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

Optimal. Leaf size=128 \[ \frac {4 g \sqrt {d+e x} \sqrt {f+g x}}{3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)^2}-\frac {2 (d+e x)^{3/2} \sqrt {f+g x}}{3 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 128, 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 = {868, 860} \[ \frac {4 g \sqrt {d+e x} \sqrt {f+g x}}{3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)^2}-\frac {2 (d+e x)^{3/2} \sqrt {f+g x}}{3 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 868

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))/((p + 1)*(c*e*f + c*d*g - b*e*g)), x]
+ Dist[(e^2*g*(m - n - 2))/((p + 1)*(c*e*f + c*d*g - b*e*g)), Int[(d + e*x)^(m - 1)*(f + g*x)^n*(a + b*x + c*
x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, 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[p, -1] && RationalQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 68, normalized size = 0.53 \[ \frac {2 (d+e x)^{3/2} \sqrt {f+g x} (3 a e g-c d (f-2 g x))}{3 ((d+e x) (a e+c d x))^{3/2} (c d f-a e g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.08, size = 318, normalized size = 2.48 \[ \frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d g x - c d f + 3 \, a e g\right )} \sqrt {e x + d} \sqrt {g x + f}}{3 \, {\left (a^{2} c^{2} d^{3} e^{2} f^{2} - 2 \, a^{3} c d^{2} e^{3} f g + a^{4} d e^{4} g^{2} + {\left (c^{4} d^{4} e f^{2} - 2 \, a c^{3} d^{3} e^{2} f g + a^{2} c^{2} d^{2} e^{3} g^{2}\right )} x^{3} + {\left ({\left (c^{4} d^{5} + 2 \, a c^{3} d^{3} e^{2}\right )} f^{2} - 2 \, {\left (a c^{3} d^{4} e + 2 \, a^{2} c^{2} d^{2} e^{3}\right )} f g + {\left (a^{2} c^{2} d^{3} e^{2} + 2 \, a^{3} c d e^{4}\right )} g^{2}\right )} x^{2} + {\left ({\left (2 \, a c^{3} d^{4} e + a^{2} c^{2} d^{2} e^{3}\right )} f^{2} - 2 \, {\left (2 \, a^{2} c^{2} d^{3} e^{2} + a^{3} c d e^{4}\right )} f g + {\left (2 \, a^{3} c d^{2} e^{3} + a^{4} e^{5}\right )} g^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(g*x+f)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),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 - c*d*f + 3*a*e*g)*sqrt(e*x + d)*sqrt(g*x + f)/(a^2
*c^2*d^3*e^2*f^2 - 2*a^3*c*d^2*e^3*f*g + a^4*d*e^4*g^2 + (c^4*d^4*e*f^2 - 2*a*c^3*d^3*e^2*f*g + a^2*c^2*d^2*e^
3*g^2)*x^3 + ((c^4*d^5 + 2*a*c^3*d^3*e^2)*f^2 - 2*(a*c^3*d^4*e + 2*a^2*c^2*d^2*e^3)*f*g + (a^2*c^2*d^3*e^2 + 2
*a^3*c*d*e^4)*g^2)*x^2 + ((2*a*c^3*d^4*e + a^2*c^2*d^2*e^3)*f^2 - 2*(2*a^2*c^2*d^3*e^2 + a^3*c*d*e^4)*f*g + (2
*a^3*c*d^2*e^3 + a^4*e^5)*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)^(5/2)/(g*x+f)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 99, normalized size = 0.77 \[ \frac {2 \sqrt {g x +f}\, \left (c d x +a e \right ) \left (2 c d g x +3 a e g -c d f \right ) \left (e x +d \right )^{\frac {5}{2}}}{3 \left (a^{2} e^{2} g^{2}-2 a c d e f g +f^{2} c^{2} d^{2}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.06, size = 246, normalized size = 1.92 \[ \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {4\,g^2\,x^2\,\sqrt {d+e\,x}}{3\,c\,d\,e\,{\left (a\,e\,g-c\,d\,f\right )}^2}-\frac {\left (2\,c\,d\,f^2-6\,a\,e\,f\,g\right )\,\sqrt {d+e\,x}}{3\,c^2\,d^2\,e\,{\left (a\,e\,g-c\,d\,f\right )}^2}+\frac {x\,\left (6\,a\,e\,g^2+2\,c\,d\,f\,g\right )\,\sqrt {d+e\,x}}{3\,c^2\,d^2\,e\,{\left (a\,e\,g-c\,d\,f\right )}^2}\right )}{x^3\,\sqrt {f+g\,x}+\frac {a^2\,e\,\sqrt {f+g\,x}}{c^2\,d}+\frac {x^2\,\sqrt {f+g\,x}\,\left (c\,d^2+2\,a\,e^2\right )}{c\,d\,e}+\frac {a\,x\,\sqrt {f+g\,x}\,\left (2\,c\,d^2+a\,e^2\right )}{c^2\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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