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

Optimal. Leaf size=519 \[ -\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a g^2-b f g+c f^2\right ) \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} (b e g-10 c d g+8 c e f) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^2 g^3 \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (c g (-6 a e g-5 b d g+3 b e f)+2 b^2 e g^2-2 c^2 f (4 e f-5 d g)\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^2 g^3 \sqrt {a+b x+c x^2} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 \sqrt {f+g x} \sqrt {a+b x+c x^2} (-b e g-5 c d g+4 c e f-3 c e g x)}{15 c g^2} \]

[Out]

-2/15*(-3*c*e*g*x-b*e*g-5*c*d*g+4*c*e*f)*(g*x+f)^(1/2)*(c*x^2+b*x+a)^(1/2)/c/g^2-1/15*(2*b^2*e*g^2-2*c^2*f*(-5
*d*g+4*e*f)+c*g*(-6*a*e*g-5*b*d*g+3*b*e*f))*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^2/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*(b*e*g-10*c*d*g+8*c*e*f)*(a*g^2-b*f*g+c*f^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*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^2/
g^3/(g*x+f)^(1/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.54, antiderivative size = 519, 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 = {814, 843, 718, 424, 419} \[ -\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a g^2-b f g+c f^2\right ) \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} (b e g-10 c d g+8 c e f) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^2 g^3 \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (c g (-6 a e g-5 b d g+3 b e f)+2 b^2 e g^2-2 c^2 f (4 e f-5 d g)\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{15 c^2 g^3 \sqrt {a+b x+c x^2} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 \sqrt {f+g x} \sqrt {a+b x+c x^2} (-b e g-5 c d g+4 c e f-3 c e g x)}{15 c g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[f + g*x]*(4*c*e*f - 5*c*d*g - b*e*g - 3*c*e*g*x)*Sqrt[a + b*x + c*x^2])/(15*c*g^2) - (Sqrt[2]*Sqrt[b^
2 - 4*a*c]*(2*b^2*e*g^2 - 2*c^2*f*(4*e*f - 5*d*g) + c*g*(3*b*e*f - 5*b*d*g - 6*a*e*g))*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]]/S
qrt[2]], (-2*Sqrt[b^2 - 4*a*c]*g)/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)])/(15*c^2*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]*(8*c*e*f - 10*c*d*g + b*e*
g)*(c*f^2 - b*f*g + a*g^2)*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 + 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^2*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 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 {(d+e x) \sqrt {a+b x+c x^2}}{\sqrt {f+g x}} \, dx &=-\frac {2 \sqrt {f+g x} (4 c e f-5 c d g-b e g-3 c e g x) \sqrt {a+b x+c x^2}}{15 c g^2}-\frac {2 \int \frac {\frac {1}{2} \left (5 c d g (b f-2 a g)-b e f (4 c f-b g)+2 a e g \left (c f+\frac {b g}{2}\right )\right )+\frac {1}{2} \left (2 b^2 e g^2-2 c^2 f (4 e f-5 d g)+c g (3 b e f-5 b d g-6 a e g)\right ) x}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{15 c g^2}\\ &=-\frac {2 \sqrt {f+g x} (4 c e f-5 c d g-b e g-3 c e g x) \sqrt {a+b x+c x^2}}{15 c g^2}-\frac {\left ((8 c e f-10 c d g+b e g) \left (c f^2-b f g+a g^2\right )\right ) \int \frac {1}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{15 c g^3}-\frac {\left (2 b^2 e g^2-2 c^2 f (4 e f-5 d g)+c g (3 b e f-5 b d g-6 a e g)\right ) \int \frac {\sqrt {f+g x}}{\sqrt {a+b x+c x^2}} \, dx}{15 c g^3}\\ &=-\frac {2 \sqrt {f+g x} (4 c e f-5 c d g-b e g-3 c e g x) \sqrt {a+b x+c x^2}}{15 c g^2}-\frac {\left (\sqrt {2} \sqrt {b^2-4 a c} \left (2 b^2 e g^2-2 c^2 f (4 e f-5 d g)+c g (3 b e f-5 b d g-6 a e g)\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^2 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} (8 c e f-10 c d g+b e g) \left (c f^2-b f g+a g^2\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^2 g^3 \sqrt {f+g x} \sqrt {a+b x+c x^2}}\\ &=-\frac {2 \sqrt {f+g x} (4 c e f-5 c d g-b e g-3 c e g x) \sqrt {a+b x+c x^2}}{15 c g^2}-\frac {\sqrt {2} \sqrt {b^2-4 a c} \left (2 b^2 e g^2-2 c^2 f (4 e f-5 d g)+c g (3 b e f-5 b d g-6 a e g)\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^2 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} (8 c e f-10 c d g+b e g) \left (c f^2-b f g+a g^2\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^2 g^3 \sqrt {f+g x} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 10.51, size = 911, normalized size = 1.76 \[ \frac {2 \sqrt {a+x (b+c x)} \left (\left (2 f (4 e f-5 d g) c^2+g (-3 b e f+5 b d g+6 a e g) c-2 b^2 e g^2\right ) \left (c \left (\frac {f}{f+g x}-1\right )^2+\frac {g \left (-\frac {f b}{f+g x}+b+\frac {a g}{f+g x}\right )}{f+g x}\right )+\frac {i \sqrt {1-\frac {2 \left (c f^2+g (a g-b f)\right )}{\left (2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}\right ) (f+g x)}} \sqrt {\frac {2 \left (c f^2+g (a g-b f)\right )}{\left (-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}\right ) (f+g x)}+1} \left (\left (2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}\right ) \left (2 f (5 d g-4 e f) c^2+g (3 b e f-5 b d g-6 a e g) c+2 b^2 e g^2\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b g f+a g^2}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}}}{\sqrt {f+g x}}\right )|-\frac {-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}}\right )+\left (2 b^3 e g^3-b^2 \left (-c f e+2 \sqrt {\left (b^2-4 a c\right ) g^2} e+5 c d g\right ) g^2+b c \left (\sqrt {\left (b^2-4 a c\right ) g^2} (5 d g-3 e f)-8 a e g^2\right ) g+2 c \left (a \left (-2 c f e+3 \sqrt {\left (b^2-4 a c\right ) g^2} e+10 c d g\right ) g^2+c f \sqrt {\left (b^2-4 a c\right ) g^2} (4 e f-5 d g)\right )\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b g f+a g^2}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}}}{\sqrt {f+g x}}\right )|-\frac {-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}}\right )\right )}{2 \sqrt {2} \sqrt {\frac {c f^2+g (a g-b f)}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}} \sqrt {f+g x}}\right ) (f+g x)^{3/2}}{15 c^2 g^4 \sqrt {c x^2+b x+a} \sqrt {\frac {(f+g x)^2 \left (c \left (\frac {f}{f+g x}-1\right )^2+\frac {g \left (-\frac {f b}{f+g x}+b+\frac {a g}{f+g x}\right )}{f+g x}\right )}{g^2}}}+\left (\frac {2 (-4 c e f+5 c d g+b e g)}{15 c g^2}+\frac {2 e x}{5 g}\right ) \sqrt {a+x (b+c x)} \sqrt {f+g x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(-4*c*e*f + 5*c*d*g + b*e*g))/(15*c*g^2) + (2*e*x)/(5*g))*Sqrt[f + g*x]*Sqrt[a + x*(b + c*x)] + (2*(f + g*
x)^(3/2)*Sqrt[a + x*(b + c*x)]*((-2*b^2*e*g^2 + 2*c^2*f*(4*e*f - 5*d*g) + c*g*(-3*b*e*f + 5*b*d*g + 6*a*e*g))*
(c*(-1 + f/(f + g*x))^2 + (g*(b - (b*f)/(f + g*x) + (a*g)/(f + g*x)))/(f + g*x)) + ((I/2)*Sqrt[1 - (2*(c*f^2 +
g*(-(b*f) + a*g)))/((2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2])*(f + g*x))]*Sqrt[1 + (2*(c*f^2 + g*(-(b*f) + a*g)
))/((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])*(f + g*x))]*((2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2])*(2*b^2*e*g^2
+ 2*c^2*f*(-4*e*f + 5*d*g) + c*g*(3*b*e*f - 5*b*d*g - 6*a*e*g))*EllipticE[I*ArcSinh[(Sqrt[2]*Sqrt[(c*f^2 - b*
f*g + a*g^2)/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])])/Sqrt[f + g*x]], -((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g
^2])/(2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2]))] + (2*b^3*e*g^3 - b^2*g^2*(-(c*e*f) + 5*c*d*g + 2*e*Sqrt[(b^2 -
4*a*c)*g^2]) + b*c*g*(-8*a*e*g^2 + Sqrt[(b^2 - 4*a*c)*g^2]*(-3*e*f + 5*d*g)) + 2*c*(c*f*Sqrt[(b^2 - 4*a*c)*g^2
]*(4*e*f - 5*d*g) + a*g^2*(-2*c*e*f + 10*c*d*g + 3*e*Sqrt[(b^2 - 4*a*c)*g^2])))*EllipticF[I*ArcSinh[(Sqrt[2]*S
qrt[(c*f^2 - b*f*g + a*g^2)/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])])/Sqrt[f + g*x]], -((-2*c*f + b*g + Sqrt[
(b^2 - 4*a*c)*g^2])/(2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2]))]))/(Sqrt[2]*Sqrt[(c*f^2 + g*(-(b*f) + a*g))/(-2*c
*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])]*Sqrt[f + g*x])))/(15*c^2*g^4*Sqrt[a + b*x + c*x^2]*Sqrt[((f + g*x)^2*(c*(
-1 + f/(f + g*x))^2 + (g*(b - (b*f)/(f + g*x) + (a*g)/(f + g*x)))/(f + g*x)))/g^2])

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fricas [F]  time = 1.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (e x + d\right )}}{\sqrt {g x + f}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 6207, normalized size = 11.96 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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