3.1647 \(\int \frac {b+2 c x}{\sqrt {d+e x} (a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=290 \[ -\frac {2 \sqrt {d+e x} \left (\left (b^2-4 a c\right ) (c d-b e)-c e x \left (b^2-4 a c\right )\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {2} e \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {822, 21, 718, 424} \[ -\frac {2 \sqrt {d+e x} \left (\left (b^2-4 a c\right ) (c d-b e)-c e x \left (b^2-4 a c\right )\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {2} e \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x]*((b^2 - 4*a*c)*(c*d - b*e) - c*(b^2 - 4*a*c)*e*x))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*Sq
rt[a + b*x + c*x^2]) - (Sqrt[2]*Sqrt[b^2 - 4*a*c]*e*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]
*EllipticE[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*e)/(
2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/((c*d^2 - b*d*e + a*e^2)*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*
c])*e)]*Sqrt[a + b*x + c*x^2])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 718

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

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*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*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

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

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Mathematica [C]  time = 1.10, size = 405, normalized size = 1.40 \[ \frac {4 \sqrt {d+e x} (b e-c d+c e x)-\frac {i \sqrt {2} \left (e \left (\sqrt {b^2-4 a c}-b\right )+2 c d\right ) \sqrt {\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{e \left (\sqrt {b^2-4 a c}+b\right )-2 c d}} \sqrt {1-\frac {2 c (d+e x)}{e \left (\sqrt {b^2-4 a c}-b\right )+2 c d}} \left (E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{\left (b+\sqrt {b^2-4 a c}\right ) e-2 c d}} \sqrt {d+e x}\right )|\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c d+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )-F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{\left (b+\sqrt {b^2-4 a c}\right ) e-2 c d}} \sqrt {d+e x}\right )|\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c d+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )\right )}{\sqrt {\frac {c}{e \left (\sqrt {b^2-4 a c}+b\right )-2 c d}}}}{2 \sqrt {a+x (b+c x)} \left (e (a e-b d)+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[d + e*x]*(-(c*d) + b*e + c*e*x) - (I*Sqrt[2]*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*Sqrt[(e*(b + Sqrt[b^
2 - 4*a*c] + 2*c*x))/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[1 - (2*c*(d + e*x))/(2*c*d + (-b + Sqrt[b^2 -
4*a*c])*e)]*(EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[d + e*x]], (2*c*d -
(b + Sqrt[b^2 - 4*a*c])*e)/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)] - EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(-2*c*d
+ (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[d + e*x]], (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*d + (-b + Sqrt[b^2 - 4
*a*c])*e)]))/Sqrt[c/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)])/(2*(c*d^2 + e*(-(b*d) + a*e))*Sqrt[a + x*(b + c*x)]
)

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {e x + d}}{c^{2} e x^{5} + {\left (c^{2} d + 2 \, b c e\right )} x^{4} + {\left (2 \, b c d + {\left (b^{2} + 2 \, a c\right )} e\right )} x^{3} + a^{2} d + {\left (2 \, a b e + {\left (b^{2} + 2 \, a c\right )} d\right )} x^{2} + {\left (2 \, a b d + a^{2} e\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(e*x + d)/(c^2*e*x^5 + (c^2*d + 2*b*c*e)*x^4 + (2*b*c*d + (b^2
+ 2*a*c)*e)*x^3 + a^2*d + (2*a*b*e + (b^2 + 2*a*c)*d)*x^2 + (2*a*b*d + a^2*e)*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.14, size = 1366, normalized size = 4.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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