3.7.82 \(\int \frac {(f+g x) \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{\sqrt {d+e x}} \, dx\) [682]

Optimal. Leaf size=125 \[ -\frac {2 \left (2 a e^2 g-c d (5 e f-3 d g)\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{15 c^2 d^2 e (d+e x)^{3/2}}+\frac {2 g \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d e \sqrt {d+e x}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {808, 662} \begin {gather*} \frac {2 g \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 c d e \sqrt {d+e x}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} \left (2 a e^2 g-c d (5 e f-3 d g)\right )}{15 c^2 d^2 e (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 54, normalized size = 0.43 \begin {gather*} \frac {2 ((a e+c d x) (d+e x))^{3/2} (-2 a e g+c d (5 f+3 g x))}{15 c^2 d^2 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [A]
time = 0.15, size = 57, normalized size = 0.46

method	result	size
default	\(-\frac {2 \left (c d x +a e \right ) \left (-3 c d g x +2 a e g -5 c d f \right ) \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}}{15 c^{2} d^{2} \sqrt {e x +d}}\)	\(57\)
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-3 c d g x +2 a e g -5 c d f \right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{15 c^{2} d^{2} \sqrt {e x +d}}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 67, normalized size = 0.54 \begin {gather*} \frac {2 \, {\left (c d x + a e\right )}^{\frac {3}{2}} f}{3 \, c d} + \frac {2 \, {\left (3 \, c^{2} d^{2} x^{2} + a c d x e - 2 \, a^{2} e^{2}\right )} \sqrt {c d x + a e} g}{15 \, c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 1.42, size = 104, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (3 \, c^{2} d^{2} g x^{2} + 5 \, c^{2} d^{2} f x - 2 \, a^{2} g e^{2} + {\left (a c d g x + 5 \, a c d f\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^{2} d^{2} x e + c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(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]

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

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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 (a e + c d 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)*(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)/sqrt(d + e*x), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (114) = 228\).
time = 1.74, size = 246, normalized size = 1.97 \begin {gather*} \frac {2}{15} \, {\left (5 \, f {\left (\frac {{\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} e^{\left (-1\right )}}{c d} + \frac {\sqrt {-c d^{2} e + a e^{3}} c d^{2} - \sqrt {-c d^{2} e + a e^{3}} a e^{2}}{c d}\right )} e^{\left (-1\right )} - g {\left (\frac {{\left (5 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} - 3 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}}\right )} e^{\left (-2\right )}}{c^{2} d^{2}} + \frac {3 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} - \sqrt {-c d^{2} e + a e^{3}} a c d^{2} e^{2} - 2 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} e^{4}}{c^{2} d^{2}}\right )} e^{\left (-2\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(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]

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

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Mupad [B]
time = 3.13, size = 93, normalized size = 0.74 \begin {gather*} \frac {\left (\frac {2\,g\,x^2}{5}-\frac {4\,a^2\,e^2\,g-10\,a\,c\,d\,e\,f}{15\,c^2\,d^2}+\frac {x\,\left (10\,f\,c^2\,d^2+2\,a\,e\,g\,c\,d\right )}{15\,c^2\,d^2}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{\sqrt {d+e\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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