3.6.67 \(\int \frac {a+b x+c x^2}{(d+e x)^2 \sqrt {f+g x}} \, dx\)

Optimal. Leaf size=140 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) (c d (4 e f-3 d g)-e (-a e g-b d g+2 b e f))}{e^{5/2} (e f-d g)^{3/2}}-\frac {\sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{(d+e x) (e f-d g)}+\frac {2 c \sqrt {f+g x}}{e^2 g} \]

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Rubi [A]  time = 0.29, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {897, 1157, 388, 208} \begin {gather*} -\frac {\sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{(d+e x) (e f-d g)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) (c d (4 e f-3 d g)-e (-a e g-b d g+2 b e f))}{e^{5/2} (e f-d g)^{3/2}}+\frac {2 c \sqrt {f+g x}}{e^2 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], 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] && IntegersQ[n,
p] && FractionQ[m]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rubi steps

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

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Mathematica [A]  time = 0.56, size = 150, normalized size = 1.07 \begin {gather*} \frac {\sqrt {f+g x} \left (e g (b d-a e)+c \left (-3 d^2 g+2 d e (f-g x)+2 e^2 f x\right )\right )}{e^2 g (d+e x) (e f-d g)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) (e (-a e g-b d g+2 b e f)+c d (3 d g-4 e f))}{e^{5/2} (e f-d g)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.55, size = 200, normalized size = 1.43 \begin {gather*} \frac {\sqrt {f+g x} \left (a e^2 g^2-b d e g^2+3 c d^2 g^2+2 c d e g (f+g x)-4 c d e f g+2 c e^2 f^2-2 c e^2 f (f+g x)\right )}{e^2 g (e f-d g) (-d g-e (f+g x)+e f)}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x} \sqrt {d g-e f}}{e f-d g}\right ) \left (-a e^2 g-b d e g+2 b e^2 f+3 c d^2 g-4 c d e f\right )}{e^{5/2} (d g-e f)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[f + g*x]*(2*c*e^2*f^2 - 4*c*d*e*f*g + 3*c*d^2*g^2 - b*d*e*g^2 + a*e^2*g^2 - 2*c*e^2*f*(f + g*x) + 2*c*d*
e*g*(f + g*x)))/(e^2*g*(e*f - d*g)*(e*f - d*g - e*(f + g*x))) + ((-4*c*d*e*f + 2*b*e^2*f + 3*c*d^2*g - b*d*e*g
- a*e^2*g)*ArcTan[(Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[f + g*x])/(e*f - d*g)])/(e^(5/2)*(-(e*f) + d*g)^(3/2))

