3.739 \(\int \frac {\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{\sqrt {d+e x} (f+g x)^{7/2}} \, dx\)

Optimal. Leaf size=129 \[ \frac {4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{15 (d+e x)^{3/2} (f+g x)^{3/2} (c d f-a e g)^2}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 (d+e x)^{3/2} (f+g x)^{5/2} (c d f-a e g)} \]

[Out]

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

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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} \[ \frac {4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{15 (d+e x)^{3/2} (f+g x)^{3/2} (c d f-a e g)^2}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 (d+e x)^{3/2} (f+g x)^{5/2} (c d f-a e g)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/(Sqrt[d + e*x]*(f + g*x)^(7/2)),x]

[Out]

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

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 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 {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x} (f+g x)^{7/2}} \, dx &=\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 (c d f-a e g) (d+e x)^{3/2} (f+g x)^{5/2}}+\frac {(2 c d) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x} (f+g x)^{5/2}} \, dx}{5 (c d f-a e g)}\\ &=\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 (c d f-a e g) (d+e x)^{3/2} (f+g x)^{5/2}}+\frac {4 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{15 (c d f-a e g)^2 (d+e x)^{3/2} (f+g x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 69, normalized size = 0.53 \[ \frac {2 ((d+e x) (a e+c d x))^{3/2} (c d (5 f+2 g x)-3 a e g)}{15 (d+e x)^{3/2} (f+g x)^{5/2} (c d f-a e g)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/(Sqrt[d + e*x]*(f + g*x)^(7/2)),x]

[Out]

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

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fricas [B]  time = 0.92, size = 402, normalized size = 3.12 \[ \frac {2 \, {\left (2 \, c^{2} d^{2} g x^{2} + 5 \, a c d e f - 3 \, a^{2} e^{2} g + {\left (5 \, c^{2} d^{2} f - a c d e g\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f}}{15 \, {\left (c^{2} d^{3} f^{5} - 2 \, a c d^{2} e f^{4} g + a^{2} d e^{2} f^{3} g^{2} + {\left (c^{2} d^{2} e f^{2} g^{3} - 2 \, a c d e^{2} f g^{4} + a^{2} e^{3} g^{5}\right )} x^{4} + {\left (3 \, c^{2} d^{2} e f^{3} g^{2} + a^{2} d e^{2} g^{5} + {\left (c^{2} d^{3} - 6 \, a c d e^{2}\right )} f^{2} g^{3} - {\left (2 \, a c d^{2} e - 3 \, a^{2} e^{3}\right )} f g^{4}\right )} x^{3} + 3 \, {\left (c^{2} d^{2} e f^{4} g + a^{2} d e^{2} f g^{4} + {\left (c^{2} d^{3} - 2 \, a c d e^{2}\right )} f^{3} g^{2} - {\left (2 \, a c d^{2} e - a^{2} e^{3}\right )} f^{2} g^{3}\right )} x^{2} + {\left (c^{2} d^{2} e f^{5} + 3 \, a^{2} d e^{2} f^{2} g^{3} + {\left (3 \, c^{2} d^{3} - 2 \, a c d e^{2}\right )} f^{4} g - {\left (6 \, a c d^{2} e - a^{2} e^{3}\right )} f^{3} g^{2}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(g*x+f)^(7/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

2/15*(2*c^2*d^2*g*x^2 + 5*a*c*d*e*f - 3*a^2*e^2*g + (5*c^2*d^2*f - a*c*d*e*g)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d
^2 + a*e^2)*x)*sqrt(e*x + d)*sqrt(g*x + f)/(c^2*d^3*f^5 - 2*a*c*d^2*e*f^4*g + a^2*d*e^2*f^3*g^2 + (c^2*d^2*e*f
^2*g^3 - 2*a*c*d*e^2*f*g^4 + a^2*e^3*g^5)*x^4 + (3*c^2*d^2*e*f^3*g^2 + a^2*d*e^2*g^5 + (c^2*d^3 - 6*a*c*d*e^2)
*f^2*g^3 - (2*a*c*d^2*e - 3*a^2*e^3)*f*g^4)*x^3 + 3*(c^2*d^2*e*f^4*g + a^2*d*e^2*f*g^4 + (c^2*d^3 - 2*a*c*d*e^
2)*f^3*g^2 - (2*a*c*d^2*e - a^2*e^3)*f^2*g^3)*x^2 + (c^2*d^2*e*f^5 + 3*a^2*d*e^2*f^2*g^3 + (3*c^2*d^3 - 2*a*c*
d*e^2)*f^4*g - (6*a*c*d^2*e - a^2*e^3)*f^3*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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(g*x+f)^(7/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 99, normalized size = 0.77 \[ -\frac {2 \left (c d x +a e \right ) \left (-2 c d g x +3 a e g -5 c d f \right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{15 \left (g x +f \right )^{\frac {5}{2}} \left (a^{2} e^{2} g^{2}-2 a c d e f g +f^{2} c^{2} d^{2}\right ) \sqrt {e x +d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)/(g*x+f)^(7/2)/(e*x+d)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{\sqrt {e x + d} {\left (g x + f\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(g*x+f)^(7/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)/(sqrt(e*x + d)*(g*x + f)^(7/2)), x)

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mupad [B]  time = 4.08, size = 187, normalized size = 1.45 \[ \frac {\left (\frac {x\,\left (10\,c^2\,d^2\,f-2\,a\,c\,d\,e\,g\right )}{15\,g^2\,{\left (a\,e\,g-c\,d\,f\right )}^2}-\frac {6\,a^2\,e^2\,g-10\,a\,c\,d\,e\,f}{15\,g^2\,{\left (a\,e\,g-c\,d\,f\right )}^2}+\frac {4\,c^2\,d^2\,x^2}{15\,g\,{\left (a\,e\,g-c\,d\,f\right )}^2}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{x^2\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}+\frac {f^2\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{g^2}+\frac {2\,f\,x\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{g}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)/((f + g*x)^(7/2)*(d + e*x)^(1/2)),x)

[Out]

(((x*(10*c^2*d^2*f - 2*a*c*d*e*g))/(15*g^2*(a*e*g - c*d*f)^2) - (6*a^2*e^2*g - 10*a*c*d*e*f)/(15*g^2*(a*e*g -
c*d*f)^2) + (4*c^2*d^2*x^2)/(15*g*(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 + e*x)^(1/2) + (f^2*(f + g*x)^(1/2)*(d + e*x)^(1/2))/g^2 + (2*f*x*(f + g*x)^(1/2)*(d + e*x)^(1/
2))/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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2)/(g*x+f)**(7/2)/(e*x+d)**(1/2),x)

[Out]

Timed out

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