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

Optimal. Leaf size=233 \[ -\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (d+e x)^{5/2} (2 c d-b e)}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+5 c d g+3 c e f)}{4 e^2 (d+e x)^{3/2} (2 c d-b e)^2}-\frac {c (-4 b e g+5 c d g+3 c e f) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{4 e^2 (2 c d-b e)^{5/2}} \]

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Rubi [A]  time = 0.37, antiderivative size = 233, 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, 672, 660, 208} \begin {gather*} -\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{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} (-4 b e g+5 c d g+3 c e f)}{4 e^2 (d+e x)^{3/2} (2 c d-b e)^2}-\frac {c (-4 b e g+5 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 )}{4 e^2 (2 c d-b e)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*f - d*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(2*e^2*(2*c*d - b*e)*(d + e*x)^(5/2)) - ((3*c*e*f + 5
*c*d*g - 4*b*e*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(4*e^2*(2*c*d - b*e)^2*(d + e*x)^(3/2)) - (c*(3*c
*e*f + 5*c*d*g - 4*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])]
)/(4*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 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + 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}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx &=-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (2 c d-b e) (d+e x)^{5/2}}+\frac {(3 c e f+5 c d g-4 b e g) \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{4 e (2 c d-b e)}\\ &=-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (2 c d-b e) (d+e x)^{5/2}}-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 e^2 (2 c d-b e)^2 (d+e x)^{3/2}}+\frac {(c (3 c e f+5 c d g-4 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}{8 e (2 c d-b e)^2}\\ &=-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (2 c d-b e) (d+e x)^{5/2}}-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 e^2 (2 c d-b e)^2 (d+e x)^{3/2}}+\frac {(c (3 c e f+5 c d g-4 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 )}{4 (2 c d-b e)^2}\\ &=-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (2 c d-b e) (d+e x)^{5/2}}-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 e^2 (2 c d-b e)^2 (d+e x)^{3/2}}-\frac {c (3 c e f+5 c d g-4 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 )}{4 e^2 (2 c d-b e)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 209, normalized size = 0.90 \begin {gather*} \frac {\sqrt {(d+e x) (c (d-e x)-b e)} \left (-\frac {(d+e x) (-4 b e g+5 c d g+3 c e f) \left (\sqrt {e} (2 c d-b e) \sqrt {c (d-e x)-b e}+c (d+e x) \sqrt {e (b e-2 c d)} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {c (d-e x)-b e}}{\sqrt {e (b e-2 c d)}}\right )\right )}{2 e^{3/2} (b e-2 c d)^2 \sqrt {c (d-e x)-b e}}+\frac {d g}{e}-f\right )}{2 e (d+e x)^{5/2} (2 c d-b e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.96, size = 238, normalized size = 1.02 \begin {gather*} \frac {\left (-4 b c e g+5 c^2 d g+3 c^2 e f\right ) \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 )}{4 e^2 (2 c d-b e)^2 \sqrt {b e-2 c d}}+\frac {\sqrt {-b e (d+e x)-c (d+e x)^2+2 c d (d+e x)} \left (4 b e g (d+e x)-2 b d e g+2 b e^2 f+4 c d^2 g-3 c e f (d+e x)-4 c d e f-5 c d g (d+e x)\right )}{4 e^2 (d+e x)^{5/2} (b e-2 c d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 1168, normalized size = 5.01

