*b-1/8/c^(1/2)*e*a^3*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/48/c*e*b^2*(c*x^2+b*
x+a)^(3/2)*x-1/128/c^2*e*b^4*(c*x^2+b*x+a)^(1/2)*x+1/32/c^2*e*b^3*(c*x^2+b*x+a)^(1/2)*a-1/30/c*e*b*(c*x^2+b*x+
a)^(5/2)+1/96/c^2*e*b^3*(c*x^2+b*x+a)^(3/2)-1/256/c^3*e*b^5*(c*x^2+b*x+a)^(1/2)-1/12*e*a*(c*x^2+b*x+a)^(3/2)*x
-1/8*e*a^2*(c*x^2+b*x+a)^(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((2*c*x+b)*(e*x+d)*(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.18425, size = 1316, normalized size = 8.22 \begin{align*} \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 (1280 \, c^{6} e x^{5} + 1536 \, a^{2} c^{4} d + 128 \,{\left (12 \, c^{6} d + 19 \, b c^{5} e\right )} x^{4} + 16 \,{\left (192 \, b c^{5} d +{\left (69 \, b^{2} c^{4} + 140 \, a c^{5}\right )} e\right )} x^{3} + 8 \,{\left (192 \,{\left (b^{2} c^{4} + 2 \, a c^{5}\right )} d -{\left (b^{3} c^{3} - 228 \, a b c^{4}\right )} e\right )} x^{2} -{\left (15 \, b^{5} c - 160 \, a b^{3} c^{2} + 528 \, a^{2} b c^{3}\right )} e + 2 \,{\left (1536 \, a b c^{4} d +{\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} e\right )} x\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} e \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 ) - 2 \,{\left (1280 \, c^{6} e x^{5} + 1536 \, a^{2} c^{4} d + 128 \,{\left (12 \, c^{6} d + 19 \, b c^{5} e\right )} x^{4} + 16 \,{\left (192 \, b c^{5} d +{\left (69 \, b^{2} c^{4} + 140 \, a c^{5}\right )} e\right )} x^{3} + 8 \,{\left (192 \,{\left (b^{2} c^{4} + 2 \, a c^{5}\right )} d -{\left (b^{3} c^{3} - 228 \, a b c^{4}\right )} e\right )} x^{2} -{\left (15 \, b^{5} c - 160 \, a b^{3} c^{2} + 528 \, a^{2} b c^{3}\right )} e + 2 \,{\left (1536 \, a b c^{4} d +{\left (5 \, b^{4} c^{2} - 48 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} e\right )} x\right )} \sqrt{c x^{2} + b x + a}}{7680 \, c^{4}}\right ] \end{align*}

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*(1280*c^6*e*x^5 + 1536*a^2*c^4*d + 128*(12*c^6*d + 19*b*c
^5*e)*x^4 + 16*(192*b*c^5*d + (69*b^2*c^4 + 140*a*c^5)*e)*x^3 + 8*(192*(b^2*c^4 + 2*a*c^5)*d - (b^3*c^3 - 228*
a*b*c^4)*e)*x^2 - (15*b^5*c - 160*a*b^3*c^2 + 528*a^2*b*c^3)*e + 2*(1536*a*b*c^4*d + (5*b^4*c^2 - 48*a*b^2*c^3
+ 240*a^2*c^4)*e)*x)*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)
*sqrt(-c)*e*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)) - 2*(1280*c^6*e*x^5
+ 1536*a^2*c^4*d + 128*(12*c^6*d + 19*b*c^5*e)*x^4 + 16*(192*b*c^5*d + (69*b^2*c^4 + 140*a*c^5)*e)*x^3 + 8*(1
92*(b^2*c^4 + 2*a*c^5)*d - (b^3*c^3 - 228*a*b*c^4)*e)*x^2 - (15*b^5*c - 160*a*b^3*c^2 + 528*a^2*b*c^3)*e + 2*(
1536*a*b*c^4*d + (5*b^4*c^2 - 48*a*b^2*c^3 + 240*a^2*c^4)*e)*x)*sqrt(c*x^2 + b*x + a))/c^4]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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]  time = 1.18267, size = 397, normalized size = 2.48 \begin{align*} \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{align*}

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)