3.758 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{5/2} (f+g x)^{11/2}} \, dx\)

Optimal. Leaf size=129 \[ \frac {4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{63 (d+e x)^{7/2} (f+g x)^{7/2} (c d f-a e g)^2}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{9 (d+e x)^{7/2} (f+g x)^{9/2} (c d f-a e g)} \]

[Out]

2/9*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(7/2)/(-a*e*g+c*d*f)/(e*x+d)^(7/2)/(g*x+f)^(9/2)+4/63*c*d*(a*d*e+(a*e^2+
c*d^2)*x+c*d*e*x^2)^(7/2)/(-a*e*g+c*d*f)^2/(e*x+d)^(7/2)/(g*x+f)^(7/2)

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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} \[ \frac {4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{63 (d+e x)^{7/2} (f+g x)^{7/2} (c d f-a e g)^2}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{9 (d+e x)^{7/2} (f+g x)^{9/2} (c d f-a e g)} \]

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*(f + g*x)^(11/2)),x]

[Out]

(2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(9*(c*d*f - a*e*g)*(d + e*x)^(7/2)*(f + g*x)^(9/2)) + (4*c*d
*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(63*(c*d*f - a*e*g)^2*(d + e*x)^(7/2)*(f + g*x)^(7/2))

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 )^{5/2}}{(d+e x)^{5/2} (f+g x)^{11/2}} \, dx &=\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{9 (c d f-a e g) (d+e x)^{7/2} (f+g x)^{9/2}}+\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 )^{5/2}}{(d+e x)^{5/2} (f+g x)^{9/2}} \, dx}{9 (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 )^{7/2}}{9 (c d f-a e g) (d+e x)^{7/2} (f+g x)^{9/2}}+\frac {4 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{63 (c d f-a e g)^2 (d+e x)^{7/2} (f+g x)^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 79, normalized size = 0.61 \[ \frac {2 (a e+c d x)^3 \sqrt {(d+e x) (a e+c d x)} (c d (9 f+2 g x)-7 a e g)}{63 \sqrt {d+e x} (f+g x)^{9/2} (c d f-a e g)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*(f + g*x)^(11/2)),x]

[Out]

(2*(a*e + c*d*x)^3*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-7*a*e*g + c*d*(9*f + 2*g*x)))/(63*(c*d*f - a*e*g)^2*Sqrt[d
+ e*x]*(f + g*x)^(9/2))

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fricas [B]  time = 0.91, size = 639, normalized size = 4.95 \[ \frac {2 \, {\left (2 \, c^{4} d^{4} g x^{4} + 9 \, a^{3} c d e^{3} f - 7 \, a^{4} e^{4} g + {\left (9 \, c^{4} d^{4} f - a c^{3} d^{3} e g\right )} x^{3} + 3 \, {\left (9 \, a c^{3} d^{3} e f - 5 \, a^{2} c^{2} d^{2} e^{2} g\right )} x^{2} + {\left (27 \, a^{2} c^{2} d^{2} e^{2} f - 19 \, a^{3} c d e^{3} g\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f}}{63 \, {\left (c^{2} d^{3} f^{7} - 2 \, a c d^{2} e f^{6} g + a^{2} d e^{2} f^{5} g^{2} + {\left (c^{2} d^{2} e f^{2} g^{5} - 2 \, a c d e^{2} f g^{6} + a^{2} e^{3} g^{7}\right )} x^{6} + {\left (5 \, c^{2} d^{2} e f^{3} g^{4} + a^{2} d e^{2} g^{7} + {\left (c^{2} d^{3} - 10 \, a c d e^{2}\right )} f^{2} g^{5} - {\left (2 \, a c d^{2} e - 5 \, a^{2} e^{3}\right )} f g^{6}\right )} x^{5} + 5 \, {\left (2 \, c^{2} d^{2} e f^{4} g^{3} + a^{2} d e^{2} f g^{6} + {\left (c^{2} d^{3} - 4 \, a c d e^{2}\right )} f^{3} g^{4} - 2 \, {\left (a c d^{2} e - a^{2} e^{3}\right )} f^{2} g^{5}\right )} x^{4} + 10 \, {\left (c^{2} d^{2} e f^{5} g^{2} + a^{2} d e^{2} f^{2} g^{5} + {\left (c^{2} d^{3} - 2 \, a c d e^{2}\right )} f^{4} g^{3} - {\left (2 \, a c d^{2} e - a^{2} e^{3}\right )} f^{3} g^{4}\right )} x^{3} + 5 \, {\left (c^{2} d^{2} e f^{6} g + 2 \, a^{2} d e^{2} f^{3} g^{4} + 2 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} f^{5} g^{2} - {\left (4 \, a c d^{2} e - a^{2} e^{3}\right )} f^{4} g^{3}\right )} x^{2} + {\left (c^{2} d^{2} e f^{7} + 5 \, a^{2} d e^{2} f^{4} g^{3} + {\left (5 \, c^{2} d^{3} - 2 \, a c d e^{2}\right )} f^{6} g - {\left (10 \, a c d^{2} e - a^{2} e^{3}\right )} f^{5} g^{2}\right )} x\right )}} \]

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)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(11/2),x, algorithm="fricas")

[Out]

