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

Optimal. Leaf size=364 \[ \frac {\sqrt {c+d x} \sqrt {e+f x} \left (2 a^2 C d f-a b (B d f+c C f+C d e)+b^2 (A d f+c C e)\right )}{b^2 f (b c-a d) (b e-a f)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {b e-a f}}{\sqrt {e+f x} \sqrt {b c-a d}}\right ) \left (4 a^3 C d f-a^2 b (2 B d f+3 c C f+5 C d e)+a b^2 (B c f+3 B d e+4 c C e)-b^3 (-A c f+A d e+2 B c e)\right )}{b^3 \sqrt {b c-a d} (b e-a f)^{3/2}}-\frac {(c+d x)^{3/2} \sqrt {e+f x} \left (A b^2-a (b B-a C)\right )}{b (a+b x) (b c-a d) (b e-a f)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) (4 a C d f+b (-2 B d f-c C f+C d e))}{b^3 \sqrt {d} f^{3/2}} \]

[Out]

-(4*a*C*d*f+b*(-2*B*d*f-C*c*f+C*d*e))*arctanh(f^(1/2)*(d*x+c)^(1/2)/d^(1/2)/(f*x+e)^(1/2))/b^3/f^(3/2)/d^(1/2)
+(4*a^3*C*d*f-b^3*(-A*c*f+A*d*e+2*B*c*e)+a*b^2*(B*c*f+3*B*d*e+4*C*c*e)-a^2*b*(2*B*d*f+3*C*c*f+5*C*d*e))*arctan
h((-a*f+b*e)^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2)/(f*x+e)^(1/2))/b^3/(-a*f+b*e)^(3/2)/(-a*d+b*c)^(1/2)-(A*b^2-
a*(B*b-C*a))*(d*x+c)^(3/2)*(f*x+e)^(1/2)/b/(-a*d+b*c)/(-a*f+b*e)/(b*x+a)+(2*a^2*C*d*f+b^2*(A*d*f+C*c*e)-a*b*(B
*d*f+C*c*f+C*d*e))*(d*x+c)^(1/2)*(f*x+e)^(1/2)/b^2/(-a*d+b*c)/f/(-a*f+b*e)

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Rubi [A]  time = 1.10, antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1613, 154, 157, 63, 217, 206, 93, 208} \[ \frac {\sqrt {c+d x} \sqrt {e+f x} \left (2 a^2 C d f-a b (B d f+c C f+C d e)+b^2 (A d f+c C e)\right )}{b^2 f (b c-a d) (b e-a f)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {b e-a f}}{\sqrt {e+f x} \sqrt {b c-a d}}\right ) \left (-a^2 b (2 B d f+3 c C f+5 C d e)+4 a^3 C d f+a b^2 (B c f+3 B d e+4 c C e)-b^3 (-A c f+A d e+2 B c e)\right )}{b^3 \sqrt {b c-a d} (b e-a f)^{3/2}}-\frac {(c+d x)^{3/2} \sqrt {e+f x} \left (A b^2-a (b B-a C)\right )}{b (a+b x) (b c-a d) (b e-a f)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) (4 a C d f+b (-2 B d f-c C f+C d e))}{b^3 \sqrt {d} f^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*C*d*f + b^2*(c*C*e + A*d*f) - a*b*(C*d*e + c*C*f + B*d*f))*Sqrt[c + d*x]*Sqrt[e + f*x])/(b^2*(b*c - a*
d)*f*(b*e - a*f)) - ((A*b^2 - a*(b*B - a*C))*(c + d*x)^(3/2)*Sqrt[e + f*x])/(b*(b*c - a*d)*(b*e - a*f)*(a + b*
x)) - ((4*a*C*d*f + b*(C*d*e - c*C*f - 2*B*d*f))*ArcTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/(b^
3*Sqrt[d]*f^(3/2)) + ((4*a^3*C*d*f - b^3*(2*B*c*e + A*d*e - A*c*f) + a*b^2*(4*c*C*e + 3*B*d*e + B*c*f) - a^2*b
*(5*C*d*e + 3*c*C*f + 2*B*d*f))*ArcTanh[(Sqrt[b*e - a*f]*Sqrt[c + d*x])/(Sqrt[b*c - a*d]*Sqrt[e + f*x])])/(b^3
*Sqrt[b*c - a*d]*(b*e - a*f)^(3/2))

