3.7.58 \(\int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [658]

Optimal. Leaf size=200 \[ -\frac {8 (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^3 d^3 e \sqrt {d+e x}}+\frac {8 g (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2 e}+\frac {2 (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d \sqrt {d+e x}} \]

[Out]

-8/15*(-a*e*g+c*d*f)*(2*a*e^2*g-c*d*(-d*g+3*e*f))*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^3/d^3/e/(e*x+d)^(1
/2)+2/5*(g*x+f)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/(e*x+d)^(1/2)+8/15*g*(-a*e*g+c*d*f)*(e*x+d)^(1/2
)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^2/d^2/e

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Rubi [A]
time = 0.15, 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 = {884, 808, 662} \begin {gather*} -\frac {8 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right )}{15 c^3 d^3 e \sqrt {d+e x}}+\frac {8 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)}{15 c^2 d^2 e}+\frac {2 (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(c*d*f - a*e*g)*(2*a*e^2*g - c*d*(3*e*f - d*g))*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(15*c^3*d^3*e
*Sqrt[d + e*x]) + (8*g*(c*d*f - a*e*g)*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(15*c^2*d^2*
e) + (2*(f + g*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*c*d*Sqrt[d + e*x])

Rule 662

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 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])

Rule 884

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] &&  !Int
egerQ[p] && EqQ[m + p, 0] && GtQ[n, 0] && NeQ[m - n - 1, 0] && (IntegerQ[2*p] || IntegerQ[n])

Rubi steps

\begin {align*} \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac {2 (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d \sqrt {d+e x}}+\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 {\sqrt {d+e x} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{5 c d e^2}\\ &=\frac {8 g (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2 e}+\frac {2 (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d \sqrt {d+e x}}-\frac {\left (4 (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{15 c^2 d^2 e}\\ &=-\frac {8 (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^3 d^3 e \sqrt {d+e x}}+\frac {8 g (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2 e}+\frac {2 (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 89, normalized size = 0.44 \begin {gather*} \frac {2 \sqrt {(a e+c d x) (d+e x)} \left (8 a^2 e^2 g^2-4 a c d e g (5 f+g x)+c^2 d^2 \left (15 f^2+10 f g x+3 g^2 x^2\right )\right )}{15 c^3 d^3 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(8*a^2*e^2*g^2 - 4*a*c*d*e*g*(5*f + g*x) + c^2*d^2*(15*f^2 + 10*f*g*x + 3*g^2
*x^2)))/(15*c^3*d^3*Sqrt[d + e*x])

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Maple [A]
time = 0.14, size = 98, normalized size = 0.49

method	result	size
default	\(\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (3 g^{2} x^{2} c^{2} d^{2}-4 a c d e \,g^{2} x +10 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-20 a c d e f g +15 f^{2} c^{2} d^{2}\right )}{15 \sqrt {e x +d}\, c^{3} d^{3}}\)	\(98\)
gosper	\(\frac {2 \left (c d x +a e \right ) \left (3 g^{2} x^{2} c^{2} d^{2}-4 a c d e \,g^{2} x +10 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-20 a c d e f g +15 f^{2} c^{2} d^{2}\right ) \sqrt {e x +d}}{15 c^{3} d^{3} \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/(e*x+d)^(1/2)*((c*d*x+a*e)*(e*x+d))^(1/2)*(3*c^2*d^2*g^2*x^2-4*a*c*d*e*g^2*x+10*c^2*d^2*f*g*x+8*a^2*e^2*g
^2-20*a*c*d*e*f*g+15*c^2*d^2*f^2)/c^3/d^3

