3.2345 \(\int (d+e x) (a+b x+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=161 \[ -\frac{3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2} (2 c d-b e)}{128 c^3}+\frac{3 \left (b^2-4 a c\right )^2 (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{7/2}}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{16 c^2}+\frac{e \left (a+b x+c x^2\right )^{5/2}}{5 c} \]

[Out]

(-3*(b^2 - 4*a*c)*(2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^3) + ((2*c*d - b*e)*(b + 2*c*x)*(a +
b*x + c*x^2)^(3/2))/(16*c^2) + (e*(a + b*x + c*x^2)^(5/2))/(5*c) + (3*(b^2 - 4*a*c)^2*(2*c*d - b*e)*ArcTanh[(
b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(7/2))

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Rubi [A]  time = 0.0623502, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {640, 612, 621, 206} \[ -\frac{3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2} (2 c d-b e)}{128 c^3}+\frac{3 \left (b^2-4 a c\right )^2 (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{7/2}}+\frac{(b+2 c x) \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{16 c^2}+\frac{e \left (a+b x+c x^2\right )^{5/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b^2 - 4*a*c)*(2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^3) + ((2*c*d - b*e)*(b + 2*c*x)*(a +
b*x + c*x^2)^(3/2))/(16*c^2) + (e*(a + b*x + c*x^2)^(5/2))/(5*c) + (3*(b^2 - 4*a*c)^2*(2*c*d - b*e)*ArcTanh[(
b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(7/2))

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[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 (d+e x) \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac{e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac{(2 c d-b e) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{2 c}\\ &=\frac{(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac{e \left (a+b x+c x^2\right )^{5/2}}{5 c}-\frac{\left (3 \left (b^2-4 a c\right ) (2 c d-b e)\right ) \int \sqrt{a+b x+c x^2} \, dx}{32 c^2}\\ &=-\frac{3 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \sqrt{a+b x+c x^2}}{128 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac{e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac{\left (3 \left (b^2-4 a c\right )^2 (2 c d-b e)\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{256 c^3}\\ &=-\frac{3 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \sqrt{a+b x+c x^2}}{128 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac{e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac{\left (3 \left (b^2-4 a c\right )^2 (2 c d-b 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 )}{128 c^3}\\ &=-\frac{3 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \sqrt{a+b x+c x^2}}{128 c^3}+\frac{(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac{e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac{3 \left (b^2-4 a c\right )^2 (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.297728, size = 144, normalized size = 0.89 \[ \frac{5 (2 c d-b e) \left (\frac{3 \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )-2 \sqrt{c} (b+2 c x) \sqrt{a+x (b+c x)}\right )}{128 c^{5/2}}+\frac{(b+2 c x) (a+x (b+c x))^{3/2}}{8 c}\right )+2 e (a+x (b+c x))^{5/2}}{10 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(a + x*(b + c*x))^(5/2) + 5*(2*c*d - b*e)*(((b + 2*c*x)*(a + x*(b + c*x))^(3/2))/(8*c) + (3*(b^2 - 4*a*c)
*(-2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)] + (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b +
c*x)])]))/(128*c^(5/2))))/(10*c)