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

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])

Rule 208

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

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 1613

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[{
Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[(b*R*(a + b*x)^(m + 1)
*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e
- a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*ExpandToSum[(m + 1)*(b*c - a*d)*(b*e - a*f)*Qx + a*d*f
*R*(m + 1) - b*R*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*R*(m + n + p + 3)*x, x], x], x]] /; FreeQ[{a, b,
c, d, e, f, n, p}, x] && PolyQ[Px, x] && ILtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{(a+b x)^2 \sqrt {e+f x}} \, dx &=-\frac {\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt {e+f x}}{b (b c-a d) (b e-a f) (a+b x)}-\frac {\int \frac {\sqrt {c+d x} \left (-\frac {a^2 C (3 d e+c f)+b^2 (2 B c e+A d e-A c f)-a b (2 c C e+3 B d e+B c f-2 A d f)}{2 b}+\left (-\frac {2 a^2 C d f}{b}-b (c C e+A d f)+a (C d e+c C f+B d f)\right ) x\right )}{(a+b x) \sqrt {e+f x}} \, dx}{(b c-a d) (b e-a f)}\\ &=\frac {\left (2 a^2 C d f+b^2 (c C e+A d f)-a b (C d e+c C f+B d f)\right ) \sqrt {c+d x} \sqrt {e+f x}}{b^2 (b c-a d) f (b e-a f)}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt {e+f x}}{b (b c-a d) (b e-a f) (a+b x)}-\frac {\int \frac {-\frac {(b c-a d) \left (2 a^2 C f (d e+c f)+b^2 f (2 B c e+A d e-A c f)-a b (B f (d e+c f)+C e (d e+3 c f))\right )}{2 b}+\frac {(b c-a d) (b e-a f) (4 a C d f+b (C d e-c C f-2 B d f)) x}{2 b}}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx}{b (b c-a d) f (b e-a f)}\\ &=\frac {\left (2 a^2 C d f+b^2 (c C e+A d f)-a b (C d e+c C f+B d f)\right ) \sqrt {c+d x} \sqrt {e+f x}}{b^2 (b c-a d) f (b e-a f)}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt {e+f x}}{b (b c-a d) (b e-a f) (a+b x)}-\frac {(4 a C d f+b (C d e-c C f-2 B d f)) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}} \, dx}{2 b^3 f}-\frac {\left (4 a^3 C d f-b^3 (2 B c e+A d e-A c f)+a b^2 (4 c C e+3 B d e+B c f)-a^2 b (5 C d e+3 c C f+2 B d f)\right ) \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx}{2 b^3 (b e-a f)}\\ &=\frac {\left (2 a^2 C d f+b^2 (c C e+A d f)-a b (C d e+c C f+B d f)\right ) \sqrt {c+d x} \sqrt {e+f x}}{b^2 (b c-a d) f (b e-a f)}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt {e+f x}}{b (b c-a d) (b e-a f) (a+b x)}-\frac {(4 a C d f+b (C d e-c C f-2 B d f)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {e-\frac {c f}{d}+\frac {f x^2}{d}}} \, dx,x,\sqrt {c+d x}\right )}{b^3 d f}-\frac {\left (4 a^3 C d f-b^3 (2 B c e+A d e-A c f)+a b^2 (4 c C e+3 B d e+B c f)-a^2 b (5 C d e+3 c C f+2 B d f)\right ) \operatorname {Subst}\left (\int \frac {1}{-b c+a d-(-b e+a f) x^2} \, dx,x,\frac {\sqrt {c+d x}}{\sqrt {e+f x}}\right )}{b^3 (b e-a f)}\\ &=\frac {\left (2 a^2 C d f+b^2 (c C e+A d f)-a b (C d e+c C f+B d f)\right ) \sqrt {c+d x} \sqrt {e+f x}}{b^2 (b c-a d) f (b e-a f)}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt {e+f x}}{b (b c-a d) (b e-a f) (a+b x)}+\frac {\left (4 a^3 C d f-b^3 (2 B c e+A d e-A c f)+a b^2 (4 c C e+3 B d e+B c f)-a^2 b (5 C d e+3 c C f+2 B d f)\right ) \tanh ^{-1}\left (\frac {\sqrt {b e-a f} \sqrt {c+d x}}{\sqrt {b c-a d} \sqrt {e+f x}}\right )}{b^3 \sqrt {b c-a d} (b e-a f)^{3/2}}-\frac {(4 a C d f+b (C d e-c C f-2 B d f)) \operatorname {Subst}\left (\int \frac {1}{1-\frac {f x^2}{d}} \, dx,x,\frac {\sqrt {c+d x}}{\sqrt {e+f x}}\right )}{b^3 d f}\\ &=\frac {\left (2 a^2 C d f+b^2 (c C e+A d f)-a b (C d e+c C f+B d f)\right ) \sqrt {c+d x} \sqrt {e+f x}}{b^2 (b c-a d) f (b e-a f)}-\frac {\left (A b^2-a (b B-a C)\right ) (c+d x)^{3/2} \sqrt {e+f x}}{b (b c-a d) (b e-a f) (a+b x)}-\frac {(4 a C d f+b (C d e-c C f-2 B d f)) \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{b^3 \sqrt {d} f^{3/2}}+\frac {\left (4 a^3 C d f-b^3 (2 B c e+A d e-A c f)+a b^2 (4 c C e+3 B d e+B c f)-a^2 b (5 C d e+3 c C f+2 B d f)\right ) \tanh ^{-1}\left (\frac {\sqrt {b e-a f} \sqrt {c+d x}}{\sqrt {b c-a d} \sqrt {e+f x}}\right )}{b^3 \sqrt {b c-a d} (b e-a f)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 2.40, size = 417, normalized size = 1.15 \[ \frac {-\frac {2 b \sqrt {c+d x} \sqrt {e+f x} \left (a (a C-b B)+A b^2\right )}{(a+b x) (b e-a f)}+\frac {2 b (c f-d e) \left (a (a C-b B)+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {a f-b e}}{\sqrt {e+f x} \sqrt {a d-b c}}\right )}{\sqrt {a d-b c} (a f-b e)^{3/2}}+\frac {4 (b B-2 a C) \sqrt {d e-c f} \sqrt {\frac {d (e+f x)}{d e-c f}} \sinh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d e-c f}}\right )}{\sqrt {f} \sqrt {e+f x}}-\frac {4 (b B-2 a C) \sqrt {a d-b c} \tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {a f-b e}}{\sqrt {e+f x} \sqrt {a d-b c}}\right )}{\sqrt {a f-b e}}+\frac {2 b C \sqrt {e+f x} \left (\sqrt {f} \sqrt {c+d x}-\frac {\sqrt {d e-c f} \sinh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d e-c f}}\right )}{\sqrt {\frac {d (e+f x)}{d e-c f}}}\right )}{f^{3/2}}}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*(A*b^2 + a*(-(b*B) + a*C))*Sqrt[c + d*x]*Sqrt[e + f*x])/((b*e - a*f)*(a + b*x)) + (4*(b*B - 2*a*C)*Sqrt
[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]])/(Sqrt[f]*Sqrt[e
+ f*x]) + (2*b*C*Sqrt[e + f*x]*(Sqrt[f]*Sqrt[c + d*x] - (Sqrt[d*e - c*f]*ArcSinh[(Sqrt[f]*Sqrt[c + d*x])/Sqrt[
d*e - c*f]])/Sqrt[(d*(e + f*x))/(d*e - c*f)]))/f^(3/2) - (4*(b*B - 2*a*C)*Sqrt[-(b*c) + a*d]*ArcTanh[(Sqrt[-(b
*e) + a*f]*Sqrt[c + d*x])/(Sqrt[-(b*c) + a*d]*Sqrt[e + f*x])])/Sqrt[-(b*e) + a*f] + (2*b*(A*b^2 + a*(-(b*B) +
a*C))*(-(d*e) + c*f)*ArcTanh[(Sqrt[-(b*e) + a*f]*Sqrt[c + d*x])/(Sqrt[-(b*c) + a*d]*Sqrt[e + f*x])])/(Sqrt[-(b
*c) + a*d]*(-(b*e) + a*f)^(3/2)))/(2*b^3)

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

[Out]

