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

Optimal. Leaf size=164 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )}{4 d^{5/2} f^{5/2}}-\frac {\sqrt {c+d x} \sqrt {e+f x} (-4 B d f+5 c C f+3 C d e)}{4 d^2 f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f} \]

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {951, 80, 63, 217, 206} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )}{4 d^{5/2} f^{5/2}}-\frac {\sqrt {c+d x} \sqrt {e+f x} (-4 B d f+5 c C f+3 C d e)}{4 d^2 f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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])

Rubi steps

\begin {align*} \int \frac {A+B x+C x^2}{\sqrt {c+d x} \sqrt {e+f x}} \, dx &=\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}+\frac {\int \frac {\frac {1}{2} \left (-3 c C d e-c^2 C f+4 A d^2 f\right )-\frac {1}{2} d (3 C d e+5 c C f-4 B d f) x}{\sqrt {c+d x} \sqrt {e+f x}} \, dx}{2 d^2 f}\\ &=-\frac {(3 C d e+5 c C f-4 B d f) \sqrt {c+d x} \sqrt {e+f x}}{4 d^2 f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}+\frac {\left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}} \, dx}{8 d^2 f^2}\\ &=-\frac {(3 C d e+5 c C f-4 B d f) \sqrt {c+d x} \sqrt {e+f x}}{4 d^2 f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}+\frac {\left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\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 )}{4 d^3 f^2}\\ &=-\frac {(3 C d e+5 c C f-4 B d f) \sqrt {c+d x} \sqrt {e+f x}}{4 d^2 f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}+\frac {\left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 d f (2 A d f-B (d e+c f))\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 )}{4 d^3 f^2}\\ &=-\frac {(3 C d e+5 c C f-4 B d f) \sqrt {c+d x} \sqrt {e+f x}}{4 d^2 f^2}+\frac {C (c+d x)^{3/2} \sqrt {e+f x}}{2 d^2 f}+\frac {\left (C \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )+4 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 )}{4 d^{5/2} f^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.79, size = 173, normalized size = 1.05 \begin {gather*} \frac {\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 ) \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+d \sqrt {f} \sqrt {c+d x} (e+f x) (4 B d f+C (-3 c f-3 d e+2 d f x))}{4 d^3 f^{5/2} \sqrt {e+f x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[f]*Sqrt[c + d*x]*(e + f*x)*(4*B*d*f + C*(-3*d*e - 3*c*f + 2*d*f*x)) + Sqrt[d*e - c*f]*(C*(3*d^2*e^2 +
2*c*d*e*f + 3*c^2*f^2) + 4*d*f*(2*A*d*f - B*(d*e + c*f)))*Sqrt[(d*(e + f*x))/(d*e - c*f)]*ArcSinh[(Sqrt[f]*Sqr
t[c + d*x])/Sqrt[d*e - c*f]])/(4*d^3*f^(5/2)*Sqrt[e + f*x])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.38, size = 229, normalized size = 1.40 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {f} \sqrt {c+d x}}\right ) \left (8 A d^2 f^2-4 B c d f^2-4 B d^2 e f+3 c^2 C f^2+2 c C d e f+3 C d^2 e^2\right )}{4 d^{5/2} f^{5/2}}+\frac {\sqrt {e+f x} (d e-c f) \left (\frac {4 B d^2 f (e+f x)}{c+d x}-4 B d f^2-\frac {3 C d^2 e (e+f x)}{c+d x}-\frac {5 c C d f (e+f x)}{c+d x}+3 c C f^2+5 C d e f\right )}{4 d^2 f^2 \sqrt {c+d x} \left (\frac {d (e+f x)}{c+d x}-f\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.57, size = 380, normalized size = 2.32 \begin {gather*} \left [\frac {{\left (3 \, C d^{2} e^{2} + 2 \, {\left (C c d - 2 \, B d^{2}\right )} e f + {\left (3 \, C c^{2} - 4 \, B c d + 8 \, A d^{2}\right )} f^{2}\right )} \sqrt {d f} \log \left (8 \, d^{2} f^{2} x^{2} + d^{2} e^{2} + 6 \, c d e f + c^{2} f^{2} + 4 \, {\left (2 \, d f x + d e + c f\right )} \sqrt {d f} \sqrt {d x + c} \sqrt {f x + e} + 8 \, {\left (d^{2} e f + c d f^{2}\right )} x\right ) + 4 \, {\left (2 \, C d^{2} f^{2} x - 3 \, C d^{2} e f - {\left (3 \, C c d - 4 \, B d^{2}\right )} f^{2}\right )} \sqrt {d x + c} \sqrt {f x + e}}{16 \, d^{3} f^{3}}, -\frac {{\left (3 \, C d^{2} e^{2} + 2 \, {\left (C c d - 2 \, B d^{2}\right )} e f + {\left (3 \, C c^{2} - 4 \, B c d + 8 \, A d^{2}\right )} f^{2}\right )} \sqrt {-d f} \arctan \left (\frac {{\left (2 \, d f x + d e + c f\right )} \sqrt {-d f} \sqrt {d x + c} \sqrt {f x + e}}{2 \, {\left (d^{2} f^{2} x^{2} + c d e f + {\left (d^{2} e f + c d f^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, C d^{2} f^{2} x - 3 \, C d^{2} e f - {\left (3 \, C c d - 4 \, B d^{2}\right )} f^{2}\right )} \sqrt {d x + c} \sqrt {f x + e}}{8 \, d^{3} f^{3}}\right ] \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)/(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

