3.43 \(\int \sqrt{c+d x} \sqrt{e+f x} (A+B x+C x^2) \, dx\)

Optimal. Leaf size=330 \[ \frac{(c+d x)^{3/2} \sqrt{e+f x} \left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right )}{32 d^3 f^2}+\frac{\sqrt{c+d x} \sqrt{e+f x} (d e-c f) \left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right )}{64 d^3 f^3}-\frac{(d e-c f)^2 \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right ) \left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right )}{64 d^{7/2} f^{7/2}}-\frac{(c+d x)^{3/2} (e+f x)^{3/2} (-8 B d f+11 c C f+5 C d e)}{24 d^2 f^2}+\frac{C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f} \]

[Out]

((d*e - c*f)*(C*(5*d^2*e^2 + 6*c*d*e*f + 5*c^2*f^2) + 8*d*f*(2*A*d*f - B*(d*e + c*f)))*Sqrt[c + d*x]*Sqrt[e +
f*x])/(64*d^3*f^3) + ((C*(5*d^2*e^2 + 6*c*d*e*f + 5*c^2*f^2) + 8*d*f*(2*A*d*f - B*(d*e + c*f)))*(c + d*x)^(3/2
)*Sqrt[e + f*x])/(32*d^3*f^2) - ((5*C*d*e + 11*c*C*f - 8*B*d*f)*(c + d*x)^(3/2)*(e + f*x)^(3/2))/(24*d^2*f^2)
+ (C*(c + d*x)^(5/2)*(e + f*x)^(3/2))/(4*d^2*f) - ((d*e - c*f)^2*(C*(5*d^2*e^2 + 6*c*d*e*f + 5*c^2*f^2) + 8*d*
f*(2*A*d*f - B*(d*e + c*f)))*ArcTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/(64*d^(7/2)*f^(7/2))

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Rubi [A]  time = 0.297592, antiderivative size = 330, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {951, 80, 50, 63, 217, 206} \[ \frac{(c+d x)^{3/2} \sqrt{e+f x} \left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right )}{32 d^3 f^2}+\frac{\sqrt{c+d x} \sqrt{e+f x} (d e-c f) \left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right )}{64 d^3 f^3}-\frac{(d e-c f)^2 \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right ) \left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right )}{64 d^{7/2} f^{7/2}}-\frac{(c+d x)^{3/2} (e+f x)^{3/2} (-8 B d f+11 c C f+5 C d e)}{24 d^2 f^2}+\frac{C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Sqrt[e + f*x]*(A + B*x + C*x^2),x]

[Out]

((d*e - c*f)*(C*(5*d^2*e^2 + 6*c*d*e*f + 5*c^2*f^2) + 8*d*f*(2*A*d*f - B*(d*e + c*f)))*Sqrt[c + d*x]*Sqrt[e +
f*x])/(64*d^3*f^3) + ((C*(5*d^2*e^2 + 6*c*d*e*f + 5*c^2*f^2) + 8*d*f*(2*A*d*f - B*(d*e + c*f)))*(c + d*x)^(3/2
)*Sqrt[e + f*x])/(32*d^3*f^2) - ((5*C*d*e + 11*c*C*f - 8*B*d*f)*(c + d*x)^(3/2)*(e + f*x)^(3/2))/(24*d^2*f^2)
+ (C*(c + d*x)^(5/2)*(e + f*x)^(3/2))/(4*d^2*f) - ((d*e - c*f)^2*(C*(5*d^2*e^2 + 6*c*d*e*f + 5*c^2*f^2) + 8*d*
f*(2*A*d*f - B*(d*e + c*f)))*ArcTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/(64*d^(7/2)*f^(7/2))

