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

Optimal. Leaf size=153 \[ -\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (d+e x)^{3/2} (2 c d-b e)}-\frac {(-2 b e g+3 c d g+c e f) \tanh ^{-1}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e^2 (2 c d-b e)^{3/2}} \]

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Rubi [A]  time = 0.23, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {792, 660, 208} \begin {gather*} -\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (d+e x)^{3/2} (2 c d-b e)}-\frac {(-2 b e g+3 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 (2 c d-b e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f + g*x)/((d + e*x)^(3/2)*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])/(e^2*(2*c*d - b*e)*(d + e*x)^(3/2))) - ((c*e*f + 3*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)^(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 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 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)^{3/2} \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}}{e^2 (2 c d-b e) (d+e x)^{3/2}}+\frac {(c e f+3 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 {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (2 c d-b e) (d+e x)^{3/2}}+\frac {(c e f+3 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}\\ &=-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (2 c d-b e) (d+e x)^{3/2}}-\frac {(c e f+3 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)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 153, normalized size = 1.00 \begin {gather*} \frac {(e f-d g) (b e-c d+c e x)-\frac {(d+e x) \sqrt {c (d-e x)-b e} (-2 b e g+3 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 )}{\sqrt {2 c d-b e}}}{e^2 \sqrt {d+e x} (2 c d-b e) \sqrt {(d+e x) (c (d-e x)-b e)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.73, size = 167, normalized size = 1.09 \begin {gather*} \frac {(e f-d g) \sqrt {-b e (d+e x)-c (d+e x)^2+2 c d (d+e x)}}{e^2 (d+e x)^{3/2} (b e-2 c d)}+\frac {(2 b e g-3 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 (b e-2 c d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 698, normalized size = 4.56 \begin {gather*} \left [\frac {{\left (c d^{2} e f + {\left (c e^{3} f + {\left (3 \, c d e^{2} - 2 \, b e^{3}\right )} g\right )} x^{2} + {\left (3 \, c d^{3} - 2 \, b d^{2} e\right )} g + 2 \, {\left (c d e^{2} f + {\left (3 \, 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 ({\left (2 \, c d e - b e^{2}\right )} f - {\left (2 \, c d^{2} - b d e\right )} g\right )} \sqrt {e x + d}}{2 \, {\left (4 \, c^{2} d^{4} e^{2} - 4 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4} + {\left (4 \, c^{2} d^{2} e^{4} - 4 \, b c d e^{5} + b^{2} e^{6}\right )} x^{2} + 2 \, {\left (4 \, c^{2} d^{3} e^{3} - 4 \, b c d^{2} e^{4} + b^{2} d e^{5}\right )} x\right )}}, -\frac {{\left (c d^{2} e f + {\left (c e^{3} f + {\left (3 \, c d e^{2} - 2 \, b e^{3}\right )} g\right )} x^{2} + {\left (3 \, c d^{3} - 2 \, b d^{2} e\right )} g + 2 \, {\left (c d e^{2} f + {\left (3 \, 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 ({\left (2 \, c d e - b e^{2}\right )} f - {\left (2 \, c d^{2} - b d e\right )} g\right )} \sqrt {e x + d}}{4 \, c^{2} d^{4} e^{2} - 4 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4} + {\left (4 \, c^{2} d^{2} e^{4} - 4 \, b c d e^{5} + b^{2} e^{6}\right )} x^{2} + 2 \, {\left (4 \, c^{2} d^{3} e^{3} - 4 \, b c d^{2} e^{4} + b^{2} d e^{5}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [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 {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.08, size = 328, normalized size = 2.14 \begin {gather*} \frac {\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 )-3 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 )-3 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 )-\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 ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}{\left (b e -2 c d \right )^{\frac {3}{2}} \sqrt {-c e x -b e +c d}\, \left (e x +d \right )^{\frac {3}{2}} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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