3.2449 \(\int \frac {(a+b x+c x^2)^{3/2}}{\sqrt {d+e x}} \, dx\)

Optimal. Leaf size=579 \[ \frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \left (-4 c e (4 b d-5 a e)-b^2 e^2+16 c^2 d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{35 c^2 e^4 \sqrt {d+e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{35 c^2 e^4 \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {2 \sqrt {d+e x} \sqrt {a+b x+c x^2} \left (-c e (11 b d-10 a e)+b^2 e^2-3 c e x (2 c d-b e)+8 c^2 d^2\right )}{35 c e^3}+\frac {2 \sqrt {d+e x} \left (a+b x+c x^2\right )^{3/2}}{7 e} \]

[Out]

2/7*(c*x^2+b*x+a)^(3/2)*(e*x+d)^(1/2)/e+2/35*(8*c^2*d^2+b^2*e^2-c*e*(-10*a*e+11*b*d)-3*c*e*(-b*e+2*c*d)*x)*(e*
x+d)^(1/2)*(c*x^2+b*x+a)^(1/2)/c/e^3-2/35*(-b*e+2*c*d)*(4*c^2*d^2-b^2*e^2-4*c*e*(-2*a*e+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^2/e^4/(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)+2/35*(a*e^2-b*d*e+c*d^2)*(16*c^2*d^2-b^2*e^2
-4*c*e*(-5*a*e+4*b*d))*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^2/e^4/(e*x+d)^(1/2)/(c*x^2+b*x+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.67, antiderivative size = 579, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {734, 814, 843, 718, 424, 419} \[ \frac {2 \sqrt {d+e x} \sqrt {a+b x+c x^2} \left (-c e (11 b d-10 a e)+b^2 e^2-3 c e x (2 c d-b e)+8 c^2 d^2\right )}{35 c e^3}+\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \left (-4 c e (4 b d-5 a e)-b^2 e^2+16 c^2 d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{35 c^2 e^4 \sqrt {d+e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{35 c^2 e^4 \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {2 \sqrt {d+e x} \left (a+b x+c x^2\right )^{3/2}}{7 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(8*c^2*d^2 + b^2*e^2 - c*e*(11*b*d - 10*a*e) - 3*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/
(35*c*e^3) + (2*Sqrt[d + e*x]*(a + b*x + c*x^2)^(3/2))/(7*e) - (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(2*c*d - b*e)*(4*c
^2*d^2 - b^2*e^2 - 4*c*e*(b*d - 2*a*e))*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[A
rcSin[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)])/(35*c^2*e^4*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]*(c*d^2 - b*d*e + a*e^2)*(16*c^2*d^2 - b^2*e^2 - 4*c*e*(4*b*d - 5*a*e))*Sq
rt[(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[A
rcSin[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)])/(35*c^2*e^4*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 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (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 {\left (a+b x+c x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx &=\frac {2 \sqrt {d+e x} \left (a+b x+c x^2\right )^{3/2}}{7 e}-\frac {3 \int \frac {(b d-2 a e+(2 c d-b e) x) \sqrt {a+b x+c x^2}}{\sqrt {d+e x}} \, dx}{7 e}\\ &=\frac {2 \sqrt {d+e x} \left (8 c^2 d^2+b^2 e^2-c e (11 b d-10 a e)-3 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{35 c e^3}+\frac {2 \sqrt {d+e x} \left (a+b x+c x^2\right )^{3/2}}{7 e}+\frac {2 \int \frac {\frac {1}{2} \left (5 c e (b d-2 a e)^2-2 (2 c d-b e) \left (\frac {1}{2} b d (4 c d-b e)-a e \left (c d+\frac {b e}{2}\right )\right )\right )-(2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) x}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{35 c e^3}\\ &=\frac {2 \sqrt {d+e x} \left (8 c^2 d^2+b^2 e^2-c e (11 b d-10 a e)-3 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{35 c e^3}+\frac {2 \sqrt {d+e x} \left (a+b x+c x^2\right )^{3/2}}{7 e}-\frac {\left (2 (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx}{35 c e^4}+\frac {\left (2 \left (d (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right )+\frac {1}{2} e \left (5 c e (b d-2 a e)^2-2 (2 c d-b e) \left (\frac {1}{2} b d (4 c d-b e)-a e \left (c d+\frac {b e}{2}\right )\right )\right )\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{35 c e^4}\\ &=\frac {2 \sqrt {d+e x} \left (8 c^2 d^2+b^2 e^2-c e (11 b d-10 a e)-3 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{35 c e^3}+\frac {2 \sqrt {d+e x} \left (a+b x+c x^2\right )^{3/2}}{7 e}-\frac {\left (2 \sqrt {2} \sqrt {b^2-4 a c} (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\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 )}{35 c^2 e^4 \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 (4 \sqrt {2} \sqrt {b^2-4 a c} \left (d (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right )+\frac {1}{2} e \left (5 c e (b d-2 a e)^2-2 (2 c d-b e) \left (\frac {1}{2} b d (4 c d-b e)-a e \left (c d+\frac {b e}{2}\right )\right )\right )\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 )}{35 c^2 e^4 \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=\frac {2 \sqrt {d+e x} \left (8 c^2 d^2+b^2 e^2-c e (11 b d-10 a e)-3 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{35 c e^3}+\frac {2 \sqrt {d+e x} \left (a+b x+c x^2\right )^{3/2}}{7 e}-\frac {2 \sqrt {2} \sqrt {b^2-4 a c} (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\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 )}{35 c^2 e^4 \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} \left (c d^2-b d e+a e^2\right ) \left (16 c^2 d^2-16 b c d e-b^2 e^2+20 a c 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 )}{35 c^2 e^4 \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 13.29, size = 5338, normalized size = 9.22 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

________________________________________________________________________________________

fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{\sqrt {e x + d}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.12, size = 6516, normalized size = 11.25 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________