Timed out

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giac [B]  time = 10.82, size = 1388, normalized size = 3.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*sqrt(d*f)*C*a^2*b*c*d^2*f - sqrt(d*f)*B*a*b^2*c*d^2*f - sqrt(d*f)*A*b^3*c*d^2*f - 4*sqrt(d*f)*C*a^3*d^3*f +
2*sqrt(d*f)*B*a^2*b*d^3*f - 4*sqrt(d*f)*C*a*b^2*c*d^2*e + 2*sqrt(d*f)*B*b^3*c*d^2*e + 5*sqrt(d*f)*C*a^2*b*d^3
*e - 3*sqrt(d*f)*B*a*b^2*d^3*e + sqrt(d*f)*A*b^3*d^3*e)*arctan(-1/2*(b*c*d*f - 2*a*d^2*f + b*d^2*e - (sqrt(d*f
)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^2*b)/(sqrt(a*b*c*d*f^2 - a^2*d^2*f^2 - b^2*c*d*f*e + a*
b*d^2*f*e)*d))/(sqrt(a*b*c*d*f^2 - a^2*d^2*f^2 - b^2*c*d*f*e + a*b*d^2*f*e)*(a*b^3*f*abs(d) - b^4*abs(d)*e)*d)
+ 2*(sqrt(d*f)*C*a^2*b*c^2*d^3*f^2 - sqrt(d*f)*B*a*b^2*c^2*d^3*f^2 + sqrt(d*f)*A*b^3*c^2*d^3*f^2 - 2*sqrt(d*f
)*C*a^2*b*c*d^4*f*e + 2*sqrt(d*f)*B*a*b^2*c*d^4*f*e - 2*sqrt(d*f)*A*b^3*c*d^4*f*e - sqrt(d*f)*(sqrt(d*f)*sqrt(
d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^2*C*a^2*b*c*d^2*f + sqrt(d*f)*(sqrt(d*f)*sqrt(d*x + c) - sqrt(
(d*x + c)*d*f - c*d*f + d^2*e))^2*B*a*b^2*c*d^2*f - sqrt(d*f)*(sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f -
c*d*f + d^2*e))^2*A*b^3*c*d^2*f + 2*sqrt(d*f)*(sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^
2*C*a^3*d^3*f - 2*sqrt(d*f)*(sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^2*B*a^2*b*d^3*f +
2*sqrt(d*f)*(sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^2*A*a*b^2*d^3*f + sqrt(d*f)*C*a^2*
b*d^5*e^2 - sqrt(d*f)*B*a*b^2*d^5*e^2 + sqrt(d*f)*A*b^3*d^5*e^2 - sqrt(d*f)*(sqrt(d*f)*sqrt(d*x + c) - sqrt((d
*x + c)*d*f - c*d*f + d^2*e))^2*C*a^2*b*d^3*e + sqrt(d*f)*(sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*
f + d^2*e))^2*B*a*b^2*d^3*e - sqrt(d*f)*(sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^2*A*b^
3*d^3*e)/((b*c^2*d^2*f^2 - 2*b*c*d^3*f*e - 2*(sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^2
*b*c*d*f + 4*(sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^2*a*d^2*f + b*d^4*e^2 - 2*(sqrt(d
*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^2*b*d^2*e + (sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)
*d*f - c*d*f + d^2*e))^4*b)*(a*b^3*f*abs(d) - b^4*abs(d)*e)) + sqrt((d*x + c)*d*f - c*d*f + d^2*e)*sqrt(d*x +
c)*C*abs(d)/(b^2*d^2*f) - 1/2*(sqrt(d*f)*C*b*c*f - 4*sqrt(d*f)*C*a*d*f + 2*sqrt(d*f)*B*b*d*f - sqrt(d*f)*C*b*d
*e)*log((sqrt(d*f)*sqrt(d*x + c) - sqrt((d*x + c)*d*f - c*d*f + d^2*e))^2)/(b^3*f^2*abs(d))

