3.681 \(\int \frac {(f+g x)^2 \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{\sqrt {d+e x}} \, dx\)

Optimal. Leaf size=200 \[ -\frac {8 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g) \left (2 a e^2 g-c d (5 e f-3 d g)\right )}{105 c^3 d^3 e (d+e x)^{3/2}}+\frac {8 g \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)}{35 c^2 d^2 e \sqrt {d+e x}}+\frac {2 (f+g x)^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{7 c d (d+e x)^{3/2}} \]

[Out]

-8/105*(-a*e*g+c*d*f)*(2*a*e^2*g-c*d*(-3*d*g+5*e*f))*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c^3/d^3/e/(e*x+d)
^(3/2)+2/7*(g*x+f)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c/d/(e*x+d)^(3/2)+8/35*g*(-a*e*g+c*d*f)*(a*d*e+(a
*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c^2/d^2/e/(e*x+d)^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 200, 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 = {870, 794, 648} \[ \frac {8 g \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)}{35 c^2 d^2 e \sqrt {d+e x}}-\frac {8 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g) \left (2 a e^2 g-c d (5 e f-3 d g)\right )}{105 c^3 d^3 e (d+e x)^{3/2}}+\frac {2 (f+g x)^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{7 c d (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(c*d*f - a*e*g)*(2*a*e^2*g - c*d*(5*e*f - 3*d*g))*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(105*c^3*
d^3*e*(d + e*x)^(3/2)) + (8*g*(c*d*f - a*e*g)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(35*c^2*d^2*e*Sqr
t[d + e*x]) + (2*(f + g*x)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(7*c*d*(d + e*x)^(3/2))

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])

Rule 870

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

Rubi steps

\begin {align*} \int \frac {(f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}} \, dx &=\frac {2 (f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{7 c d (d+e x)^{3/2}}+\frac {\left (4 \left (c d e^2 f+c d^2 e g-e \left (c d^2+a e^2\right ) g\right )\right ) \int \frac {(f+g x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}} \, dx}{7 c d e^2}\\ &=\frac {8 g (c d f-a e g) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{35 c^2 d^2 e \sqrt {d+e x}}+\frac {2 (f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{7 c d (d+e x)^{3/2}}-\frac {\left (4 (c d f-a e g) \left (2 a e^2 g-c d (5 e f-3 d g)\right )\right ) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}} \, dx}{35 c^2 d^2 e}\\ &=-\frac {8 (c d f-a e g) \left (2 a e^2 g-c d (5 e f-3 d g)\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{105 c^3 d^3 e (d+e x)^{3/2}}+\frac {8 g (c d f-a e g) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{35 c^2 d^2 e \sqrt {d+e x}}+\frac {2 (f+g x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{7 c d (d+e x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 90, normalized size = 0.45 \[ \frac {2 ((d+e x) (a e+c d x))^{3/2} \left (8 a^2 e^2 g^2-4 a c d e g (7 f+3 g x)+c^2 d^2 \left (35 f^2+42 f g x+15 g^2 x^2\right )\right )}{105 c^3 d^3 (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(3/2)*(8*a^2*e^2*g^2 - 4*a*c*d*e*g*(7*f + 3*g*x) + c^2*d^2*(35*f^2 + 42*f*g*x + 1
5*g^2*x^2)))/(105*c^3*d^3*(d + e*x)^(3/2))