[1/16*((3*C*d^2*e^2 + 2*(C*c*d - 2*B*d^2)*e*f + (3*C*c^2 - 4*B*c*d + 8*A*d^2)*f^2)*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*(2*C*d^2*f^2*x - 3*C*d^2*e*f - (3*C*c*d - 4*B*d^2)*f^2)*sqrt(d*x + c)*sqrt(f*x + e))/(d^3*f^
3), -1/8*((3*C*d^2*e^2 + 2*(C*c*d - 2*B*d^2)*e*f + (3*C*c^2 - 4*B*c*d + 8*A*d^2)*f^2)*sqrt(-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*(2*C*d^2*f^2*x - 3*C*d^2*e*f - (3*C*c*d - 4*B*d^2)*f^2)*sqrt(d*x + c)*sqrt(f*x + e))/(d^3*f^3)]

________________________________________________________________________________________

giac [A]  time = 1.22, size = 194, normalized size = 1.18 \begin {gather*} \frac {{\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 )} C}{d^{3} f} - \frac {5 \, C c d^{5} f^{2} - 4 \, B d^{6} f^{2} + 3 \, C d^{6} f e}{d^{8} f^{3}}\right )} - \frac {{\left (3 \, C c^{2} f^{2} - 4 \, B c d f^{2} + 8 \, A d^{2} f^{2} + 2 \, C c d f e - 4 \, B d^{2} f e + 3 \, C 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^{2} f^{2}}\right )} d}{4 \, {\left | d \right |}} \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)/(f*x+e)^(1/2),x, algorithm="giac")

[Out]

1/4*(sqrt((d*x + c)*d*f - c*d*f + d^2*e)*sqrt(d*x + c)*(2*(d*x + c)*C/(d^3*f) - (5*C*c*d^5*f^2 - 4*B*d^6*f^2 +
3*C*d^6*f*e)/(d^8*f^3)) - (3*C*c^2*f^2 - 4*B*c*d*f^2 + 8*A*d^2*f^2 + 2*C*c*d*f*e - 4*B*d^2*f*e + 3*C*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^2*f^2))*d/abs(d)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 425, normalized size = 2.59 \begin {gather*} \frac {\left (8 A \,d^{2} f^{2} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )-4 B c d \,f^{2} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )-4 B \,d^{2} e f \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )+3 C \,c^{2} f^{2} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )+2 C c d e f \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )+3 C \,d^{2} e^{2} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )+4 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, C d f x +8 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, B d f -6 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, C c f -6 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, C d e \right ) \sqrt {d x +c}\, \sqrt {f x +e}}{8 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, d^{2} f^{2}} \end {gather*}