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maple [B]  time = 0.05, size = 3670, normalized size = 10.08 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d*x+c)^(1/2)*(f*x+e)^(1/2)*(-2*A*b^4*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(
1/2)*(d*f)^(1/2)+4*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b
^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*a^4*d*f^2*(d*f)^(1/2)+B*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d
*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*a^2*b^2*c*f^2*(
d*f)^(1/2)-2*B*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*a^2*b^2*d*f^2*((a^2
*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)-3*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*
b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*a^3*b*c*f^2*(d*f)^(1/2)+4*C*ln(1/2*(2*d*
f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*a^3*b*d*f^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)
/b^2)^(1/2)-C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*a^2*b^2*c*f^2*((a^2*
d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)-C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)
^(1/2))*a*b^3*d*e^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+2*B*a*b^3*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*
e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+2*C*x*b^4*e*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((
d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)-4*C*a^2*b^2*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e)
)^(1/2)*(d*f)^(1/2)+2*C*a*b^3*e*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1
/2)+A*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x
+c)*(f*x+e))^(1/2)*b)/(b*x+a))*x*b^4*c*f^2*(d*f)^(1/2)-C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*
f)^(1/2))/(d*f)^(1/2))*x*b^4*d*e^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+A*ln((-2*a*d*f*x+b*c*f*x+b*d*
e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*a*
b^3*c*f^2*(d*f)^(1/2)-2*B*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c
*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*a^3*b*d*f^2*(d*f)^(1/2)-2*B*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-
a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*x*b^4*
c*e*f*(d*f)^(1/2)+4*C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*x*a^2*b^2*d*
f^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)-C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1
/2))/(d*f)^(1/2))*x*a*b^3*c*f^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*
x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*x*b^4*c*e*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)-A*ln((
-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e
))^(1/2)*b)/(b*x+a))*a*b^3*d*e*f*(d*f)^(1/2)+3*B*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*
f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*a^2*b^2*d*e*f*(d*f)^(1/2)-2*B*ln((-2
*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))
^(1/2)*b)/(b*x+a))*a*b^3*c*e*f*(d*f)^(1/2)+2*B*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/
(d*f)^(1/2))*a*b^3*d*e*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)-5*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*
f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*a^3*b*d*e*
f*(d*f)^(1/2)+4*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)
^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*a^2*b^2*c*e*f*(d*f)^(1/2)-3*C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f
*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*a^2*b^2*d*e*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+C*ln(1/2*(2
*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*a*b^3*c*e*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c
*e)/b^2)^(1/2)-2*C*x*a*b^3*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)
-A*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)
*(f*x+e))^(1/2)*b)/(b*x+a))*x*b^4*d*e*f*(d*f)^(1/2)-2*B*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*(
(a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*x*a^2*b^2*d*f^2*(d*f)^(1/2)+B
*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(
f*x+e))^(1/2)*b)/(b*x+a))*x*a*b^3*c*f^2*(d*f)^(1/2)-2*B*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f
)^(1/2))/(d*f)^(1/2))*x*a*b^3*d*f^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+2*B*ln(1/2*(2*d*f*x+c*f+d*e+
2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*x*b^4*d*e*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+
4*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c
)*(f*x+e))^(1/2)*b)/(b*x+a))*x*a^3*b*d*f^2*(d*f)^(1/2)-3*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+
2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*x*a^2*b^2*c*f^2*(d*f)^(1/2
)+3*B*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x
+c)*(f*x+e))^(1/2)*b)/(b*x+a))*x*a*b^3*d*e*f*(d*f)^(1/2)-5*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*
e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*x*a^2*b^2*d*e*f*(d*f)^(1
/2)+4*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x-a*c*f-a*d*e+2*b*c*e+2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)*((d
*x+c)*(f*x+e))^(1/2)*b)/(b*x+a))*x*a*b^3*c*e*f*(d*f)^(1/2)-3*C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/
2)*(d*f)^(1/2))/(d*f)^(1/2))*x*a*b^3*d*e*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2))/((d*x+c)*(f*x+e))^(1
/2)/(a*f-b*e)/(b*x+a)/(d*f)^(1/2)/((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)/f/b^4

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((-(2*a*d*f)/b^2)>0)', see `as
sume?` for more details)Is ((-(2*a*d*f)/b^2)    +(c*f)/b    +(d*e)/b)    ^2    -(4*d*f       *((a^2*d*f)/b^2
-(a*c*f)/b        -(a*d*e)/b        +c*e))     /b^2 zero or nonzero?

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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

[Out]

Timed out

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