3.2374 $$\int \frac{(d+e x)^3}{\sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=174 $\frac{e \sqrt{a+b x+c x^2} \left (-2 c e (8 a e+27 b d)+15 b^2 e^2+10 c e x (2 c d-b e)+64 c^2 d^2\right )}{24 c^3}+\frac{(2 c d-b e) \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{7/2}}+\frac{e (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}$

[Out]

(e*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(3*c) + (e*(64*c^2*d^2 + 15*b^2*e^2 - 2*c*e*(27*b*d + 8*a*e) + 10*c*e*(2
*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(24*c^3) + ((2*c*d - b*e)*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e)
)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(7/2))

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Rubi [A]  time = 0.151003, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {742, 779, 621, 206} $\frac{e \sqrt{a+b x+c x^2} \left (-2 c e (8 a e+27 b d)+15 b^2 e^2+10 c e x (2 c d-b e)+64 c^2 d^2\right )}{24 c^3}+\frac{(2 c d-b e) \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{7/2}}+\frac{e (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(e*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(3*c) + (e*(64*c^2*d^2 + 15*b^2*e^2 - 2*c*e*(27*b*d + 8*a*e) + 10*c*e*(2
*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(24*c^3) + ((2*c*d - b*e)*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e)
)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(7/2))

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 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 \frac{(d+e x)^3}{\sqrt{a+b x+c x^2}} \, dx &=\frac{e (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{\int \frac{(d+e x) \left (\frac{1}{2} \left (6 c d^2-e (b d+4 a e)\right )+\frac{5}{2} e (2 c d-b e) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{3 c}\\ &=\frac{e (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{e \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)+10 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{24 c^3}+\frac{\left ((2 c d-b e) \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{16 c^3}\\ &=\frac{e (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{e \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)+10 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{24 c^3}+\frac{\left ((2 c d-b e) \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right )\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 )}{8 c^3}\\ &=\frac{e (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{e \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)+10 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{24 c^3}+\frac{(2 c d-b e) \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.224165, size = 210, normalized size = 1.21 $\frac{e \left (-16 a^2 c e^2+a \left (15 b^2 e^2-2 b c e (27 d+13 e x)+4 c^2 \left (18 d^2+9 d e x-2 e^2 x^2\right )\right )+x (b+c x) \left (15 b^2 e^2-2 b c e (27 d+5 e x)+4 c^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )\right )}{24 c^3 \sqrt{a+x (b+c x)}}+\frac{(2 c d-b e) \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{16 c^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(e*(-16*a^2*c*e^2 + a*(15*b^2*e^2 - 2*b*c*e*(27*d + 13*e*x) + 4*c^2*(18*d^2 + 9*d*e*x - 2*e^2*x^2)) + x*(b + c
*x)*(15*b^2*e^2 - 2*b*c*e*(27*d + 5*e*x) + 4*c^2*(18*d^2 + 9*d*e*x + 2*e^2*x^2))))/(24*c^3*Sqrt[a + x*(b + c*x
)]) + ((2*c*d - b*e)*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x
*(b + c*x)])])/(16*c^(7/2))

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Maple [B]  time = 0.05, size = 366, normalized size = 2.1 \begin{align*}{\frac{{e}^{3}{x}^{2}}{3\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,b{e}^{3}x}{12\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,{b}^{2}{e}^{3}}{8\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,{b}^{3}{e}^{3}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}+{\frac{3\,ab{e}^{3}}{4}\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\,a{e}^{3}}{3\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,d{e}^{2}x}{2\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{9\,d{e}^{2}b}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{9\,d{e}^{2}{b}^{2}}{8}\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\,ad{e}^{2}}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+3\,{\frac{{d}^{2}e\sqrt{c{x}^{2}+bx+a}}{c}}-{\frac{3\,{d}^{2}eb}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{{d}^{3}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*e^3*x^2/c*(c*x^2+b*x+a)^(1/2)-5/12*e^3*b/c^2*x*(c*x^2+b*x+a)^(1/2)+5/8*e^3*b^2/c^3*(c*x^2+b*x+a)^(1/2)-5/1
6*e^3*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/4*e^3*b/c^(5/2)*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^
2+b*x+a)^(1/2))-2/3*e^3*a/c^2*(c*x^2+b*x+a)^(1/2)+3/2*d*e^2*x/c*(c*x^2+b*x+a)^(1/2)-9/4*d*e^2*b/c^2*(c*x^2+b*x
+a)^(1/2)+9/8*d*e^2*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3/2*d*e^2*a/c^(3/2)*ln((1/2*b+c*x)
/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3*d^2*e/c*(c*x^2+b*x+a)^(1/2)-3/2*d^2*e*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+
b*x+a)^(1/2))+d^3*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)

