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

Optimal. Leaf size=290 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) (2 b d f (4 A b d f-a C (c f+3 d e))+(2 a d f-b c f+b d e) (4 a C d f+b (-4 B d f+c C f+3 C d e)))}{4 b^3 d^{3/2} f^{5/2}}-\frac {2 \sqrt {b c-a d} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {b e-a f}}{\sqrt {e+f x} \sqrt {b c-a d}}\right )}{b^3 \sqrt {b e-a f}}-\frac {\sqrt {c+d x} \sqrt {e+f x} (4 a C d f+b (-4 B d f+c C f+3 C d e))}{4 b^2 d f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f} \]

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Rubi [A]  time = 0.67, antiderivative size = 290, 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 = {1615, 154, 157, 63, 217, 206, 93, 208} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) (2 b d f (4 A b d f-a C (c f+3 d e))+(2 a d f-b c f+b d e) (4 a C d f+b (-4 B d f+c C f+3 C d e)))}{4 b^3 d^{3/2} f^{5/2}}-\frac {2 \sqrt {b c-a d} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {b e-a f}}{\sqrt {e+f x} \sqrt {b c-a d}}\right )}{b^3 \sqrt {b e-a f}}-\frac {\sqrt {c+d x} \sqrt {e+f x} (4 a C d f+b (-4 B d f+c C f+3 C d e))}{4 b^2 d f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1615

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[
{q = Expon[Px, x], k = Coeff[Px, x, Expon[Px, x]]}, Simp[(k*(a + b*x)^(m + q - 1)*(c + d*x)^(n + 1)*(e + f*x)^
(p + 1))/(d*f*b^(q - 1)*(m + n + p + q + 1)), x] + Dist[1/(d*f*b^q*(m + n + p + q + 1)), Int[(a + b*x)^m*(c +
d*x)^n*(e + f*x)^p*ExpandToSum[d*f*b^q*(m + n + p + q + 1)*Px - d*f*k*(m + n + p + q + 1)*(a + b*x)^q + k*(a +
b*x)^(q - 2)*(a^2*d*f*(m + n + p + q + 1) - b*(b*c*e*(m + q - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*
(2*(m + q) + n + p) - b*(d*e*(m + q + n) + c*f*(m + q + p)))*x), x], x], x] /; NeQ[m + n + p + q + 1, 0]] /; F
reeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && 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) \sqrt {e+f x}} \, dx &=\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}+\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} b (4 A b d f-a C (3 d e+c f))-\frac {1}{2} b (4 a C d f+b (3 C d e+c C f-4 B d f)) x\right )}{(a+b x) \sqrt {e+f x}} \, dx}{2 b^2 d f}\\ &=-\frac {(4 a C d f+b (3 C d e+c C f-4 B d f)) \sqrt {c+d x} \sqrt {e+f x}}{4 b^2 d f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}+\frac {\int \frac {\frac {1}{4} b (2 b c f (4 A b d f-a C (3 d e+c f))+a (d e+c f) (4 a C d f+b (3 C d e+c C f-4 B d f)))+\frac {1}{4} b (2 b d f (4 A b d f-a C (3 d e+c f))+(b d e-b c f+2 a d f) (4 a C d f+b (3 C d e+c C f-4 B d f))) x}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx}{2 b^3 d f^2}\\ &=-\frac {(4 a C d f+b (3 C d e+c C f-4 B d f)) \sqrt {c+d x} \sqrt {e+f x}}{4 b^2 d f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}+\frac {\left (\left (A b^2-a (b B-a C)\right ) (b c-a d)\right ) \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx}{b^3}+\frac {(2 b d f (4 A b d f-a C (3 d e+c f))+(b d e-b c f+2 a d f) (4 a C d f+b (3 C d e+c C f-4 B d f))) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}} \, dx}{8 b^3 d f^2}\\ &=-\frac {(4 a C d f+b (3 C d e+c C f-4 B d f)) \sqrt {c+d x} \sqrt {e+f x}}{4 b^2 d f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}+\frac {\left (2 \left (A b^2-a (b B-a C)\right ) (b c-a d)\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}+\frac {(2 b d f (4 A b d f-a C (3 d e+c f))+(b d e-b c f+2 a d f) (4 a C d f+b (3 C d e+c C f-4 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 )}{4 b^3 d^2 f^2}\\ &=-\frac {(4 a C d f+b (3 C d e+c C f-4 B d f)) \sqrt {c+d x} \sqrt {e+f x}}{4 b^2 d f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {b c-a d} \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 e-a f}}+\frac {(2 b d f (4 A b d f-a C (3 d e+c f))+(b d e-b c f+2 a d f) (4 a C d f+b (3 C d e+c C f-4 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 )}{4 b^3 d^2 f^2}\\ &=-\frac {(4 a C d f+b (3 C d e+c C f-4 B d f)) \sqrt {c+d x} \sqrt {e+f x}}{4 b^2 d f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 b d f}+\frac {(2 b d f (4 A b d f-a C (3 d e+c f))+(b d e-b c f+2 a d f) (4 a C d f+b (3 C d e+c C f-4 B d f))) \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{4 b^3 d^{3/2} f^{5/2}}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {b c-a d} \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 e-a f}}\\ \end {align*}

