3.21.6 \(\int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^{5/2}} \, dx\)

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

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Rubi [A]  time = 0.36, 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, 664, 660, 208} \begin {gather*} -\frac {(e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{e^2 (d+e x)^{5/2} (2 c d-b e)}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (2 b e g-5 c d g+c e f)}{e^2 \sqrt {d+e x} (2 c d-b e)}+\frac {(2 b e g-5 c d g+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 \sqrt {2 c d-b e}} \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)^(5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.23, size = 173, normalized size = 0.78 \begin {gather*} \frac {\sqrt {(d+e x) (c (d-e x)-b e)} \left ((d+e x) \sqrt {2 c d-b e} (-2 b e g+5 c d g-c e f) \tanh ^{-1}\left (\frac {\sqrt {-b e+c d-c e x}}{\sqrt {2 c d-b e}}\right )-(2 c d-b e) \sqrt {c (d-e x)-b e} (3 d g-e f+2 e g x)\right )}{e^2 (d+e x)^{3/2} (b e-2 c d) \sqrt {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)^(5/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.87, size = 165, normalized size = 0.74 \begin {gather*} \frac {\sqrt {-b e (d+e x)-c (d+e x)^2+2 c d (d+e x)} (2 g (d+e x)+d g-e f)}{e^2 (d+e x)^{3/2}}+\frac {(-2 b e g+5 c d g-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 \sqrt {b e-2 c d}} \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)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.55, size = 659, normalized size = 2.96 \begin {gather*} \left [\frac {{\left (c d^{2} e f + {\left (c e^{3} f - {\left (5 \, c d e^{2} - 2 \, b e^{3}\right )} g\right )} x^{2} - {\left (5 \, c d^{3} - 2 \, b d^{2} e\right )} g + 2 \, {\left (c d e^{2} f - {\left (5 \, c d^{2} e - 2 \, b d e^{2}\right )} 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 (2 \, {\left (2 \, c d e - b e^{2}\right )} g x - {\left (2 \, c d e - b e^{2}\right )} f + 3 \, {\left (2 \, c d^{2} - b d e\right )} g\right )} \sqrt {e x + d}}{2 \, {\left (2 \, c d^{3} e^{2} - b d^{2} e^{3} + {\left (2 \, c d e^{4} - b e^{5}\right )} x^{2} + 2 \, {\left (2 \, c d^{2} e^{3} - b d e^{4}\right )} x\right )}}, \frac {{\left (c d^{2} e f + {\left (c e^{3} f - {\left (5 \, c d e^{2} - 2 \, b e^{3}\right )} g\right )} x^{2} - {\left (5 \, c d^{3} - 2 \, b d^{2} e\right )} g + 2 \, {\left (c d e^{2} f - {\left (5 \, c d^{2} e - 2 \, b d e^{2}\right )} 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 (2 \, {\left (2 \, c d e - b e^{2}\right )} g x - {\left (2 \, c d e - b e^{2}\right )} f + 3 \, {\left (2 \, c d^{2} - b d e\right )} g\right )} \sqrt {e x + d}}{2 \, c d^{3} e^{2} - b d^{2} e^{3} + {\left (2 \, c d e^{4} - b e^{5}\right )} x^{2} + 2 \, {\left (2 \, c d^{2} e^{3} - b d e^{4}\right )} x}\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)^(5/2),x, algorithm="fricas")

[Out]

[1/2*((c*d^2*e*f + (c*e^3*f - (5*c*d*e^2 - 2*b*e^3)*g)*x^2 - (5*c*d^3 - 2*b*d^2*e)*g + 2*(c*d*e^2*f - (5*c*d^2
*e - 2*b*d*e^2)*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*(2*c*d*e - b*e^2)*g*x - (2*c*d*e - b*e^2)*f + 3*(2*c*d^2 - b*d*e)*g)*sqrt(e
*x + d))/(2*c*d^3*e^2 - b*d^2*e^3 + (2*c*d*e^4 - b*e^5)*x^2 + 2*(2*c*d^2*e^3 - b*d*e^4)*x), ((c*d^2*e*f + (c*e
^3*f - (5*c*d*e^2 - 2*b*e^3)*g)*x^2 - (5*c*d^3 - 2*b*d^2*e)*g + 2*(c*d*e^2*f - (5*c*d^2*e - 2*b*d*e^2)*g)*x)*s
qrt(-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*(2*c*d*e - b*e^2)*g*x - (2*c*d*e
- b*e^2)*f + 3*(2*c*d^2 - b*d*e)*g)*sqrt(e*x + d))/(2*c*d^3*e^2 - b*d^2*e^3 + (2*c*d*e^4 - b*e^5)*x^2 + 2*(2*
c*d^2*e^3 - b*d*e^4)*x)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: AttributeError} \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)^(5/2),x, algorithm="giac")

[Out]

Exception raised: AttributeError >> type

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maple [A]  time = 0.07, size = 359, normalized size = 1.61 \begin {gather*} \frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, \left (-2 b \,e^{2} g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+5 c d e g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-c \,e^{2} f x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-2 b d e g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+5 c \,d^{2} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-c d e f \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}\, \sqrt {-c e x -b e +c d}\, e g x +3 \sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}\, d g -\sqrt {b e -2 c d}\, \sqrt {-c e x -b e +c d}\, e f \right )}{\left (e x +d \right )^{\frac {3}{2}} \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, 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)^(5/2),x)

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \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 )}^{5/2}} \,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)^(5/2),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)^(5/2), 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 {5}{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)**(5/2),x)

[Out]

Integral(sqrt(-(d + e*x)*(b*e - c*d + c*e*x))*(f + g*x)/(d + e*x)**(5/2), x)

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