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

Optimal. Leaf size=118 \[ -\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2} (-2 b e g-c d g+5 c e f)}{15 c^2 e^2 (d+e x)^{3/2}}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{5 c e^2 \sqrt {d+e x}} \]

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Rubi [A]  time = 0.17, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {794, 648} \begin {gather*} -\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2} (-2 b e g-c d g+5 c e f)}{15 c^2 e^2 (d+e x)^{3/2}}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{5 c e^2 \sqrt {d+e x}} \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])/Sqrt[d + e*x],x]

[Out]

(-2*(5*c*e*f - c*d*g - 2*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(15*c^2*e^2*(d + e*x)^(3/2)) - (2
*g*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(5*c*e^2*Sqrt[d + e*x])

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

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}}{\sqrt {d+e x}} \, dx &=-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{5 c e^2 \sqrt {d+e x}}-\frac {\left (2 \left (\frac {3}{2} e \left (-2 c e^2 f+b e^2 g\right )+\frac {1}{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}}{\sqrt {d+e x}} \, dx}{5 c e^3}\\ &=-\frac {2 (5 c e f-c d g-2 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{15 c^2 e^2 (d+e x)^{3/2}}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{5 c e^2 \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 76, normalized size = 0.64 \begin {gather*} \frac {2 (b e-c d+c e x) \sqrt {(d+e x) (c (d-e x)-b e)} (c (2 d g+5 e f+3 e g x)-2 b e g)}{15 c^2 e^2 \sqrt {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])/Sqrt[d + e*x],x]

[Out]

(2*(-(c*d) + b*e + c*e*x)*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))]*(-2*b*e*g + c*(5*e*f + 2*d*g + 3*e*g*x)))/(15
*c^2*e^2*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 0.25, size = 74, normalized size = 0.63 \begin {gather*} -\frac {2 \left ((d+e x) (2 c d-b e)-c (d+e x)^2\right )^{3/2} (-2 b e g+3 c g (d+e x)-c d g+5 c e f)}{15 c^2 e^2 (d+e x)^{3/2}} \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])/Sqrt[d + e*x],x]

[Out]

(-2*(5*c*e*f - c*d*g - 2*b*e*g + 3*c*g*(d + e*x))*((2*c*d - b*e)*(d + e*x) - c*(d + e*x)^2)^(3/2))/(15*c^2*e^2
*(d + e*x)^(3/2))

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fricas [A]  time = 0.57, size = 140, normalized size = 1.19 \begin {gather*} \frac {2 \, {\left (3 \, c^{2} e^{2} g x^{2} - 5 \, {\left (c^{2} d e - b c e^{2}\right )} f - 2 \, {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} g + {\left (5 \, c^{2} e^{2} f - {\left (c^{2} d e - b c e^{2}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}{15 \, {\left (c^{2} e^{3} x + c^{2} d e^{2}\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)^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*c^2*e^2*g*x^2 - 5*(c^2*d*e - b*c*e^2)*f - 2*(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*g + (5*c^2*e^2*f - (c^2*d*
e - b*c*e^2)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(e*x + d)/(c^2*e^3*x + c^2*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 )}}{\sqrt {e x + d}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 79, normalized size = 0.67 \begin {gather*} -\frac {2 \left (c e x +b e -c d \right ) \left (-3 c e g x +2 b e g -2 c d g -5 c e f \right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}{15 \sqrt {e x +d}\, c^{2} 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)^(1/2),x)

[Out]

-2/15*(c*e*x+b*e-c*d)*(-3*c*e*g*x+2*b*e*g-2*c*d*g-5*c*e*f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)/c^2/e^2/(e*x
+d)^(1/2)

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maxima [A]  time = 0.67, size = 112, normalized size = 0.95 \begin {gather*} \frac {2 \, {\left (c e x - c d + b e\right )} \sqrt {-c e x + c d - b e} f}{3 \, c e} + \frac {2 \, {\left (3 \, c^{2} e^{2} x^{2} - 2 \, c^{2} d^{2} + 4 \, b c d e - 2 \, b^{2} e^{2} - {\left (c^{2} d e - b c e^{2}\right )} x\right )} \sqrt {-c e x + c d - b e} g}{15 \, c^{2} e^{2}} \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)^(1/2),x, algorithm="maxima")

[Out]

2/3*(c*e*x - c*d + b*e)*sqrt(-c*e*x + c*d - b*e)*f/(c*e) + 2/15*(3*c^2*e^2*x^2 - 2*c^2*d^2 + 4*b*c*d*e - 2*b^2
*e^2 - (c^2*d*e - b*c*e^2)*x)*sqrt(-c*e*x + c*d - b*e)*g/(c^2*e^2)

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mupad [B]  time = 2.50, size = 100, normalized size = 0.85 \begin {gather*} \frac {\left (\frac {2\,g\,x^2}{5}+\frac {2\,x\,\left (b\,e\,g-c\,d\,g+5\,c\,e\,f\right )}{15\,c\,e}+\frac {2\,\left (b\,e-c\,d\right )\,\left (2\,c\,d\,g-2\,b\,e\,g+5\,c\,e\,f\right )}{15\,c^2\,e^2}\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}}{\sqrt {d+e\,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)^(1/2),x)

[Out]

(((2*g*x^2)/5 + (2*x*(b*e*g - c*d*g + 5*c*e*f))/(15*c*e) + (2*(b*e - c*d)*(2*c*d*g - 2*b*e*g + 5*c*e*f))/(15*c
^2*e^2))*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2))/(d + e*x)^(1/2)

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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 )}{\sqrt {d + e x}}\, 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)**(1/2),x)

[Out]

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

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