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

Optimal. Leaf size=186 \[ \frac {2 (e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 \sqrt {d+e x}}-\frac {2 \sqrt {2 c d-b e} (e f-d g) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 c e^2 (d+e x)^{3/2}} \]

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Rubi [A]  time = 0.35, antiderivative size = 186, 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 = {794, 664, 660, 208} \begin {gather*} \frac {2 (e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 \sqrt {d+e x}}-\frac {2 \sqrt {2 c d-b e} (e f-d g) \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}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 c e^2 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 664

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

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 154, normalized size = 0.83 \begin {gather*} \frac {2 \sqrt {d+e x} \sqrt {c (d-e x)-b e} \left (\sqrt {c (d-e x)-b e} (b e g+c (-4 d g+3 e f+e g x))+3 c \sqrt {2 c d-b e} (d g-e f) \tanh ^{-1}\left (\frac {\sqrt {-b e+c d-c e x}}{\sqrt {2 c d-b e}}\right )\right )}{3 c e^2 \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.89, size = 171, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {-b e (d+e x)-c (d+e x)^2+2 c d (d+e x)} (b e g+c g (d+e x)-5 c d g+3 c e f)}{3 c e^2 \sqrt {d+e x}}+\frac {2 \sqrt {b e-2 c d} (d g-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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 419, normalized size = 2.25 \begin {gather*} \left [-\frac {3 \, {\left (c d e f - c d^{2} g + {\left (c e^{2} f - c d e g\right )} x\right )} \sqrt {2 \, c d - b e} \log \left (-\frac {c e^{2} x^{2} - 3 \, c d^{2} + 2 \, b d e - 2 \, {\left (c d e - b e^{2}\right )} 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}}\right ) - 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (c e g x + 3 \, c e f - {\left (4 \, c d - b e\right )} g\right )} \sqrt {e x + d}}{3 \, {\left (c e^{3} x + c d e^{2}\right )}}, -\frac {2 \, {\left (3 \, {\left (c d e f - c d^{2} g + {\left (c e^{2} f - c d e g\right )} x\right )} \sqrt {-2 \, c d + b e} \arctan \left (\frac {\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}\right ) - \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (c e g x + 3 \, c e f - {\left (4 \, c d - b e\right )} g\right )} \sqrt {e x + d}\right )}}{3 \, {\left (c e^{3} x + c d e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 330, normalized size = 1.77 \begin {gather*} \frac {2 \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, \left (3 b c d e g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-3 b c \,e^{2} f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-6 c^{2} d^{2} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+6 c^{2} d e f \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}\, \sqrt {b e -2 c d}\, c e g x +\sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, b e g -4 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c d g +3 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c e f \right )}{3 \sqrt {e x +d}\, \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*(3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c*d*e*g-3*arc
tan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c*e^2*f-6*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2
*d^2*g+6*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d*e*f+x*c*e*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)
^(1/2)+b*e*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-4*c*d*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+3*c*e*f
*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2))/(e*x+d)^(1/2)/(-c*e*x-b*e+c*d)^(1/2)/c/e^2/(b*e-2*c*d)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (f+g\,x\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}}{{\left (d+e\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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