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(e*x+d)^(5/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*((3*c^2*d^3*e*f + (3*c^2*e^4*f + (5*c^2*d*e^3 - 4*b*c*e^4)*g)*x^3 + 3*(3*c^2*d*e^3*f + (5*c^2*d^2*e^2 -
4*b*c*d*e^3)*g)*x^2 + (5*c^2*d^4 - 4*b*c*d^3*e)*g + 3*(3*c^2*d^2*e^2*f + (5*c^2*d^3*e - 4*b*c*d^2*e^2)*g)*x)*s
qrt(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)*((14*c^2*d^2*e - 11*b*c*d*e^2 + 2*b^2*e^3)*f + (2*c^2*d^3 - 5*b*c*d^2*e + 2*b^2*d*e^2)*g + (3*(2*c^2*
d*e^2 - b*c*e^3)*f + (10*c^2*d^2*e - 13*b*c*d*e^2 + 4*b^2*e^3)*g)*x)*sqrt(e*x + d))/(8*c^3*d^6*e^2 - 12*b*c^2*
d^5*e^3 + 6*b^2*c*d^4*e^4 - b^3*d^3*e^5 + (8*c^3*d^3*e^5 - 12*b*c^2*d^2*e^6 + 6*b^2*c*d*e^7 - b^3*e^8)*x^3 + 3
*(8*c^3*d^4*e^4 - 12*b*c^2*d^3*e^5 + 6*b^2*c*d^2*e^6 - b^3*d*e^7)*x^2 + 3*(8*c^3*d^5*e^3 - 12*b*c^2*d^4*e^4 +
6*b^2*c*d^3*e^5 - b^3*d^2*e^6)*x), -1/4*((3*c^2*d^3*e*f + (3*c^2*e^4*f + (5*c^2*d*e^3 - 4*b*c*e^4)*g)*x^3 + 3*
(3*c^2*d*e^3*f + (5*c^2*d^2*e^2 - 4*b*c*d*e^3)*g)*x^2 + (5*c^2*d^4 - 4*b*c*d^3*e)*g + 3*(3*c^2*d^2*e^2*f + (5*
c^2*d^3*e - 4*b*c*d^2*e^2)*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)*(
(14*c^2*d^2*e - 11*b*c*d*e^2 + 2*b^2*e^3)*f + (2*c^2*d^3 - 5*b*c*d^2*e + 2*b^2*d*e^2)*g + (3*(2*c^2*d*e^2 - b*
c*e^3)*f + (10*c^2*d^2*e - 13*b*c*d*e^2 + 4*b^2*e^3)*g)*x)*sqrt(e*x + d))/(8*c^3*d^6*e^2 - 12*b*c^2*d^5*e^3 +
6*b^2*c*d^4*e^4 - b^3*d^3*e^5 + (8*c^3*d^3*e^5 - 12*b*c^2*d^2*e^6 + 6*b^2*c*d*e^7 - b^3*e^8)*x^3 + 3*(8*c^3*d^
4*e^4 - 12*b*c^2*d^3*e^5 + 6*b^2*c*d^2*e^6 - b^3*d*e^7)*x^2 + 3*(8*c^3*d^5*e^3 - 12*b*c^2*d^4*e^4 + 6*b^2*c*d^
3*e^5 - b^3*d^2*e^6)*x)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)/(e*x+d)^(5/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.09, size = 630, normalized size = 2.70 \begin {gather*} -\frac {\sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}\, \left (4 b c \,e^{3} g \,x^{2} \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-5 c^{2} d \,e^{2} g \,x^{2} \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-3 c^{2} e^{3} f \,x^{2} \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+8 b c d \,e^{2} g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-10 c^{2} d^{2} e g x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-6 c^{2} d \,e^{2} f x \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )+4 b c \,d^{2} e g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-5 c^{2} d^{3} g \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-3 c^{2} d^{2} e f \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right )-4 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, b \,e^{2} g x +5 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c d e g x +3 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c \,e^{2} f x -2 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, b d e g -2 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, b \,e^{2} f +\sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c \,d^{2} g +7 \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, c d e f \right )}{4 \left (e x +d \right )^{\frac {5}{2}} \left (b e -2 c d \right )^{\frac {5}{2}} \sqrt {-c e x -b e +c d}\, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)*(4*b*c*e^3*g*x^2*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))-
5*c^2*d*e^2*g*x^2*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))-3*c^2*e^3*f*x^2*arctan((-c*e*x-b*e+c*d)^(1/
2)/(b*e-2*c*d)^(1/2))+8*b*c*d*e^2*g*x*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))-10*c^2*d^2*e*g*x*arctan
((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))-6*c^2*d*e^2*f*x*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))+4*
b*c*d^2*e*g*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))-5*c^2*d^3*g*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*
c*d)^(1/2))-3*c^2*d^2*e*f*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))-4*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d
)^(1/2)*b*e^2*g*x+5*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)*c*d*e*g*x+3*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1
/2)*c*e^2*f*x-2*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)*b*d*e*g-2*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)*b*
e^2*f+(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)*c*d^2*g+7*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)*c*d*e*f)/(e*
x+d)^(5/2)/(b*e-2*c*d)^(5/2)/e^2/(-c*e*x-b*e+c*d)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {g x + f}{\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\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)/(e*x+d)^(5/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((g*x + f)/(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(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 {f+g\,x}{{\left (d+e\,x\right )}^{5/2}\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {f + g x}{\sqrt {- \left (d + e x\right ) \left (b e - c d + c e 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)/(e*x+d)**(5/2)/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(1/2),x)

[Out]

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

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