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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+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.60892, size = 890, normalized size = 5.11 \begin{align*} \left [\frac{3 \,{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \,{\left (3 \, b^{2} c - 4 \, a c^{2}\right )} d e^{2} -{\left (5 \, b^{3} - 12 \, a b c\right )} e^{3}\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 (8 \, c^{3} e^{3} x^{2} + 72 \, c^{3} d^{2} e - 54 \, b c^{2} d e^{2} +{\left (15 \, b^{2} c - 16 \, a c^{2}\right )} e^{3} + 2 \,{\left (18 \, c^{3} d e^{2} - 5 \, b c^{2} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{96 \, c^{4}}, -\frac{3 \,{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \,{\left (3 \, b^{2} c - 4 \, a c^{2}\right )} d e^{2} -{\left (5 \, b^{3} - 12 \, a b c\right )} e^{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 ) - 2 \,{\left (8 \, c^{3} e^{3} x^{2} + 72 \, c^{3} d^{2} e - 54 \, b c^{2} d e^{2} +{\left (15 \, b^{2} c - 16 \, a c^{2}\right )} e^{3} + 2 \,{\left (18 \, c^{3} d e^{2} - 5 \, b c^{2} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{48 \, c^{4}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*(3*b^2*c - 4*a*c^2)*d*e^2 - (5*b^3 - 12*a*b*c)*e^3)*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*(8*c^3*e^3*x^2 + 72*c^3*d^2*
e - 54*b*c^2*d*e^2 + (15*b^2*c - 16*a*c^2)*e^3 + 2*(18*c^3*d*e^2 - 5*b*c^2*e^3)*x)*sqrt(c*x^2 + b*x + a))/c^4,
-1/48*(3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 6*(3*b^2*c - 4*a*c^2)*d*e^2 - (5*b^3 - 12*a*b*c)*e^3)*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*(8*c^3*e^3*x^2 + 72*c^3*d^2*e - 5
4*b*c^2*d*e^2 + (15*b^2*c - 16*a*c^2)*e^3 + 2*(18*c^3*d*e^2 - 5*b*c^2*e^3)*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 \frac{\left (d + e x\right )^{3}}{\sqrt{a + b x + c x^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+b*x+a)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.13047, size = 230, normalized size = 1.32 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + b x + a}{\left (2 \, x{\left (\frac{4 \, x e^{3}}{c} + \frac{18 \, c^{2} d e^{2} - 5 \, b c e^{3}}{c^{3}}\right )} + \frac{72 \, c^{2} d^{2} e - 54 \, b c d e^{2} + 15 \, b^{2} e^{3} - 16 \, a c e^{3}}{c^{3}}\right )} - \frac{{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 18 \, b^{2} c d e^{2} - 24 \, a c^{2} d e^{2} - 5 \, b^{3} e^{3} + 12 \, a b c e^{3}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{16 \, c^{\frac{7}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + b*x + a)*(2*x*(4*x*e^3/c + (18*c^2*d*e^2 - 5*b*c*e^3)/c^3) + (72*c^2*d^2*e - 54*b*c*d*e^2 +
15*b^2*e^3 - 16*a*c*e^3)/c^3) - 1/16*(16*c^3*d^3 - 24*b*c^2*d^2*e + 18*b^2*c*d*e^2 - 24*a*c^2*d*e^2 - 5*b^3*e^
3 + 12*a*b*c*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)