3.7.74 \(\int \frac {(d+e x)^{5/2} (f+g x)}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\) [674]

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

[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 = 0.09, 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 = {802, 662} \begin {gather*} \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}} \end {gather*}

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 662

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 802

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 \begin {gather*} -\frac {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}} \end {gather*}

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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Maple [A]
time = 0.14, size = 58, normalized size = 0.38

method	result	size
default	\(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (3 c d g x +2 a e g +c d f \right )}{3 \sqrt {e x +d}\, \left (c d x +a e \right )^{2} c^{2} d^{2}}\)	\(58\)
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (3 c d g x +2 a e g +c d f \right ) \left (e x +d \right )^{\frac {5}{2}}}{3 c^{2} d^{2} \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.32, size = 78, normalized size = 0.51 \begin {gather*} -\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}} \end {gather*}

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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Fricas [A]
time = 2.40, size = 130, normalized size = 0.84 \begin {gather*} -\frac {2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (3 \, c d g x + c d f + 2 \, a g e\right )} \sqrt {x e + d}}{3 \, {\left (c^{4} d^{5} x^{2} + a^{2} c^{2} d^{2} x e^{3} + {\left (2 \, a c^{3} d^{3} x^{2} + a^{2} c^{2} d^{3}\right )} e^{2} + {\left (c^{4} d^{4} x^{3} + 2 \, a c^{3} d^{4} x\right )} e\right )}} \end {gather*}

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

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Sympy [F(-1)] Timed out
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((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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Giac [A]
time = 2.38, size = 153, normalized size = 0.99 \begin {gather*} \frac {2 \, {\left (3 \, c d^{2} g e - c d f e^{2} - 2 \, a g e^{3}\right )}}{3 \, {\left (\sqrt {-c d^{2} e + a e^{3}} c^{3} d^{4} - \sqrt {-c d^{2} e + a e^{3}} a c^{2} d^{2} e^{2}\right )}} - \frac {2 \, {\left (c d f e^{3} - a g e^{4} + 3 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )} g e\right )}}{3 \, {\left ({\left (x e + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{2} d^{2}} \end {gather*}

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]

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

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Mupad [B]
time = 3.50, size = 149, normalized size = 0.97 \begin {gather*} -\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}} \end {gather*}

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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