3.59 \(\int \frac {A+B x+C x^2}{(a+b x)^3 \sqrt {c+d x} \sqrt {e+f x}} \, dx\)

Optimal. Leaf size=424 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {b e-a f}}{\sqrt {e+f x} \sqrt {b c-a d}}\right ) \left (a^2 \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+a b \left (-2 c d \left (4 A f^2-7 B e f+4 C e^2\right )+d^2 e (B e-8 A f)+c^2 (-f) (8 C e-B f)\right )+b^2 \left (c^2 \left (3 A f^2-4 B e f+8 C e^2\right )-2 c d e (2 B e-A f)+3 A d^2 e^2\right )\right )}{4 (b c-a d)^{5/2} (b e-a f)^{5/2}}+\frac {\sqrt {c+d x} \sqrt {e+f x} \left (2 a^3 C d f+a^2 b (2 B d f-5 C (c f+d e))+a b^2 (-6 A d f+B c f+B d e+8 c C e)-b^3 (4 B c e-3 A (c f+d e))\right )}{4 b (a+b x) (b c-a d)^2 (b e-a f)^2}-\frac {\sqrt {c+d x} \sqrt {e+f x} \left (A b^2-a (b B-a C)\right )}{2 b (a+b x)^2 (b c-a d) (b e-a f)} \]

[Out]

-1/4*(b^2*(3*A*d^2*e^2-2*c*d*e*(-A*f+2*B*e)+c^2*(3*A*f^2-4*B*e*f+8*C*e^2))+a*b*(d^2*e*(-8*A*f+B*e)-c^2*f*(-B*f
+8*C*e)-2*c*d*(4*A*f^2-7*B*e*f+4*C*e^2))+a^2*(C*(3*c^2*f^2+2*c*d*e*f+3*d^2*e^2)+4*d*f*(2*A*d*f-B*(c*f+d*e))))*
arctanh((-a*f+b*e)^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2)/(f*x+e)^(1/2))/(-a*d+b*c)^(5/2)/(-a*f+b*e)^(5/2)-1/2*(
A*b^2-a*(B*b-C*a))*(d*x+c)^(1/2)*(f*x+e)^(1/2)/b/(-a*d+b*c)/(-a*f+b*e)/(b*x+a)^2+1/4*(2*a^3*C*d*f+a*b^2*(-6*A*
d*f+B*c*f+B*d*e+8*C*c*e)-b^3*(4*B*c*e-3*A*(c*f+d*e))+a^2*b*(2*B*d*f-5*C*(c*f+d*e)))*(d*x+c)^(1/2)*(f*x+e)^(1/2
)/b/(-a*d+b*c)^2/(-a*f+b*e)^2/(b*x+a)

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Rubi [A]  time = 0.97, antiderivative size = 424, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {1613, 151, 12, 93, 208} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {b e-a f}}{\sqrt {e+f x} \sqrt {b c-a d}}\right ) \left (a^2 \left (4 d f (2 A d f-B (c f+d e))+C \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )+a b \left (-2 c d \left (4 A f^2-7 B e f+4 C e^2\right )+d^2 e (B e-8 A f)+c^2 (-f) (8 C e-B f)\right )+b^2 \left (c^2 \left (3 A f^2-4 B e f+8 C e^2\right )-2 c d e (2 B e-A f)+3 A d^2 e^2\right )\right )}{4 (b c-a d)^{5/2} (b e-a f)^{5/2}}+\frac {\sqrt {c+d x} \sqrt {e+f x} \left (a^2 b (2 B d f-5 C (c f+d e))+2 a^3 C d f+a b^2 (-6 A d f+B c f+B d e+8 c C e)-b^3 (4 B c e-3 A (c f+d e))\right )}{4 b (a+b x) (b c-a d)^2 (b e-a f)^2}-\frac {\sqrt {c+d x} \sqrt {e+f x} \left (A b^2-a (b B-a C)\right )}{2 b (a+b x)^2 (b c-a d) (b e-a f)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/((a + b*x)^3*Sqrt[c + d*x]*Sqrt[e + f*x]),x]

