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*(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])

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Rubi [A]  time = 0.133725, antiderivative size = 154, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.0498132, 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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Maple [A]  time = 0.046, size = 66, normalized size = 0.4 \begin{align*} -{\frac{ \left ( 2\,cdx+2\,ae \right ) \left ( 3\,xcdg+2\,aeg+cdf \right ) }{3\,{c}^{2}{d}^{2}} \left ( ex+d \right ) ^{{\frac{5}{2}}} \left ( cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade \right ) ^{-{\frac{5}{2}}}} \end{align*}

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)

[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 = 1.17725, size = 99, normalized size = 0.64 \begin{align*} -\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{align*}

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 = 1.56673, size = 270, normalized size = 1.75 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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]

sage0*x