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Maple [B]  time = 0.045, size = 469, normalized size = 2.9 \begin{align*}{\frac{e}{5\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}}-{\frac{bxe}{8\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}e}{16\,{c}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,bxea}{16\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,e{b}^{3}x}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{b}^{2}ea}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,e{b}^{4}}{128\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{a}^{2}be}{16}\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}^{3}a}{32}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{3\,e{b}^{5}}{256}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}+{\frac{dx}{4} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{bd}{8\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,adx}{8}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,dx{b}^{2}}{32\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,abd}{16\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,d{b}^{3}}{64\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{a}^{2}d}{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{3\,a{b}^{2}d}{16}\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\,d{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.06053, size = 1215, normalized size = 7.55 \begin{align*} \left [-\frac{15 \,{\left (2 \,{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d -{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e\right )} \sqrt{c} \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 (128 \, c^{5} e x^{4} + 16 \,{\left (10 \, c^{5} d + 11 \, b c^{4} e\right )} x^{3} + 8 \,{\left (30 \, b c^{4} d +{\left (b^{2} c^{3} + 32 \, a c^{4}\right )} e\right )} x^{2} - 10 \,{\left (3 \, b^{3} c^{2} - 20 \, a b c^{3}\right )} d +{\left (15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3}\right )} e + 2 \,{\left (10 \,{\left (b^{2} c^{3} + 20 \, a c^{4}\right )} d -{\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} e\right )} x\right )} \sqrt{c x^{2} + b x + a}}{2560 \, c^{4}}, -\frac{15 \,{\left (2 \,{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d -{\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e\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 ) - 2 \,{\left (128 \, c^{5} e x^{4} + 16 \,{\left (10 \, c^{5} d + 11 \, b c^{4} e\right )} x^{3} + 8 \,{\left (30 \, b c^{4} d +{\left (b^{2} c^{3} + 32 \, a c^{4}\right )} e\right )} x^{2} - 10 \,{\left (3 \, b^{3} c^{2} - 20 \, a b c^{3}\right )} d +{\left (15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3}\right )} e + 2 \,{\left (10 \,{\left (b^{2} c^{3} + 20 \, a c^{4}\right )} d -{\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} e\right )} x\right )} \sqrt{c x^{2} + b x + a}}{1280 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2560*(15*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e)*sqrt(c)*log(-8*c^2*
x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(128*c^5*e*x^4 + 16*(10*c^5*d +
11*b*c^4*e)*x^3 + 8*(30*b*c^4*d + (b^2*c^3 + 32*a*c^4)*e)*x^2 - 10*(3*b^3*c^2 - 20*a*b*c^3)*d + (15*b^4*c - 1
00*a*b^2*c^2 + 128*a^2*c^3)*e + 2*(10*(b^2*c^3 + 20*a*c^4)*d - (5*b^3*c^2 - 28*a*b*c^3)*e)*x)*sqrt(c*x^2 + b*x
+ a))/c^4, -1/1280*(15*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e)*sqrt(-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)) - 2*(128*c^5*e*x^4 + 16*(10*c^
5*d + 11*b*c^4*e)*x^3 + 8*(30*b*c^4*d + (b^2*c^3 + 32*a*c^4)*e)*x^2 - 10*(3*b^3*c^2 - 20*a*b*c^3)*d + (15*b^4*
c - 100*a*b^2*c^2 + 128*a^2*c^3)*e + 2*(10*(b^2*c^3 + 20*a*c^4)*d - (5*b^3*c^2 - 28*a*b*c^3)*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 (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((e*x+d)*(c*x**2+b*x+a)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.12276, size = 356, normalized size = 2.21 \begin{align*} \frac{1}{640} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \, c x e + \frac{10 \, c^{5} d + 11 \, b c^{4} e}{c^{4}}\right )} x + \frac{30 \, b c^{4} d + b^{2} c^{3} e + 32 \, a c^{4} e}{c^{4}}\right )} x + \frac{10 \, b^{2} c^{3} d + 200 \, a c^{4} d - 5 \, b^{3} c^{2} e + 28 \, a b c^{3} e}{c^{4}}\right )} x - \frac{30 \, b^{3} c^{2} d - 200 \, a b c^{3} d - 15 \, b^{4} c e + 100 \, a b^{2} c^{2} e - 128 \, a^{2} c^{3} e}{c^{4}}\right )} - \frac{3 \,{\left (2 \, b^{4} c d - 16 \, a b^{2} c^{2} d + 32 \, a^{2} c^{3} d - b^{5} e + 8 \, a b^{3} c e - 16 \, a^{2} b c^{2} e\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/640*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*c*x*e + (10*c^5*d + 11*b*c^4*e)/c^4)*x + (30*b*c^4*d + b^2*c^3*e + 32*
a*c^4*e)/c^4)*x + (10*b^2*c^3*d + 200*a*c^4*d - 5*b^3*c^2*e + 28*a*b*c^3*e)/c^4)*x - (30*b^3*c^2*d - 200*a*b*c
^3*d - 15*b^4*c*e + 100*a*b^2*c^2*e - 128*a^2*c^3*e)/c^4) - 3/256*(2*b^4*c*d - 16*a*b^2*c^2*d + 32*a^2*c^3*d -
b^5*e + 8*a*b^3*c*e - 16*a^2*b*c^2*e)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)