3.8.17 \(\int \frac {\sqrt {d+e x}}{(f+g x)^{5/2} \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [717]

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

[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 = 0.10, 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 = {886, 874} \begin {gather*} \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)} \end {gather*}

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 874

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 886

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 = 0.12, size = 69, normalized size = 0.53 \begin {gather*} \frac {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}} \end {gather*}

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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Maple [A]
time = 0.16, size = 61, normalized size = 0.47

method	result	size
default	\(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (-2 c d g x +a e g -3 c d f \right )}{3 \sqrt {e x +d}\, \left (g x +f \right )^{\frac {3}{2}} \left (a e g -c d f \right )^{2}}\)	\(61\)
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-2 c d g x +a e g -3 c d f \right ) \sqrt {e x +d}}{3 \left (g x +f \right )^{\frac {3}{2}} \left (a^{2} e^{2} g^{2}-2 a c d e f g +f^{2} c^{2} d^{2}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (119) = 238\).
time = 3.21, size = 302, normalized size = 2.34 \begin {gather*} \frac {2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d g x + 3 \, c d f - a g e\right )} \sqrt {g x + f} \sqrt {x e + d}}{3 \, {\left (c^{2} d^{3} f^{2} g^{2} x^{2} + 2 \, c^{2} d^{3} f^{3} g x + c^{2} d^{3} f^{4} + {\left (a^{2} g^{4} x^{3} + 2 \, a^{2} f g^{3} x^{2} + a^{2} f^{2} g^{2} x\right )} e^{3} - {\left (2 \, a c d f g^{3} x^{3} - a^{2} d f^{2} g^{2} + {\left (4 \, a c d f^{2} g^{2} - a^{2} d g^{4}\right )} x^{2} + 2 \, {\left (a c d f^{3} g - a^{2} d f g^{3}\right )} x\right )} e^{2} + {\left (c^{2} d^{2} f^{2} g^{2} x^{3} - 2 \, a c d^{2} f^{3} g + 2 \, {\left (c^{2} d^{2} f^{3} g - a c d^{2} f g^{3}\right )} x^{2} + {\left (c^{2} d^{2} f^{4} - 4 \, a c d^{2} f^{2} g^{2}\right )} x\right )} e\right )}} \end {gather*}

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

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

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]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Mupad [B]
time = 4.90, size = 147, normalized size = 1.14 \begin {gather*} -\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}} \end {gather*}

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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