[Out]

-((A*b^2 - a*(b*B - a*C))*Sqrt[c + d*x]*Sqrt[e + f*x])/(2*b*(b*c - a*d)*(b*e - a*f)*(a + b*x)^2) + ((2*a^3*C*d
*f + a*b^2*(8*c*C*e + B*d*e + B*c*f - 6*A*d*f) - b^3*(4*B*c*e - 3*A*(d*e + c*f)) + a^2*b*(2*B*d*f - 5*C*(d*e +
c*f)))*Sqrt[c + d*x]*Sqrt[e + f*x])/(4*b*(b*c - a*d)^2*(b*e - a*f)^2*(a + b*x)) - ((b^2*(3*A*d^2*e^2 - 2*c*d*
e*(2*B*e - A*f) + c^2*(8*C*e^2 - 4*B*e*f + 3*A*f^2)) + a*b*(d^2*e*(B*e - 8*A*f) - c^2*f*(8*C*e - B*f) - 2*c*d*
(4*C*e^2 - 7*B*e*f + 4*A*f^2)) + a^2*(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[b*e - a*f]*Sqrt[c + d*x])/(Sqrt[b*c - a*d]*Sqrt[e + f*x])])/(4*(b*c - a*d)^(5/2)*(b*e - a*f)^(
5/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

