3.23.35 \(\int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^{3/2}} \, dx\) [2235]

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

[Out]

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

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Rubi [A]
time = 0.22, 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 = {808, 678, 674, 214} \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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 674

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 678

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 808

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)) \text {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 = 0.24, size = 152, normalized size = 0.82 \begin {gather*} \frac {2 \sqrt {d+e x} \sqrt {-b e+c (d-e x)} \left (\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) \tan ^{-1}\left (\frac {\sqrt {-b e+c (d-e x)}}{\sqrt {-2 c d+b e}}\right )\right )}{3 c e^2 \sqrt {(d+e x) (-b e+c (d-e x))}} \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)*ArcTan[Sqrt[-(b*e) + c*(d - 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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Maple [A]
time = 0.04, size = 322, normalized size = 1.73

method	result	size
default	\(\frac {2 \sqrt {-\left (e x +d \right ) \left (c e x +b e -c d \right )}\, \left (3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c d e g -3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c \,e^{2} f -6 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{2} g +6 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d e f +c e g x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+b e g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-4 c d g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+3 c e f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\right )}{3 \sqrt {e x +d}\, \sqrt {-c e x -b e +c d}\, c \,e^{2} \sqrt {b e -2 c d}}\)	\(322\)

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,method=_RETURNVERBOSE)

[Out]

2/3*(-(e*x+d)*(c*e*x+b*e-c*d))^(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*arctan((-
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+c*e*g*x*(-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*} \text {Failed to integrate} \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*x^2*e^2 + c*d^2 - b*x*e^2 - b*d*e)*(g*x + f)/(x*e + d)^(3/2), x)

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Fricas [A]
time = 3.33, size = 386, normalized size = 2.08 \begin {gather*} \left [-\frac {3 \, {\left (c d^{2} g - c f x e^{2} + {\left (c d g x - c d f\right )} e\right )} \sqrt {2 \, c d - b e} \log \left (\frac {3 \, c d^{2} - {\left (c x^{2} + 2 \, b x\right )} e^{2} + 2 \, {\left (c d x - b d\right )} e - 2 \, \sqrt {c d^{2} - b d e - {\left (c x^{2} + b x\right )} e^{2}} \sqrt {2 \, c d - b e} \sqrt {x e + d}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 2 \, \sqrt {c d^{2} - b d e - {\left (c x^{2} + b x\right )} e^{2}} {\left (4 \, c d g - {\left (c g x + 3 \, c f + b g\right )} e\right )} \sqrt {x e + d}}{3 \, {\left (c x e^{3} + c d e^{2}\right )}}, \frac {2 \, {\left (3 \, {\left (c d^{2} g - c f x e^{2} + {\left (c d g x - c d f\right )} e\right )} \sqrt {-2 \, c d + b e} \arctan \left (-\frac {\sqrt {-2 \, c d + b e} \sqrt {x e + d}}{\sqrt {c d^{2} - b d e - {\left (c x^{2} + b x\right )} e^{2}}}\right ) - \sqrt {c d^{2} - b d e - {\left (c x^{2} + b x\right )} e^{2}} {\left (4 \, c d g - {\left (c g x + 3 \, c f + b g\right )} e\right )} \sqrt {x e + d}\right )}}{3 \, {\left (c x e^{3} + 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^2*g - c*f*x*e^2 + (c*d*g*x - c*d*f)*e)*sqrt(2*c*d - b*e)*log((3*c*d^2 - (c*x^2 + 2*b*x)*e^2 + 2*
(c*d*x - b*d)*e - 2*sqrt(c*d^2 - b*d*e - (c*x^2 + b*x)*e^2)*sqrt(2*c*d - b*e)*sqrt(x*e + d))/(x^2*e^2 + 2*d*x*
e + d^2)) + 2*sqrt(c*d^2 - b*d*e - (c*x^2 + b*x)*e^2)*(4*c*d*g - (c*g*x + 3*c*f + b*g)*e)*sqrt(x*e + d))/(c*x*
e^3 + c*d*e^2), 2/3*(3*(c*d^2*g - c*f*x*e^2 + (c*d*g*x - c*d*f)*e)*sqrt(-2*c*d + b*e)*arctan(-sqrt(-2*c*d + b*
e)*sqrt(x*e + d)/sqrt(c*d^2 - b*d*e - (c*x^2 + b*x)*e^2)) - sqrt(c*d^2 - b*d*e - (c*x^2 + b*x)*e^2)*(4*c*d*g -
(c*g*x + 3*c*f + b*g)*e)*sqrt(x*e + d))/(c*x*e^3 + c*d*e^2)]

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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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (169) = 338\).
time = 1.42, size = 406, normalized size = 2.18 \begin {gather*} -\frac {2}{3} \, {\left (\frac {3 \, {\left (2 \, c d^{2} g - 2 \, c d f e - b d g e + b f e^{2}\right )} \arctan \left (\frac {\sqrt {-{\left (x e + d\right )} c + 2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right )}{\sqrt {-2 \, c d + b e}} + \frac {3 \, \sqrt {-{\left (x e + d\right )} c + 2 \, c d - b e} c^{3} d g - 3 \, \sqrt {-{\left (x e + d\right )} c + 2 \, c d - b e} c^{3} f e + {\left (-{\left (x e + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{2} g}{c^{3}}\right )} e^{\left (-2\right )} + \frac {2 \, {\left (6 \, c^{2} d^{2} g \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) - 6 \, c^{2} d f \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) e - 3 \, b c d g \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) e + 3 \, b c f \arctan \left (\frac {\sqrt {2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right ) e^{2} + 5 \, \sqrt {2 \, c d - b e} \sqrt {-2 \, c d + b e} c d g - 3 \, \sqrt {2 \, c d - b e} \sqrt {-2 \, c d + b e} c f e - \sqrt {2 \, c d - b e} \sqrt {-2 \, c d + b e} b g e\right )} e^{\left (-2\right )}}{3 \, \sqrt {-2 \, c d + b e} c} \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]

-2/3*(3*(2*c*d^2*g - 2*c*d*f*e - b*d*g*e + b*f*e^2)*arctan(sqrt(-(x*e + d)*c + 2*c*d - b*e)/sqrt(-2*c*d + b*e)
)/sqrt(-2*c*d + b*e) + (3*sqrt(-(x*e + d)*c + 2*c*d - b*e)*c^3*d*g - 3*sqrt(-(x*e + d)*c + 2*c*d - b*e)*c^3*f*
e + (-(x*e + d)*c + 2*c*d - b*e)^(3/2)*c^2*g)/c^3)*e^(-2) + 2/3*(6*c^2*d^2*g*arctan(sqrt(2*c*d - b*e)/sqrt(-2*
c*d + b*e)) - 6*c^2*d*f*arctan(sqrt(2*c*d - b*e)/sqrt(-2*c*d + b*e))*e - 3*b*c*d*g*arctan(sqrt(2*c*d - b*e)/sq
rt(-2*c*d + b*e))*e + 3*b*c*f*arctan(sqrt(2*c*d - b*e)/sqrt(-2*c*d + b*e))*e^2 + 5*sqrt(2*c*d - b*e)*sqrt(-2*c
*d + b*e)*c*d*g - 3*sqrt(2*c*d - b*e)*sqrt(-2*c*d + b*e)*c*f*e - sqrt(2*c*d - b*e)*sqrt(-2*c*d + b*e)*b*g*e)*e
^(-2)/(sqrt(-2*c*d + b*e)*c)

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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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