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fricas [B]  time = 0.65, size = 637, normalized size = 4.55 \begin {gather*} \left [-\frac {\sqrt {e^{2} f - d e g} {\left (2 \, {\left (2 \, c d^{2} e - b d e^{2}\right )} f g - {\left (3 \, c d^{3} - b d^{2} e - a d e^{2}\right )} g^{2} + {\left (2 \, {\left (2 \, c d e^{2} - b e^{3}\right )} f g - {\left (3 \, c d^{2} e - b d e^{2} - a e^{3}\right )} g^{2}\right )} x\right )} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {e^{2} f - d e g} \sqrt {g x + f}}{e x + d}\right ) - 2 \, {\left (2 \, c d e^{3} f^{2} - {\left (5 \, c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} f g + {\left (3 \, c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} g^{2} + 2 \, {\left (c e^{4} f^{2} - 2 \, c d e^{3} f g + c d^{2} e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}}{2 \, {\left (d e^{5} f^{2} g - 2 \, d^{2} e^{4} f g^{2} + d^{3} e^{3} g^{3} + {\left (e^{6} f^{2} g - 2 \, d e^{5} f g^{2} + d^{2} e^{4} g^{3}\right )} x\right )}}, -\frac {\sqrt {-e^{2} f + d e g} {\left (2 \, {\left (2 \, c d^{2} e - b d e^{2}\right )} f g - {\left (3 \, c d^{3} - b d^{2} e - a d e^{2}\right )} g^{2} + {\left (2 \, {\left (2 \, c d e^{2} - b e^{3}\right )} f g - {\left (3 \, c d^{2} e - b d e^{2} - a e^{3}\right )} g^{2}\right )} x\right )} \arctan \left (\frac {\sqrt {-e^{2} f + d e g} \sqrt {g x + f}}{e g x + e f}\right ) - {\left (2 \, c d e^{3} f^{2} - {\left (5 \, c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} f g + {\left (3 \, c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} g^{2} + 2 \, {\left (c e^{4} f^{2} - 2 \, c d e^{3} f g + c d^{2} e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}}{d e^{5} f^{2} g - 2 \, d^{2} e^{4} f g^{2} + d^{3} e^{3} g^{3} + {\left (e^{6} f^{2} g - 2 \, d e^{5} f g^{2} + d^{2} e^{4} g^{3}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(e^2*f - d*e*g)*(2*(2*c*d^2*e - b*d*e^2)*f*g - (3*c*d^3 - b*d^2*e - a*d*e^2)*g^2 + (2*(2*c*d*e^2 -
b*e^3)*f*g - (3*c*d^2*e - b*d*e^2 - a*e^3)*g^2)*x)*log((e*g*x + 2*e*f - d*g - 2*sqrt(e^2*f - d*e*g)*sqrt(g*x +
f))/(e*x + d)) - 2*(2*c*d*e^3*f^2 - (5*c*d^2*e^2 - b*d*e^3 + a*e^4)*f*g + (3*c*d^3*e - b*d^2*e^2 + a*d*e^3)*g
^2 + 2*(c*e^4*f^2 - 2*c*d*e^3*f*g + c*d^2*e^2*g^2)*x)*sqrt(g*x + f))/(d*e^5*f^2*g - 2*d^2*e^4*f*g^2 + d^3*e^3*
g^3 + (e^6*f^2*g - 2*d*e^5*f*g^2 + d^2*e^4*g^3)*x), -(sqrt(-e^2*f + d*e*g)*(2*(2*c*d^2*e - b*d*e^2)*f*g - (3*c
*d^3 - b*d^2*e - a*d*e^2)*g^2 + (2*(2*c*d*e^2 - b*e^3)*f*g - (3*c*d^2*e - b*d*e^2 - a*e^3)*g^2)*x)*arctan(sqrt
(-e^2*f + d*e*g)*sqrt(g*x + f)/(e*g*x + e*f)) - (2*c*d*e^3*f^2 - (5*c*d^2*e^2 - b*d*e^3 + a*e^4)*f*g + (3*c*d^
3*e - b*d^2*e^2 + a*d*e^3)*g^2 + 2*(c*e^4*f^2 - 2*c*d*e^3*f*g + c*d^2*e^2*g^2)*x)*sqrt(g*x + f))/(d*e^5*f^2*g
- 2*d^2*e^4*f*g^2 + d^3*e^3*g^3 + (e^6*f^2*g - 2*d*e^5*f*g^2 + d^2*e^4*g^3)*x)]

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giac [A]  time = 0.18, size = 175, normalized size = 1.25 \begin {gather*} \frac {2 \, \sqrt {g x + f} c e^{\left (-2\right )}}{g} - \frac {{\left (3 \, c d^{2} g - 4 \, c d f e - b d g e + 2 \, b f e^{2} - a g e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right )}{{\left (d g e^{2} - f e^{3}\right )} \sqrt {d g e - f e^{2}}} + \frac {\sqrt {g x + f} c d^{2} g - \sqrt {g x + f} b d g e + \sqrt {g x + f} a g e^{2}}{{\left (d g e^{2} - f e^{3}\right )} {\left (d g + {\left (g x + f\right )} e - f e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(g*x + f)*c*e^(-2)/g - (3*c*d^2*g - 4*c*d*f*e - b*d*g*e + 2*b*f*e^2 - a*g*e^2)*arctan(sqrt(g*x + f)*e/sq
rt(d*g*e - f*e^2))/((d*g*e^2 - f*e^3)*sqrt(d*g*e - f*e^2)) + (sqrt(g*x + f)*c*d^2*g - sqrt(g*x + f)*b*d*g*e +
sqrt(g*x + f)*a*g*e^2)/((d*g*e^2 - f*e^3)*(d*g + (g*x + f)*e - f*e))

