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

Optimal. Leaf size=631 \[ \frac {\sqrt {2} e \sqrt {b^2-4 a c} \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (c e g (-9 a e g-30 b d g+7 b e f)+8 b^2 e^2 g^2+c^2 \left (45 d^2 g^2-30 d e f g+8 e^2 f^2\right )\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^3 g^3 \sqrt {a+b x+c x^2} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} \left (-c e^2 g (a g (7 e f-15 d g)-3 b f (e f-5 d g))+4 b e^3 g^2 (b f-a g)+c^2 \left (-15 d^3 g^3+45 d^2 e f g^2-30 d e^2 f^2 g+8 e^3 f^3\right )\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^3 g^3 \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {8 e^2 \sqrt {f+g x} \sqrt {a+b x+c x^2} (b e g-3 c d g+c e f)}{15 c^2 g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{5 c g} \]

[Out]

-8/15*e^2*(b*e*g-3*c*d*g+c*e*f)*(g*x+f)^(1/2)*(c*x^2+b*x+a)^(1/2)/c^2/g^2+2/5*e^2*(e*x+d)*(g*x+f)^(1/2)*(c*x^2
+b*x+a)^(1/2)/c/g+1/15*e*(8*b^2*e^2*g^2+c*e*g*(-9*a*e*g-30*b*d*g+7*b*e*f)+c^2*(45*d^2*g^2-30*d*e*f*g+8*e^2*f^2
))*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*g*(-4*a*c+b^2)^(1/2)/(2*c
*f-g*(b+(-4*a*c+b^2)^(1/2))))^(1/2))*2^(1/2)*(-4*a*c+b^2)^(1/2)*(g*x+f)^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^
(1/2)/c^3/g^3/(c*x^2+b*x+a)^(1/2)/(c*(g*x+f)/(2*c*f-g*(b+(-4*a*c+b^2)^(1/2))))^(1/2)-2/15*(4*b*e^3*g^2*(-a*g+b
*f)+c^2*(-15*d^3*g^3+45*d^2*e*f*g^2-30*d*e^2*f^2*g+8*e^3*f^3)-c*e^2*g*(a*g*(-15*d*g+7*e*f)-3*b*f*(-5*d*g+e*f))
)*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*g*(-4*a*c+b^2)^(1/2)/(2*c*
f-g*(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*(g*x+f
)/(2*c*f-g*(b+(-4*a*c+b^2)^(1/2))))^(1/2)/c^3/g^3/(g*x+f)^(1/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 1.09, antiderivative size = 631, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {930, 1653, 843, 718, 424, 419} \[ -\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} \left (-c e^2 g (a g (7 e f-15 d g)-3 b f (e f-5 d g))+4 b e^3 g^2 (b f-a g)+c^2 \left (45 d^2 e f g^2-15 d^3 g^3-30 d e^2 f^2 g+8 e^3 f^3\right )\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^3 g^3 \sqrt {f+g x} \sqrt {a+b x+c x^2}}+\frac {\sqrt {2} e \sqrt {b^2-4 a c} \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (c e g (-9 a e g-30 b d g+7 b e f)+8 b^2 e^2 g^2+c^2 \left (45 d^2 g^2-30 d e f g+8 e^2 f^2\right )\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^3 g^3 \sqrt {a+b x+c x^2} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {8 e^2 \sqrt {f+g x} \sqrt {a+b x+c x^2} (b e g-3 c d g+c e f)}{15 c^2 g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{5 c g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*e^2*(c*e*f - 3*c*d*g + b*e*g)*Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2])/(15*c^2*g^2) + (2*e^2*(d + e*x)*Sqrt[f
+ g*x]*Sqrt[a + b*x + c*x^2])/(5*c*g) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*e*(8*b^2*e^2*g^2 + c*e*g*(7*b*e*f - 30*b*d*
g - 9*a*e*g) + c^2*(8*e^2*f^2 - 30*d*e*f*g + 45*d^2*g^2))*Sqrt[f + g*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
]*g)/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)])/(15*c^3*g^3*Sqrt[(c*(f + g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]
*Sqrt[a + b*x + c*x^2]) - (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(4*b*e^3*g^2*(b*f - a*g) + c^2*(8*e^3*f^3 - 30*d*e^2*f^
2*g + 45*d^2*e*f*g^2 - 15*d^3*g^3) - c*e^2*g*(a*g*(7*e*f - 15*d*g) - 3*b*f*(e*f - 5*d*g)))*Sqrt[(c*(f + g*x))/
(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqrt[(b + S
qrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*g)/(2*c*f - (b + Sqrt[b^2 - 4*a*c
])*g)])/(15*c^3*g^3*Sqrt[f + g*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 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]

