3.21.43 \(\int \frac {f+g x}{\sqrt {d+e x} (c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=223 \[ -\frac {e f-d g}{e^2 \sqrt {d+e x} (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {\sqrt {d+e x} (-2 b e g+c d g+3 c e f)}{e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(-2 b e g+c d g+3 c e f) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{5/2}} \]

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Rubi [A]  time = 0.35, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {792, 666, 660, 208} \begin {gather*} -\frac {e f-d g}{e^2 \sqrt {d+e x} (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {\sqrt {d+e x} (-2 b e g+c d g+3 c e f)}{e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(-2 b e g+c d g+3 c e f) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f + g*x)/(Sqrt[d + e*x]*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(3/2)),x]

[Out]

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

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 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 666

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((2*c*d - b*e)*(d +
e*x)^m*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*c*d - b*e)*(m + 2*p + 2))/((p + 1)*
(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

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

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Mathematica [C]  time = 0.07, size = 117, normalized size = 0.52 \begin {gather*} \frac {(d+e x) (-2 b e g+c d g+3 c e f) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {-c d+b e+c e x}{b e-2 c d}\right )+(2 c d-b e) (d g-e f)}{e^2 \sqrt {d+e x} (b e-2 c d)^2 \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)/(Sqrt[d + e*x]*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(3/2)),x]

[Out]

((2*c*d - b*e)*(-(e*f) + d*g) + (3*c*e*f + c*d*g - 2*b*e*g)*(d + e*x)*Hypergeometric2F1[-1/2, 1, 1/2, (-(c*d)
+ b*e + c*e*x)/(-2*c*d + b*e)])/(e^2*(-2*c*d + b*e)^2*Sqrt[d + e*x]*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])

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IntegrateAlgebraic [A]  time = 3.64, size = 231, normalized size = 1.04 \begin {gather*} \frac {\sqrt {-b e (d+e x)-c (d+e x)^2+2 c d (d+e x)} \left (2 b e g (d+e x)+b d e g-b e^2 f-2 c d^2 g-3 c e f (d+e x)+2 c d e f-c d g (d+e x)\right )}{e^2 (d+e x)^{3/2} (b e-2 c d)^2 (b e+c (d+e x)-2 c d)}+\frac {(-2 b e g+c d g+3 c e f) \tan ^{-1}\left (\frac {\sqrt {b e-2 c d} \sqrt {(d+e x) (2 c d-b e)-c (d+e x)^2}}{\sqrt {d+e x} (b e+c (d+e x)-2 c d)}\right )}{e^2 (b e-2 c d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(f + g*x)/(Sqrt[d + e*x]*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(3/2)),x]

[Out]

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

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fricas [B]  time = 0.46, size = 1351, normalized size = 6.06

