3.2387 \(\int \frac {1}{(d+e x) (a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=155 \[ \frac {e^2 \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{\left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {2 \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 155, 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 = {740, 12, 724, 206} \[ \frac {e^2 \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{\left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {2 \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

Rule 724

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

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}-\frac {2 \int -\frac {\left (b^2-4 a c\right ) e^2}{2 (d+e x) \sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}+\frac {e^2 \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{c d^2-b d e+a e^2}\\ &=-\frac {2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}-\frac {\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{c d^2-b d e+a e^2}\\ &=-\frac {2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}+\frac {e^2 \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{\left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 162, normalized size = 1.05 \[ \frac {\frac {e^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )}{\left (e (a e-b d)+c d^2\right )^{3/2}}+\frac {4 c (a e+c d x)-2 b^2 e+2 b c (d-e x)}{\sqrt {a+x (b+c x)} \left (e (a e-b d)+c d^2\right )}}{4 a c-b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 2.04, size = 1349, normalized size = 8.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((b^2*c - 4*a*c^2)*e^2*x^2 + (b^3 - 4*a*b*c)*e^2*x + (a*b^2 - 4*a^2*c)*e^2)*sqrt(c*d^2 - b*d*e + a*e^2)*
log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 4*sqrt(c*d^
2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 +
4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2)) - 4*(b*c^2*d^3 - 2*(b^2*c - a*c^2)*d^2*e + (b^3 - a*b*c)*d*e^2 - (a
*b^2 - 2*a^2*c)*e^3 + (2*c^3*d^3 - 3*b*c^2*d^2*e - a*b*c*e^3 + (b^2*c + 2*a*c^2)*d*e^2)*x)*sqrt(c*x^2 + b*x +
a))/((a*b^2*c^2 - 4*a^2*c^3)*d^4 - 2*(a*b^3*c - 4*a^2*b*c^2)*d^3*e + (a*b^4 - 2*a^2*b^2*c - 8*a^3*c^2)*d^2*e^2
- 2*(a^2*b^3 - 4*a^3*b*c)*d*e^3 + (a^3*b^2 - 4*a^4*c)*e^4 + ((b^2*c^3 - 4*a*c^4)*d^4 - 2*(b^3*c^2 - 4*a*b*c^3
)*d^3*e + (b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*d^2*e^2 - 2*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 + (a^2*b^2*c - 4*a^3*c^2
)*e^4)*x^2 + ((b^3*c^2 - 4*a*b*c^3)*d^4 - 2*(b^4*c - 4*a*b^2*c^2)*d^3*e + (b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*d^2*
e^2 - 2*(a*b^4 - 4*a^2*b^2*c)*d*e^3 + (a^2*b^3 - 4*a^3*b*c)*e^4)*x), (((b^2*c - 4*a*c^2)*e^2*x^2 + (b^3 - 4*a*
b*c)*e^2*x + (a*b^2 - 4*a^2*c)*e^2)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt
(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e^2)
*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) - 2*(b*c^2*d^3 - 2*(b^2*c - a*c^2)*d^2*e + (b^3 - a*b*c)*d*e^2 - (a*b
^2 - 2*a^2*c)*e^3 + (2*c^3*d^3 - 3*b*c^2*d^2*e - a*b*c*e^3 + (b^2*c + 2*a*c^2)*d*e^2)*x)*sqrt(c*x^2 + b*x + a)
)/((a*b^2*c^2 - 4*a^2*c^3)*d^4 - 2*(a*b^3*c - 4*a^2*b*c^2)*d^3*e + (a*b^4 - 2*a^2*b^2*c - 8*a^3*c^2)*d^2*e^2 -
2*(a^2*b^3 - 4*a^3*b*c)*d*e^3 + (a^3*b^2 - 4*a^4*c)*e^4 + ((b^2*c^3 - 4*a*c^4)*d^4 - 2*(b^3*c^2 - 4*a*b*c^3)*
d^3*e + (b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*d^2*e^2 - 2*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 + (a^2*b^2*c - 4*a^3*c^2)*
e^4)*x^2 + ((b^3*c^2 - 4*a*b*c^3)*d^4 - 2*(b^4*c - 4*a*b^2*c^2)*d^3*e + (b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*d^2*e^
2 - 2*(a*b^4 - 4*a^2*b^2*c)*d*e^3 + (a^2*b^3 - 4*a^3*b*c)*e^4)*x)]

