3.2630 \(\int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=545 \[ -\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} (5 A c e-4 b B e+3 B c d) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{15 c^3 e \sqrt {d+e x} \sqrt {a+b x+c x^2}}+\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (B \left (-c e (9 a e+13 b d)+8 b^2 e^2+3 c^2 d^2\right )+10 A c e (2 c d-b e)\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{15 c^3 e \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {2 \sqrt {d+e x} \sqrt {a+b x+c x^2} (5 A c e-4 b B e+3 B c d)}{15 c^2}+\frac {2 B (d+e x)^{3/2} \sqrt {a+b x+c x^2}}{5 c} \]

[Out]

2/5*B*(e*x+d)^(3/2)*(c*x^2+b*x+a)^(1/2)/c+2/15*(5*A*c*e-4*B*b*e+3*B*c*d)*(e*x+d)^(1/2)*(c*x^2+b*x+a)^(1/2)/c^2
+1/15*(10*A*c*e*(-b*e+2*c*d)+B*(3*c^2*d^2+8*b^2*e^2-c*e*(9*a*e+13*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)*(-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)/c^3/e/(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)-2/15*(5*A*c*e-4*B*b*e+3*B*c*d)*(a*e^2-b*d*e+c*d^2)*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)*(-4*a*c+b^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^3/e/(e*x+d)^(1/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.78, antiderivative size = 545, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {832, 843, 718, 424, 419} \[ -\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} (5 A c e-4 b B e+3 B c d) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{15 c^3 e \sqrt {d+e x} \sqrt {a+b x+c x^2}}+\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (B \left (-c e (9 a e+13 b d)+8 b^2 e^2+3 c^2 d^2\right )+10 A c e (2 c d-b e)\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{15 c^3 e \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {2 \sqrt {d+e x} \sqrt {a+b x+c x^2} (5 A c e-4 b B e+3 B c d)}{15 c^2}+\frac {2 B (d+e x)^{3/2} \sqrt {a+b x+c x^2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*B*c*d - 4*b*B*e + 5*A*c*e)*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])/(15*c^2) + (2*B*(d + e*x)^(3/2)*Sqrt[a +
b*x + c*x^2])/(5*c) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(10*A*c*e*(2*c*d - b*e) + B*(3*c^2*d^2 + 8*b^2*e^2 - c*e*(13
*b*d + 9*a*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)])
/(15*c^3*e*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]*Sqrt[b^
2 - 4*a*c]*(3*B*c*d - 4*b*B*e + 5*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[-((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)])/(15*c^3*e*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 832

