3.6.15 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{3/2} (f+g x)^{9/2}} \, dx\)

Optimal. Leaf size=129 \[ \frac {4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{35 (d+e x)^{5/2} (f+g x)^{5/2} (c d f-a e g)^2}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{7 (d+e x)^{5/2} (f+g x)^{7/2} (c d f-a e g)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.15, 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} \begin {gather*} \frac {4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{35 (d+e x)^{5/2} (f+g x)^{5/2} (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 )^{5/2}}{7 (d+e x)^{5/2} (f+g x)^{7/2} (c d f-a e g)} \end {gather*}

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)^(9/2)),x]

[Out]

(2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(7*(c*d*f - a*e*g)*(d + e*x)^(5/2)*(f + g*x)^(7/2)) + (4*c*d
*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(35*(c*d*f - a*e*g)^2*(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]

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 {\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)^{9/2}} \, dx &=\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{7 (c d f-a e g) (d+e x)^{5/2} (f+g x)^{7/2}}+\frac {(2 c d) \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}{7 (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 )^{5/2}}{7 (c d f-a e g) (d+e x)^{5/2} (f+g x)^{7/2}}+\frac {4 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{35 (c d f-a e g)^2 (d+e x)^{5/2} (f+g x)^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 69, normalized size = 0.53 \begin {gather*} \frac {2 ((d+e x) (a e+c d x))^{5/2} (c d (7 f+2 g x)-5 a e g)}{35 (d+e x)^{5/2} (f+g x)^{7/2} (c d f-a e g)^2} \end {gather*}

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)^(9/2)),x]

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(5/2)*(-5*a*e*g + c*d*(7*f + 2*g*x)))/(35*(c*d*f - a*e*g)^2*(d + e*x)^(5/2)*(f +
g*x)^(7/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.71, size = 249, normalized size = 1.93 \begin {gather*} \frac {2 \left (\frac {(d g+e g x) (a e g+c d g x)}{g^2}\right )^{3/2} \sqrt {a e g+c d (f+g x)-c d f} \left (-5 a^3 e^3 g^3-8 a^2 c d e^2 g^2 (f+g x)+15 a^2 c d e^2 f g^2-15 a c^2 d^2 e f^2 g-a c^2 d^2 e g (f+g x)^2+16 a c^2 d^2 e f g (f+g x)+5 c^3 d^3 f^3-8 c^3 d^3 f^2 (f+g x)+2 c^3 d^3 (f+g x)^3+c^3 d^3 f (f+g x)^2\right )}{35 g (d+e x)^{3/2} (f+g x)^{7/2} (a e g-c d f)^2 (a e g+c d g x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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)^(9/2)),x]

[Out]

(2*(((a*e*g + c*d*g*x)*(d*g + e*g*x))/g^2)^(3/2)*Sqrt[-(c*d*f) + a*e*g + c*d*(f + g*x)]*(5*c^3*d^3*f^3 - 15*a*
c^2*d^2*e*f^2*g + 15*a^2*c*d*e^2*f*g^2 - 5*a^3*e^3*g^3 - 8*c^3*d^3*f^2*(f + g*x) + 16*a*c^2*d^2*e*f*g*(f + g*x
) - 8*a^2*c*d*e^2*g^2*(f + g*x) + c^3*d^3*f*(f + g*x)^2 - a*c^2*d^2*e*g*(f + g*x)^2 + 2*c^3*d^3*(f + g*x)^3))/
(35*g*(-(c*d*f) + a*e*g)^2*(d + e*x)^(3/2)*(f + g*x)^(7/2)*(a*e*g + c*d*g*x)^(3/2))

________________________________________________________________________________________

fricas [B]  time = 0.47, size = 526, normalized size = 4.08 \begin {gather*} \frac {2 \, {\left (2 \, c^{3} d^{3} g x^{3} + 7 \, a^{2} c d e^{2} f - 5 \, a^{3} e^{3} g + {\left (7 \, c^{3} d^{3} f - a c^{2} d^{2} e g\right )} x^{2} + 2 \, {\left (7 \, a c^{2} d^{2} e f - 4 \, a^{2} c d e^{2} 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}}{35 \, {\left (c^{2} d^{3} f^{6} - 2 \, a c d^{2} e f^{5} g + a^{2} d e^{2} f^{4} g^{2} + {\left (c^{2} d^{2} e f^{2} g^{4} - 2 \, a c d e^{2} f g^{5} + a^{2} e^{3} g^{6}\right )} x^{5} + {\left (4 \, c^{2} d^{2} e f^{3} g^{3} + a^{2} d e^{2} g^{6} + {\left (c^{2} d^{3} - 8 \, a c d e^{2}\right )} f^{2} g^{4} - 2 \, {\left (a c d^{2} e - 2 \, a^{2} e^{3}\right )} f g^{5}\right )} x^{4} + 2 \, {\left (3 \, c^{2} d^{2} e f^{4} g^{2} + 2 \, a^{2} d e^{2} f g^{5} + 2 \, {\left (c^{2} d^{3} - 3 \, a c d e^{2}\right )} f^{3} g^{3} - {\left (4 \, a c d^{2} e - 3 \, a^{2} e^{3}\right )} f^{2} g^{4}\right )} x^{3} + 2 \, {\left (2 \, c^{2} d^{2} e f^{5} g + 3 \, a^{2} d e^{2} f^{2} g^{4} + {\left (3 \, c^{2} d^{3} - 4 \, a c d e^{2}\right )} f^{4} g^{2} - 2 \, {\left (3 \, a c d^{2} e - a^{2} e^{3}\right )} f^{3} g^{3}\right )} x^{2} + {\left (c^{2} d^{2} e f^{6} + 4 \, a^{2} d e^{2} f^{3} g^{3} + 2 \, {\left (2 \, c^{2} d^{3} - a c d e^{2}\right )} f^{5} g - {\left (8 \, a c d^{2} e - a^{2} e^{3}\right )} f^{4} g^{2}\right )} x\right )}} \end {gather*}