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Mathematica [A]  time = 3.45, size = 465, normalized size = 1.60 \begin {gather*} \frac {\frac {8 \sqrt {d e-c f} \left (a (a C-b B)+A b^2\right ) \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 {8 \sqrt {a d-b c} \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 f-b e}}+\frac {4 b \sqrt {e+f x} (a C f-b B f+b C e) \left (\sqrt {c+d x} (d e-c f) \sinh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d e-c f}}\right )-\sqrt {f} (c+d x) \sqrt {d e-c f} \sqrt {\frac {d (e+f x)}{d e-c f}}\right )}{f^{5/2} \sqrt {c+d x} \sqrt {d e-c f} \sqrt {\frac {d (e+f x)}{d e-c f}}}+\frac {b^2 C \sqrt {e+f x} \left (\sqrt {f} \sqrt {c+d x} (c f+d (e+2 f x))-\frac {(d e-c f)^{3/2} \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 )}{d f^{5/2}}}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*(A*b^2 + a*(-(b*B) + 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]) + (4*b*(b*C*e - b*B*f + a*C*f)*Sqrt[e + f*x]*(-(Sqrt[f]*Sqrt[d*e -
c*f]*(c + d*x)*Sqrt[(d*(e + f*x))/(d*e - c*f)]) + (d*e - c*f)*Sqrt[c + d*x]*ArcSinh[(Sqrt[f]*Sqrt[c + d*x])/Sq
rt[d*e - c*f]]))/(f^(5/2)*Sqrt[d*e - c*f]*Sqrt[c + d*x]*Sqrt[(d*(e + f*x))/(d*e - c*f)]) + (b^2*C*Sqrt[e + f*x
]*(Sqrt[f]*Sqrt[c + d*x]*(c*f + d*(e + 2*f*x)) - ((d*e - c*f)^(3/2)*ArcSinh[(Sqrt[f]*Sqrt[c + d*x])/Sqrt[d*e -
c*f]])/Sqrt[(d*(e + f*x))/(d*e - c*f)]))/(d*f^(5/2)) - (8*(A*b^2 + a*(-(b*B) + a*C))*Sqrt[-(b*c) + a*d]*ArcTa
nh[(Sqrt[-(b*e) + a*f]*Sqrt[c + d*x])/(Sqrt[-(b*c) + a*d]*Sqrt[e + f*x])])/Sqrt[-(b*e) + a*f])/(4*b^3)

