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

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

[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))

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Rubi [A]  time = 0.149335, antiderivative size = 164, normalized size of antiderivative = 1., 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} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d} \sqrt{e+f x}}\right ) \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )}{4 d^{5/2} f^{5/2}}-\frac{\sqrt{c+d x} \sqrt{e+f x} (-4 B d f+5 c C f+3 C d e)}{4 d^2 f^2}+\frac{C (c+d x)^{3/2} \sqrt{e+f x}}{2 d^2 f} \]

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 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 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 \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 = 0.765654, size = 173, normalized size = 1.05 \[ \frac{\sqrt{d e-c f} \sqrt{\frac{d (e+f x)}{d e-c f}} \sinh ^{-1}\left (\frac{\sqrt{f} \sqrt{c+d x}}{\sqrt{d e-c f}}\right ) \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+d \sqrt{f} \sqrt{c+d x} (e+f x) (4 B d f+C (-3 c f-3 d e+2 d f x))}{4 d^3 f^{5/2} \sqrt{e+f x}} \]

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])

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Maple [B]  time = 0.02, size = 425, normalized size = 2.6 \begin{align*}{\frac{1}{8\,{d}^{2}{f}^{2}} \left ( 8\,A\ln \left ( 1/2\,{\frac{2\,dfx+2\,\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }\sqrt{df}+cf+de}{\sqrt{df}}} \right ){d}^{2}{f}^{2}-4\,B\ln \left ( 1/2\,{\frac{2\,dfx+2\,\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }\sqrt{df}+cf+de}{\sqrt{df}}} \right ) cd{f}^{2}-4\,B\ln \left ( 1/2\,{\frac{2\,dfx+2\,\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }\sqrt{df}+cf+de}{\sqrt{df}}} \right ){d}^{2}ef+3\,C\ln \left ( 1/2\,{\frac{2\,dfx+2\,\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }\sqrt{df}+cf+de}{\sqrt{df}}} \right ){c}^{2}{f}^{2}+2\,C\ln \left ( 1/2\,{\frac{2\,dfx+2\,\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }\sqrt{df}+cf+de}{\sqrt{df}}} \right ) cdef+3\,C\ln \left ( 1/2\,{\frac{2\,dfx+2\,\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }\sqrt{df}+cf+de}{\sqrt{df}}} \right ){d}^{2}{e}^{2}+4\,C\sqrt{df}\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }xdf+8\,B\sqrt{df}\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }df-6\,C\sqrt{df}\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }cf-6\,C\sqrt{df}\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }de \right ) \sqrt{dx+c}\sqrt{fx+e}{\frac{1}{\sqrt{ \left ( dx+c \right ) \left ( fx+e \right ) }}}{\frac{1}{\sqrt{df}}}} \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]

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

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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 = 2.51424, size = 879, normalized size = 5.36 \begin{align*} \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{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]

[1/16*((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)]

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

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Giac [A]  time = 3.6856, size = 262, normalized size = 1.6 \begin{align*} \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{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]

1/4*(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)