3.747 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{3/2} (f+g x)^{7/2}} \, dx\)

Optimal. Leaf size=63 \[ \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 (d+e x)^{5/2} (f+g x)^{5/2} (c d f-a e g)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {860} \[ \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 (d+e x)^{5/2} (f+g x)^{5/2} (c d f-a e g)} \]

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/((d + e*x)^(3/2)*(f + g*x)^(7/2)),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 52, normalized size = 0.83 \[ \frac {2 ((d+e x) (a e+c d x))^{5/2}}{5 (d+e x)^{5/2} (f+g x)^{5/2} (c d f-a e g)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/((d + e*x)^(3/2)*(f + g*x)^(7/2)),x]

[Out]

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

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fricas [B]  time = 0.87, size = 232, normalized size = 3.68 \[ \frac {2 \, {\left (c^{2} d^{2} x^{2} + 2 \, a c d e x + a^{2} e^{2}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f}}{5 \, {\left (c d^{2} f^{4} - a d e f^{3} g + {\left (c d e f g^{3} - a e^{2} g^{4}\right )} x^{4} + {\left (3 \, c d e f^{2} g^{2} - a d e g^{4} + {\left (c d^{2} - 3 \, a e^{2}\right )} f g^{3}\right )} x^{3} + 3 \, {\left (c d e f^{3} g - a d e f g^{3} + {\left (c d^{2} - a e^{2}\right )} f^{2} g^{2}\right )} x^{2} + {\left (c d e f^{4} - 3 \, a d e f^{2} g^{2} + {\left (3 \, c d^{2} - a e^{2}\right )} f^{3} g\right )} x\right )}} \]

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)^(3/2)/(e*x+d)^(3/2)/(g*x+f)^(7/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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maple [A]  time = 0.01, size = 63, normalized size = 1.00 \[ -\frac {2 \left (c d x +a e \right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}{5 \left (g x +f \right )^{\frac {5}{2}} \left (a e g -c d f \right ) \left (e x +d \right )^{\frac {3}{2}}} \]

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)^(3/2)/(e*x+d)^(3/2)/(g*x+f)^(7/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {3}{2}}}{{\left (e x + d\right )}^{\frac {3}{2}} {\left (g x + f\right )}^{\frac {7}{2}}}\,{d x} \]

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)^(3/2)/(e*x+d)^(3/2)/(g*x+f)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)/((e*x + d)^(3/2)*(g*x + f)^(7/2)), x)

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mupad [B]  time = 4.07, size = 232, normalized size = 3.68 \[ -\frac {\left (\frac {2\,a^2\,e^2}{5\,a\,e\,g^3-5\,c\,d\,f\,g^2}+\frac {2\,c^2\,d^2\,x^2}{5\,a\,e\,g^3-5\,c\,d\,f\,g^2}+\frac {4\,a\,c\,d\,e\,x}{5\,a\,e\,g^3-5\,c\,d\,f\,g^2}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{x^2\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}-\frac {\sqrt {f+g\,x}\,\left (5\,c\,d\,f^3-5\,a\,e\,f^2\,g\right )\,\sqrt {d+e\,x}}{5\,a\,e\,g^3-5\,c\,d\,f\,g^2}+\frac {x\,\sqrt {f+g\,x}\,\left (10\,a\,e\,f\,g^2-10\,c\,d\,f^2\,g\right )\,\sqrt {d+e\,x}}{5\,a\,e\,g^3-5\,c\,d\,f\,g^2}} \]

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)^(3/2)/((f + g*x)^(7/2)*(d + e*x)^(3/2)),x)

[Out]

-(((2*a^2*e^2)/(5*a*e*g^3 - 5*c*d*f*g^2) + (2*c^2*d^2*x^2)/(5*a*e*g^3 - 5*c*d*f*g^2) + (4*a*c*d*e*x)/(5*a*e*g^
3 - 5*c*d*f*g^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 +
g*x)^(1/2)*(5*c*d*f^3 - 5*a*e*f^2*g)*(d + e*x)^(1/2))/(5*a*e*g^3 - 5*c*d*f*g^2) + (x*(f + g*x)^(1/2)*(10*a*e*f
*g^2 - 10*c*d*f^2*g)*(d + e*x)^(1/2))/(5*a*e*g^3 - 5*c*d*f*g^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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2)/(e*x+d)**(3/2)/(g*x+f)**(7/2),x)

[Out]

Timed out

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