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IntegrateAlgebraic [B]  time = 34.33, size = 1375, normalized size = 4.74 \begin {gather*} \frac {C \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right ) c^2}{4 b \left (\frac {d}{f}\right )^{3/2} f^2}-\frac {B d \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right ) c}{b \left (\frac {d}{f}\right )^{3/2} f^2}+\frac {a C d \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right ) c}{b^2 \left (\frac {d}{f}\right )^{3/2} f^2}+\frac {C d e \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right ) c}{2 b \left (\frac {d}{f}\right )^{3/2} f^3}+\frac {\sqrt {\frac {d}{f}} \sqrt {e+f x} (-5 b C d e+b c C f+4 b B d f-4 a C d f+2 b C d (e+f x)) \sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}} \left (c^2 f^3-2 c d e f^2+8 c d (e+f x) f^2+d^2 e^2 f+8 d^2 (e+f x)^2 f-8 d^2 e (e+f x) f\right )+\sqrt {e+f x} (-5 b C d e+b c C f+4 b B d f-4 a C d f+2 b C d (e+f x)) \left (-8 (e+f x)^{5/2} d^3+12 e (e+f x)^{3/2} d^3-4 e^2 \sqrt {e+f x} d^3-12 c f (e+f x)^{3/2} d^2+8 c e f \sqrt {e+f x} d^2-4 c^2 f^2 \sqrt {e+f x} d\right )}{4 b^2 d \sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}} \left (-\frac {8 (e+f x)^{3/2} d^2}{f^2}+\frac {4 e \sqrt {e+f x} d^2}{f^2}-\frac {4 c \sqrt {e+f x} d}{f}\right ) f^5+4 b^2 d \sqrt {\frac {d}{f}} \left (c^2+\frac {8 d (e+f x) c}{f}-\frac {2 d e c}{f}+\frac {8 d^2 (e+f x)^2}{f^2}-\frac {8 d^2 e (e+f x)}{f^2}+\frac {d^2 e^2}{f^2}\right ) f^5}+\left (\frac {2 C \sqrt {d} \sqrt {b c-a d} a^2}{b^3 \sqrt {\frac {d}{f}} \sqrt {f} \sqrt {b e-a f}}-\frac {2 B \sqrt {d} \sqrt {b c-a d} a}{b^2 \sqrt {\frac {d}{f}} \sqrt {f} \sqrt {b e-a f}}+\frac {2 A \sqrt {d} \sqrt {b c-a d}}{b \sqrt {\frac {d}{f}} \sqrt {f} \sqrt {b e-a f}}\right ) \tanh ^{-1}\left (\frac {-b d e+a d f+b d (e+f x)-b \sqrt {\frac {d}{f}} f \sqrt {e+f x} \sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}}{\sqrt {d} \sqrt {b c-a d} \sqrt {f} \sqrt {b e-a f}}\right )+\frac {2 a B d^2 \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right )}{b^2 \left (\frac {d}{f}\right )^{3/2} f^2}-\frac {2 a^2 C d^2 \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right )}{b^3 \left (\frac {d}{f}\right )^{3/2} f^2}-\frac {2 A d^2 \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right )}{b \left (\frac {d}{f}\right )^{3/2} f^2}+\frac {B d^2 e \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right )}{b \left (\frac {d}{f}\right )^{3/2} f^3}-\frac {a C d^2 e \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right )}{b^2 \left (\frac {d}{f}\right )^{3/2} f^3}-\frac {3 C d^2 e^2 \log \left (\sqrt {c+\frac {d (e+f x)}{f}-\frac {d e}{f}}-\sqrt {\frac {d}{f}} \sqrt {e+f x}\right )}{4 b \left (\frac {d}{f}\right )^{3/2} f^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d/f]*Sqrt[e + f*x]*(-5*b*C*d*e + b*c*C*f + 4*b*B*d*f - 4*a*C*d*f + 2*b*C*d*(e + f*x))*Sqrt[c - (d*e)/f +
(d*(e + f*x))/f]*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3 - 8*d^2*e*f*(e + f*x) + 8*c*d*f^2*(e + f*x) + 8*d^2*f*(e
+ f*x)^2) + Sqrt[e + f*x]*(-5*b*C*d*e + b*c*C*f + 4*b*B*d*f - 4*a*C*d*f + 2*b*C*d*(e + f*x))*(-4*d^3*e^2*Sqrt[
e + f*x] + 8*c*d^2*e*f*Sqrt[e + f*x] - 4*c^2*d*f^2*Sqrt[e + f*x] + 12*d^3*e*(e + f*x)^(3/2) - 12*c*d^2*f*(e +
f*x)^(3/2) - 8*d^3*(e + f*x)^(5/2)))/(4*b^2*d*f^5*Sqrt[c - (d*e)/f + (d*(e + f*x))/f]*((4*d^2*e*Sqrt[e + f*x])
/f^2 - (4*c*d*Sqrt[e + f*x])/f - (8*d^2*(e + f*x)^(3/2))/f^2) + 4*b^2*d*Sqrt[d/f]*f^5*(c^2 + (d^2*e^2)/f^2 - (
2*c*d*e)/f - (8*d^2*e*(e + f*x))/f^2 + (8*c*d*(e + f*x))/f + (8*d^2*(e + f*x)^2)/f^2)) + ((2*A*Sqrt[d]*Sqrt[b*
c - a*d])/(b*Sqrt[d/f]*Sqrt[f]*Sqrt[b*e - a*f]) - (2*a*B*Sqrt[d]*Sqrt[b*c - a*d])/(b^2*Sqrt[d/f]*Sqrt[f]*Sqrt[
b*e - a*f]) + (2*a^2*C*Sqrt[d]*Sqrt[b*c - a*d])/(b^3*Sqrt[d/f]*Sqrt[f]*Sqrt[b*e - a*f]))*ArcTanh[(-(b*d*e) + a
*d*f + b*d*(e + f*x) - b*Sqrt[d/f]*f*Sqrt[e + f*x]*Sqrt[c - (d*e)/f + (d*(e + f*x))/f])/(Sqrt[d]*Sqrt[b*c - a*
d]*Sqrt[f]*Sqrt[b*e - a*f])] - (3*C*d^2*e^2*Log[-(Sqrt[d/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/
f]])/(4*b*(d/f)^(3/2)*f^4) + (c*C*d*e*Log[-(Sqrt[d/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/f]])/(
2*b*(d/f)^(3/2)*f^3) + (B*d^2*e*Log[-(Sqrt[d/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/f]])/(b*(d/f
)^(3/2)*f^3) - (a*C*d^2*e*Log[-(Sqrt[d/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/f]])/(b^2*(d/f)^(3
/2)*f^3) + (c^2*C*Log[-(Sqrt[d/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/f]])/(4*b*(d/f)^(3/2)*f^2)
- (B*c*d*Log[-(Sqrt[d/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/f]])/(b*(d/f)^(3/2)*f^2) + (a*c*C*
d*Log[-(Sqrt[d/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/f]])/(b^2*(d/f)^(3/2)*f^2) - (2*A*d^2*Log[
-(Sqrt[d/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/f]])/(b*(d/f)^(3/2)*f^2) + (2*a*B*d^2*Log[-(Sqrt
[d/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/f]])/(b^2*(d/f)^(3/2)*f^2) - (2*a^2*C*d^2*Log[-(Sqrt[d
/f]*Sqrt[e + f*x]) + Sqrt[c - (d*e)/f + (d*(e + f*x))/f]])/(b^3*(d/f)^(3/2)*f^2)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)/(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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)/(f*x+e)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 0.47