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giac [B]  time = 0.27, size = 447, normalized size = 2.88 \[ -\frac {2 \, {\left (\frac {{\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + b^{2} c d e^{2} + 2 \, a c^{2} d e^{2} - a b c e^{3}\right )} x}{b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4} - 2 \, b^{3} c d^{3} e + 8 \, a b c^{2} d^{3} e + b^{4} d^{2} e^{2} - 2 \, a b^{2} c d^{2} e^{2} - 8 \, a^{2} c^{2} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 8 \, a^{2} b c d e^{3} + a^{2} b^{2} e^{4} - 4 \, a^{3} c e^{4}} + \frac {b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + 2 \, a c^{2} d^{2} e + b^{3} d e^{2} - a b c d e^{2} - a b^{2} e^{3} + 2 \, a^{2} c e^{3}}{b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4} - 2 \, b^{3} c d^{3} e + 8 \, a b c^{2} d^{3} e + b^{4} d^{2} e^{2} - 2 \, a b^{2} c d^{2} e^{2} - 8 \, a^{2} c^{2} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 8 \, a^{2} b c d e^{3} + a^{2} b^{2} e^{4} - 4 \, a^{3} c e^{4}}\right )}}{\sqrt {c x^{2} + b x + a}} + \frac {2 \, \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e - a e^{2}}}\right ) e^{2}}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-c d^{2} + b d e - a e^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((2*c^3*d^3 - 3*b*c^2*d^2*e + b^2*c*d*e^2 + 2*a*c^2*d*e^2 - a*b*c*e^3)*x/(b^2*c^2*d^4 - 4*a*c^3*d^4 - 2*b^3
*c*d^3*e + 8*a*b*c^2*d^3*e + b^4*d^2*e^2 - 2*a*b^2*c*d^2*e^2 - 8*a^2*c^2*d^2*e^2 - 2*a*b^3*d*e^3 + 8*a^2*b*c*d
*e^3 + a^2*b^2*e^4 - 4*a^3*c*e^4) + (b*c^2*d^3 - 2*b^2*c*d^2*e + 2*a*c^2*d^2*e + b^3*d*e^2 - a*b*c*d*e^2 - a*b
^2*e^3 + 2*a^2*c*e^3)/(b^2*c^2*d^4 - 4*a*c^3*d^4 - 2*b^3*c*d^3*e + 8*a*b*c^2*d^3*e + b^4*d^2*e^2 - 2*a*b^2*c*d
^2*e^2 - 8*a^2*c^2*d^2*e^2 - 2*a*b^3*d*e^3 + 8*a^2*b*c*d*e^3 + a^2*b^2*e^4 - 4*a^3*c*e^4))/sqrt(c*x^2 + b*x +
a) + 2*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*x + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e - a*e^2))*e^2/((c*d^2 -
b*d*e + a*e^2)*sqrt(-c*d^2 + b*d*e - a*e^2))

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maple [B]  time = 0.06, size = 603, normalized size = 3.89 \[ -\frac {2 b c e x}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (4 a c -b^{2}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}+\frac {4 c^{2} d x}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (4 a c -b^{2}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {b^{2} e}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (4 a c -b^{2}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}+\frac {2 b c d}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (4 a c -b^{2}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {e \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}+\frac {e}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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(a*e^2-b*d*e>0)', see `assume?`
for more details)Is a*e^2-b*d*e                            +c*d^2    positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x\right ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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