2/63*(2*c^4*d^4*g*x^4 + 9*a^3*c*d*e^3*f - 7*a^4*e^4*g + (9*c^4*d^4*f - a*c^3*d^3*e*g)*x^3 + 3*(9*a*c^3*d^3*e*f
- 5*a^2*c^2*d^2*e^2*g)*x^2 + (27*a^2*c^2*d^2*e^2*f - 19*a^3*c*d*e^3*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^7 - 2*a*c*d^2*e*f^6*g + a^2*d*e^2*f^5*g^2 + (c^2*d^2*e*f^2*g^5
- 2*a*c*d*e^2*f*g^6 + a^2*e^3*g^7)*x^6 + (5*c^2*d^2*e*f^3*g^4 + a^2*d*e^2*g^7 + (c^2*d^3 - 10*a*c*d*e^2)*f^2*
g^5 - (2*a*c*d^2*e - 5*a^2*e^3)*f*g^6)*x^5 + 5*(2*c^2*d^2*e*f^4*g^3 + a^2*d*e^2*f*g^6 + (c^2*d^3 - 4*a*c*d*e^2
)*f^3*g^4 - 2*(a*c*d^2*e - a^2*e^3)*f^2*g^5)*x^4 + 10*(c^2*d^2*e*f^5*g^2 + a^2*d*e^2*f^2*g^5 + (c^2*d^3 - 2*a*
c*d*e^2)*f^4*g^3 - (2*a*c*d^2*e - a^2*e^3)*f^3*g^4)*x^3 + 5*(c^2*d^2*e*f^6*g + 2*a^2*d*e^2*f^3*g^4 + 2*(c^2*d^
3 - a*c*d*e^2)*f^5*g^2 - (4*a*c*d^2*e - a^2*e^3)*f^4*g^3)*x^2 + (c^2*d^2*e*f^7 + 5*a^2*d*e^2*f^4*g^3 + (5*c^2*
d^3 - 2*a*c*d*e^2)*f^6*g - (10*a*c*d^2*e - a^2*e^3)*f^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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(11/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 99, normalized size = 0.77 \[ -\frac {2 \left (c d x +a e \right ) \left (-2 c d g x +7 a e g -9 c d f \right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}}}{63 \left (g x +f \right )^{\frac {9}{2}} \left (a^{2} e^{2} g^{2}-2 a c d e f g +f^{2} c^{2} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}} \]

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)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(11/2),x)

[Out]

-2/63*(c*d*x+a*e)*(-2*c*d*g*x+7*a*e*g-9*c*d*f)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(5/2)/(g*x+f)^(9/2)/(a^2*e^2*
g^2-2*a*c*d*e*f*g+c^2*d^2*f^2)/(e*x+d)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}^{\frac {11}{2}}}\,{d x} \]

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)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(11/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/((e*x + d)^(5/2)*(g*x + f)^(11/2)), x)

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mupad [B]  time = 4.54, size = 315, normalized size = 2.44 \[ -\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {2\,a^3\,e^3\,\left (7\,a\,e\,g-9\,c\,d\,f\right )}{63\,g^4\,{\left (a\,e\,g-c\,d\,f\right )}^2}-\frac {4\,c^4\,d^4\,x^4}{63\,g^3\,{\left (a\,e\,g-c\,d\,f\right )}^2}+\frac {2\,c^3\,d^3\,x^3\,\left (a\,e\,g-9\,c\,d\,f\right )}{63\,g^4\,{\left (a\,e\,g-c\,d\,f\right )}^2}+\frac {2\,a^2\,c\,d\,e^2\,x\,\left (19\,a\,e\,g-27\,c\,d\,f\right )}{63\,g^4\,{\left (a\,e\,g-c\,d\,f\right )}^2}+\frac {2\,a\,c^2\,d^2\,e\,x^2\,\left (5\,a\,e\,g-9\,c\,d\,f\right )}{21\,g^4\,{\left (a\,e\,g-c\,d\,f\right )}^2}\right )}{x^4\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}+\frac {f^4\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{g^4}+\frac {4\,f\,x^3\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{g}+\frac {4\,f^3\,x\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{g^3}+\frac {6\,f^2\,x^2\,\sqrt {f+g\,x}\,\sqrt {d+e\,x}}{g^2}} \]

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)^(5/2)/((f + g*x)^(11/2)*(d + e*x)^(5/2)),x)

[Out]

-((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((2*a^3*e^3*(7*a*e*g - 9*c*d*f))/(63*g^4*(a*e*g - c*d*f)^2) -
(4*c^4*d^4*x^4)/(63*g^3*(a*e*g - c*d*f)^2) + (2*c^3*d^3*x^3*(a*e*g - 9*c*d*f))/(63*g^4*(a*e*g - c*d*f)^2) + (2
*a^2*c*d*e^2*x*(19*a*e*g - 27*c*d*f))/(63*g^4*(a*e*g - c*d*f)^2) + (2*a*c^2*d^2*e*x^2*(5*a*e*g - 9*c*d*f))/(21
*g^4*(a*e*g - c*d*f)^2)))/(x^4*(f + g*x)^(1/2)*(d + e*x)^(1/2) + (f^4*(f + g*x)^(1/2)*(d + e*x)^(1/2))/g^4 + (
4*f*x^3*(f + g*x)^(1/2)*(d + e*x)^(1/2))/g + (4*f^3*x*(f + g*x)^(1/2)*(d + e*x)^(1/2))/g^3 + (6*f^2*x^2*(f + g
*x)^(1/2)*(d + e*x)^(1/2))/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)**(5/2)/(e*x+d)**(5/2)/(g*x+f)**(11/2),x)

[Out]

Timed out

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