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

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) (d+e x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx &=\frac {2 B (d+e x)^{3/2} \sqrt {a+b x+c x^2}}{5 c}+\frac {2 \int \frac {\sqrt {d+e x} \left (\frac {1}{2} (-b B d+5 A c d-3 a B e)+\frac {1}{2} (3 B c d-4 b B e+5 A c e) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{5 c}\\ &=\frac {2 (3 B c d-4 b B e+5 A c e) \sqrt {d+e x} \sqrt {a+b x+c x^2}}{15 c^2}+\frac {2 B (d+e x)^{3/2} \sqrt {a+b x+c x^2}}{5 c}+\frac {4 \int \frac {\frac {1}{4} \left (4 b^2 B d e+c \left (15 A c d^2-12 a B d e-5 a A e^2\right )-b \left (6 B c d^2+5 A c d e-4 a B e^2\right )\right )+\frac {1}{4} \left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2+8 b^2 e^2-c e (13 b d+9 a e)\right )\right ) x}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{15 c^2}\\ &=\frac {2 (3 B c d-4 b B e+5 A c e) \sqrt {d+e x} \sqrt {a+b x+c x^2}}{15 c^2}+\frac {2 B (d+e x)^{3/2} \sqrt {a+b x+c x^2}}{5 c}-\frac {\left ((3 B c d-4 b B e+5 A c e) \left (c d^2-b d e+a e^2\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{15 c^2 e}+\frac {\left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2+8 b^2 e^2-c e (13 b d+9 a e)\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx}{15 c^2 e}\\ &=\frac {2 (3 B c d-4 b B e+5 A c e) \sqrt {d+e x} \sqrt {a+b x+c x^2}}{15 c^2}+\frac {2 B (d+e x)^{3/2} \sqrt {a+b x+c x^2}}{5 c}+\frac {\left (\sqrt {2} \sqrt {b^2-4 a c} \left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2+8 b^2 e^2-c e (13 b d+9 a e)\right )\right ) \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 )}{15 c^3 e \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} \sqrt {b^2-4 a c} (3 B c d-4 b B e+5 A c e) \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 {-\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 )}{15 c^3 e \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=\frac {2 (3 B c d-4 b B e+5 A c e) \sqrt {d+e x} \sqrt {a+b x+c x^2}}{15 c^2}+\frac {2 B (d+e x)^{3/2} \sqrt {a+b x+c x^2}}{5 c}+\frac {\sqrt {2} \sqrt {b^2-4 a c} \left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2+8 b^2 e^2-c e (13 b d+9 a e)\right )\right ) \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 )}{15 c^3 e \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} \sqrt {b^2-4 a c} (3 B c d-4 b B e+5 A c e) \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 {-\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 )}{15 c^3 e \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 10.58, size = 978, normalized size = 1.79 \[ \frac {2 \sqrt {c x^2+b x+a} \left (\left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2+8 b^2 e^2-c e (13 b d+9 a e)\right )\right ) \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 )-\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 (\left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) \left (10 A c e (2 c d-b e)+B \left (3 c^2 d^2+8 b^2 e^2-c e (13 b d+9 a e)\right )\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 (8 b^3 B e^3-b^2 \left (21 c d B+8 \sqrt {\left (b^2-4 a c\right ) e^2} B+10 A c e\right ) e^2+b c \left (10 A e \left (3 c d+\sqrt {\left (b^2-4 a c\right ) e^2}\right )+B \left (15 c d^2+13 \sqrt {\left (b^2-4 a c\right ) e^2} d-17 a e^2\right )\right ) e+c \left (-3 B c \sqrt {\left (b^2-4 a c\right ) e^2} d^2+3 a B e^2 \left (8 c d+3 \sqrt {\left (b^2-4 a c\right ) e^2}\right )-10 A c e \left (3 c d^2+2 \sqrt {\left (b^2-4 a c\right ) e^2} d-a e^2\right )\right )\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}}\right ) (d+e x)^{3/2}}{15 c^3 e^2 \sqrt {a+x (b+c x)} \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}}}+\frac {\left (\frac {2 (6 B c d-4 b B e+5 A c e)}{15 c^2}+\frac {2 B e x}{5 c}\right ) \left (c x^2+b x+a\right ) \sqrt {d+e x}}{\sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*((2*(6*B*c*d - 4*b*B*e + 5*A*c*e))/(15*c^2) + (2*B*e*x)/(5*c))*(a + b*x + c*x^2))/Sqrt[a + x*(b
+ c*x)] + (2*(d + e*x)^(3/2)*Sqrt[a + b*x + c*x^2]*((10*A*c*e*(2*c*d - b*e) + B*(3*c^2*d^2 + 8*b^2*e^2 - c*e*
(13*b*d + 9*a*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)*S
qrt[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))]*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))]*((2*c*d - b*e + Sqrt[(b^2 - 4*a*c)
*e^2])*(10*A*c*e*(2*c*d - b*e) + B*(3*c^2*d^2 + 8*b^2*e^2 - c*e*(13*b*d + 9*a*e)))*EllipticE[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 + Sq
rt[(b^2 - 4*a*c)*e^2])/(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2]))] + (8*b^3*B*e^3 - b^2*e^2*(21*B*c*d + 10*A*c*e
+ 8*B*Sqrt[(b^2 - 4*a*c)*e^2]) + c*(-3*B*c*d^2*Sqrt[(b^2 - 4*a*c)*e^2] + 3*a*B*e^2*(8*c*d + 3*Sqrt[(b^2 - 4*a
*c)*e^2]) - 10*A*c*e*(3*c*d^2 - a*e^2 + 2*d*Sqrt[(b^2 - 4*a*c)*e^2])) + b*c*e*(10*A*e*(3*c*d + Sqrt[(b^2 - 4*a
*c)*e^2]) + B*(15*c*d^2 - 17*a*e^2 + 13*d*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])))/(15*c^3*e^2*Sqrt[a + x*(b + c*x)]*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 = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e x^{2} + A d + {\left (B d + A e\right )} x\right )} \sqrt {e x + d}}{\sqrt {c x^{2} + b x + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.15, size = 7523, normalized size = 13.80 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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