3.2638 \(\int \frac {A+B x}{\sqrt {d+e x} (a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=528 \[ \frac {2 \sqrt {d+e x} \left (-A \left (2 a c e+b^2 (-e)+b c d\right )+c x (-2 a B e+A b e-2 A c d+b B d)+a B (2 c d-b e)\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (-2 a B e+A b e-2 A c d+b B d) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (b B-2 A c) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 528, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {822, 843, 718, 424, 419} \[ \frac {2 \sqrt {d+e x} \left (-A \left (2 a c e+b^2 (-e)+b c d\right )+c x (-2 a B e+A b e-2 A c d+b B d)+a B (2 c d-b e)\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (-2 a B e+A b e-2 A c d+b B d) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (b B-2 A c) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(a*B*(2*c*d - b*e) - A*(b*c*d - b^2*e + 2*a*c*e) + c*(b*B*d - 2*A*c*d + A*b*e - 2*a*B*e)*x))/
((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*Sqrt[a + b*x + c*x^2]) - (Sqrt[2]*(b*B*d - 2*A*c*d + A*b*e - 2*a*B*e)*S
qrt[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)])/(Sqrt[b^2 - 4*a*
c]*(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]) + (2
*Sqrt[2]*(b*B - 2*A*c)*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[-((c*(a + b*x + c*x^2))/(b
^2 - 4*a*c))]*EllipticF[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*Sqrt[b^2 - 4*a*c]*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

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])

Rule 843

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

Rubi steps

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

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Mathematica [C]  time = 9.00, size = 893, normalized size = 1.69 \[ \frac {2 \sqrt {d+e x} \left (-A e b^2+A c d b+a B e b-B c d x b-A c e x b-2 a B c d+2 a A c e+2 A c^2 d x+2 a B c e x\right ) \left (c x^2+b x+a\right )}{\left (4 a c-b^2\right ) \left (c d^2-b e d+a e^2\right ) (a+x (b+c x))^{3/2}}-\frac {2 (d+e x)^{3/2} \left (c x^2+b x+a\right )^{3/2} \left (\frac {i \sqrt {1-\frac {2 \left (c d^2+e (a e-b d)\right )}{\left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} \sqrt {\frac {2 \left (c d^2+e (a e-b d)\right )}{\left (-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}+1} \left ((b B d-2 A c d+A b e-2 a B e) \left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )-\left (e (B d-A e) b^2+\sqrt {\left (b^2-4 a c\right ) e^2} (B d+A e) b-2 a e \left (2 c d B+\sqrt {\left (b^2-4 a c\right ) e^2} B-2 A c e\right )-2 A c d \sqrt {\left (b^2-4 a c\right ) e^2}\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )\right )}{2 \sqrt {2} \sqrt {\frac {c d^2+e (a e-b d)}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}} \sqrt {d+e x}}-(b B d-2 A c d+A b e-2 a B e) \left (c \left (\frac {d}{d+e x}-1\right )^2+\frac {e \left (-\frac {d b}{d+e x}+b+\frac {a e}{d+e x}\right )}{d+e x}\right )\right )}{\left (4 a c-b^2\right ) e \left (c d^2-b e d+a e^2\right ) (a+x (b+c x))^{3/2} \sqrt {\frac {(d+e x)^2 \left (c \left (\frac {d}{d+e x}-1\right )^2+\frac {e \left (-\frac {d b}{d+e x}+b+\frac {a e}{d+e x}\right )}{d+e x}\right )}{e^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(A*b*c*d - 2*a*B*c*d - A*b^2*e + a*b*B*e + 2*a*A*c*e - b*B*c*d*x + 2*A*c^2*d*x - A*b*c*e*x +
2*a*B*c*e*x)*(a + b*x + c*x^2))/((-b^2 + 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(a + x*(b + c*x))^(3/2)) - (2*(d + e*x
)^(3/2)*(a + b*x + c*x^2)^(3/2)*(-((b*B*d - 2*A*c*d + A*b*e - 2*a*B*e)*(c*(-1 + d/(d + e*x))^2 + (e*(b - (b*d)
/(d + e*x) + (a*e)/(d + e*x)))/(d + e*x))) + ((I/2)*Sqrt[1 - (2*(c*d^2 + e*(-(b*d) + a*e)))/((2*c*d - b*e + Sq
rt[(b^2 - 4*a*c)*e^2])*(d + e*x))]*Sqrt[1 + (2*(c*d^2 + e*(-(b*d) + a*e)))/((-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)
*e^2])*(d + e*x))]*((b*B*d - 2*A*c*d + A*b*e - 2*a*B*e)*(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])*EllipticE[I*Ar
cSinh[(Sqrt[2]*Sqrt[(c*d^2 - b*d*e + a*e^2)/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])])/Sqrt[d + e*x]], -((-2*c
*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])/(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2]))] - (-2*A*c*d*Sqrt[(b^2 - 4*a*c)*e
^2] + b^2*e*(B*d - A*e) + b*Sqrt[(b^2 - 4*a*c)*e^2]*(B*d + A*e) - 2*a*e*(2*B*c*d - 2*A*c*e + B*Sqrt[(b^2 - 4*a
*c)*e^2]))*EllipticF[I*ArcSinh[(Sqrt[2]*Sqrt[(c*d^2 - b*d*e + a*e^2)/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])]
)/Sqrt[d + e*x]], -((-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])/(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2]))]))/(Sqrt
[2]*Sqrt[(c*d^2 + e*(-(b*d) + a*e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*Sqrt[d + e*x])))/((-b^2 + 4*a*c)
*e*(c*d^2 - b*d*e + a*e^2)*(a + x*(b + c*x))^(3/2)*Sqrt[((d + e*x)^2*(c*(-1 + d/(d + e*x))^2 + (e*(b - (b*d)/(
d + e*x) + (a*e)/(d + e*x)))/(d + e*x)))/e^2])

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (B x + A\right )} \sqrt {e x + d}}{c^{2} e x^{5} + {\left (c^{2} d + 2 \, b c e\right )} x^{4} + {\left (2 \, b c d + {\left (b^{2} + 2 \, a c\right )} e\right )} x^{3} + a^{2} d + {\left (2 \, a b e + {\left (b^{2} + 2 \, a c\right )} d\right )} x^{2} + {\left (2 \, a b d + a^{2} e\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + b*x + a)*(B*x + A)*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 \[ \int \frac {B x + A}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} \sqrt {e x + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.10, size = 5059, normalized size = 9.58 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)/((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 \[ \int \frac {A+B\,x}{\sqrt {d+e\,x}\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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