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)^(9/2),x, algorithm="fricas")

[Out]

2/35*(2*c^3*d^3*g*x^3 + 7*a^2*c*d*e^2*f - 5*a^3*e^3*g + (7*c^3*d^3*f - a*c^2*d^2*e*g)*x^2 + 2*(7*a*c^2*d^2*e*f
- 4*a^2*c*d*e^2*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^6 -
2*a*c*d^2*e*f^5*g + a^2*d*e^2*f^4*g^2 + (c^2*d^2*e*f^2*g^4 - 2*a*c*d*e^2*f*g^5 + a^2*e^3*g^6)*x^5 + (4*c^2*d^2
*e*f^3*g^3 + a^2*d*e^2*g^6 + (c^2*d^3 - 8*a*c*d*e^2)*f^2*g^4 - 2*(a*c*d^2*e - 2*a^2*e^3)*f*g^5)*x^4 + 2*(3*c^2
*d^2*e*f^4*g^2 + 2*a^2*d*e^2*f*g^5 + 2*(c^2*d^3 - 3*a*c*d*e^2)*f^3*g^3 - (4*a*c*d^2*e - 3*a^2*e^3)*f^2*g^4)*x^
3 + 2*(2*c^2*d^2*e*f^5*g + 3*a^2*d*e^2*f^2*g^4 + (3*c^2*d^3 - 4*a*c*d*e^2)*f^4*g^2 - 2*(3*a*c*d^2*e - a^2*e^3)
*f^3*g^3)*x^2 + (c^2*d^2*e*f^6 + 4*a^2*d*e^2*f^3*g^3 + 2*(2*c^2*d^3 - a*c*d*e^2)*f^5*g - (8*a*c*d^2*e - a^2*e^
3)*f^4*g^2)*x)

________________________________________________________________________________________

giac [F(-1)]  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((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)^(9/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.01, size = 99, normalized size = 0.77 \begin {gather*} -\frac {2 \left (c d x +a e \right ) \left (-2 c d g x +5 a e g -7 c d f \right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}{35 \left (g x +f \right )^{\frac {7}{2}} \left (a^{2} e^{2} g^{2}-2 a c d e f g +f^{2} c^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}} \end {gather*}

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)^(9/2),x)

[Out]

-2/35*(c*d*x+a*e)*(-2*c*d*g*x+5*a*e*g-7*c*d*f)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(3/2)/(g*x+f)^(7/2)/(a^2*e^2*
g^2-2*a*c*d*e*f*g+c^2*d^2*f^2)/(e*x+d)^(3/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {9}{2}}}\,{d x} \end {gather*}

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)^(9/2),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)^(9/2)), x)

________________________________________________________________________________________

mupad [B]  time = 4.31, size = 247, normalized size = 1.91 \begin {gather*} -\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {2\,a^2\,e^2\,\left (5\,a\,e\,g-7\,c\,d\,f\right )}{35\,g^3\,{\left (a\,e\,g-c\,d\,f\right )}^2}-\frac {4\,c^3\,d^3\,x^3}{35\,g^2\,{\left (a\,e\,g-c\,d\,f\right )}^2}+\frac {2\,c^2\,d^2\,x^2\,\left (a\,e\,g-7\,c\,d\,f\right )}{35\,g^3\,{\left (a\,e\,g-c\,d\,f\right )}^2}+\frac {4\,a\,c\,d\,e\,x\,\left (4\,a\,e\,g-7\,c\,d\,f\right )}{35\,g^3\,{\left (a\,e\,g-c\,d\,f\right )}^2}\right )}{x^3\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}+\frac {f^3\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{g^3}+\frac {3\,f\,x^2\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{g}+\frac {3\,f^2\,x\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{g^2}} \end {gather*}

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)^(9/2)*(d + e*x)^(3/2)),x)

[Out]

-((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((2*a^2*e^2*(5*a*e*g - 7*c*d*f))/(35*g^3*(a*e*g - c*d*f)^2) -
(4*c^3*d^3*x^3)/(35*g^2*(a*e*g - c*d*f)^2) + (2*c^2*d^2*x^2*(a*e*g - 7*c*d*f))/(35*g^3*(a*e*g - c*d*f)^2) + (4
*a*c*d*e*x*(4*a*e*g - 7*c*d*f))/(35*g^3*(a*e*g - c*d*f)^2)))/(x^3*(f + g*x)^(1/2)*(d + e*x)^(1/2) + (f^3*(f +
g*x)^(1/2)*(d + e*x)^(1/2))/g^3 + (3*f*x^2*(f + g*x)^(1/2)*(d + e*x)^(1/2))/g + (3*f^2*x*(f + g*x)^(1/2)*(d +
e*x)^(1/2))/g^2)

________________________________________________________________________________________

sympy [F(-1)]  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((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)**(9/2),x)

[Out]

Timed out

________________________________________________________________________________________