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]

1/8*(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))*d^2*f^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))*c*d*f^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))*d^2*e*f+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))*c^2*f^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))*c*d
*e*f+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))*d^2*e^2+4*C*(d*f)^(1/2)*(
(d*x+c)*(f*x+e))^(1/2)*x*d*f+8*B*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*d*f-6*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/
2)*c*f-6*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*d*e)*(d*x+c)^(1/2)*(f*x+e)^(1/2)/(d*f)^(1/2)/f^2/d^2/((d*x+c)*(
f*x+e))^(1/2)

________________________________________________________________________________________

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)/(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(c*f-d*e>0)', see `assume?` for
more details)Is c*f-d*e zero or nonzero?

________________________________________________________________________________________

mupad [B]  time = 25.89, size = 833, normalized size = 5.08 \begin {gather*} \frac {\frac {\left (2\,B\,c\,f+2\,B\,d\,e\right )\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{f^3\,\left (\sqrt {e+f\,x}-\sqrt {e}\right )}+\frac {\left (2\,B\,c\,f+2\,B\,d\,e\right )\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}{d\,f^2\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^3}-\frac {8\,B\,\sqrt {c}\,\sqrt {e}\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{f^2\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^2}}{\frac {{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}{{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^4}+\frac {d^2}{f^2}-\frac {2\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{f\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^2}}-\frac {\frac {\left (\sqrt {c+d\,x}-\sqrt {c}\right )\,\left (\frac {3\,C\,c^2\,d\,f^2}{2}+C\,c\,d^2\,e\,f+\frac {3\,C\,d^3\,e^2}{2}\right )}{f^6\,\left (\sqrt {e+f\,x}-\sqrt {e}\right )}-\frac {{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3\,\left (\frac {11\,C\,c^2\,f^2}{2}+25\,C\,c\,d\,e\,f+\frac {11\,C\,d^2\,e^2}{2}\right )}{f^5\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^3}+\frac {{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7\,\left (\frac {3\,C\,c^2\,f^2}{2}+C\,c\,d\,e\,f+\frac {3\,C\,d^2\,e^2}{2}\right )}{d^2\,f^3\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^7}-\frac {{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5\,\left (\frac {11\,C\,c^2\,f^2}{2}+25\,C\,c\,d\,e\,f+\frac {11\,C\,d^2\,e^2}{2}\right )}{d\,f^4\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^5}+\frac {\sqrt {c}\,\sqrt {e}\,\left (32\,C\,c\,f+32\,C\,d\,e\right )\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}{f^4\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^4}}{\frac {{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}{{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^8}+\frac {d^4}{f^4}-\frac {4\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}{f\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^6}-\frac {4\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{f^3\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^2}+\frac {6\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}{f^2\,{\left (\sqrt {e+f\,x}-\sqrt {e}\right )}^4}}-\frac {4\,A\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {e+f\,x}-\sqrt {e}\right )}{\sqrt {-d\,f}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )}{\sqrt {-d\,f}}-\frac {2\,B\,\mathrm {atanh}\left (\frac {\sqrt {f}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{\sqrt {d}\,\left (\sqrt {e+f\,x}-\sqrt {e}\right )}\right )\,\left (c\,f+d\,e\right )}{d^{3/2}\,f^{3/2}}+\frac {C\,\mathrm {atanh}\left (\frac {\sqrt {f}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{\sqrt {d}\,\left (\sqrt {e+f\,x}-\sqrt {e}\right )}\right )\,\left (3\,c^2\,f^2+2\,c\,d\,e\,f+3\,d^2\,e^2\right )}{2\,d^{5/2}\,f^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((2*B*c*f + 2*B*d*e)*((c + d*x)^(1/2) - c^(1/2)))/(f^3*((e + f*x)^(1/2) - e^(1/2))) + ((2*B*c*f + 2*B*d*e)*((
c + d*x)^(1/2) - c^(1/2))^3)/(d*f^2*((e + f*x)^(1/2) - e^(1/2))^3) - (8*B*c^(1/2)*e^(1/2)*((c + d*x)^(1/2) - c
^(1/2))^2)/(f^2*((e + f*x)^(1/2) - e^(1/2))^2))/(((c + d*x)^(1/2) - c^(1/2))^4/((e + f*x)^(1/2) - e^(1/2))^4 +
d^2/f^2 - (2*d*((c + d*x)^(1/2) - c^(1/2))^2)/(f*((e + f*x)^(1/2) - e^(1/2))^2)) - ((((c + d*x)^(1/2) - c^(1/
2))*((3*C*d^3*e^2)/2 + (3*C*c^2*d*f^2)/2 + C*c*d^2*e*f))/(f^6*((e + f*x)^(1/2) - e^(1/2))) - (((c + d*x)^(1/2)
- c^(1/2))^3*((11*C*c^2*f^2)/2 + (11*C*d^2*e^2)/2 + 25*C*c*d*e*f))/(f^5*((e + f*x)^(1/2) - e^(1/2))^3) + (((c
+ d*x)^(1/2) - c^(1/2))^7*((3*C*c^2*f^2)/2 + (3*C*d^2*e^2)/2 + C*c*d*e*f))/(d^2*f^3*((e + f*x)^(1/2) - e^(1/2
))^7) - (((c + d*x)^(1/2) - c^(1/2))^5*((11*C*c^2*f^2)/2 + (11*C*d^2*e^2)/2 + 25*C*c*d*e*f))/(d*f^4*((e + f*x)
^(1/2) - e^(1/2))^5) + (c^(1/2)*e^(1/2)*(32*C*c*f + 32*C*d*e)*((c + d*x)^(1/2) - c^(1/2))^4)/(f^4*((e + f*x)^(
1/2) - e^(1/2))^4))/(((c + d*x)^(1/2) - c^(1/2))^8/((e + f*x)^(1/2) - e^(1/2))^8 + d^4/f^4 - (4*d*((c + d*x)^(
1/2) - c^(1/2))^6)/(f*((e + f*x)^(1/2) - e^(1/2))^6) - (4*d^3*((c + d*x)^(1/2) - c^(1/2))^2)/(f^3*((e + f*x)^(
1/2) - e^(1/2))^2) + (6*d^2*((c + d*x)^(1/2) - c^(1/2))^4)/(f^2*((e + f*x)^(1/2) - e^(1/2))^4)) - (4*A*atan((d
*((e + f*x)^(1/2) - e^(1/2)))/((-d*f)^(1/2)*((c + d*x)^(1/2) - c^(1/2)))))/(-d*f)^(1/2) - (2*B*atanh((f^(1/2)*
((c + d*x)^(1/2) - c^(1/2)))/(d^(1/2)*((e + f*x)^(1/2) - e^(1/2))))*(c*f + d*e))/(d^(3/2)*f^(3/2)) + (C*atanh(
(f^(1/2)*((c + d*x)^(1/2) - c^(1/2)))/(d^(1/2)*((e + f*x)^(1/2) - e^(1/2))))*(3*c^2*f^2 + 3*d^2*e^2 + 2*c*d*e*
f))/(2*d^(5/2)*f^(5/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x + C x^{2}}{\sqrt {c + d x} \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)/(f*x+e)**(1/2),x)

[Out]

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

________________________________________________________________________________________