3.717 \(\int \frac {\sqrt {d+e x}}{(f+g x)^{5/2} \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\)

Optimal. Leaf size=129 \[ \frac {4 c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 \sqrt {d+e x} \sqrt {f+g x} (c d f-a e g)^2}+\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 \sqrt {d+e x} (f+g x)^{3/2} (c d f-a e g)} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {872, 860} \[ \frac {4 c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 \sqrt {d+e x} \sqrt {f+g x} (c d f-a e g)^2}+\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 \sqrt {d+e x} (f+g x)^{3/2} (c d f-a e g)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + e*x]/((f + g*x)^(5/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

(2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*(c*d*f - a*e*g)*Sqrt[d + e*x]*(f + g*x)^(3/2)) + (4*c*d*Sqr
t[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*(c*d*f - a*e*g)^2*Sqrt[d + e*x]*Sqrt[f + g*x])

Rule 860

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

Rule 872

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 69, normalized size = 0.53 \[ \frac {2 \sqrt {(d+e x) (a e+c d x)} (c d (3 f+2 g x)-a e g)}{3 \sqrt {d+e x} (f+g x)^{3/2} (c d f-a e g)^2} \]

Antiderivative was successfully verified.

[In]

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

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fricas [B]  time = 0.71, size = 288, normalized size = 2.23 \[ \frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d g x + 3 \, c d f - a e g\right )} \sqrt {e x + d} \sqrt {g x + f}}{3 \, {\left (c^{2} d^{3} f^{4} - 2 \, a c d^{2} e f^{3} g + a^{2} d e^{2} f^{2} g^{2} + {\left (c^{2} d^{2} e f^{2} g^{2} - 2 \, a c d e^{2} f g^{3} + a^{2} e^{3} g^{4}\right )} x^{3} + {\left (2 \, c^{2} d^{2} e f^{3} g + a^{2} d e^{2} g^{4} + {\left (c^{2} d^{3} - 4 \, a c d e^{2}\right )} f^{2} g^{2} - 2 \, {\left (a c d^{2} e - a^{2} e^{3}\right )} f g^{3}\right )} x^{2} + {\left (c^{2} d^{2} e f^{4} + 2 \, a^{2} d e^{2} f g^{3} + 2 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} f^{3} g - {\left (4 \, a c d^{2} e - a^{2} e^{3}\right )} f^{2} g^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(g*x+f)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(g*x+f)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 98, normalized size = 0.76 \[ -\frac {2 \left (c d x +a e \right ) \left (-2 c d g x +a e g -3 c d f \right ) \sqrt {e x +d}}{3 \left (g x +f \right )^{\frac {3}{2}} \left (a^{2} e^{2} g^{2}-2 a c d e f g +f^{2} c^{2} d^{2}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)/(g*x+f)^(5/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e x + d}}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (g x + f\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(g*x+f)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(e*x + d)/(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(g*x + f)^(5/2)), x)

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mupad [B]  time = 4.90, size = 147, normalized size = 1.14 \[ -\frac {\left (\frac {\left (2\,a\,e\,g-6\,c\,d\,f\right )\,\sqrt {d+e\,x}}{3\,e\,g\,{\left (a\,e\,g-c\,d\,f\right )}^2}-\frac {4\,c\,d\,x\,\sqrt {d+e\,x}}{3\,e\,{\left (a\,e\,g-c\,d\,f\right )}^2}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{x^2\,\sqrt {f+g\,x}+\frac {d\,f\,\sqrt {f+g\,x}}{e\,g}+\frac {x\,\sqrt {f+g\,x}\,\left (d\,g+e\,f\right )}{e\,g}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(1/2)/((f + g*x)^(5/2)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)),x)

[Out]

-((((2*a*e*g - 6*c*d*f)*(d + e*x)^(1/2))/(3*e*g*(a*e*g - c*d*f)^2) - (4*c*d*x*(d + e*x)^(1/2))/(3*e*(a*e*g - c
*d*f)^2))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(x^2*(f + g*x)^(1/2) + (d*f*(f + g*x)^(1/2))/(e*g) +
(x*(f + g*x)^(1/2)*(d*g + e*f))/(e*g))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)/(g*x+f)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

Timed out

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