3.24.74 \(\int \frac {(d+e x)^3}{\sqrt {a+b x+c x^2}} \, dx\) [2374]

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

[Out]

1/16*(-b*e+2*c*d)*(8*c^2*d^2+5*b^2*e^2-4*c*e*(3*a*e+2*b*d))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))
/c^(7/2)+1/3*e*(e*x+d)^2*(c*x^2+b*x+a)^(1/2)/c+1/24*e*(64*c^2*d^2+15*b^2*e^2-2*c*e*(8*a*e+27*b*d)+10*c*e*(-b*e
+2*c*d)*x)*(c*x^2+b*x+a)^(1/2)/c^3

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 174, normalized size of antiderivative = 1.00, 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 = {756, 793, 635, 212} \begin {gather*} \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 \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 {e (d+e x)^2 \sqrt {a+b x+c x^2}}{3 c} \end {gather*}

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 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 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 756

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 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 \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 ) \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 )}{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 = 0.53, size = 150, normalized size = 0.86 \begin {gather*} \frac {2 \sqrt {c} e \sqrt {a+x (b+c x)} \left (15 b^2 e^2-2 c e (27 b d+8 a e+5 b e x)+4 c^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )-3 (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 ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{48 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*e*Sqrt[a + x*(b + c*x)]*(15*b^2*e^2 - 2*c*e*(27*b*d + 8*a*e + 5*b*e*x) + 4*c^2*(18*d^2 + 9*d*e*x +
2*e^2*x^2)) - 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))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a
+ x*(b + c*x)]])/(48*c^(7/2))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(386\) vs. \(2(156)=312\).
time = 0.82, size = 387, normalized size = 2.22

method	result	size
risch	\(-\frac {e \left (-8 c^{2} e^{2} x^{2}+10 b c \,e^{2} x -36 c^{2} d e x +16 a c \,e^{2}-15 b^{2} e^{2}+54 b c d e -72 c^{2} d^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{24 c^{3}}+\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a b \,e^{3}}{4 c^{\frac {5}{2}}}-\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) d \,e^{2} a}{2 c^{\frac {3}{2}}}-\frac {5 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{3} e^{3}}{16 c^{\frac {7}{2}}}+\frac {9 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{2} d \,e^{2}}{8 c^{\frac {5}{2}}}-\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b \,d^{2} e}{2 c^{\frac {3}{2}}}+\frac {d^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}\)	\(287\)
default	\(e^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )+3 d \,e^{2} \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+3 d^{2} e \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+\frac {d^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}\)	\(387\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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

________________________________________________________________________________________

Fricas [A]
time = 3.05, size = 371, normalized size = 2.13 \begin {gather*} \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 (72 \, c^{3} d^{2} e + {\left (8 \, c^{3} x^{2} - 10 \, b c^{2} x + 15 \, b^{2} c - 16 \, a c^{2}\right )} e^{3} + 18 \, {\left (2 \, c^{3} d x - 3 \, b c^{2} d\right )} e^{2}\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 (72 \, c^{3} d^{2} e + {\left (8 \, c^{3} x^{2} - 10 \, b c^{2} x + 15 \, b^{2} c - 16 \, a c^{2}\right )} e^{3} + 18 \, {\left (2 \, c^{3} d x - 3 \, b c^{2} d\right )} e^{2}\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{4}}\right ] \end {gather*}

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*(72*c^3*d^2*e + (8*c^3*x^2 -
10*b*c^2*x + 15*b^2*c - 16*a*c^2)*e^3 + 18*(2*c^3*d*x - 3*b*c^2*d)*e^2)*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*(72*c^3*d^2*e + (8*c^3*x^2 - 10*b*c^2*x + 1
5*b^2*c - 16*a*c^2)*e^3 + 18*(2*c^3*d*x - 3*b*c^2*d)*e^2)*sqrt(c*x^2 + b*x + a))/c^4]

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [A]
time = 1.25, size = 170, normalized size = 0.98 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________