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

Optimal. Leaf size=154 \[ \frac {2 \sqrt {d+e x} \left (2 a e^2 g+c d (e f-3 d g)\right )}{3 c^2 d^2 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)^{5/2} (c d f-a e g)}{3 c d \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

[Out]

-2/3*(-a*e*g+c*d*f)*(e*x+d)^(5/2)/c/d/(-a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+2/3*(2*a*e^2*g+c*
d*(-3*d*g+e*f))*(e*x+d)^(1/2)/c^2/d^2/(-a*e^2+c*d^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 = 0.13, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {788, 648} \[ \frac {2 \sqrt {d+e x} \left (2 a e^2 g+c d (e f-3 d g)\right )}{3 c^2 d^2 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)^{5/2} (c d f-a e g)}{3 c d \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 788

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((g*(c*d - b*e) + c*e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)*(2*c*d - b*e)), x] - Dist[(e*(m*(g
*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g)))/(c*(p + 1)*(2*c*d - b*e)), Int[(d + e*x)^(m - 1)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2,
0] && LtQ[p, -1] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(d+e x)^{5/2} (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 (c d f-a e g) (d+e x)^{5/2}}{3 c d \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {\left (2 a e^2 g+c d (e f-3 d g)\right ) \int \frac {(d+e x)^{3/2}}{\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 \left (c d^2-a e^2\right )}\\ &=-\frac {2 (c d f-a e g) (d+e x)^{5/2}}{3 c d \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {2 \left (2 a e^2 g+c d (e f-3 d g)\right ) \sqrt {d+e x}}{3 c^2 d^2 \left (c d^2-a e^2\right ) \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 = 0.05, size = 52, normalized size = 0.34 \[ -\frac {2 (d+e x)^{3/2} (2 a e g+c d (f+3 g x))}{3 c^2 d^2 ((d+e x) (a e+c d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 1.20, size = 129, normalized size = 0.84 \[ -\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (3 \, c d g x + c d f + 2 \, a e g\right )} \sqrt {e x + d}}{3 \, {\left (c^{4} d^{4} e x^{3} + a^{2} c^{2} d^{3} e^{2} + {\left (c^{4} d^{5} + 2 \, a c^{3} d^{3} e^{2}\right )} x^{2} + {\left (2 \, a c^{3} d^{4} e + a^{2} c^{2} d^{2} e^{3}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 3.14Unable to transpose Error: Bad Argument Value

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maple [A]  time = 0.01, size = 66, normalized size = 0.43 \[ -\frac {2 \left (c d x +a e \right ) \left (3 c d g x +2 a e g +c d f \right ) \left (e x +d \right )^{\frac {5}{2}}}{3 \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}} c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.60, size = 73, normalized size = 0.47 \[ -\frac {2 \, {\left (3 \, c d x + 2 \, a e\right )} g}{3 \, {\left (c^{3} d^{3} x + a c^{2} d^{2} e\right )} \sqrt {c d x + a e}} - \frac {2 \, f}{3 \, {\left (c^{2} d^{2} x + a c d e\right )} \sqrt {c d x + a e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.50, size = 149, normalized size = 0.97 \[ -\frac {\left (\frac {\left (\frac {4\,a\,e\,g}{3}+\frac {2\,c\,d\,f}{3}\right )\,\sqrt {d+e\,x}}{c^4\,d^4\,e}+\frac {2\,g\,x\,\sqrt {d+e\,x}}{c^3\,d^3\,e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{x^3+\frac {a^2\,e}{c^2\,d}+\frac {a\,x\,\left (2\,c\,d^2+a\,e^2\right )}{c^2\,d^2}+\frac {x^2\,\left (c^4\,d^5+2\,a\,c^3\,d^3\,e^2\right )}{c^4\,d^4\,e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(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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