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fricas [A]  time = 1.14, size = 173, normalized size = 0.86 \[ \frac {2 \, {\left (15 \, c^{3} d^{3} g^{2} x^{3} + 35 \, a c^{2} d^{2} e f^{2} - 28 \, a^{2} c d e^{2} f g + 8 \, a^{3} e^{3} g^{2} + 3 \, {\left (14 \, c^{3} d^{3} f g + a c^{2} d^{2} e g^{2}\right )} x^{2} + {\left (35 \, c^{3} d^{3} f^{2} + 14 \, a c^{2} d^{2} e f g - 4 \, a^{2} c d e^{2} g^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{105 \, {\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*c^3*d^3*g^2*x^3 + 35*a*c^2*d^2*e*f^2 - 28*a^2*c*d*e^2*f*g + 8*a^3*e^3*g^2 + 3*(14*c^3*d^3*f*g + a*c^
2*d^2*e*g^2)*x^2 + (35*c^3*d^3*f^2 + 14*a*c^2*d^2*e*f*g - 4*a^2*c*d*e^2*g^2)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^
2 + a*e^2)*x)*sqrt(e*x + d)/(c^3*d^3*e*x + c^3*d^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (g x + f\right )}^{2}}{\sqrt {e x + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 116, normalized size = 0.58 \[ \frac {2 \left (c d x +a e \right ) \left (15 g^{2} x^{2} c^{2} d^{2}-12 a c d e \,g^{2} x +42 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-28 a c d e f g +35 f^{2} c^{2} d^{2}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{105 \sqrt {e x +d}\, c^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(c*d*x+a*e)*(15*c^2*d^2*g^2*x^2-12*a*c*d*e*g^2*x+42*c^2*d^2*f*g*x+8*a^2*e^2*g^2-28*a*c*d*e*f*g+35*c^2*d^
2*f^2)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)/c^3/d^3/(e*x+d)^(1/2)

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maxima [A]  time = 0.59, size = 133, normalized size = 0.66 \[ \frac {2 \, {\left (c d x + a e\right )}^{\frac {3}{2}} f^{2}}{3 \, c d} + \frac {4 \, {\left (3 \, c^{2} d^{2} x^{2} + a c d e x - 2 \, a^{2} e^{2}\right )} \sqrt {c d x + a e} f g}{15 \, c^{2} d^{2}} + \frac {2 \, {\left (15 \, c^{3} d^{3} x^{3} + 3 \, a c^{2} d^{2} e x^{2} - 4 \, a^{2} c d e^{2} x + 8 \, a^{3} e^{3}\right )} \sqrt {c d x + a e} g^{2}}{105 \, c^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

2/3*(c*d*x + a*e)^(3/2)*f^2/(c*d) + 4/15*(3*c^2*d^2*x^2 + a*c*d*e*x - 2*a^2*e^2)*sqrt(c*d*x + a*e)*f*g/(c^2*d^
2) + 2/105*(15*c^3*d^3*x^3 + 3*a*c^2*d^2*e*x^2 - 4*a^2*c*d*e^2*x + 8*a^3*e^3)*sqrt(c*d*x + a*e)*g^2/(c^3*d^3)

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mupad [B]  time = 3.25, size = 157, normalized size = 0.78 \[ \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {2\,g^2\,x^3}{7}+\frac {16\,a^3\,e^3\,g^2-56\,a^2\,c\,d\,e^2\,f\,g+70\,a\,c^2\,d^2\,e\,f^2}{105\,c^3\,d^3}+\frac {x\,\left (-8\,a^2\,c\,d\,e^2\,g^2+28\,a\,c^2\,d^2\,e\,f\,g+70\,c^3\,d^3\,f^2\right )}{105\,c^3\,d^3}+\frac {2\,g\,x^2\,\left (a\,e\,g+14\,c\,d\,f\right )}{35\,c\,d}\right )}{\sqrt {d+e\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((2*g^2*x^3)/7 + (16*a^3*e^3*g^2 + 70*a*c^2*d^2*e*f^2 - 56*a^2*
c*d*e^2*f*g)/(105*c^3*d^3) + (x*(70*c^3*d^3*f^2 - 8*a^2*c*d*e^2*g^2 + 28*a*c^2*d^2*e*f*g))/(105*c^3*d^3) + (2*
g*x^2*(a*e*g + 14*c*d*f))/(35*c*d)))/(d + e*x)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\left (d + e x\right ) \left (a e + c d x\right )} \left (f + g x\right )^{2}}{\sqrt {d + e x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**2*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2)/(e*x+d)**(1/2),x)

[Out]

Integral(sqrt((d + e*x)*(a*e + c*d*x))*(f + g*x)**2/sqrt(d + e*x), x)

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