Rule 951

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[(c^p*(d + e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], 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] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sqrt{c+d x} \sqrt{e+f x} \left (A+B x+C x^2\right ) \, dx &=\frac{C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}+\frac{\int \sqrt{c+d x} \sqrt{e+f x} \left (\frac{1}{2} \left (-5 c C d e-3 c^2 C f+8 A d^2 f\right )-\frac{1}{2} d (5 C d e+11 c C f-8 B d f) x\right ) \, dx}{4 d^2 f}\\ &=-\frac{(5 C d e+11 c C f-8 B d f) (c+d x)^{3/2} (e+f x)^{3/2}}{24 d^2 f^2}+\frac{C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}+\frac{\left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) \int \sqrt{c+d x} \sqrt{e+f x} \, dx}{16 d^2 f^2}\\ &=\frac{\left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) (c+d x)^{3/2} \sqrt{e+f x}}{32 d^3 f^2}-\frac{(5 C d e+11 c C f-8 B d f) (c+d x)^{3/2} (e+f x)^{3/2}}{24 d^2 f^2}+\frac{C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}+\frac{\left ((d e-c f) \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right )\right ) \int \frac{\sqrt{c+d x}}{\sqrt{e+f x}} \, dx}{64 d^3 f^2}\\ &=\frac{(d e-c f) \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) \sqrt{c+d x} \sqrt{e+f x}}{64 d^3 f^3}+\frac{\left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) (c+d x)^{3/2} \sqrt{e+f x}}{32 d^3 f^2}-\frac{(5 C d e+11 c C f-8 B d f) (c+d x)^{3/2} (e+f x)^{3/2}}{24 d^2 f^2}+\frac{C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac{\left ((d e-c f)^2 \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right )\right ) \int \frac{1}{\sqrt{c+d x} \sqrt{e+f x}} \, dx}{128 d^3 f^3}\\ &=\frac{(d e-c f) \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) \sqrt{c+d x} \sqrt{e+f x}}{64 d^3 f^3}+\frac{\left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) (c+d x)^{3/2} \sqrt{e+f x}}{32 d^3 f^2}-\frac{(5 C d e+11 c C f-8 B d f) (c+d x)^{3/2} (e+f x)^{3/2}}{24 d^2 f^2}+\frac{C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac{\left ((d e-c f)^2 \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{e-\frac{c f}{d}+\frac{f x^2}{d}}} \, dx,x,\sqrt{c+d x}\right )}{64 d^4 f^3}\\ &=\frac{(d e-c f) \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) \sqrt{c+d x} \sqrt{e+f x}}{64 d^3 f^3}+\frac{\left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) (c+d x)^{3/2} \sqrt{e+f x}}{32 d^3 f^2}-\frac{(5 C d e+11 c C f-8 B d f) (c+d x)^{3/2} (e+f x)^{3/2}}{24 d^2 f^2}+\frac{C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac{\left ((d e-c f)^2 \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{f x^2}{d}} \, dx,x,\frac{\sqrt{c+d x}}{\sqrt{e+f x}}\right )}{64 d^4 f^3}\\ &=\frac{(d e-c f) \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) \sqrt{c+d x} \sqrt{e+f x}}{64 d^3 f^3}+\frac{\left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) (c+d x)^{3/2} \sqrt{e+f x}}{32 d^3 f^2}-\frac{(5 C d e+11 c C f-8 B d f) (c+d x)^{3/2} (e+f x)^{3/2}}{24 d^2 f^2}+\frac{C (c+d x)^{5/2} (e+f x)^{3/2}}{4 d^2 f}-\frac{(d e-c f)^2 \left (C \left (5 d^2 e^2+6 c d e f+5 c^2 f^2\right )+8 d f (2 A d f-B (d e+c f))\right ) \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right )}{64 d^{7/2} f^{7/2}}\\ \end{align*}

Mathematica [A]  time = 1.84877, size = 306, normalized size = 0.93 \[ \frac{d \sqrt{f} \sqrt{c+d x} (e+f x) \left (8 d f \left (6 A d f (c f+d (e+2 f x))+B \left (-3 c^2 f^2+2 c d f (e+f x)+d^2 \left (-3 e^2+2 e f x+8 f^2 x^2\right )\right )\right )+C \left (-c^2 d f^2 (7 e+10 f x)+15 c^3 f^3+c d^2 f \left (-7 e^2+4 e f x+8 f^2 x^2\right )+d^3 \left (-10 e^2 f x+15 e^3+8 e f^2 x^2+48 f^3 x^3\right )\right )\right )-3 (d e-c f)^{5/2} \sqrt{\frac{d (e+f x)}{d e-c f}} \sinh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d e-c f}}\right ) \left (8 d f (2 A d f-B (c f+d e))+C \left (5 c^2 f^2+6 c d e f+5 d^2 e^2\right )\right )}{192 d^4 f^{7/2} \sqrt{e+f x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]*Sqrt[e + f*x]*(A + B*x + C*x^2),x]

[Out]