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 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 {A+B x+C x^2}{(a+b x)^3 \sqrt {c+d x} \sqrt {e+f x}} \, dx &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d x} \sqrt {e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}-\frac {\int \frac {-\frac {a^2 C (d e+c f)-a b (4 c C e+B d e+B c f-4 A d f)+b^2 (4 B c e-3 A (d e+c f))}{2 b}+\left (-2 b c C e+2 a C d e+2 a c C f+A b d f-a B d f-\frac {a^2 C d f}{b}\right ) x}{(a+b x)^2 \sqrt {c+d x} \sqrt {e+f x}} \, dx}{2 (b c-a d) (b e-a f)}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d x} \sqrt {e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}+\frac {\left (2 a^3 C d f+a b^2 (8 c C e+B d e+B c f-6 A d f)-b^3 (4 B c e-3 A (d e+c f))+a^2 b (2 B d f-5 C (d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{4 b (b c-a d)^2 (b e-a f)^2 (a+b x)}+\frac {\int \frac {b^2 \left (3 A d^2 e^2-2 c d e (2 B e-A f)+c^2 \left (8 C e^2-4 B e f+3 A f^2\right )\right )+a b \left (d^2 e (B e-8 A f)-c^2 f (8 C e-B f)-2 c d \left (4 C e^2-7 B e f+4 A f^2\right )\right )+a^2 \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 )}{4 (a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx}{2 (b c-a d)^2 (b e-a f)^2}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d x} \sqrt {e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}+\frac {\left (2 a^3 C d f+a b^2 (8 c C e+B d e+B c f-6 A d f)-b^3 (4 B c e-3 A (d e+c f))+a^2 b (2 B d f-5 C (d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{4 b (b c-a d)^2 (b e-a f)^2 (a+b x)}+\frac {\left (b^2 \left (3 A d^2 e^2-2 c d e (2 B e-A f)+c^2 \left (8 C e^2-4 B e f+3 A f^2\right )\right )+a b \left (d^2 e (B e-8 A f)-c^2 f (8 C e-B f)-2 c d \left (4 C e^2-7 B e f+4 A f^2\right )\right )+a^2 \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 )\right ) \int \frac {1}{(a+b x) \sqrt {c+d x} \sqrt {e+f x}} \, dx}{8 (b c-a d)^2 (b e-a f)^2}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d x} \sqrt {e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}+\frac {\left (2 a^3 C d f+a b^2 (8 c C e+B d e+B c f-6 A d f)-b^3 (4 B c e-3 A (d e+c f))+a^2 b (2 B d f-5 C (d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{4 b (b c-a d)^2 (b e-a f)^2 (a+b x)}+\frac {\left (b^2 \left (3 A d^2 e^2-2 c d e (2 B e-A f)+c^2 \left (8 C e^2-4 B e f+3 A f^2\right )\right )+a b \left (d^2 e (B e-8 A f)-c^2 f (8 C e-B f)-2 c d \left (4 C e^2-7 B e f+4 A f^2\right )\right )+a^2 \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 )\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 )}{4 (b c-a d)^2 (b e-a f)^2}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {c+d x} \sqrt {e+f x}}{2 b (b c-a d) (b e-a f) (a+b x)^2}+\frac {\left (2 a^3 C d f+a b^2 (8 c C e+B d e+B c f-6 A d f)-b^3 (4 B c e-3 A (d e+c f))+a^2 b (2 B d f-5 C (d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{4 b (b c-a d)^2 (b e-a f)^2 (a+b x)}-\frac {\left (b^2 \left (3 A d^2 e^2-2 c d e (2 B e-A f)+c^2 \left (8 C e^2-4 B e f+3 A f^2\right )\right )+a b \left (d^2 e (B e-8 A f)-c^2 f (8 C e-B f)-2 c d \left (4 C e^2-7 B e f+4 A f^2\right )\right )+a^2 \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 )\right ) \tanh ^{-1}\left (\frac {\sqrt {b e-a f} \sqrt {c+d x}}{\sqrt {b c-a d} \sqrt {e+f x}}\right )}{4 (b c-a d)^{5/2} (b e-a f)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 2.09, size = 512, normalized size = 1.21 \[ \frac {\frac {\left (a (a C-b B)+A b^2\right ) \left (\frac {\left (8 a^2 d^2 f^2-8 a b d f (c f+d e)+b^2 \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {a f-b e}}{\sqrt {e+f x} \sqrt {a d-b c}}\right )}{(a d-b c)^{3/2} (a f-b e)^{3/2}}+\frac {3 b \sqrt {c+d x} \sqrt {e+f x} (-2 a d f+b c f+b d e)}{(a+b x) (b c-a d) (b e-a f)}\right )}{(b c-a d) (b e-a f)}-\frac {2 b \sqrt {c+d x} \sqrt {e+f x} \left (a (a C-b B)+A b^2\right )}{(a+b x)^2 (b c-a d) (b e-a f)}-\frac {4 b \sqrt {c+d x} \sqrt {e+f x} (b B-2 a C)}{(a+b x) (b c-a d) (b e-a f)}-\frac {4 (b B-2 a C) (-2 a d f+b c f+b d e) \tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {a f-b e}}{\sqrt {e+f x} \sqrt {a d-b c}}\right )}{(a d-b c)^{3/2} (a f-b e)^{3/2}}+\frac {8 C \tanh ^{-1}\left (\frac {\sqrt {c+d x} \sqrt {a f-b e}}{\sqrt {e+f x} \sqrt {a d-b c}}\right )}{\sqrt {a d-b c} \sqrt {a f-b e}}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

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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)/(b*x+a)^3/(d*x+c)^(1/2)/(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [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)/(b*x+a)^3/(d*x+c)^(1/2)/(f*x+e)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.13, size = 7119, normalized size = 16.79 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^(1/2)/(f*x+e)^(1/2),x)

[Out]

result too large to display

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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)/(b*x+a)^3/(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((a*d-b*c)>0)', see `assume?` f
or more details)Is (a*d-b*c)    *(a*f-b*e)    positive, negative or zero?

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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((A + B*x + C*x^2)/((e + f*x)^(1/2)*(a + b*x)^3*(c + d*x)^(1/2)),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)/(b*x+a)**3/(d*x+c)**(1/2)/(f*x+e)**(1/2),x)

[Out]

Timed out

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