Rule 930

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

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx &=\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{5 c g}-\frac {\int \frac {b d e^2 f-5 c d^3 g+a e^2 (2 e f+d g)+e (c d (2 e f-15 d g)+e (3 b e f+2 b d g+3 a e g)) x+4 e^2 (c e f-3 c d g+b e g) x^2}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{5 c g}\\ &=-\frac {8 e^2 (c e f-3 c d g+b e g) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{15 c^2 g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{5 c g}-\frac {2 \int \frac {-\frac {1}{2} g \left (4 b^2 e^3 f g+b e^2 \left (4 a e g^2+c f (4 e f-15 d g)\right )+c g \left (15 c d^3 g-a e^2 (2 e f+15 d g)\right )\right )-\frac {1}{2} e g \left (8 b^2 e^2 g^2+c e g (7 b e f-30 b d g-9 a e g)+c^2 \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right ) x}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{15 c^2 g^3}\\ &=-\frac {8 e^2 (c e f-3 c d g+b e g) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{15 c^2 g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{5 c g}+\frac {\left (e \left (8 b^2 e^2 g^2+c e g (7 b e f-30 b d g-9 a e g)+c^2 \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right )\right ) \int \frac {\sqrt {f+g x}}{\sqrt {a+b x+c x^2}} \, dx}{15 c^2 g^3}-\frac {\left (4 b e^3 g^2 (b f-a g)+c^2 \left (8 e^3 f^3-30 d e^2 f^2 g+45 d^2 e f g^2-15 d^3 g^3\right )-c e^2 g (a g (7 e f-15 d g)-3 b f (e f-5 d g))\right ) \int \frac {1}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{15 c^2 g^3}\\ &=-\frac {8 e^2 (c e f-3 c d g+b e g) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{15 c^2 g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{5 c g}+\frac {\left (\sqrt {2} \sqrt {b^2-4 a c} e \left (8 b^2 e^2 g^2+c e g (7 b e f-30 b d g-9 a e g)+c^2 \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right ) \sqrt {f+g 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} g x^2}{2 c f-b g-\sqrt {b^2-4 a c} g}}}{\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 g^3 \sqrt {\frac {c (f+g x)}{2 c f-b g-\sqrt {b^2-4 a c} g}} \sqrt {a+b x+c x^2}}-\frac {\left (2 \sqrt {2} \sqrt {b^2-4 a c} \left (4 b e^3 g^2 (b f-a g)+c^2 \left (8 e^3 f^3-30 d e^2 f^2 g+45 d^2 e f g^2-15 d^3 g^3\right )-c e^2 g (a g (7 e f-15 d g)-3 b f (e f-5 d g))\right ) \sqrt {\frac {c (f+g x)}{2 c f-b g-\sqrt {b^2-4 a c} g}} \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} g x^2}{2 c f-b g-\sqrt {b^2-4 a c} g}}} \, 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 g^3 \sqrt {f+g x} \sqrt {a+b x+c x^2}}\\ &=-\frac {8 e^2 (c e f-3 c d g+b e g) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{15 c^2 g^2}+\frac {2 e^2 (d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}}{5 c g}+\frac {\sqrt {2} \sqrt {b^2-4 a c} e \left (8 b^2 e^2 g^2+c e g (7 b e f-30 b d g-9 a e g)+c^2 \left (8 e^2 f^2-30 d e f g+45 d^2 g^2\right )\right ) \sqrt {f+g 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^3 g^3 \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \left (4 b e^3 g^2 (b f-a g)+c^2 \left (8 e^3 f^3-30 d e^2 f^2 g+45 d^2 e f g^2-15 d^3 g^3\right )-c e^2 g (a g (7 e f-15 d g)-3 b f (e f-5 d g))\right ) \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^3 g^3 \sqrt {f+g x} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 13.69, size = 12746, normalized size = 20.20 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [F]  time = 1.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt {c x^{2} + b x + a} \sqrt {g x + f}}{c g x^{3} + {\left (c f + b g\right )} x^{2} + a f + {\left (b f + a g\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.08, size = 8755, normalized size = 13.87 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(g*x+f)^(1/2)/(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 \[ \int \frac {{\left (e x + d\right )}^{3}}{\sqrt {c x^{2} + b x + a} \sqrt {g x + f}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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