(d*Sqrt[f]*Sqrt[c + d*x]*(e + f*x)*(C*(15*c^3*f^3 - c^2*d*f^2*(7*e + 10*f*x) + c*d^2*f*(-7*e^2 + 4*e*f*x + 8*f
^2*x^2) + d^3*(15*e^3 - 10*e^2*f*x + 8*e*f^2*x^2 + 48*f^3*x^3)) + 8*d*f*(6*A*d*f*(c*f + d*(e + 2*f*x)) + B*(-3
*c^2*f^2 + 2*c*d*f*(e + f*x) + d^2*(-3*e^2 + 2*e*f*x + 8*f^2*x^2)))) - 3*(d*e - c*f)^(5/2)*(C*(5*d^2*e^2 + 6*c
*d*e*f + 5*c^2*f^2) + 8*d*f*(2*A*d*f - B*(d*e + c*f)))*Sqrt[(d*(e + f*x))/(d*e - c*f)]*ArcSinh[(Sqrt[f]*Sqrt[c
+ d*x])/Sqrt[d*e - c*f]])/(192*d^4*f^(7/2)*Sqrt[e + f*x])

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Maple [B]  time = 0.016, size = 1431, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)*(d*x+c)^(1/2)*(f*x+e)^(1/2),x)

[Out]

-1/384*(d*x+c)^(1/2)*(f*x+e)^(1/2)*(48*B*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*d^3*e^2*f+48*B*(d*f)^(1/2
)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*c^2*d*f^3-96*A*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*d^3*e*f^2-96*A*(d
*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*c*d^2*f^3-6*C*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f
)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c^2*d^2*e^2*f^2-12*C*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f)^(1/
2)+c*f+d*e)/(d*f)^(1/2))*c*d^3*e^3*f-96*A*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f)^(1/2)+c*f+d*
e)/(d*f)^(1/2))*c*d^3*e*f^3-192*A*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*x*d^3*f^3+24*B*ln(1/2*(2*d*f*x+2
*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c^2*d^2*e*f^3+24*B*ln(1/2*(2*d*f*x+2*(d*f*x
^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c*d^3*e^2*f^2-12*C*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*
x+d*e*x+c*e)^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c^3*d*e*f^3-96*C*x^3*d^3*f^3*(d*f)^(1/2)*(d*f*x^2+c*f*x+d
*e*x+c*e)^(1/2)-128*B*x^2*d^3*f^3*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)-30*C*(d*f)^(1/2)*(d*f*x^2+c*f*x+
d*e*x+c*e)^(1/2)*d^3*e^3-24*B*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/
2))*c^3*d*f^4-24*B*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*d^4*e^3
*f-30*C*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*c^3*f^3+48*A*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*x+d*e*x+c*e)^(
1/2)*(d*f)^(1/2)+c*f+d*e)/(d*f)^(1/2))*c^2*d^2*f^4+48*A*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f
)^(1/2)+c*f+d*e)/(d*f)^(1/2))*d^4*e^2*f^2-32*B*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*x*c*d^2*f^3-32*B*(d