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((3*c^2*e^4*f + (c^2*d*e^3 - 2*b*c*e^4)*g)*x^3 + (3*(c^2*d*e^3 + b*c*e^4)*f + (c^2*d^2*e^2 - b*c*d*e^3 -
2*b^2*e^4)*g)*x^2 - 3*(c^2*d^3*e - b*c*d^2*e^2)*f - (c^2*d^4 - 3*b*c*d^3*e + 2*b^2*d^2*e^2)*g - (3*(c^2*d^2*e
^2 - 2*b*c*d*e^3)*f + (c^2*d^3*e - 4*b*c*d^2*e^2 + 4*b^2*d*e^3)*g)*x)*sqrt(2*c*d - b*e)*log(-(c*e^2*x^2 - 3*c*
d^2 + 2*b*d*e - 2*(c*d*e - b*e^2)*x - 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(2*c*d - b*e)*sqrt(e*x
+ d))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*((2*c^2*d^2*e + b*c*d*e^2 - b^
2*e^3)*f + 3*(2*c^2*d^3 - 3*b*c*d^2*e + b^2*d*e^2)*g + (3*(2*c^2*d*e^2 - b*c*e^3)*f + (2*c^2*d^2*e - 5*b*c*d*e
^2 + 2*b^2*e^3)*g)*x)*sqrt(e*x + d))/(8*c^4*d^6*e^2 - 20*b*c^3*d^5*e^3 + 18*b^2*c^2*d^4*e^4 - 7*b^3*c*d^3*e^5
+ b^4*d^2*e^6 - (8*c^4*d^3*e^5 - 12*b*c^3*d^2*e^6 + 6*b^2*c^2*d*e^7 - b^3*c*e^8)*x^3 - (8*c^4*d^4*e^4 - 4*b*c^
3*d^3*e^5 - 6*b^2*c^2*d^2*e^6 + 5*b^3*c*d*e^7 - b^4*e^8)*x^2 + (8*c^4*d^5*e^3 - 28*b*c^3*d^4*e^4 + 30*b^2*c^2*
d^3*e^5 - 13*b^3*c*d^2*e^6 + 2*b^4*d*e^7)*x), (((3*c^2*e^4*f + (c^2*d*e^3 - 2*b*c*e^4)*g)*x^3 + (3*(c^2*d*e^3
+ b*c*e^4)*f + (c^2*d^2*e^2 - b*c*d*e^3 - 2*b^2*e^4)*g)*x^2 - 3*(c^2*d^3*e - b*c*d^2*e^2)*f - (c^2*d^4 - 3*b*c
*d^3*e + 2*b^2*d^2*e^2)*g - (3*(c^2*d^2*e^2 - 2*b*c*d*e^3)*f + (c^2*d^3*e - 4*b*c*d^2*e^2 + 4*b^2*d*e^3)*g)*x)
*sqrt(-2*c*d + b*e)*arctan(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(-2*c*d + b*e)*sqrt(e*x + d)/(c*e^2*
x^2 + b*e^2*x - c*d^2 + b*d*e)) + sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*((2*c^2*d^2*e + b*c*d*e^2 - b^2*e
^3)*f + 3*(2*c^2*d^3 - 3*b*c*d^2*e + b^2*d*e^2)*g + (3*(2*c^2*d*e^2 - b*c*e^3)*f + (2*c^2*d^2*e - 5*b*c*d*e^2
+ 2*b^2*e^3)*g)*x)*sqrt(e*x + d))/(8*c^4*d^6*e^2 - 20*b*c^3*d^5*e^3 + 18*b^2*c^2*d^4*e^4 - 7*b^3*c*d^3*e^5 + b
^4*d^2*e^6 - (8*c^4*d^3*e^5 - 12*b*c^3*d^2*e^6 + 6*b^2*c^2*d*e^7 - b^3*c*e^8)*x^3 - (8*c^4*d^4*e^4 - 4*b*c^3*d
^3*e^5 - 6*b^2*c^2*d^2*e^6 + 5*b^3*c*d*e^7 - b^4*e^8)*x^2 + (8*c^4*d^5*e^3 - 28*b*c^3*d^4*e^4 + 30*b^2*c^2*d^3
*e^5 - 13*b^3*c*d^2*e^6 + 2*b^4*d*e^7)*x)]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [b,c,d,exp(1),exp(2)]=[72,91,-18,-31,46]Warning, need to choose a branch for the root of a polyn
omial with parameters. This might be wrong.The choice was done assuming [b,c,d,exp(1),exp(2)]=[-76,-17,63,68,9
8]Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choic
e was done assuming [b,c,d,exp(1),exp(2)]=[-95,-95,-57,19,-77]Warning, need to choose a branch for the root of
a polynomial with parameters. This might be wrong.The choice was done assuming [b,c,d,exp(1),exp(2)]=[-29,-98
,-9,8,96]Evaluation time: 119.66Unable to transpose Error: Bad Argument Value

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maple [B]  time = 0.07, size = 479, normalized size = 2.15 \begin {gather*} \frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, \left (2 \sqrt {-c e x -b e +c d}\, b \,e^{2} g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-\sqrt {-c e x -b e +c d}\, c d e g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-3 \sqrt {-c e x -b e +c d}\, c \,e^{2} f x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+2 \sqrt {-c e x -b e +c d}\, b d e g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+2 \sqrt {b e -2 c d}\, b \,e^{2} g x -\sqrt {-c e x -b e +c d}\, c \,d^{2} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-3 \sqrt {-c e x -b e +c d}\, c d e f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-\sqrt {b e -2 c d}\, c d e g x -3 \sqrt {b e -2 c d}\, c \,e^{2} f x +3 \sqrt {b e -2 c d}\, b d e g -\sqrt {b e -2 c d}\, b \,e^{2} f -3 \sqrt {b e -2 c d}\, c \,d^{2} g -\sqrt {b e -2 c d}\, c d e f \right )}{\left (e x +d \right )^{\frac {3}{2}} \left (c e x +b e -c d \right ) \left (b e -2 c d \right )^{\frac {5}{2}} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)/(e*x+d)^(1/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {g x + f}{{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {3}{2}} \sqrt {e x + d}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(e*x+d)^(1/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((g*x + f)/((-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)^(3/2)*sqrt(e*x + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {f+g\,x}{\sqrt {d+e\,x}\,{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)/((d + e*x)^(1/2)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(3/2)),x)

[Out]

int((f + g*x)/((d + e*x)^(1/2)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {f + g x}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}} \sqrt {d + e x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(e*x+d)**(1/2)/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(3/2),x)

[Out]

Integral((f + g*x)/((-(d + e*x)*(b*e - c*d + c*e*x))**(3/2)*sqrt(d + e*x)), x)

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