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Maxima [A]
time = 0.33, size = 135, normalized size = 0.68 \begin {gather*} \frac {2 \, \sqrt {c d x + a e} f^{2}}{c d} + \frac {4 \, {\left (c^{2} d^{2} x^{2} - a c d x e - 2 \, a^{2} e^{2}\right )} f g}{3 \, \sqrt {c d x + a e} c^{2} d^{2}} + \frac {2 \, {\left (3 \, c^{3} d^{3} x^{3} - a c^{2} d^{2} x^{2} e + 4 \, a^{2} c d x e^{2} + 8 \, a^{3} e^{3}\right )} g^{2}}{15 \, \sqrt {c d x + a e} c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.34, size = 124, normalized size = 0.62 \begin {gather*} \frac {2 \, {\left (3 \, c^{2} d^{2} g^{2} x^{2} + 10 \, c^{2} d^{2} f g x + 15 \, c^{2} d^{2} f^{2} + 8 \, a^{2} g^{2} e^{2} - 4 \, {\left (a c d g^{2} x + 5 \, a c d f g\right )} e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{15 \, {\left (c^{3} d^{3} x e + c^{3} d^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*c^2*d^2*g^2*x^2 + 10*c^2*d^2*f*g*x + 15*c^2*d^2*f^2 + 8*a^2*g^2*e^2 - 4*(a*c*d*g^2*x + 5*a*c*d*f*g)*e)
*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(x*e + d)/(c^3*d^3*x*e + c^3*d^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.16, size = 344, normalized size = 1.72 \begin {gather*} \frac {2 \, {\left (c^{2} d^{2} f^{2} - 2 \, a c d f g e + a^{2} g^{2} e^{2}\right )} \sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} e^{\left (-1\right )}}{c^{3} d^{3}} - \frac {2 \, {\left (3 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} g^{2} - 10 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{3} f g e + 15 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{2} f^{2} e^{2} + 4 \, \sqrt {-c d^{2} e + a e^{3}} a c d^{2} g^{2} e^{2} - 20 \, \sqrt {-c d^{2} e + a e^{3}} a c d f g e^{3} + 8 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} g^{2} e^{4}\right )} e^{\left (-3\right )}}{15 \, c^{3} d^{3}} + \frac {2 \, {\left (10 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c d f g e^{2} - 10 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a g^{2} e^{3} + 3 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} g^{2}\right )} e^{\left (-5\right )}}{15 \, c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(c^2*d^2*f^2 - 2*a*c*d*f*g*e + a^2*g^2*e^2)*sqrt((x*e + d)*c*d*e - c*d^2*e + a*e^3)*e^(-1)/(c^3*d^3) - 2/15*
(3*sqrt(-c*d^2*e + a*e^3)*c^2*d^4*g^2 - 10*sqrt(-c*d^2*e + a*e^3)*c^2*d^3*f*g*e + 15*sqrt(-c*d^2*e + a*e^3)*c^
2*d^2*f^2*e^2 + 4*sqrt(-c*d^2*e + a*e^3)*a*c*d^2*g^2*e^2 - 20*sqrt(-c*d^2*e + a*e^3)*a*c*d*f*g*e^3 + 8*sqrt(-c
*d^2*e + a*e^3)*a^2*g^2*e^4)*e^(-3)/(c^3*d^3) + 2/15*(10*((x*e + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c*d*f*g*e^2
- 10*((x*e + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*g^2*e^3 + 3*((x*e + d)*c*d*e - c*d^2*e + a*e^3)^(5/2)*g^2)*e
^(-5)/(c^3*d^3)

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Mupad [B]
time = 3.40, size = 142, normalized size = 0.71 \begin {gather*} \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {\sqrt {d+e\,x}\,\left (16\,a^2\,e^2\,g^2-40\,a\,c\,d\,e\,f\,g+30\,c^2\,d^2\,f^2\right )}{15\,c^3\,d^3\,e}+\frac {2\,g^2\,x^2\,\sqrt {d+e\,x}}{5\,c\,d\,e}-\frac {4\,g\,x\,\left (2\,a\,e\,g-5\,c\,d\,f\right )\,\sqrt {d+e\,x}}{15\,c^2\,d^2\,e}\right )}{x+\frac {d}{e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*(((d + e*x)^(1/2)*(16*a^2*e^2*g^2 + 30*c^2*d^2*f^2 - 40*a*c*d*e
*f*g))/(15*c^3*d^3*e) + (2*g^2*x^2*(d + e*x)^(1/2))/(5*c*d*e) - (4*g*x*(2*a*e*g - 5*c*d*f)*(d + e*x)^(1/2))/(1
5*c^2*d^2*e)))/(x + d/e)

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