*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*x*d^3*e*f^2+20*C*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*x*c^2*d
*f^3+20*C*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*x*d^3*e^2*f-32*B*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(
1/2)*c*d^2*e*f^2+14*C*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*c^2*d*e*f^2+14*C*(d*f)^(1/2)*(d*f*x^2+c*f*x+
d*e*x+c*e)^(1/2)*c*d^2*e^2*f-16*C*x^2*c*d^2*f^3*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)-16*C*x^2*d^3*e*f^2
*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)+15*C*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f)^(1/2
)+c*f+d*e)/(d*f)^(1/2))*c^4*f^4+15*C*ln(1/2*(2*d*f*x+2*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*(d*f)^(1/2)+c*f+d*e)/(d
*f)^(1/2))*d^4*e^4-8*C*(d*f)^(1/2)*(d*f*x^2+c*f*x+d*e*x+c*e)^(1/2)*x*c*d^2*e*f^2)/(d*f*x^2+c*f*x+d*e*x+c*e)^(1
/2)/d^3/f^3/(d*f)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54313, size = 1843, normalized size = 5.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*(5*C*d^4*e^4 - 4*(C*c*d^3 + 2*B*d^4)*e^3*f - 2*(C*c^2*d^2 - 4*B*c*d^3 - 8*A*d^4)*e^2*f^2 - 4*(C*c^3*
d - 2*B*c^2*d^2 + 8*A*c*d^3)*e*f^3 + (5*C*c^4 - 8*B*c^3*d + 16*A*c^2*d^2)*f^4)*sqrt(d*f)*log(8*d^2*f^2*x^2 + d
^2*e^2 + 6*c*d*e*f + c^2*f^2 - 4*(2*d*f*x + d*e + c*f)*sqrt(d*f)*sqrt(d*x + c)*sqrt(f*x + e) + 8*(d^2*e*f + c*
d*f^2)*x) + 4*(48*C*d^4*f^4*x^3 + 15*C*d^4*e^3*f - (7*C*c*d^3 + 24*B*d^4)*e^2*f^2 - (7*C*c^2*d^2 - 16*B*c*d^3
- 48*A*d^4)*e*f^3 + 3*(5*C*c^3*d - 8*B*c^2*d^2 + 16*A*c*d^3)*f^4 + 8*(C*d^4*e*f^3 + (C*c*d^3 + 8*B*d^4)*f^4)*x
^2 - 2*(5*C*d^4*e^2*f^2 - 2*(C*c*d^3 + 4*B*d^4)*e*f^3 + (5*C*c^2*d^2 - 8*B*c*d^3 - 48*A*d^4)*f^4)*x)*sqrt(d*x
+ c)*sqrt(f*x + e))/(d^4*f^4), 1/384*(3*(5*C*d^4*e^4 - 4*(C*c*d^3 + 2*B*d^4)*e^3*f - 2*(C*c^2*d^2 - 4*B*c*d^3
- 8*A*d^4)*e^2*f^2 - 4*(C*c^3*d - 2*B*c^2*d^2 + 8*A*c*d^3)*e*f^3 + (5*C*c^4 - 8*B*c^3*d + 16*A*c^2*d^2)*f^4)*s
qrt(-d*f)*arctan(1/2*(2*d*f*x + d*e + c*f)*sqrt(-d*f)*sqrt(d*x + c)*sqrt(f*x + e)/(d^2*f^2*x^2 + c*d*e*f + (d^
2*e*f + c*d*f^2)*x)) + 2*(48*C*d^4*f^4*x^3 + 15*C*d^4*e^3*f - (7*C*c*d^3 + 24*B*d^4)*e^2*f^2 - (7*C*c^2*d^2 -
16*B*c*d^3 - 48*A*d^4)*e*f^3 + 3*(5*C*c^3*d - 8*B*c^2*d^2 + 16*A*c*d^3)*f^4 + 8*(C*d^4*e*f^3 + (C*c*d^3 + 8*B*
d^4)*f^4)*x^2 - 2*(5*C*d^4*e^2*f^2 - 2*(C*c*d^3 + 4*B*d^4)*e*f^3 + (5*C*c^2*d^2 - 8*B*c*d^3 - 48*A*d^4)*f^4)*x
)*sqrt(d*x + c)*sqrt(f*x + e))/(d^4*f^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c + d x} \sqrt{e + f x} \left (A + B x + C x^{2}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)*(d*x+c)**(1/2)*(f*x+e)**(1/2),x)