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maple [B]  time = 0.04, size = 1822, normalized size = 6.28

result 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)/(f*x+e)^(1/2),x)

[Out]

1/8*(8*A*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))*b^3*d^2*f^2*((a^2*d*f-a*b
*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+8*A*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x+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-a*c*f-a*d*e+2*b*c*e)/(b*x+a))*a*b^2*d^2*f^2*(d*f)^(1/2)-8*A*ln((-2*a*d*f*x+b*c
*f*x+b*d*e*x+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-a*c*f-a*d*e+2*b*c*e)/(b
*x+a))*b^3*c*d*f^2*(d*f)^(1/2)-8*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^2*d^2*f^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+4*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))*b^3*c*d*f^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)-4*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))*b^3*d^2*e*f*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)
^(1/2)-8*B*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x+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-a*c*f-a*d*e+2*b*c*e)/(b*x+a))*a^2*b*d^2*f^2*(d*f)^(1/2)+8*B*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x+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-a*c*f-a*d*e+2*b*c*e)/(b*x+a))*a*b^2*c*d*f^2*(d*f)
^(1/2)+8*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*d^2*f^2*((a^2*d*f
-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(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*b^2*c*d*f^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(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*b^2*d^2*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))*b^3*c^2*f^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e
)/b^2)^(1/2)-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))*b^3*c*d*e*f*((a^2
*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(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))*b^3*d^2*e^2*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+8*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x+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-a*c*f-a*d*e+2*b*c*e)/(b*x+a))*a^3*d^2*f^2*
(d*f)^(1/2)-8*C*ln((-2*a*d*f*x+b*c*f*x+b*d*e*x+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-a*c*f-a*d*e+2*b*c*e)/(b*x+a))*a^2*b*c*d*f^2*(d*f)^(1/2)+4*C*x*b^3*d*f*((d*x+c)*(f*x+e))^(1/2)*(d*f)
^(1/2)*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)+8*B*b^3*d*f*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)*((a^2*d*f
-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)-8*C*a*b^2*d*f*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)*((a^2*d*f-a*b*c*f-a*b*d
*e+b^2*c*e)/b^2)^(1/2)+2*C*b^3*c*f*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)*((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)
^(1/2)-6*C*b^3*d*e*((d*x+c)*(f*x+e))^(1/2)*(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*x+e)^
(1/2)*(d*x+c)^(1/2)/((a^2*d*f-a*b*c*f-a*b*d*e+b^2*c*e)/b^2)^(1/2)/(d*f)^(1/2)/d/f^2/b^4/((d*x+c)*(f*x+e))^(1/2
)

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

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)/(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 \begin {gather*} \text {Hanged} \end {gather*}

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)),x)

[Out]

\text{Hanged}

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

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)/(f*x+e)**(1/2),x)

[Out]

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

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