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maple [B]  time = 0.02, size = 371, normalized size = 2.65 \begin {gather*} \frac {a g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}+\frac {b d g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}\, e}-\frac {2 b f \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {3 c \,d^{2} g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}\, e^{2}}+\frac {4 c d f \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}\, e}+\frac {\sqrt {g x +f}\, a g}{\left (d g -e f \right ) \left (e g x +d g \right )}-\frac {\sqrt {g x +f}\, b d g}{\left (d g -e f \right ) \left (e g x +d g \right ) e}+\frac {\sqrt {g x +f}\, c \,d^{2} g}{\left (d g -e f \right ) \left (e g x +d g \right ) e^{2}}+\frac {2 \sqrt {g x +f}\, c}{e^{2} g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(g*x+f)^(1/2)*c/e^2/g+g/(d*g-e*f)*(g*x+f)^(1/2)/(e*g*x+d*g)*a-g/e/(d*g-e*f)*(g*x+f)^(1/2)/(e*g*x+d*g)*b*d+g/
e^2/(d*g-e*f)*(g*x+f)^(1/2)/(e*g*x+d*g)*c*d^2+g/(d*g-e*f)/((d*g-e*f)*e)^(1/2)*arctan((g*x+f)^(1/2)/((d*g-e*f)*
e)^(1/2)*e)*a+g/e/(d*g-e*f)/((d*g-e*f)*e)^(1/2)*arctan((g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*b*d-2/(d*g-e*f)/((
d*g-e*f)*e)^(1/2)*arctan((g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*b*f-3*g/e^2/(d*g-e*f)/((d*g-e*f)*e)^(1/2)*arctan
((g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)*e)*c*d^2+4/e/(d*g-e*f)/((d*g-e*f)*e)^(1/2)*arctan((g*x+f)^(1/2)/((d*g-e*f)*
e)^(1/2)*e)*d*c*f

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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)/(e*x+d)^2/(g*x+f)^(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(d*g-e*f>0)', see `assume?` for
more details)Is d*g-e*f positive or negative?

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mupad [B]  time = 0.23, size = 146, normalized size = 1.04 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (a\,e^2\,g-2\,b\,e^2\,f-3\,c\,d^2\,g+b\,d\,e\,g+4\,c\,d\,e\,f\right )}{e^{5/2}\,{\left (d\,g-e\,f\right )}^{3/2}}+\frac {\sqrt {f+g\,x}\,\left (c\,g\,d^2-b\,g\,d\,e+a\,g\,e^2\right )}{\left (d\,g-e\,f\right )\,\left (e^3\,\left (f+g\,x\right )-e^3\,f+d\,e^2\,g\right )}+\frac {2\,c\,\sqrt {f+g\,x}}{e^2\,g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((e^(1/2)*(f + g*x)^(1/2))/(d*g - e*f)^(1/2))*(a*e^2*g - 2*b*e^2*f - 3*c*d^2*g + b*d*e*g + 4*c*d*e*f))/(e
^(5/2)*(d*g - e*f)^(3/2)) + ((f + g*x)^(1/2)*(a*e^2*g + c*d^2*g - b*d*e*g))/((d*g - e*f)*(e^3*(f + g*x) - e^3*
f + d*e^2*g)) + (2*c*(f + g*x)^(1/2))/(e^2*g)

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sympy [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)/(e*x+d)**2/(g*x+f)**(1/2),x)

[Out]

Timed out

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