[Out]

Integral(sqrt(c + d*x)*sqrt(e + f*x)*(A + B*x + C*x**2), x)

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Giac [B]  time = 1.58253, size = 856, normalized size = 2.59 \begin{align*} \frac{\frac{20 \,{\left (\sqrt{{\left (d x + c\right )} d f - c d f + d^{2} e} \sqrt{d x + c}{\left (\frac{2 \,{\left (d x + c\right )}}{d^{4} f^{2}} - \frac{c f^{2} - d f e}{d^{4} f^{4}}\right )} + \frac{{\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} \log \left ({\left | -\sqrt{d f} \sqrt{d x + c} + \sqrt{{\left (d x + c\right )} d f - c d f + d^{2} e} \right |}\right )}{\sqrt{d f} d^{3} f^{3}}\right )} A{\left | d \right |}}{d^{2}} + \frac{10 \,{\left (\sqrt{{\left (d x + c\right )} d f - c d f + d^{2} e}{\left (2 \,{\left (d x + c\right )}{\left (4 \,{\left (d x + c\right )}{\left (\frac{6 \,{\left (d x + c\right )}}{d^{2}} - \frac{17 \, c d^{6} f^{6} - d^{7} f^{5} e}{d^{8} f^{6}}\right )} + \frac{59 \, c^{2} d^{6} f^{6} - 6 \, c d^{7} f^{5} e - 5 \, d^{8} f^{4} e^{2}}{d^{8} f^{6}}\right )} - \frac{3 \,{\left (5 \, c^{3} d^{6} f^{6} + c^{2} d^{7} f^{5} e - c d^{8} f^{4} e^{2} - 5 \, d^{9} f^{3} e^{3}\right )}}{d^{8} f^{6}}\right )} \sqrt{d x + c} + \frac{3 \,{\left (5 \, c^{4} f^{4} - 4 \, c^{3} d f^{3} e - 2 \, c^{2} d^{2} f^{2} e^{2} - 4 \, c d^{3} f e^{3} + 5 \, d^{4} e^{4}\right )} \log \left ({\left | -\sqrt{d f} \sqrt{d x + c} + \sqrt{{\left (d x + c\right )} d f - c d f + d^{2} e} \right |}\right )}{\sqrt{d f} d f^{3}}\right )} C{\left | d \right |}}{d^{2}} + \frac{{\left (\sqrt{{\left (d x + c\right )} d f - c d f + d^{2} e} \sqrt{d x + c}{\left (2 \,{\left (d x + c\right )}{\left (\frac{4 \,{\left (d x + c\right )}}{d^{6} f^{2}} - \frac{7 \, c f^{4} - d f^{3} e}{d^{6} f^{6}}\right )} + \frac{3 \,{\left (c^{2} f^{4} - d^{2} f^{2} e^{2}\right )}}{d^{6} f^{6}}\right )} - \frac{3 \,{\left (c^{3} f^{3} - c^{2} d f^{2} e - c d^{2} f e^{2} + d^{3} e^{3}\right )} \log \left ({\left | -\sqrt{d f} \sqrt{d x + c} + \sqrt{{\left (d x + c\right )} d f - c d f + d^{2} e} \right |}\right )}{\sqrt{d f} d^{5} f^{4}}\right )} B{\left | d \right |}}{d^{3}}}{1920 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*(20*(sqrt((d*x + c)*d*f - c*d*f + d^2*e)*sqrt(d*x + c)*(2*(d*x + c)/(d^4*f^2) - (c*f^2 - d*f*e)/(d^4*f^
4)) + (c^2*f^2 - 2*c*d*f*e + d^2*e^2)*log(abs(-sqrt(d*f)*sqrt(d*x + c) + sqrt((d*x + c)*d*f - c*d*f + d^2*e)))
/(sqrt(d*f)*d^3*f^3))*A*abs(d)/d^2 + 10*(sqrt((d*x + c)*d*f - c*d*f + d^2*e)*(2*(d*x + c)*(4*(d*x + c)*(6*(d*x
+ c)/d^2 - (17*c*d^6*f^6 - d^7*f^5*e)/(d^8*f^6)) + (59*c^2*d^6*f^6 - 6*c*d^7*f^5*e - 5*d^8*f^4*e^2)/(d^8*f^6)
) - 3*(5*c^3*d^6*f^6 + c^2*d^7*f^5*e - c*d^8*f^4*e^2 - 5*d^9*f^3*e^3)/(d^8*f^6))*sqrt(d*x + c) + 3*(5*c^4*f^4
- 4*c^3*d*f^3*e - 2*c^2*d^2*f^2*e^2 - 4*c*d^3*f*e^3 + 5*d^4*e^4)*log(abs(-sqrt(d*f)*sqrt(d*x + c) + sqrt((d*x
+ c)*d*f - c*d*f + d^2*e)))/(sqrt(d*f)*d*f^3))*C*abs(d)/d^2 + (sqrt((d*x + c)*d*f - c*d*f + d^2*e)*sqrt(d*x +
c)*(2*(d*x + c)*(4*(d*x + c)/(d^6*f^2) - (7*c*f^4 - d*f^3*e)/(d^6*f^6)) + 3*(c^2*f^4 - d^2*f^2*e^2)/(d^6*f^6))
- 3*(c^3*f^3 - c^2*d*f^2*e - c*d^2*f*e^2 + d^3*e^3)*log(abs(-sqrt(d*f)*sqrt(d*x + c) + sqrt((d*x + c)*d*f - c
*d*f + d^2*e)))/(sqrt(d*f)*d^5*f^4))*B*abs(d)/d^3)/d